Dynamics Pytel Kiusalaas 4th Solutions

Chapter 11 11.1 24 lb

2 = 4:511 slugs J 5:32 ft/s = mg = 4:511(32:2) = 145:3 lb J

(a) m = (b) W

11.2 W

1 2 R h g = (0:0752 )(0:125)(2700)(9:81) = 19:503 N 3 3 0:2284 lb (19:503 N) = 4:45 lb J 1:0 N

= =

11.3 (a) 100 kN/m2 =

(b) 30 m/s =

100

30 m s

103 N m2

0:2248 lb 1:0 N

3:281 ft 1:0 m

1:0 mi 5280 ft

20 lb ft2

3600 s = 67:1 mi/h J 1:0 h

14:593 kg = 11:67 1:0 slug

(c) 800 slugs = 800 slugs

(d) 20 lb/ft2 =

1:0 m2 = 14:50 lb/in.2 J 1550 in.2

4:448 N 1:0 lb

103 kg = 11:67 Mg J

1:0 ft2 = 958 N/m2 J 0:092 903 04 m2

11.4 I = 20 kg m2 = 20 kg m2

0:06853 slugs 1:0 kg

10:764 ft2 = 14:75 slugs ft2 1:0 m2

But 1:0 slug = 1:0 lb s2 =ft I = 14:75

lb s2 2 ft = 14:75 lb ft s2 J ft

11.5 1 1 mv 2 + mk 2 ! 2 2 2 Since the dimensions of each term must be the same, we have KE =

[KE] = [M ]

L2 = [M ] k 2 T2

1 T2

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Therefore, [k] = [L] (a) In the SI system kg m2 L2 = J T2 s2

[KE] = [M ]

[k] = m J

(b) In the US system [KE]

= [M ]

FT2 L2 = T2 L

L2 = [F L] = lb ft J T2

= ft J

[k]

11.6 1 W

[g] [k] [x]

=

L T2

1 F [L] L F

=

L = [a] Q.E.D. T2

11.7 (a) [ ]=

PL EA

[L] =

1 E

FL L2

[E] =

F L2

J

(b) Substituting [F ] = M L=T 2 into the result of part (a): [E] =

ML T2

1 M = 2 L T 2L

J

11.8 (a)

mv 2 =

FT2 L

L2 = [F L] J T2

(b) [mv] =

FT2 L

L = [F T ] J T

(c) [ma] =

FT2 L

L = [F ] J T2


11.9 Rewrite the equation as y = 1:0 x2 [y] = [1:0] x2

[L] = [1:0] L2

1 L

[1:0] =

y = x2 can be dimensionally correct only if the units of the implied constant 1:0 are in. 1 . J

11.10 FT2 L

(a) [I] = mR2 =

L2 = F LT 2 J

(b) [I] = mR2 = M L2 J

11.11 (a)

v 3 = [A] x2 + [B] [v] t2 L J T3

[A] =

(b)

x2 = [A] t2

h

2 e[B][t ]

[A] =

i

L2 T2

L3 L = [A] L2 + [B] T3 T L2 [B] = J T4

L2 = [A] T 2 [1] J

[B] =

T2

[B] T 2 = [1]

1 T2

J

11.12 dx d2 x +c + kx = P0 sin !t 2 dt dt d2 x FT2 L [m] = = [F ] 2 dt L T2 m

Therefore, the dimension of each term in the expression is [F ]. [c]

dx L = [c] = [F ] dt T [k] [x] = [k] [L] = [F ]

[P0 ] [sin !t] = [P0 ] [1] = [F ] [!] [t] = [!] [T ] = [1]

FT L

[c] =

[k] =

J

F J L

[P0 ] = [F ] J [!] =

1 T

J


11.13 F =G

(a) [G] =

(b) [G] =

mA mB R2

[F ] L2 2

[F T 2 =L]

=

G=

L4 FT4

F R2 mA mb

[G] =

[F ] L2 [M 2 ]

J

M L=T 2 L2 L3 = 2 [M ] MT2

J

11.14 Using the base dimensions of an absolute [MLT] system: ML M = [C] 3 2 T L

[F ] = [C][ ][v 2 ][A]

L2 [L2 ] T2

[C] = [1] Q.E.D J

11.15 F W F W

100%

82 m2 = (6:67 10 11 ) 2 = 2:668 10 8 N 2 R 0:4 = mg = 8(9:81) = 78:48 N 2:668 10 8 = 100% = 3:40 10 8 % J 78:48 = G

11.16 F =G

2

m2 = 3:44 R2

10

8

(2=32:2) = 7:46 (16=12)2

10

11

lb J

11.17 m=

W R2 (3000) (6378 + 1600)2 106 = = 479 kg J GMe (6:67 10 11 ) (5:9742 1024 )

11.18 GMm 2 Rm

ge =

0:073483 5:9742

6378 1738

gm = gm Mm = ge Me

Re Rm

2

=

GMe Re2 2

= 0:1656 t

1 Q.E.D 6


11.19 Me = 5:9742

Me m

= 3:44

10

Ms m = 6:67 R2

10

2

(2Re )

3:281 ft = 20:93 1:0 m

103 m

Re = 6378 W =G

0:06853 slugs = 0:4094 1:0 kg

1024 kg

(2

106 ft

1024 )(150=32:2)

8 (0:4094

2

106 )

20:93

1024 slugs

= 37:4 lb J

11.20 F =G

11

1:9891

1030 (1:0)

(149:6

2

109 )

= 0:00593 N J

11.21 Me m r2 Me Ms 5:9742 1024 1:9891 1030

=

0

=

G

Ms m Me Ms = (R r)2 r2 (R r)2 r2 2 R 2Rr + r2 r2 9 2 (149:6 10 ) 2(149:6 109 )r + r2

= G

=

2:238

r = 259

1022

2:992

106 m = 259

1011 r

3:329 4

105 r2

103 km J


Chapter 12 12.1 y v

= 0:16t4 + 4:9t3 + 0:14t2 ft = y_ = 0:64t3 + 14:7t2 + 0:28t ft/s

a = v_ =

1:92t2 + 29:4t + 0:28 ft/s

2

At maximum velocity (a = 0): 1:92t2 + 29:4t + 0:28 = 0 vmax y

t = 15:322 s

= 0:64(15:3223 ) + 14:7(15:3222 ) + 0:28(15:322) = 1153 ft/s J = 0:16(15:3224 ) + 4:9(15:3223 ) + 0:14(15:3222 ) = 8840 ft J

12.2


12.65 y = 64:4(4

x_ =

y_ =

128:8t ft/s

y• =

128:8 ft/s

x •=

Fx Fy

t2 ) ft

x = 20 cos

t ft 2 10 sin t ft/s 2 2 5 2 cos t ft/s 2

2

4 5 2 cos t = 6:130 cos t lb 32:2 2 2 4 = m• y= ( 128:8) = 16:00 lb 32:2 = m• x=

t (s) 0 1 2

Fx (lb) 6:13 0 6:13

Fy (lb) 16:0 16:0 16:0

J

12.66

12.67 x = b sin

2 t t0

y=

2 b 2 t cos t0 t0 2 t 4 2b sin ax = t20 t0 vx =

b 4

vy = ay =

4 t t0 b 4 t sin t0 t0 2 4 b 4 t cos t20 t0 1 + cos

At point B: x = 0 ) t = 0 ) ax = 0

) ay =

4

2

t20

b

=

4

2

(1:2) = 0:82

74:02 m/s

2


N

0.2N

y x F = 0.5(74.02) N MAD

0.5(9.81) N FBD Fy = may

+"

N

0:5(9:81) =

0:5(74:02)

N = 32:11 N Fx = 0

+ !

F

0:2N = 0

F = 0:2N = 0:2(32:11) = 6:42 N J

12.68


12.69

x

y g

20o

From the acceleration diagram of a water droplet: ax

=

g sin 20 =

32:2 sin 20 =

11:013 ft/s

ay

=

g cos 20 =

32:2 cos 20 =

30:26 ft/s

2

2

Initial conditions at t = 0 : x = y=0 vx = 22 cos 30 = 19:053 ft/s

vy = 22 sin 30 = 11:0 ft/s

Integrating and using initial conditions: vx = x =

11:013t + 19:053 ft/s 5:507t2 + 19:053t ft

vy = 30:26t + 11:0 ft/s y = 15:13t2 + 11:0t ft

Droplet lands when y = 0: y R

= 15:13t2 + 11:0t = 0 t = 0:7270 s = xjt=0:7270s = 5:507(0:72702 ) + 19:053(0:7270) = 10:94 ft J


12.70 From Eqs. (e) of Sample Problem 2.11: x = y

= =

(v0 cos )t = (65 cos 55 ) t = 37:28t ft 1 2 32:2 2 gt + (v0 sin )t = t + (65 sin 55 )t 2 2 16:1t2 + 53:24t ft

At point B: x = 60 ft 60 = 37:28t t = 1:6094 s h = yjt=1:6094s = 16:1(1:60942 ) + 53:24(1:6094) = 44:0 ft J

*12.71 From Sample Problem 12.12: x = C1 e

c m

C1

ct=m

c e m

mgt + C4 c mg c C3 e ct=m m c

+ C2

y = C3 e

ct=m

vy =

vx

=

=

0:0025 = 0:06708 s 1:2=32:2

1

mg 1:2 = = 480:0 ft/s c 0:0025

x = C1 e 0:06708t + C2 y = C3 e 0:06708t vx = C1 0:06708e vy

=

C3 0:06708e v0 sin v0 cos

0:06708t

= =

ct=m

0:06708t

480t + C4

480:0

70 sin 65 = 63:44 ft/s 70 cos 65 = 29:58 ft/s

Initial conditions at t = 0: x y vx vy

= = = =

0 ) C2 0 ) C4 v0 cos v0 sin

= C1 = C3 ) C1 (0:06708) = 29:58 C1 = ) C3 (0:06708) 480:0 = 63:44

441:0 ft C3 = 8101 ft

When x = 60 ft: 60 = 441:0e 0:06708t + 441:0 t = 2:180 s (0:06708)(2:180) h = yjt=2:180 = 8101e 480(2:180) + (8101) = 55:7 ft J


12.72 y

x g

Acceleration diagram At t = 0 (initial conditions): x = y =

0 vx = 200 sin 30 = 100 m/s 1200 m vy = 200 cos 30 = 173:21 m/s

Integrating acceleration and applying initial conditions: ax = 0 vx = 100 m/s x = 100t m

2

ay = 9:81 m/s vy = 9:81t 173:21 m/s y = 4:905t2 173:21t + 1200 m

When y = 0: 4:905t2

173:21t + 1200 = 0

t = 5:932 s

x = 100(5:932) = 593:2 m 593:2 = 99:6 m J

d = 1200 tan 30

12.73 Eqs. (d) and (e) of Sample Problem 12.11: x = v0 t cos = 2500t cos ft 1 2 y = v0 t sin gt = 2500t sin 2 Setting x = R = 5280 ft and solving for t: 5280 = 2500t cos

t=

5280 2500 cos

16:1t2 ft

=

2:112 s cos

Setting y = 0, we get after dividing by t: 2500 sin

16:1t =

0

16:1

2:112 2500

= 0:013 601

sin cos

=

16:1

1 sin 2 2

=

0:013 601

=

1 sin 2

2:112 cos

2500 sin

1

=0

sin 2 = 2(0:013 601) = 0:02720

(0:02720) = 0:779

J


12.74 From Eqs. (e) of Sample Problem 12.11: x = (v0 cos ) t

1 2 gt + (v0 sin ) t 2

y=

(a) x = y

(42 cos 28 ) t = 37:08t ft

)t=

1 (32:2)t2 + (42 sin 28 ) t = 2

=

x s 37:08

16:1t2 + 19:718t

Substituting for t: y

= =

2 x x + 19:718 37:08 37:08 0:01171x2 + 0:5318x ft J

16:1

(b) Check if ball hits the ceiling. dy dx

=

ymax

=

0:02342x + 0:5318 = 0

x = 22:71 ft

2

0:01171(22:71) + 0:5318(22:71) = 6:04 ft

Since ymax < 25 ft, the ball will not hit the ceiling. Check if ball clears the net. When x = 22 ft: y=

0:01171(22)2 + 0:5318(22) = 6:03 ft

Since y > 5 ft, the ball clears the net. J When x = 42 ft: y=

0:01171(42)2 + 0:5318(42) = 1:679 ft

Since y > 0, the ball lands behind the baseline. J

12.75 From Eqs. (e) of Sample Problem 12.11: x = (v0 cos ) t y vy

1 2 gt + (v0 sin ) t 2

y=

1 (32:2)t2 + (v0 sin 70 )t = 2 = y_ = 32:2t + 0:9397v0 ft/s

=

16:1t2 + 0:9397v0 t ft

When y = ymax vy = 0 ymax = 27 ft

32:2t + 0:9397v0 = 0

t = 0:02918v0 s

2

16:1(0:02918v0 ) + 0:9397v0 (0:02918v0 ) = 27 v0 = 44:4 ft/s J

0:013712v02 = 27


12.76 Equations (e) of Sample Problem 12.10: x = y

= =

(v0 cos )t = (30 cos 60 )t = 15t ft 1 2 1 gt + (v0 sin )t = (32:2)t2 + (30 sin 60 )t 2 2 16:1t2 + 25:98t ft

vx = x_ = 15 ft/s

vy = y_ =

v 30o

B

y

30o

32:2t + 25:98 ft/s

h y

30o

x tan 30o

30o

x

x

At point B: v is parallel to the inclined surface 32:2t + 25:98 = tan 30 15

)

vy = tan 30 vx

t = 0:5379 s

x = 15(0:5379) = 8:069 ft y = 16:1(0:53792 ) + 25:98(0:5379) = 9:316 ft h = (y x tan 30 ) cos 30 = (9:316 8:069 tan 30 ) cos 30 = 4:03 ft J


12.77

*12.78


12.79 (a)

FD(vx /v) FD

y mg vx v vy

x = max FD(vy /v) MAD FBD

Fx = max ax

= =

= =

+ !

FD

vx = max v

0:0005v 2 vx FD vx = = 0:005vvx m v q 0:1 v 2 0:005vx vx2 + vy2 m/s J

Fy = may ay

may

+"

FD

vy v

mg = may

FD vy 0:0005v 2 vy g= 9:81 = m v q 0:1 v 2 0:005vy vx2 + vy2 9:81 m/s J

0:005vvy

9:81


(b) Letting x1 = x, x2 = y, x3 = vx and x4 = vy , the equivalent …rst-order equations are x_ 1 = x3 x_ 2 = x4 q q 2 2 x_ 4 = 0:005x4 x23 + x24 9:81 x_ 3 = 0:005x3 x3 + x4 The initial conditions are

x1 (0) = 0

x2 (0) = 2 m/s

x3 (0) = 30 cos 50 = 19:284 m/s

x4 (0) = 30 sin 50 = 22:981 m/s

The following MATLAB program was used to integrate the equations: function problem12_79 [t,x] = ode45(@f,(0:0.05:2),[0 2 19.284 22.981]); printSol(t,x) function dxdt = f(t,x) v = sqrt(x(3)^2 + x(4)^2); dxdt = [x(3) x(4) -0.005*x(3)*v -0.005*x(4)*v-9.81]; end end The two lines of output that span x = 30 m are t 1.7000e+000 1.7500e+000

x1 2.9607e+001 3.0405e+001

x2 2.3944e+001 2.4117e+001

x3 1.5990e+001 1.5925e+001

x4 3.6997e+000 3.1952e+000

Linear interpolation for h: 30:405 24:117

29:607 30 = 23:944 h

29:607 23:944

h = 24:0 m J

Linear interpolation for vx and vy : 30:405 15:925

29:607 30 = 15:990 vx

29:607 15:990

vx = 15:958 m/s

30:405 3:1952

29:607 30 29:607 = vy = 3:451 m/s 3:6997 vy 3:6997 p ) v = 15:9582 + 3:4512 = 16:33 m/s J


12.80 (a)

F y

F(y/d)

may

x

max

F(x/d) = FBD Fx = max ax =

+ !

x = max d 0:5x 2 = m/s J 3=2 2 2 (x + y ) y F = may d 0:5y 2 m/s J = 3=2 2 2 (x + y )

F

1 x 1 0:005 x x F = = 0:5 3 m d 0:01 d2 d d Fy = may

ay =

MAD

+"

1 y 1 0:005 y y F = = 0:5 3 m d 0:01 d2 d d

The initial conditions are: x = 0:3 m y = 0:4 m vx = 0

vy =

2 m/s at t = 0 J

(b) Letting x1 = x, x2 = y, x3 = vx and x4 = vy , the equivalent …rst-order equations are x_ 1 = x3

x_ 2 = x4

x_ 3 =

0:5x1 (x21

+

3=2 x22 )

x_ 4 =

0:5x2 (x21

3=2

+ x22 )

The MATLAB program for solving the equations is function problem12_80 [t,x] = ode45(@f,(0:0.005:0.25),[0.3 0.4 0 -2]); printSol(t,x) function dxdt = f(t,x) d3 = (sqrt(x(1)^2 + x(2)^2))^3; dxdt = [x(3) x(4) 0.5*x(1)/d3 0.5*x(2)/d3]; end end The two output lines spanning y = 0 are shown below. t 2.2000e-001 2.2500e-001

x1 x2 3.5879e-001 3.0450e-003 3.6213e-001 -5.2884e-003

x3 x4 6.5959e-001 -1.6667e+000 6.7883e-001 -1.6668e+000


Linear interpolation for x at y = 0: 0:36213 0:35879 x 0:35879 = 0:0052884 0:0030450 0 0:0030450

x = 0:360 m J

Linear interpolation for vx : vx :

0:67883 0:65959 vx 0:65959 = 0:0052884 0:0030450 0 0:0030450

vx = 0:6666 m/s

By inspection vy =

1:6667 m/s. p ) v = 0:66662 + 1:66672 = 1:795 m/s J

12.81

(a) The signs of ax and ay in the solution of Prob. 12.80 must be reversed. ax =

0:5x (x2

+

3=2 y2 )

m/s

2

J

ay =

0:5y (x2

+

3=2 y2 )

m/s

2

J

The initial conditions are the same as in Problem 12.80: x = 0:3 m y = 0:4 m vx = 0 vy =

2 m/s

at t = 0: J

(b) MATLAB program: function problem12_81 [t,x] = ode45(@f,(0:0.005:0.2),[0.3 0.4 0 -2]); printSol(t,x) function dxdt = f(t,x) d3 = (sqrt(x(1)^2 + x(2)^2))^3; dxdt = [x(3) x(4) -0.5*x(1)/d3 -0.5*x(2)/d3]; end end The two lines of output spanning y = 0 are: t 1.8000e-001 1.8500e-001

x1 x2 x3 x4 2.5952e-001 8.5117e-003 -6.3935e-001 -2.3329e+000 2.5623e-001 -3.1544e-003 -6.7692e-001 -2.3333e+000


x = 0:257 m J


Linear interpolation for vx : 0:67692 ( 0:63935) vx = 0:0031544 0:0085117 0 By inspection, vy =

12.82

( 0:63935) 0:0085177

vx =

0:6668 m/s

2:3332 m/s. p v = 0:66682 + 2:33322 = 2:43 m/s J

FD (vx/v) FD

mg vx v y vy

may

x

max

=

FD (vy/v) FBD

MAD

vx vx = max cD v 1:5 = max v v cD p ) ax = vx v m 0:0012 cD 0:5 = = 0:06869 (ft s) m (9=16)(32:2)

Fx = max

+ !

FD

) ax = Fy = may

+"

0:06869vx vx2 + vy2 FD

vy y

cD p vy v g = m The initial conditions are: ) ay =

x=0

y = 6 ft

0:25

ft/s

J

cD v 1:5

mg = may 0:06869vy vx2 + vy2

vx = 120 ft/s

2

0:25

vy = 0

vy y

mg = may

32:2 ft/s

2

J

at t = 0 J

(b) Letting x1 = x, x2 = y, x3 = vx and x4 = vy , the equivalent …rst-order equations are x_ 1

= x3

x_ 2 = x4

x_ 3

=

0:06869x3 x23 + x24

0:25

x4

=

0:06869x4 x23 + x24

0:25

32:2

The MATLAB program is


function problem12_82 [t,x] = ode45(@f,(0:0.02:0.7),[0 6 120 0]); printSol(t,x) function dxdt = f(t,x) v25 = sqrt((sqrt(x(3)^2 + x(4)^2))); dxdt = [x(3) x(4) -0.06869*x(3)*v25 -0.06869*x(4)*v25-32.2]; end end The two lines of output that span y = 0 are: t 6.4000e-001 6.6000e-001

x1 x2 6.1878e+001 2.6073e-001 6.3426e+001 -8.0650e-002

x3 x4 7.7840e+001 -1.6851e+001 7.6894e+001 -1.7286e+001

Linear interpolation for x at y = 0: R 61:878 63:426 61:878 = 0:080650 0:26073 0 0:26072

R = 63:1 ft J

Linear interpolation for t at y = 0: 0:66 0:64 t 0:64 = 0:080650 0:26073 0 0:26072

t = 0:655 s J

12.83 ax =

10

0:5vx m/s

2

ay =

9:81

0:5vy m/s

2

(a) Letting x1 = x, x2 = y, x3 = vx and x4 = vy , the equivalent …rst-order equations are x_ 1 = x3

x_ 2 = x4

x_ 3 =

10

0:5x3

x_ 4 =

9:81

0:5x4

At t = 0 (initial conditions): x1 = x2 = 0

x3 = 30 cos 40 = 22:98 m/s

x4 = 30 sin 40 = 19:284 m/s

MATLAB program: function problem12_83 [t,x] = ode45(@f,(0:0.05:3.5),[0 0 22.98 19.284]); printSol(t,x) axes(’fontsize’,14) plot(x(:,1),x(:,2),’linewidth’,1.5)


grid on xlabel(’x (ft)’); ylabel(’y (ft)’) function dxdt = f(t,x) dxdt = [x(3) x(4) -10-0.5*x(3) -9.81-0.5*x(4)]; end end Two lines of output that span y = 0: t 3.1000e+000 3.1500e+000

x1 x2 x3 x4 5.7152e+000 4.7141e-001 -1.0878e+001 -1.1363e+001 5.1656e+000 -1.0184e-001 -1.1103e+001 -1.1567e+001

Linear interpolation for x at y = 0: b 5:1656 5:7152 = 0:10184 0:47141 0

5:7152 0:47141

b = 5:26 m J

3:15 3:10 t 3:10 = 0:10184 0:47141 0 0:47141

t = 3:14 s J

Linear interpolation for t at y = 0:

(b) 15

y (ft)

10

5

0 0

5

10 x (ft)

15

20


12.84 (a)

y

mg

x R

y

θ

Spring force is F = k(R

x

F

MAD

FBD

+ !

p x2 + y 2 :

F cos = max

k(R L0 ) x F cos = = m m R 10 0:5 1 x = 40 1 0:25 R

= =

Fy = may ay

max

=

L0 ), where R =

Fx = max

ax

θ

may

=

F sin m

=

40 1

+" k(R

g= 0:5 R

y

F sin L0 ) y m R

9:81 m/s

2

k L0 1 x m R 0:5 2 x m/s J R mg = may

g=

k m

1

L0 R

y

g

J

The initial conditions are: x = 0:5 m y =

0:5 m vx = vy = 0 at t = 0 J

(b) Letting x1 = x, x2 = y, x3 = vx and x4 = vy , the equivalent …rst-order equations are x_ 1

= x3

x_ 2 = x4 0:5 x1 40 1 R

x_ 3

=

x_ 4 =

40 1

0:5 R

x2

9:81

p where R = x21 + x22 . The MATLAB program is: function problem12_84 [t,x] = ode45(@f,(0:0.02:2),[0.5 -0.5 0 0]); axes(’fontsize’,14) plot(x(:,1),x(:,2),’linewidth’,1.5) grid on xlabel(’x (m)’); ylabel(’y (m)’) function dxdt = f(t,x)


rr = 1-0.5/sqrt(x(1)^2 + x(2)^2); dxdt = [x(3) x(4) -40*rr*x(1) -40*rr*x(2)-9.81]; end end

-0.4 -0.5

y (m)

-0.6 -0.7 -0.8 -0.9 -1 -1.1 -0.5 -0.4 -0.3 -0.2 -0.1

0 0.1 0.2 0.3 0.4 0.5 x (m)

12.85 (a) The expressions for the accelerations in Prob. 12.84 are now valid only when the spring is in tension. If the spring is not in tension, the spring force is zero. Therefore, we have 8 0:5 < 2 40 1 x m/s if R > 0:5 m ax = J R : 0 if R 0:5 m 8 < 40 1 0:5 y 9:81 m/s2 if R > 0:5 m ay = J R : 2 9:81 m/s if R 0:5 m The initial conditions are:

x = y = 0:5m

vy = vy = 0 at t = 0 J

(b) MATLAB program: function problem12_85 [t,x] = ode45(@f,(0:0.02:2),[0.5 0.5 0 0]);


axes(’fontsize’,14) plot(x(:,1),x(:,2),’linewidth’,1.5) grid on xlabel(’x (m)’); ylabel(’y (m)’) function dxdt = f(t,x) rr = 1-0.5/sqrt(x(1)^2 + x(2)^2); if rr < 0; rr = 0; end dxdt = [x(3) x(4) -40*rr*x(1) -40*rr*x(2)-9.81]; end end 1

0.5

y (m)

0

-0.5

-1

-1.5 -0.4

-0.2

0

0.2

0.4

0.6

x (m)

12.86 (a) ax =

aD cos + aL sin

ay =

aD sin

aL cos

g

Substitute sin aL where v =

vx2

+

= =

vy vx cos = aD = 0:05v 2 v v 0:16!v = 0:16(10)v = 1:6v

vy2 .

vx vy + 1:6v = v v vy vx ay = 0:05v 2 1:6v 32:2 = v v The initial conditions at t = 0 are: ax =

x=y=0

0:05v 2

0:05vvx + 1:6vy ft/s 0:05vvy

vx = 60 cos 60 = 30 ft/s

1:6vx

2

J

32:2 ft/s

2

J

vy = 60 sin 60 = 51:96 ft/s


(b) Letting x1 = x, x2 = y, x3 = vx and x4 = vy , the equivalent …rst-order equations are x_ 1 x_ 3

= x3 x_ 2 = x4 = 0:05vx3 + 16x4

x_ 4 =

0:05vx4

16x3

32:2

MATLAB program: function problem12_86 [t,x] = ode45(@f,(0:0.02:1.2),[0 0 30 51.96]); printSol(t,x) axes(’fontsize’,14) plot(x(:,1),x(:,2),’linewidth’,1.5) grid on xlabel(’x (ft)’); ylabel(’y (ft)’) function dxdt = f(t,x) v = sqrt(x(3)^2 + x(4)^2); dxdt = [x(3) x(4) -0.05*v*x(3)+1.6*x(4) -0.05*v*x(4)-1.6*x(3)-32.2]; end end The two lines of output that span y = 0 are t 1.0400e+000 1.0600e+000

x1 x2 1.9377e+001 2.5868e-001 1.9388e+001 -1.9335e-001

x3 x4 9.6888e-001 -2.2523e+001 2.3210e-001 -2.2675e+001

Linear interpolation for t at y = 0: t 1:04 1:06 1:04 = 0:19335 0:25868 0 0:25868

t = 1:051 s J


x = 19:38 ft J


(c) 10

8

y (ft)

6

4

2

0 0

5

10 x (ft)

15

20

12.87 a (ft/s2) 3 60 150 170 190 t (s) -20 -40

0 0 20 -1 -2 v (ft/s) 60 40 7800

1000

600 400

0

t (s)

x (ft) 9800 9400 8400 600

t (s)

0 d = 9800 ft J


12.88 x (m) x = 10t 2 4000 0 0 v (m/s) 400

t (s)

20

0 0

t (s)

20

a (m/s2 ) 20 0 0

v

t (s)

20

= slope of x diagram

a = slope of v diagram

vj20s = a=

dx dy

= 400 m/s 20s

400 2 = 20 m/s 20

12.89 It is su¢ cient to consider vertical motion only: a=

g=

32:2 ft/s

2

Initial conditions: vjt=0 = v0 sin 60 = 0:8660v0 ft/s

yjt=0 = 0


a (ft/s2) 0 0

t1

t (s)

-32.2t1 -32.2 v (ft/s) 0.8660v0 0

0.4330v0t1 0

t1

t (s)

y (ft) 27 0 0

t1 t (s)

End conditions (t1 is the time when the ball is at its maximum height): vjt=t1 yjt=t1

= 0 0:8660v0 32:2t = 0 t1 = 0:026 89v0 2 = 27 ft 0:4330(0:026 89)v0 = 27 v0 = 48:2 ft/s J


12.90

a (ft/s2 ) 00

t1

6

−24

t (s)

−8 (t1 −6)

-8 v (ft/s) 88 64

96 384

42

0

6

x (ft) 480

t1 = 14 s t (s) 522

0 0

6

14

t (s)

From a and v diagrams 8(t1 6) = 64 t1 = 14:0 s J From x diagram xjt1 = 522 ft J


12.91 Horizontal motion (ax = 0)

Vertical motion (ay =

ax (m/s2) 0

t (s)

0

vx (m/s) 240 0

ay (m/s2) 0 0 -9.81

9:81 m/s2 )

t1 -9.81t1

t (s)

vy (m/s) 240t1 t1

0

5

00

t (s)

t (s)

1 9.81

−49.05 y (m) h

x (m) 1200 0 0

0 0

t1 t (s)

t (s)

5

End condition: xjt=t1

=

h

=

1200 m

240t1 = 1200

t1 = 5:0 s 1 [area under vy diagram] = (49:05)(5) = 122:6 m J 2

12.92 Horizontal motion: ax = 0 Vertical motion: ay =

32:2 ft/s

2


ax (ft/s2) 0 0

10

ay (ft/s2) 00

t (s)

-32.2 vx (ft/s) v0 0

t (s)

vy (ft/s) 10v0

0

10

10

00

t (s)

-1610

t (s)

-322 y (ft) 1610

x (ft) 10v0 0 0

vy vx

10 -322

0 0

10 t (s)

= tan 20 t=10

10

322 = tan 20 v0

R = xjt=10 = 10v0 = 8850 ft J

t (s)

v0 = 885 ft/s J

h = yjt=0 = 1610 ft J

12.93

y 22.6o

g

x

Acceleration diagram

ax

= g sin 22:6 = 32:2 sin 22:6 = 12:374 ft/s

ay

=

g cos 22:6 =

32:2 cos 22:6 =

2

29:73 ft/s

2

At t = 0 (initial conditions): x = 0 y = 0

vx = 260 cos 22:6 = 240:0 ft/s vy = 260 sin 22:6 = 99:92 ft/s


ax (ft/s) 12.374 0 0

t1 = 3.361 s 6.722 t (s)

ay (ft/s) 83.18 6.722

0

t (s)

-29.73 t1 = -99.92 -29.73

vx (ft/s) 323.2 240.0 0

vy (ft/s) 167.92

99.92 t (s) 0 -99.92

1892.8

x (ft) 1892.8

t (s)

y (ft) 167.92

0

t (s)

vy = 0 at t = t1 . h = 167:9 ft J

) 99:92

t (s)

0 29:73t1 = 0

R = 1893 ft J

) t1 = 3:361 s

time of ‡ight = 6:72 s J

12.94 ax (m/s2) 0

0

16.091

ay (m/s2) 00

t (s)

-9.81 vx (m/s) 190.52 0

0

t (s)

vy (m/s) 110 3066 16.091

x (m) 3066 0 0

t1 -9.81t1

16.091

110 - 9.81t1 t (s) 0

(110 - 4.905t1)t1 0 t1 t (s) y (m) (110 - 4.905t1)t1 0 0

t (s)

End condition: yjt t1 = 500 m. ) (110 The larger root is t1 = 16:091 s J R = xjt=16:091 = 3066 m J

t1

t (s)

4:905t1 )t1 = 500


12.95


12.96 2 mi = 2(5280) = 10 560 ft

45 mi/h = 45

5280 = 66 ft/s 3600

Accelerate at the maximum rate (6.6 ft/s2 ) until maximum allowable speed (66 ft/s) is reached at time t1 . Then maintain this speed until time t2 . Finally, decelerate at the maximum rate (5.5 ft/s2 ) until the train stops at time t3 . The distance traveled during this time must be 10 560 ft.

a (ft/s2) 6.6 0 6.6t1 0

t2 t1

t3

t (s)

-5.5(t3 - t2)

-5.5 v (ft/s) 66 0

0

330 66(t2 - 10) 396 t (s) 10 t2 + 12 t2

x (ft) 10 560

0 0

10

159

171

t (s)

From a and v diagrams: 6:6t1 = 66 ft/s 5:5(t3 t2 ) = 66 ft/s

) t1 = 10 s ) t3 t2 = 12 s

From v and x diagrams: 330 + 66(t2

10) + 396 = 10 560

) t2 = 159:0 s

t3 = t2 + 12 = 159:0 + 12 = 171:0 s J


12.97


12.98 a (m/s2) 8 + t1 = 11 8 t (s) 00 3 -1.8t1 -1.8 -9.6 -25 t1 -3.2 -5.0 v (m/s) 40 30.4 105.6 89.5 5.4 0

8.1 t (s)

x (m) 203.2 195.1 105.6 t (s)

v = 0 when 40

9:6

25

1:8t1 = 0

) t1 = 3:0s

After touchdown, the plane travels 203 m J

12.99 a (m/s2) 12 0.6

0.6 0 0

0.1

0.2

0.3

0.4

0.3

0.4

t (s)

v (m/s) 1.2 0.6 00 From the v diagram:

0.1

0.2

t (s)

vj0:4s = 1:2 m/s J


xj0:4s

= area under v diagram 1 = 2 (0:6)(0:1) + 4(0:6)(0:1) = 0:28 m J 3

12.100


12.101


12.102 a (m/s2) 0.38

0.10182

00

t1 = 0.5359 1.1 t (s)

−0.11282 −0.40 v (m/s) 0.10182 0.03638 0 −0.011

0.5359

0.03638 1.0718 t (s)

y (m) 0.07276 0.03638 t (s)

0 From similar triangles on the a-diagram: t1 0:38 vmax

= =

1:1 t1 = 0:5359 s 0:78 0:1018 m/s J ymax = 0:0728 m J


12.103 a (ft/s2 ) 60 t1 t2 14 t (s) 0 0273.3 -32.2( t 2 - 14) -32.2 -78.7 = -194.6 v (ft/s) 273.3 194.6 1660 1208 588 0 t (s) 20.04 9.111 14 0 y (ft) 3456 2868 1660 t (s) 0 From similar triangles on a-diagram: 14 t1 = 60 60 + 32:2

) t1 = 9:111 s

Let t2 be the time when v = 0: Therefore, 194:6 vmax = 273 ft/s J

32:2(t2

14) = 0

) t2 = 20:04 s

ymax = 3460 ft J occurring at t = 20:0 s J

12.104 v = 2x3

8x2 + 12x mm/s

dv = 2x3 8x2 + 12x 6x2 16x + 12 dx 2 = (2(8) 8(4) + 12(2)) (6(4) 16(2) + 12) = 32:0 mm/s J a=v

ajx=2

12.105 a = At + B When t = 0: a = 0 ) B = 0 4 When t = 6 ft/s: a = 8 ft/s ) 8 = A(6) ) A = ft/s3 3 Z 4 2 2 ) a = t ft/s v = a dt + C = t2 + C 3 3

When t = 0: v = v0 ) C = v0 When t = 6 s: v = 16 ft/s )

2 (36) + v0 = 16 3

) v0 =

8:0 ft/s J


12.106 Car A a = 4 ft/s2 v = 4t + C1 ft/s vjt=0 = 30 ft/s ) C1 = 30 ft/s v = 4t + 30 ft/s x = 2t2 + 30t + C2 ft xjt=0 = 400 ft ) C2 = 400 ft x = 2t2 + 30t 400 ft

Car B a = 2 ft/s2 v = 2t + C3 ft/s vjt=0 = 60 ft/s ) C3 = 60 ft/s v = 2t + 60 ft/s x = t2 + 60t + C4 ft xjt=0 = 0 ) C4 = 0 x = t2 + 60t ft

Car A overtakes car B when xA = xB : 2t2 + 30t 3t2 30t

400 = t2 + 60t 400 = 0 t = 17:58 s J

12.107 (a) x = 3t3 9t + 4 in. x = xjt=2 xjt=0 = [3(8) (b) v = x_ = 9t2

-2

) v = 0 when t = 1:0 s

9 in./s

xjt=0 = 4 in.

xjt=1 =

0

4 = 6:0 in. J

9(2) + 4]

2 in.

xjt=2 = 10 in.

4

x (in.) 10

d = 6 + 12 = 18 in. J

12.108 Fall of the stone: a = 32:2 ft/s ) v = 32:2t + C1 ft/s ) y = 16:1t2 + C1 t + C2 ft 2

When t = 0: v = x = 0 ) C1 = C2 = 0 Let t1 be the time of fall and h the depth of the well. ) h = 16:1t21 ft Travel of the sound: Let t2 be the time for sound to travel the distance h: ) h = 1120t2 ft t 1 + t2 = 3 16:1t21 = 1120t2

) t2 = 3

16:1t21 = 1120(3

t1 )

t1 t1 = 2:881 s

h = yjt=2:881s = 16:1(2:881) = 133:6 ft J 2


12.109 2

12t 6t2 ft/s Z v = a dt = 6t2 2t3 + C1 ft/s Z x = v dt = 2t3 0:5t4 + C1 t + C2 ft a =

) C1 = C2 = 0

When t = 0: x = v = 0:

) x = 2t3

) v = 6t2

0:5t4 ft

2t3 ft/s

(a) xjt=0 = 2(5)3

x = xjt=5

0:5(5)4

62:5 ft J

0=

(b) When v = 0: 6t2 xjt=0 = 0

xjt=3 = 2(3)

2t3 = 0 3

t = 3:0 s 4

0:5(3) = 13:5 ft xjt=5 =

0

-62.5

13.5

62:5 ft

x (ft)

d = 2(13:5) + 62:5 = 89:5 ft J

12.110 dv = dt

cv

2

1 dv = v2

c dt

Z

Initial condition: vjt=0 = 800 ft/s When v = 400 ft/s: 1 v

1 800

1 dv = v2 )

1 = 400

1 = v

ct + C1

1 = v

1 = 800

C1

0:8t

t = 1:5625

)

10

ct + C1

0:8t 3

1 1 0:8t + v= 800 0:8t + 1=800 Z dt 1 1 x = + C2 = ln 0:8t + 0:8t + 1=800 0:8 800 1 1 Initial condition: xjt=0 = 0 ) 0 = ln + C2 0:8 800 1 1 C2 = ln = 8:356 ft 1 0:8 800 1 1 x = ln 0:8t + + 8:356 0:8 800

1 800

s J

=

+ C2


When v

=

x =

400 ft/s t = 1:5625 1 ln 0:8(1:5625 0:8

10 10

3

3

)+

s : 1 + 8:356 = 0:867 ft J 800

12.111 75e 0:05t ft/s Z x = 75e 0:05t dt = v

Initial condition:

=

1500e

0:05t

+ C ft

xjt=0 = 0 ) C = 1500 ft x = 1500(1 e 0:05t ) ft

Setting v = 5 ft/s and solving for t: 5 = 75e

0:05t

h

xjv=5 ft/s = 1500 1

t = 54:16 s J i e 0:05(54:16) = 1400 ft J

12.112

may max mg FBD Fx Fy

MAD

= max = may vx =

Initial condition: vx jt=0 = v0 cos 50

ax = 0 ay = g = Z

9:81 m/s

2

ax dt = C1

) C1 = v0 cos 50 = 0:6428v0 vx = 0:6428v0 Z x = vx dt = 0:6428v0 t + C2

Initial condition: xjt=0 = 0

) C2 = 0 x = 0:6428v0 t Z vy = ay dt = 9:81t + C3


Initial condition: vy jt=0 = v0 sin 50 ) C3 = v0 sin 50 = 0:7660v0 vy = 9:81t + 0:7660v0 Z y = vy dt = 4:905t2 + 0:7660v0 t + C4

Initial condition: yjt=0 = 0

) C4 = 0

4:905t2 + 0:7660v0 t

y=

At point B: x = 18 m ) 0:6428v0 t = 18 v0 t = 28:0 m y = 18 m ) 4:905t2 + 0:7660(28:0) = 18 t = 0:8384 s 28:0 = 33:4 m/s J v0 = 0:8384

12.113

12.114 F = (50

+"

a = v

=

y

=

F m Z

Z

g=

(50

t)

103 N

t) 103 1400

a dt = 25:90t v dt = 12:95t2

9:81 = 25:90

0:7143t m/s

2

0:3571t2 + C1 0:11903t3 + C1 t + C2

Initial conditions: y = v = 0 when t = 0. ) C1 = C2 = 0: yjt=20 = 12:95(20)2

0:11903(20)3 = 4230 m J


12.115

12.116

12.117 From Eqs. (d) and (e) of Sample Problem12.11: x = v0 t cos = 8t cos 30 = 6:928t 1 2 1 y = gt + v0 t sin = (9:81)t2 + 8t sin 30 = 2 2

4:905t2 + 4t


At the landing point y

=

x =

4:905t2 + 4t = 6:928t tan 20 t = 1:330 s J 9:214 6:928(1:330) = 9:214 m d= = 9:81 m J cos 20

x tan 20 4:905t = 6:522

12.118

0.3 lb y F6 FBD N

45

x

0.3 a 32.2

3

MAD

When x = 6 in., the elongation of the spring and the spring force are p = 32 + 62 3 = 3:708 in. F = k = 5(3:708) = 18:540 lb Fx

6 0:3 p F = a 32:2 45 2 1780 ft/s J

= ma

a =

6 0:3 p (18:540) = a 32:2 45

12.119


12.120

12.121 From Eqs. (d) and (e) of Sample Problem 12.11: x = v0 t cos

0

y=

1 2 gt + v0 t sin 2

(a) Let t = t1 when the ball hits the fairway at y = 1 (9:81)t21 + 45t1 sin 40 2 ) R = 45(6:162) cos 40 = 212 m J

)

8=

0

8 m, x = R. t1 = 6:162 s

(b) At t = 6:162 s: vx vy

= x_ = v0 cos 0 = 45 cos 40 = 34:47 m/s = y_ = gt + v0 sin 0 = 9:81(6:162) + 45 sin 40 = p v = 34:472 + 31:522 = 46:7 m/s J

31:52 m/s


12.122


12.3 x = v

6 1

= x_ = 6

a = v_ =

(a) xmax

=

vmax

=

e

t=2

m

1 t=2 e = 3e t=2 m/s 2 1 t=2 2 e = 1:5e t=2 m/s 3 2

6 m at t = 1 J

3 m/s and jajmax = 1:5 m/s both occurring at t = 0 J 2

(b) When x = 3 m: 3 = 6(1 e t=2 ) t = 2 ln(0:5) = 1:3863 s J v

=

3(0:5) = 1:5 m/s J

t=2

e a=

= 0:5

1:5(0:5) =

0:75 m/s

2

J

12.4 x = t3 6t2 v = x_ = 3t2 a = v_ = 6t

32t in. 12t 32 in./s 12 in./s

2

At t = 10 s: x = 103 6(102 ) 32(10) = 80 in. J v = 3(102 ) 12(10) 32 = 148 in./s J a =

6(10)

2

12 = 48 in./s

J

Reversal of velocity occurs when v = 0 (t 6= 0): v = x =

3t2 12t 32 = 0 5:8303 6(5:8302 )

t = 5:830 s 32(5:830) = 192:3 in.

At t = 10 s the distance travelled is s = 192:3 + (193:3 + 80) = 466 in. J

−192.3 in.

0

5.83 s

80 in. 10 s


12.5 (a) x = t2

t3 ft 90

(b) a = v_ = 2

v = x_ = 2t

t2 ft/s 30

xmax = 602

603 = 1200 ft J 90

t ft/s2 15

v = 0 when t = 60 s

a = 0 when t = 30 s

vmax = 2(30)

302 = 30 ft/s J 30

12.6

12.7 x = 3t2

12t in.

v = x_ = 6t

12 in./s

(a) The bead leaves the wire when x = 40 in. 3t2

t = 6:16 s J

12t = 40

(b) Reversal of velocity occurs when v = 0 (t 6= 0): v = x =

6t 10 3(2:02 )

t = 2:0 s 12(2:0) = 12:0 in.

The distance travelled is s = 2(12) + 40 = 64:0 in. J

−12 in. 0

40 in. 6.16 s

2.0 s


12.8 x = y

=

4t2 2 mm x2 16t4 16t2 + 4 4t4 = = 12 12

4t2 + 1 mm 3

When t = 2 s: vx vy v

= x_ = 8t = 8(2) = 16 mm/s 16t3 8t 16(2)3 8(2) = y_ = = = 37:33 mm/s 3 3 q p = vx2 + vy2 = 162 + 37:332 = 40:6 mm/s J 2

ax

= v_ x = 8 mm/s 48t2 8 48(2)2 8 2 ay = v_ y = = = 61:33 mm/s 3 3 q p 2 a = a2x + a2y = 82 + 61:332 = 61:9 mm/s J

12.9

12.10 y = 50 x=

2t s

6 6 = in. y 50 2t

vy = y_ = vx = x_ =

2 in./s 3

(25

t)2

ay = v_ y = 0

in./s

ax = vx =

6 (25

3

t)

At t = 20 s: vx

=

ax

=

3 20)2

(25 6 (25

3

20)

= 0:12 in./s = 0:048 in./s

vy = 2

2 in./s

ay = 0

2j in./s J

v = 0:12i 2

a = 0:048i in./s

J


12.11

12.12 x =

15

2t2 m

y

15

10t + t2 m

=

(a) At t = 0:

v=

(b) At t = 5 s:

v=

vx = x_ =

4t m/s

vy = y_ =

10j m/s J

ax = v_ x =

10 + 2t m/s a=

20i m/s J

a=

4 m/s

2

ay = v_ y = 2 m/s

2

4i + 2j m/s2 J 4i + 2j m/s2 J

12.13 x = y (a) (b) (c)

=

58t m

vx = x_ = 58 m/s

78t

2

4:91t m

vy = y_ = 78

9:82t m/s

ay = v_ y =

9:82 m/s

2

a = 9:82j m/s2 J vjt=0 = 58i + 78j m/s J y = h when vy = 0: vy h

(d)

ax = v_ x = 0

= 78 9:82t = 0 t = 7:943 s = 78(7:943) 4:91(7:9432 ) = 310 m J

x = L when y =

140 m:

y = 78t 4:91t2 = 140 t = 17:514 s L = 58(17:514) = 1016 m J


12.14 y=

x2 1000

dy x = dx 500

)2 v0 x d / (dy dy/dx 1+

1 vx

1

= v0 q

2

1 + (dy=dx)

vy

dy=dx

= v0 q

2

1 + (dy=dx)

When x = 100 m:

vx

=

vy

=

v

=

v0

=p

1+

(x=500)2

x=500

=p

1+

(x=500)2

=p

500v0 5002 + x2

=p

xv0 5002 + x2

500(6) = 5:883 m/s 5002 + 1002 100(6) p = 1:1767 m/s 5002 + 1002 5:88i + 1:177j m/s J p

ax

= v_ x =

dvx dx dvx = vx = dx dt dx

ay

= v_ y =

dvy dx dvy v = vx = 3=2 x dx dt dx (5002 + x2 )

500xv0

v 3=2 x

(5002 + x2 ) 5002 v0

When x = 100 m: ax

=

ay

=

a =

500(100)(6) (5002

3=2 1002 )

+ 5002 (6)

3=2

(5002 + 1002 )

(5:883) =

0:013 31 m/s

(5:883) = 0:0666 m/s

0:013 32i + 0:666j m/s

2

2

2

J


12.15

12.16


12.17

12.18


12.19

12.20


12.21

12.22

12.23

12.24 x = R cos y = R sin vy

v0 R cos v0 vx = ( R sin ) = R cos

= v0 )

vx = x_ = ( R sin ) _ vy = y_ = (R cos ) _

yields

_=

v0 tan


ax = v_ x =

_=

v0 sec2

With R = 6 ft, v0 = 2:5 ft/s and vy v

v0 R cos

v0 sec2

= 60 we get

= 2:5 ft/s vx = 2:5 tan 60 = = 4:33i + 2:5j ft/s J

ay

=

a =

0

v02 sec3 R

=

2:52 sec3 60 = 6 2 8:33i ft/s J ax =

4:330 ft/s

8:333 ft/s

2

12.25 _ = 1200 rev 1:0 min r

=

2 rad 1:0 rev

1:0 min = 125:66 rad/s 60 s

55 + 10 cos + 5 cos 2 dr _ v = r_ = = ( 10 sin d dv _ a = v_ = = ( 10 cos d jajmax = 30(125:66)2 = 474 000

mm 10 sin 2 ) (125:66) mm/s 20 cos 2 )(125:66)2 mm/s 2

2

mm/s = 474 m/s (at

2

= 0) J

*12.26


12.27 T

ma =

mg FBD

MAD

v = 4t m/s

a = v_ = 4 m/s

F = ma + "

T

2

mg = ma

T = m(g + a) = 50(9:81 + 4) = 691 N J

12.28 y

mg

x ma

=

F = µk N

FBD N

MAD 1000 m/km 3600 s/h

v0 = 100 km/h = 100 km/h Fy

=

0 +"

Fx

= ma v

N

kN

Z

=

x =

k gt

1 2

v dt =

kN

)a=

= ma

a dt =

Z

) N = mg

mg = 0

+ !

= 27:78 m/s

m

=

kg

+ C1

2 k gt

+ C1 t + C2

When t = 0 (initial conditions): x=0 )x=

) C2 = 0 v = v0 ) C1 = v0 1 gt2 + v0 t v= k gt + v0 2 k

When v = 0: k gt

x =

1 2

)t=

+ v0 = 0 v0 kg

kg

2

+ v0

v0 kg

v0 kg =

v02 2

kg

2

=

27:78 = 60:5 m J 2(0:65)(9:81)


12.29 y

mg 5o

x =

F = µk N FBD

ma MAD

N

v0 = 100 km/h = 27:78 m/s = 0 +" N mg cos 5 = 0 ) N = mg cos 5 + = ma ! k N + mg sin 5 = ma N k )a = + g sin 5 = (sin 5 k cos 5 )g m 2 = (sin 5 0:65 cos 5 )9:81 = 5:497 m/s Fy Fx

v

=

x =

Z Z

a dt =

5:497t + C1

v dt =

2:749t2 + C1 t + C2

When t = 0 (initial conditions): x=0

) C2 = 0

v = v0

2:749t2 + 27:78t m 5:497t + 27:78 m/s

x = v = When v = 0:

5:497t + 27:78 = 0 x=

) C1 = v0 = 27:78 m/s

) t = 5:054 s

2:749(5:054)2 + 27:78(5:054) = 70:2 m J

12.30 a = v

=

x =

F 1:2t = = 0:1 Zm Z

12t m/s

2

ax dt =

6t2 + C1 m/s

vx dt =

2t3 + C1 t + C2 m

When t = 0 (initial conditions): x = 0 ) C2 = 0

v = 64 m/s ) C1 = 64 m/s


)x=

2t3 + 64t m

v=

6t2 + 64 m/s

When t = 4 s: x=

2(4)3 + 64(4) = 128 m

When v = 0 : 6t2 + 64 = 0 x=

t = 3:266 s

2(3:266)3 + 64(3:266) = 139:35 m

Distance traveled: d = 2(139:35)

128 = 150:7 m J

139.35

0

x (m)

128

12.31 a = dt

=

p p 0:06 v F 2 = = 5 v m/s m 0:012 Z dv dv dv 2p p = = p t= v + C1 a 5 5 v 5 v

Given v = 0:25 m/s when t = 0:8 s: 0:8 =

2p 0:25 + C1 5

C1 = 0:6 s

2p v + 0:6 s 5 2 v = (2:5t 1:5) = 6:25t2 7:5t + 2:25 m/s Z 6:25 3 7:5 2 x = v dt = t t + 2:25t + C2 3 2 t

=

Initial condition: x = 0 when t = 0 xjt=1:2s =

6:25 (1:2)3 3

) C2 = 0

7:5 1:22 + 2:25(1:2) = 0:90 m J 2

12.32

T cv2 FBD

=

ma MAD


ma = T

cv 2

FD = T

a=

cv 2 m

T

Z

Z m v v dv + C = x= dv = m ln T cv 2 + C a T cv 2 2c Initial condition: v = 0 at x = 0: m m ln (T ) + C C= ln (T ) 0 = 2c 2c m m m T ) x= ln(T cv 2 ) + ln(T ) = ln 2c 2c 2c (T cv 2 ) Solve for v: T (T cv 2 ) v2

2c x m

=

exp

=

T 1 c

cv 2 = T exp s

T 2c x m

exp

v=

Terminal velocity: v1 = lim v(x) = x!1

r

2c x m T 1 c

exp

2c x m

J

T J c

12.33 4t 4 F 2 = = t 1 m/s m 4 Z 1 = a dt = t2 t + C1 2 Z 1 2 1 t + C1 t + C2 = v dt = t3 6 2

a = v y

When t = 0 (initial condition): y = 0 ) C2 = 0 )y=

1 3 t 6

When t = 8 s: y= When v = 0:

y=

1 2 t 2

1 3 (8) 6 1 2 t 2

1 3 (5:123) 6

8 m/s ) C1 =

v=

t

1 2 t 2

8t m

v=

1 2 (8) 2

8 (8) =

t

t = 5:123 s

1 2 (5:123) 2

8 (5:123) =

d = 2(31:70)

8 m/s

10:67 m

8=0

Distance traveled:

8 m/s

31:70 m

10:67 = 52:7 m J

-31.70

0

y (m)

-10.67


12.34

12.35

12.36 y mg 20o NA

ma

x =

FA = 0.4 NA FBD

MAD

Assume impending sliding (FA = 0:4NA ) Fy = 0 + "

NA cos 20

0:4NA sin 20

mg = 0

NA = 1:2455 mg Fx = max

+ !

NA sin 20 + 0:4NA cos 20 = ma

1:2455 mg(sin 20 + 0:4 cos 20 ) = ma

a = 0:894g J

12.37 Let y be measured up from the base of the cli¤.

y

ma x

=

FBD

MAD

mg


Fy = ma + " a=

dv v dy

) v dv =

1 2 v 2

At impact y = 0

)

1 2 v 2

1 2 v = 2

g dy

Initial condition: v = v0 when y = h: ) C = )

)a=

mg = ma

gy + C

1 2 v + gh 2 0

v02 = g(h )v=

v02 = gh

g

y) p

v02 + 2gh J

12.38

F

=

FBD F 1 2 v 2

= ma F0 m

=

e

x=b

Initial condition

:

1 2 v 2

=

v

vjx=1:8 ft =

s

MAD

F F0 x=b F0 = e v dv = e m m m F0 b x=b dx = e +C m

a= Z

ma

v = 0 at x = 0

)C=

x=b

dx

F0 b m

F0 b 1 e x=b m r 2F0 b 1 e x=b = m

2(1576)(2) 1 6:62 10 3 =32:2

e

1:8=2:0

= 4270 ft/s J

12.39

20o 18 lb

6t y

FBD

18 a g

= x MAD

N


Fx

= ma 32:2 (6t Z18

a = v

6t

=

Z

x =

18 sin 20 =

18 a 32:2

18 sin 20 ) = 10:733t

11:013 ft/s

a dt = 5:367t2

11:013t + C1 ft/s

v dt = 1:789t3

5:507t2 + C1 t + C2 ft

2

) C1 = C2 = 0

Initial conditions: x = v = 0 at t = 0 (a) When v = 0:

v = 5:367t2 11:013t = 0 t = 2:052 s x = 1:789 2:0523 5:507(2:0522 ) = 7:73 ft J (b) When x = 0: x = 1:789t3 5:507t2 = 0 t = 3:078 s 2 v = 5:367(3:078 ) 11:013(3:078) = 16:95 ft J

12.40 x 4

ma 3

P = 8 - 2t

FBD mg N

= 4 P 5

Fx = ma + % a=

4P 5m v

3 4 8 2t g= 5 5 5=32:2 =

x = Initial conditions: v = When x = 0 : 10:95t2

Z

Z

MAD 3 mg = ma 5

3 (32:2) = 21:90 5

a dt = 21:90t v dt = 10:95t2

10:304t ft/s

5:152t2 + C1 1:7173t3 + C1 t + C2

10 ft/s, x = 0 at t = 0. ) C1 = 1:7173t3

v = 21:90(1:1049 )

2

10t = 0

5:152(1:1049 )2

10 ft/s

C2 = 0

t = 1:1049 s J 10 = 7:91 ft/s J


12.41

(mA + mB)g A

y

θ

x

=

B

(mA + mB) a

NB FBD

MAD

Fx = ma + &

(mA + mB )g sin = (mA + mB )a a = g sin

Assume impending slipping between A and B:

mA g y

=

A

x

Fx = ma + &

F =µsNA NA FBD (mA g s

θ ma A MAD

NA ) sin +

s NA

cos = mA g sin

= tan J

12.42 (a) Assume impending sliding of crate to the left.

y 20o 60 lb 60 a g

x FA = 0.3NA NA FBD Fy

=

Fx

= max

MAD

0 NA cos 20 NA = 67:35 lb

0:3NA sin 20

NA sin 20 + 0:3NA cos 20

67:35(sin 20 + 0:3 cos 20 ) a = 11:54 ft/s

2

60 cos 20 = 0 60 sin 20 =

60 sin 20 =

60 a 32:2

60 a 32:2

J


(b) Assume impending sliding of crate to the right.

y

20 o 60 lb x 60 a g

FA = 0.3NA NA FBD Fy

=

Fx

= max

MAD

0 NA cos 20 + 0:3NA sin 20 NA = 54:09 lb NA sin 20

54:09 (sin 20

0:3NA cos 20

0:3 cos 20 )

a = 9:27 ft/s

2

60 cos 20 = 0 60 sin 20 =

60 sin 20 =

60 a 32:2

60 a 32:2

J

12.43

mg

=

ma

0.2 N N FBD

MAD

Fy = 0 N mg = 0 N = mg Fx = 0 0:2N = ma 0:2mg = ma 2 a = 0:2g = 1:962 m/s Z 1 2 v dv = a dx v = 1:962 dx = 1:962x + C 2 1 2 Initial condition: vjx=0 = 6 m/s (6 ) = C C = 18 m2 =s2 2 ) v

=

1 2 v = 1:962x + 18 2 0 when 1:962x + 18 = 0

x = 9:17 m J


12.44 W F

ma

=

µkΝ N FBD Fy Fx

= 0 = 0 F a = =

+" N W =0 N = W = 3000 lb +! F k N = ma 0:2t N 1000e 0:05(3000) k = m 3000=32:2 0:2t

10:733e

v

= =

MAD

Z

1:61 ft/s

a dt = 53:67e

Z

2

(10:733e

0:2t

0:2t

1:61)dt

1:61t + C ft/s

Initial condition: v = 0 when t = 0. ) C = 53:67 ft/s v = 53:67(1

e

0:2t

)

1:61t ft/s

Maximum velocity occurs at t = 4 s (end of powered travel) h i vmax = 53:67 1 e 0:2(4) 1:61(4) = 23:1 ft/s J

12.45

mg

kx

30o

µk N

x

y

=

N FBD Fy Fx

= 0 = ma

a =

MAD

mg cos 30 N =0 mg sin 30 kN

a = g (sin 30 9:81(sin 30

ma

k

N = mg cos 30 kx = ma

k x m 25 0:3 cos 30 ) x = 2:356 2:5 cos 30 )

10x m/s

2


v dv 1 2 v 2

Z

= a dx

1 2 v = 2

=

5x2 + C

2:356x

Initial condition:

(2:356

)C=0

vjx=0 = 0

) x = 0:471 m J

2

v = 0 when 2:356x

10x)dx

5x = 0

12.46 ma

40o P N mg µN FBD

y

= x MAD ) N = P sin 40 (P sin 40 ) = ma

= 0 +! P sin 40 N =0 = may + " P cos 40 mg

Fx Fy

P (cos 40 m When motion impends: a = 0 and = a=

0=

P (cos 40 5

sin 40 ) s

0:5 sin 40 )

= 0:5 9:81

When collar begins to slide: P = 110:31 N and a=

110:31 (cos 40 5

g

0:4 sin 40 )

P = 110:31 N =

k

= 0:4

9:81 = 1:418 m/s

2

J

12.47


12.48

12.49


12.50 v0 = 10 km/h = 10

1000 = 2:778 m/s 3600

mg kx

x = ma MAD

FBD N Fx = max

a =

dv v dx 1 2 v = 2

kx = ma

v dv = a dx

a=

v dv =

k x m k x dx m

k 2 x +C 2m ) C1 = 12 v02

Initial condition: v = v0 when x = 0: ) v 2 = v02

k 2 x m

Stopping condition: v = 0 when x = 0:5 m: 2:7782

k 18

103

(0:5)2 = 0

k = 5:56

105 N/m J

12.51


12.52

*12.53 F = ma dv = dx Z x=

a=v

T

cD v 2 = ma

1 (T + cD v 2 ) m mv dv = T + cD v 2

mv dv = dx T + cD v 2 m T + cD v 2 ln +C 2cD cD

Initial condition: v = v0 when x = 0: ) C = )x=

m T + cD v02 ln 2cD cD

m T + cD v02 ln 2cD T + cD v 2


2500 = 77:64 slugs 32:2

m=

v0 = (90)

5280 = 132 ft/s 3600

When v = 0: x=

m T + cD v02 77:64 450 + 0:006(132)2 ln = ln = 1352 ft J 2cD T 2(0:006) 450

*12.54 .

x 5 lb

30o

4 ft

θ

=

8 lb

FBD

Ν

MAD

Fx = ma + .

5 sin 30

32:2 (5 sin 30 5

8 cos ) = 16:1

)a=

cos = p a=v

5 a 32.2

dv = 16:1 dx

51:52x p x2 + 16

v dv =

Initial condition: v = 0 when x = 0: When a = 0 (v = vmax ):

1 2 v 2 max

= =

5 a 32:2

51:52 cos ft/s

2

x x =p x2 + 42 x2 + 16

1 2 v = 16:1x 2

16:1

8 cos =

2

51:52x p x2 + 16

dx

p 51:52 x2 + 16 + C

) C = 51:51(4) = 206:0 (ft/s)2

51:52x p =0 x2 + 16

16:1(1:3159)

16:1

) x = 1:3159 ft J p 51:52 (1:3159)2 + 16 + 206:0

10:241 (ft/s) p vmax = 2(10:241) = 4:53 ft/s J


*12.55 F = ma + # a= t=

1 ln cd =m

dv =g dt g

mg

cd v m

dt =

(cd =m)v +C = cd =m

Initial condition: v = 0 when t = 0: )t=

)C=

dv (cd =m) v

g

m ln cD m ln cD

mg +v +C cD mg cD

m mg=cD ln cD mg=cd v

When v = v1 (terminal velocity), a = 0: mg When v = 0:9v1 = 0:9 : cd t=

cD v = ma

) v1 =

mg cd

m 1 m m = ln ln 10 = 2:30 J cd 1 0:9 cD cd

12.56 v 7 2 + m/s 4 16 Letting x1 = x, x2 = v, the equivalent …rst-order equations and initial conditions are 7 x2 x_ 1 = x2 x_ 2 = + 4 16 x1 (0) = 0 x2 (0) = 20 m/s a=

The MATLAB program that integrates the equations is function problem12_56 [t,x] =ode45(@f,[0:0.5:25],[0 20]); printSol(t,x) function dxdt = f(t,x) dxdt = [x(2) -7/4 + x(2)/16]; end end The 2 lines of output that span the instant where v = 0 are t 2.0000e+001 2.0500e+001

x1 x2 2.4124e+002 7.7255e-002 2.4105e+002 -8.0911e-001

By inspection of output, the stopping distance is x = 241 m J


12.57 v2 10 Letting x1 = x, x2 = v, the equivalent …rst-order equations and initial conditions are x22 x_ 1 = x2 x_ 2 = 10 x1 (0) = 0 x2 (0) = 20 in. a=

The MATLAB program that integrates the equations is function problem12_57 [t,x] =ode45(@f,(0:0.01:0.51),[0 20]); printSol(t,x) function dxdt = f(t,x) dxdt = [x(2) -x(2)^2/10]; end end The 3 lines of output that span the instant where v = 10 in./s are t 4.9000e-001 5.0000e-001 5.1000e-001

x1 6.8310e+000 6.9315e+000 7.0310e+000

x2 1.0101e+001 1.0000e+001 9.9010e+000

By inspection, when v = 10 in./s, t = 0:500 s J

12.58 Letting x1 = x, x2 = v, the equivalent …rst-order equations and initial conditions are x_ 1 = x2 x_ 2 = 32:2(1 32:3 10 6 x22 ) x1 (0) = 0

x2 (0) = 0

The MATLAB program that integrates the equations is function problem12_58 [t,x] =ode45(@f,(0:0.1:7),[0 0]); printSol(t,x) function dxdt = f(t,x) dxdt = [x(2) 32.2*(1-32.3e-6*x(2)^2)]; end end The 2 lines of output that span the instant where v = 100 mi/h = 146:67 ft/s are


t 6.5000e+000 6.6000e+000

x1 5.6244e+002 5.7710e+002

x2 1.4612e+002 1.4710e+002

Linear interpolation: 6:6 147:10

6:5 t 6:5 = 146:12 146:67 146:12

t = 6:56 s J

12.59 p

5

2

x in./s +9 Letting x1 = x, x2 = v, the equivalent …rst-order equations and initial conditions are ! 5 x1 x_ 1 = x2 x_ 2 = 5796 1 p 2 x1 + 9 a=

5796 1

x2

x1 (0) = 8 in.

x2 (0) = 0

The MATLAB program that integrates the equations is function problem12_59 [t,x] =ode45(@f,(0:0.001:0.042),[8 0]); printSol(t,x) function dxdt = f(t,x) dxdt = [x(2) -5796*(1 - 5/sqrt(x(1)^2 + 9))*x(1)]; end end The 2 lines of output that span the instant where x = 0 are t x1 x2 3.9000e-002 3.7400e-002 -2.2275e+002 4.0000e-002 -1.8542e-001 -2.2304e+002 By inspection, when x = 0, v = 223 in./s J

12.60 a=

32:2 1

6:24

10

4 2

v exp( 3:211

10

5

x) ft/s

2

(a) Letting x1 = x, x2 = v, the equivalent …rst-order equations and initial conditions are x_ 1 = x2

x_ 2 =

32:2 1

6:24

x1 (0) = 30 000 ft

10

4 2 x2 e 3:211 10

5

x1

x2 (0) = 0

The MATLAB program that integrates the equations is


function problem12_60 [t,x] = ode45(@f,(0:0.2:10),[30000 0]); printSol(t,x) axes(’fontsize’,14) plot(x(:,1),x(:,2),’linewidth’,1.5) grid on xlabel(’x (ft)’); ylabel(’v (ft/s)’) function dxdt = f(t,x) dxdt = [x(2) -32.2*(1-6.24e-4*x(2)^2*exp(-3.211e-5*x(1)))]; end end The 3 lines of output that span the instant where v = vmax are t 7.4000e+000 7.6000e+000 7.8000e+000

x1 x2 2.9612e+004 -6.4384e+001 2.9599e+004 -6.4385e+001 2.9586e+004 -6.4384e+001

By inspection, vmax = 64:4 ft/s J at x = 29 600 ft J (b) 0 -10

v (ft/s)

-20 -30 -40 -50 -60 -70 2.94

2.95

2.96

2.97 x (ft)

2.98

2.99

3 x 10

4


12.61 (a)

mg y P = kx

=

v F = µN

x

N FBD

ma

MAD

The FBD shown is valid only if v > 0 (block is moving to the right.) If v < 0 (block is moving to the left), the direction of the friction force F must be reversed. Fy Fx ) =

a=

= 0 +" = ma +!

k x m

30 x 1:6

N

mg = 0 ) N = mg kx N sign(v) = ma

N sign(v) = m

0:2(9:81) sign(v) =

k m

g sign(v)

18:75x

1:962 sign(v) m/s

2

J

(b) With the notation x1 = x and x2 = v, the equivalent …rst-order equations are x_ 1 = x2 x_ 2 = 18:75x1 1:962 sign(x2 ) subject to the initial conditions x1 = 0, x2 = 6 m/s at t = 0. The corresponding MATLAB program is: function problem12_61 [t,x] =ode45(@f,[0:0.02:1.2],[0 6]); printSol(t,x) axes(’fontsize’,14) plot(x(:,1),x(:,2),’linewidth’,1.5) grid on xlabel(’x (m)’); ylabel(’v (m/s)’) function dxdt = f(t,x) dxdt = [x(2) -18.75*x(1)-1.962*sign(x(2))]; end end The block stops twice during the period 0 < t < 1:2 s. Only the lines of output that span the instant where v = 0 are shown below. t 3.4000e-001 3.6000e-001

x1 x2 1.2844e+000 1.3160e-001 1.2822e+000 -3.5272e-001


Linear interpolation: 3:6 3:4 t1 3:4 = 0:35272 0:13160 0 0:13160

t1 = 3:45 s J

1.0600e+000 -1.0745e+000 -2.1421e-001 1.0800e+000 -1.0745e+000 2.0038e-001 Linear interpolation: t2 1:06 1:08 1:06 = 0:20038 ( 0:21421) 0 ( 0:21421)

t2 = 1:070 s J

(c) 6 4

v (m/s)

2 0 -2 -4 -6 -1.5

-1

-0.5

0 x (m)

0.5

1

1.5

12.62 (a)

y

mg P(t)

kx

=

x

N FBD Fx = ma P (t)

= =

+ !

ma

MAD P (t)

kx = ma

25t N when t 25 N when t

a=

P (t) m

k x m

1s 1s

(12:5t + 12:5) + (12:5t

12:5) sgn (1

t)


) =

(12:5t + 12:5) + (12:5t 12:5) sgn (1 t) 25 x 2 2 2 6:25 [(t + 1) + (t 1) sgn (1 t)] 12:5x m/s J

a=

(b) With the notation x1 = x and x2 = v, the equivalent …rst-order equations are x_ 1 = x2 x_ 2 = 6:25 [(t + 1) + (t 1) sgn (1 t)] 12:5x1 subject to the initial conditions x1 = x2 = 0 at t = 0. The corresponding MATLAB program is: function problem12_62 [t,x] = ode45(@f,(0:0.05:3),[0 0]); printSol(t,x) axes(’fontsize’,14) plot(x(:,1),x(:,2),’linewidth’,1.5) grid on xlabel(’x (m)’); ylabel(’v (m/s)’) function dxdt = f(t,x) dxdt = [x(2) 6.25*(t+1 + (t-1)*sign(1-t)) - 12.5*x(1)]; end end Below are partial printouts that span vmax and xmax . t 8.0000e-001 9.0000e-001 1.0000e+000

x1 7.1290e-001 9.1150e-001 1.1087e+000

x2 1.9518e+000 2.0003e+000 1.9205e+000

By inspection, vmax = 2:00 m/s J 1.3000e+000 1.4000e+000 1.5000e+000

1.5271e+000 6.0188e-001 1.5535e+000 -8.0559e-002 1.5113e+000 -7.5304e-001

By inspection, xmax = 1:554 m J


(c) 2 1.5 1

v (m/s)

0.5 0 -0.5 -1 -1.5 -2 0

0.5

1 x (m)

1.5

2

12.63 a = 80

16v 1:5 ft/s

2

With the notation x1 = x and x2 = v, the equivalent …rst-order equations are x_ 1 = x2

x_ 2 = 80

16x1

subject to the initial conditions x1 = x2 = 0 at t = 0. The corresponding MATLAB program is: function problem12_63 [t,x] =ode45(@f,[0:0.005:0.15],[0 0]); printSol(t,x) function dxdt = f(t,x) dxdt = [x(2) 80 - 16*x(2)^1.5]; end end Only the 4 lines of output that span x = 0:25 ft are shown below. t 1.0500e-001 1.1000e-001 1.1500e-001 1.2000e-001

x1 2.2384e-001 2.3822e-001 2.5263e-001 2.6709e-001

x2 2.8693e+000 2.8794e+000 2.8876e+000 2.8944e+000

Use linear interpolation to …nd v at x = 0:25 ft: 2:8876 2:8794 v 2:8794 = 0:25263 0:23822 0:25 0:23822

v = 2:89 ft/s J


12.64 y=

x2 400

y_ =

x x_ = 200

x v0 200

y• =

1 v0 x_ = 200

v02 200

mg

= mv02/200

N FBD Fy = may

MAD N

mg =

m

v02 200

v2 Contact is lost when N = 0: g = 0 200 p p v0 = 200g = 200(32:2) = 80:3 ft/s J


Chapter 13 13.1 2 1

5

3 4

13.2 2

2

at = 0:8 m/s a = 1:5 m/s q p 2 an = a2 a2t = 1:52 0:82 = 1:2689 m/s an

=

v

=

v2

)v=

15:930

m s

p

an =

1 km 1000 m

p 1:2689(200) = 15:930 m/s

3600 s = 57:3 km/h J 1h

13.3

athrust g

an 30o

a

an = g cos 30 = 32:2 cos 30 = 27:89 ft/s an =

v2

)

=

2

v2 8002 = 22 900 ft J = an 27:89

13.4 At point A: at = 0

v

=

v

=

p

an = ft 109:89 s

an =

v2

p

(0:25 32:2)(1500) = 109:89 ft/s 1 mi 3600 s = 74:9 mi/h J 5280 ft 1h


13.5 At point B: v = vx = v0 cos an =

v

2

2

=

v = an

an = g 2 vA

cos2 g

J

13.6 at

dv ds dv 1 2 = v v dv = at ds v = at s + C ds dt ds 2 Initial condition: v = 0 at s = 0 ) C = 0 v2 at = 2s

= )

At point B:

a=

at

=

an

=

q

2 52 vB 2 = = 1:989 ft/s 2( R) 2(2 ) 2 vB 52 2 = = 12:5 ft/s R 2

a2n + a2t =

13.7

y

=

p 2 12:52 + 1:9892 = 12:66 ft/s J

y0 =

4 cos x 0 2

4 sin x

3=2

at

4 cos x

3=2

1 + (y )

1 + 16 sin2 x jy 00 j 4 cos x 2 2 (2 )(4 cos x) v 16 cos x = = = 3=2 2 1 + 16 sin x 1 + 16 sin2 x = v_ = 0 =

an

y 00 =

=

At A :

x=0

At B

:

x=

At C

:

x=

4 2

) a = an = 16 m/s J 16 cos( =4) ) a = an = 1 + 16 sin2 ( =4)

3=2

2

3=2

= 0:419 m/s J

) a = an = 0 J


13.8 at )

1 2 v 2

= =

dv ds dv = v ds Zds dt

v dv = at ds

at ds + C =

Z

0:05s ds + C = 0:025s2 + C

At point O: v = 20 in./s at s = 0. ) C = 200 (in./s)2 At point B: s = 80 in. 1 2 v = 0:025(80)2 + 200 2 an =

13.9

v2

=

) v 2 = 720 (in./s)

2

720 2 2 = 6:0 in./s at = 0:05s = 0:05(80) = 4:0 in./s 120 q p 2 a = a2n + a2t = 62 + 42 = 7:21 in./s J

13.10 (a)

. .

B

0 75

Dimensions in mm

A

.

O

θ

vB

300 450 mm/s


vB (an )A

=

_ = vB = 450 = 0:60 rad/s RB 750

450 mm/s 2

= RA _ = 300(0:62 ) = 108:0 mm/s

(b)

300 O

450 mm/s vA = 450 mm/s

(an )A =

2

J

.

A vA

2 vA 4502 2 = = 675 mm/s J RA 300

13.11

13.12


13.13

13.14

13.15


13.16 At point A: 1000 = 9:722 m/s 3600 v2 9:7222 2 = = = 0:9452 m/s 100 p p a2 a2n = 1:32 0:94522 = =

v

=

an at

35

1 2 v = 2

v dv = at ds

0:8925 m/s

2

0:8925s + C

Initial condition: 1 2 (9:7222 ) = 47:26 (m/s) 2 0:8925s + 47:26 s = 53:0 m J

v = 9:722 m/s when s = 0 ) C = When v = 0:

0=

13.17

13.18 y=

x2 m 80

y0 =

x 40

y 00 =

1 m 40

1

At point A (x = 10 m): y0 = 1

=h

10 = 0:25 m 40

y 00 1+

2 (y 0 )

i3=2 =

y 00 = 0:025 m 0:025 3=2

(1 + 0:252 )

1

= 0:02283 m

1


an =

v2

= 122 (0:02283) = 3:288 m/s

a=

q

a2n + a2t =

13.19

2

at = v_ = 4 m/s

p 2 3:2882 + 42 = 5:18 m/s J

at

o

20 an

g 2

at

= g sin 20 = 9:82 sin 20 = 3:359 m/s

an

= g cos 20 = 9:82 cos 20 = 9:228 m/s

an =

v2

v=

p

2

an =

2

p 9:228(4500) = 204 m/s J

*13.20 Curve Car travels at constant speed v1 , determined by an = amax . amax =

v12

v1 = t1 =

p

amax =

p 5(200) = 31:62 m/s

s1 200 = 19:871 s = v1 31:62

Straightaway Car accelerates for 500 m at the rate v_ = amax and then brakes for the next 500 m at the rate v_ = amax : . For the …rst 500 m: v

= v1 + amax t 1 s = v1 t + amax t2 2 1 500 = 31:62t2 + (5)t22 2

t2 = 9:168 s

Total time to complete the circuit is t = 2t1 + 4t2 = 2(19:871) + 4(9:168) = 76:4 s J


13.21

v

θ

v0 B

R

y

θ

O

) y_ = v0 = R cos

y = R sin

v0 )_= sec R

_

v0 v2 • = v0 sec tan _ = v0 sec tan sec = 02 sec2 tan R R R R v = v0 sec = 6 sec 60 = 12:0 in./s J 2

an = R _ = R

at

v0 sec R

= R• = R

2

=

62 v02 2 sec2 = sec2 60 = 8:0 in./s R 18

v02 sec2 tan R2

=

v02 sec2 tan R

62 2 sec2 60 tan 60 = 13:856 in./s 18 q p 2 a = a2t + a2n = 13:8562 + 8:02 = 16:0 in./s J =

13.22

x2 ft 120

y=

y0 =

x 60

y 00 =

1 ft 60

1

At point A (x = 10 ft): y0 1

an at

= = =

10 1 = 60 6 jy 00 j

y 00 =

1 ft 60 (1=60)

1

= = 0:015996 3=2 3=2 [1 + (y 0 )2 ] (1 + ( 1=6)2 ) v2 2 = (102 )(0:015996) = 1:5996 ft/s 2

= v_ = 1:2 ft/s q p 2 a = a2n + a2t = 1:59962 + 1:22 = 2:00 ft/s J


13.23

P

2 in./s

θ

v From velocity diagram: v = 2 sec R_ v_ an at When

= v = 2 sec

=

2 sec 5

= 0:4 sec rad/s

dv _ = (2 sec tan ) (0:4 sec ) = 0:8 tan sec2 d (2 sec )2 4 sec2 v2 2 = = = 0:8 sec2 in./s = R R 5 2 = v_ = 0:8 tan sec2 in./s =

in./s

= 20 : v

=

2 sec 20 = 2:13 in./s J

an

=

0:8 sec2 20 = 0:9060 in./s

2 2

0:8 tan 20 sec2 20 = 0:3298 in./s p 2 a = 0:90602 + 0:32982 = 0:964 in./s J

at

13.24

_ = 2 sec R

=

13.25


13.26

v vθ

θ

vR cot )_= R

v = R _ = vR cot • = R • R )a =

aθ

θ

aR

v = vR csc = 350 csc 40 = 545 m/s J

vR = R_ = 350 m/s

aR

a

vR

=

350 cot 40 5000

= 0:08342 rad/s

2

R _ = a sin 2 R_ 100 = sin

5000(0:08342)2 2 = 101:4 m/s J sin 40

13.27 R = 0:75 + 0:5t2 m R_ = 1:0t m/s • = 1:0 m/s2 R

t2 rad 2 _ = t rad/s • = rad/s2 =

At t = 2 s: R = 0:75 + 0:5(2)2 = 2:75 m R_ = 1:0(2) = 2 m/s • = 1:0 m/s2 R

(2)2 = 2 rad 2 _ = (2) = 2 rad/s • = rad/s2 =

vR = R_ = 2 m/s v = R _ = 2:75(2 ) = 17:279 m/s q p 2 + v2 = v = vR 22 + 17:2792 = 17:39 m/s J aR a

2

2

R _ = 1:0 2:75(2 )2 = 107:57 m/s 2 = R• + 2R_ _ = 2:75( ) + 2(2)(2 ) = 33:77 m/s

• = R


a=

q

a2R + a2 =

13.28

p 2 ( 107:57)2 + 33:772 = 112:7 m/s J

R = 4 + 2 sin ft R_ = 2 cos _ = 3 cos ft/s • = 3 sin _ = 4:5 sin ft/s2 R

_ = 1:5 rad/s •=0

(a) At point A ( = 0): R_ = 3 ft/s

R = 4 ft vR = R_ = 3 ft/s J aR a

•=0 R

v = R _ = 4(1:5) = 6 ft/s J

2

R _ = 0 4(1:5)2 = 9 ft/s J 2 = R• + 2R_ _ = 4(0) + 2(3)(1:5) = 9 ft/s J

• = R

2

At point B ( = 90 )· : R_ = 0

R = 6 ft vR = R_ = 0 J aR a

•= R

4:5 ft/s

2

v = R _ = 6(1:5) = 9 ft/s J _

• R _ 2 = 4:5 6(1:5)2 = 18 ft/s2 J = R = R• + 2R_ _ = 6(0) + 2(0)(1:5) = 0 J

13.29 R vR

= 4 + 2 sin ft = R_ = 2 cos _ ft/s

v = R _ = (4 + 2 sin ) _ ft/s

(a) At point A: = 0 ) sin = 0 cos = 1 ) vR = 2 _ ft/s v = R _ = 4 _ ft/s

v

r q 2 2 vR + v = 2_ =

v

=

p

4 ft/s

2

+ 4_

20 _ = 4

2

=

p

20 _

_ = p4 = 0:894 rad/s J 20

(b) At point B:

vR

= =

=2 0

) sin = 1 v = 6 _ ft/s

cos = 0


v

=

v

=

q 2 + v 2 = 6 _ ft/s vR 6_ = 4

4 ft/s

_ = 4 = 0:667 rad/s J 6

13.30 v = R_

vR = R_ = 600 ft/s Find _ from acceleration:

-g

aθ θ

aR

θ

_=

s

• + g sin R R

• aR = R r =

2_

R_ =

g sin

50 + 32:2 sin 30 8000

=

0:09090 rad/s

By inspection of the ‡ight path we see that _ is negative; hence _ = rad/s. ) v = 8000( 0:09090) = 727:2 ft/s q p 2 + v2 = v = vR 6002 + ( 727:2)2 = 943 ft/s J

0:09090

13.31

Di¤erentiate cam pro…le twice: R2 + 6R cos 27 = 0 2RR_ + 6R_ cos 6R sin _ = 0 • + 6R • cos 2R_ 2 + 2RR Substituting

_

6R_ sin

6R cos

_2

6R sin • =

0

(a) (b) (c)

= 0, _ = 2 rad/s and R = 3 in. into Eq. (b) yields 2(3)R_ + 6R_

0=0

) v = R_ = 0 J

Eq. (c) now becomes • + 6R • 0 6(3)(22 ) 0 = 0 0 + 2(3)R • = 6:0 in./s2 J )a=R


13.32 R = 0:3 _

R_ =

0:4

=

0:4

•= R (a) When

2

•

(b) When

•=0

=2

= 0:1 m

v = R _ = 0:1(2) = 0:2 m/s J

2_

R _ = 0 0:1(2)2 = 0:4 m/s J 2 = R• + 2R_ _ = 0 + 2( 0:2546)(2) = 1:018 m/s J

• = R

2

= =3:

vR = R_ =

a

0:4

0:255 m/s J

R = 0:3

aR

=0

_ = 2 rad/s

= =2:

vR = R_ =

a

m

0:2546 m/s

0:4

R = 0:3

aR

=

0:4

0:255 m/s J

0:4

=3

= 0:16667 m

v = R _ = 0:16667(2) = 0:333 m/s J

2_

R _ = 0 0:16667(2)2 = 0:667 m/s J 2 = R• + 2R_ _ = 0 + 2( 0:2546)(2) = 1:018 m/s J

• = R

2

13.33 • = 0 and _ = 15(2 =60) = 0:5 rad/s. Given: R = 0:7 ft, R_ = 4 ft/s, R Note that in the position = 0 we have eR = i and e = j. (a) When = =2: v

_ R + R _ e = 4eR + 0:7(0:5 )e = Re = 4eR + 1:100e = 4i + 1:100j ft/s J

(b) • R _ 2 )eR + (R• + 2R_ _ )e a = (R = 0 0:7(0:5 )2 eR + [0:7(0) + 2(4)(0:5 )] e =

1:727eR + 12:567e =

1:727i + 12:567j ft/s

2

J


13.34

13.35 _

= 2t rad = 2 rad/s

• = When R vR v v

= 2:5 mm: t = cosh 1 2:5 = 1:5668 s = R_ = sinh(1:5668) = 2:291 mm/s = R _ = 2:5(2) = 5:0 mm/s p 2:2912 + 5:02 = 5:50 mm/s J = 2

2

R _ = 2:5 2:5(22 ) = 7:50 mm/s 2 a = R• + 2R_ _ = 0 + 2 [sinh(1:5668)] (2) = 9:165 mm/s p 2 a = 7:52 + 9:1652 = 11:84 mm/s J

aR

13.36

0

R = cosh t mm R_ = sinh t mm/s • = cosh t mm/s2 = R R

• = R


13.37

13.38

13.39


13.40 R2 = b2 sin 2 2RR_ = 2b2 cos 2 _ • + R_ 2 ) = 2b2 ( 2 sin 2 _ 2 + cos 2 •) 2(RR At point A ( = 45 ): R 2RR_ • 2bR

= b = 0 ) R_ = 0 2 •= = 2b2 ( 2 _ ) ) R 2

R_ =

• aR = R v

2

2

b_ =

2

3b _

2

v0 v = )_= b b v0 2 v02 3b = 3 J b b

= R_ = b_ )

an =

2b _

2b _

aR =

aR =

v02

3

v02 v2 = 0 b

=

b J 3

13.41


13.42 h = R sin = 8200 sin 35 = 4703 ft J h_ = R_ sin + R _ cos = 0 0 = R_ sin 35 + 8200(0:03) cos 35 ) R_ = 351:3 ft/s vR v

13.43

= R_ = 351:3 ft/s v = R _ = 8200(0:03) = 246:0 ft/s q p 2 + v2 = = vR ( 351:3)2 + 246:02 = 429 ft/s J

13.44

A

5.25 m

5.25 m

θ

θ

6m

R=

C

A

28.96o

6m

28.96o B

vR

B

v

C

57.92o

vθ 57.92o

Given: R = 6 m, R_ = 1:2 m/s cos vR

5:25 = 28:96 6 = R_ = 1:2 m/s =

From velocity diagram (note that v is perpendicular to AB): v=

1:2 vR = = 1:416 m/s J sin 57:92 sin 57:92


13.45 •=0 R = 0:4 m R_ = R z = 0:2 z_ = 0:2 _ = 0:2(6) = z• = 0:2• = 0:2( 10) = 2 m/s v v

1:2 m/s

= R_ eR + R _ e + z_ ez = 0 + 0:4(6)e + ( 1:2)ez = 2:4e p = 2:42 + ( 1:2)2 = 2:68 m/s J a = =

1:2ez m/s

• R _ 2 )eR + (R• + 2R_ _ )e + z•ez (R 0 0:4(6)2 eR + [(0:4)( 10) + 0] e + 2ez 2

=

14:4eR 4e + 2ez m/s p 2 ( 14:4)2 + ( 4)2 + 22 = 15:08 m/s J a =

13.46

R _ z z_

•=0 = 4 in. (const.) ) R_ = R = 0:8 rad/s (const.) )•=0 = 0:5 sin 4 in. = 2 _ cos 4 = 2(0:8) cos 4 = 1:6 cos 4 in./s

z• = =

2

8 _ sin 4 + 2• cos 4 =

8(0:8)2 sin 4 + 0

2

5:12 sin 4 in./s

v = R _ = 4(0:8) = 3:2 in./s vz = z_ = 1:6 cos 4 in./s p n = 3:22 + 1:62 = 3:58 in./s at = , n = 0; 1; 2; : : : J 4

vR = R_ = 0 ) vmax

aR

) amax

2

2

R _ = 4(0:8)2 = 2:56 in./s 2 = R• + 2R_ _ = 0 az = z• = 5:12 sin 4 in./s

• = R

a p 2 = 2:562 + 5:122 = 5:72 in./s at

=

8

+

n , n = 0; 1; 2; : : : J 4


13.47

13.48

13.49 _ = 3 rad/s

•=0

_ = 2 rad/s

L = 0:8 m


R_ =

R = L cos z = L sin When

= 50 and

z_ = L cos

az

z• =

_2

L cos L sin

_2

= 0:8 cos 40 = 0:6128 m = 0:8(sin 40 )(2) = 1:0285 m/s = 0:8(cos 40 )(22 ) = 2:451 m/s 0:8(sin 40 )(22 ) =

z• =

a

_

•= R

= 40 : R R_ • R

aR

_

L sin

2:057 m/s

2

2

R _ = 2:451 0:6128(32 ) = 7:966 m/s 2 = R• + 2R_ _ = 0 + 2( 1:0285)(3) = 6:171 m/s

• = R

2

2

= z• = 2:057 m/s p 2 a = 7:9662 + 6:1712 + 2:0572 = 10:28 m/s J


13.50

13.51 z

ρ

mg = man

µsN FBD

Fz

=

Fn

= man

v

13.52

=

0

p

+"

s

N

N

mg = 0

MAD

) N = mg

+ s N = man s mg = m p g = 0:4(600)(32:2) = 87:9 ft/s J

v = 70 km/h =

v2

70 103 m/s = 19:444 m/s 3600


Assume impending sliding of crate.

Mg

R

y v MR

2

x

15o

µsΝ

N FBD

Fy

= may

N

= M

N

s

M g cos 15 = M

v2 sin 15 R

v2 sin 15 R 19:4442 9:81 cos 15 + sin 15 50 g cos 15 +

= M

Fx

MAD

= max

sN

M g sin 15 =

= 11:433M

M

v2 cos 15 R

v 2 =R cos 15 M g sin 15 N 19:4442 =50 cos 15 9:81 sin 15 = 0:417 J 11:433

= M =

13.53 W

T 35

N FR T cos 35

N

= maR =

maR

o

MAD

+ !

60 0 32:2

Fz 60 + 90:99 sin 35

FBD

T cos 35 =

10(2)2

= 0 = 0

W • (R g

2

R_ )

T = 90:99 lb

+ " N W + T sin 35 = 0 N = 7:81 N J


13.54 ρ =4m N

=

z mρv

mg FBD

µsN Fz

z

0

+"

MAD sN

)N =

mg = 0 v2

+! N =m ) r r g 4(9:81) = = = 8:09 m/s J 0:6 s

Fn

= man

v

2

mg

mg s

=m

v2

s

13.55 o

3200 lb 20

n

man t

F = 0.6N

mat

N FBD

MAD

Assume impending sliding between the tires and the road. = man =

Ft

3200 602 N = 4796 lb 32:2 200 = mat 3200 sin 20 0:6N = mat 3200 2 3200 sin 20 0:6(4796) = at at = 17:94 ft/s J 32:2 N

N

W cos 20 = m

v2

Fn

3200 cos 20 =

13.56

mv R

2

mg

n

. mv

t

µN

N FBD

MAD


(a) Fn

= man

N

N

= m g+

v2 R

mg = m

v2 R

= 8 9:81 +

2:82 2:5

= 103:57 N J

(b) Ft v_

= mat N = = m

N = mv_ 0:2(103:57) = 8

2:59 m/s

2

J

13.57 T FBD

=

mg

θ

• = 0, we have aR = Since R_ = R F

= ma g sin • = R

FR _2 _

maθ MAD maR

2

R _ and a = R•

+% mg sin = mR• 9:81 sin 25 2 = = 2:07 rad/s J 2:0 2

= maR + & mg cos T = mR _ 1 T 1 8:5 = g cos = 9:81 cos 25 R m 2:0 0:5 p 4:055 = 2:01 rad/s J =

2

= 4:055 (rad/s)

13.58


13.59

ω

y

ρ

x v

The bead moves on a horizonal circle of radius v = ! = !R sin .

= R sin

with the speed

y mg mv2/ρ

x N θ FBD Fx

= m

Fy

=

v2

MAD

N sin = m! 2 R sin

N = m! 2 R

0

mg N cos = 0 mg g 9:81 cos = = 2 = = 0:8741 N ! R (6:72 ) (0:25) = 29:1 J

13.60


13.61 Critical position is

= 90 .

N

FBD

MAD man

W Fn _

W _2 R = man +# W = g r r g 32:2 = = = 4:01 rad/s J R 2

Ft = mat

+

N=

W • R = 0 (since • = 0) g

Because N = 0 in the critical position, friction would not change the result.

13.62

N y F n W = 0.4 lb FBD v

=

an

=

30o t

x mat ma n

ma MAD

vx 8 = = 9:238 ft/s cos 30 cos 30 v2 9:2382 2 = = 71:12 ft/s 1:2

Because vx is constant, we have ax = 0, so that a is vertical. Therefore (see MAD), 2 at = an tan 30 = 71:12 tan 30 = 41:06 ft/s


Ft

= mat

F cos 30 + W sin 30 = mat 0:4 F cos 30 + 0:4 sin 30 = (41:06) 32:2 = 0:358 lb J

F

13.63 x

N

F

FBD

MAD

β R

man β

O Fx

= max

+

kx0

x = k

2 m_

=

F = man sin 2

k(x

2

x0 ) = m(R _ )

x R

2(0:5) = 2:24 ft J (2=32:2) (5)2

13.64 .

θ

R = 3 ft

W = 1.5 lb FBD

MAD maR

F = 2 lb

(a) Rotation in the horizontal plane: delete W from the FBD (it acts perpendicular to the plane of motion). FR

= maR 1:5 (0 2 = 32:2

+# 2

3_ )

• F = m(R

2

R_ )

_ = 3:78 rad/s J

(b) Rotation in the vertical plane: use FBD as shown. FR

= maR 1:5 (0 2 + 1:5 = 32:2

+# 2

3_ )

• F + W = m(R

2

R_ )

_ = 1:892 rad/s J


13.65 When slider is at C:

mat

F 45o man

N mg FBD F an

Ft at

MAD

p = k = 1:2(2:5 2 1:3) = 2:683 lb v2 82 2 = = = 25:60 ft/s R 2:5

= mat F cos 45 mg = mat 2:683 F cos 45 g= cos 45 = m 1:0=32:2

Fn

= man

N

= F sin 45 a=

F sin 45

a2n + a2t =

13.66 mg

1:0 (25:60) = 1:102 lb J 32:2

p 2 25:602 + 28:892 = 38:6 ft/s J

R T

maR

.

θ MAD

N FBD FR = maR

2

N = man

man = 2:683 sin 45

q

32:2 = 28:89 ft/s

+ !

• = 0. ) T = (a) If string is intact, R

T = 2:5 0

(b) After string breaks, T = 0. ) 0 = • = 560 m/s2 J )R

h • 2:5 R

• m(R

2

R_ )

1:4(20)2 = 1400 N J 1:4(20)2

i


13.67

13.68 Assume impending sliding of the block up the cone.

mg

µsN

y

mv2 /ρ

x

β

β N FBD

Fy

=

MAD

0

N cos

N=

Fx

cos

= max mg

sN

sin

mg = 0

mg s

N sin

sin + cos

sin

+

cos s sin

s

s N cos

=m

=m

v2

v2

Substituting v = !: ) !2

= =

sin + cos g sin + cos

g

cos = !2 sin s 32:2 sin 35 + 0:28 cos 35 s cos = 1:2 cos 35 0:28 sin 35 s sin

s

2

32:7 (rad/s)

! = 5:72 rad/s


Assume impending sliding of the block down the cone. The direction of the friction force s N must be reversed on the FBD. This is equivalent to reversing the sign of s . The result is !2

=

!

=

32:2 sin 35 0:28 cos 35 = 9:427 1:2 cos 35 + 0:28 sin 35 3:07 rad/s

) The block will not slide if 3:07 rad/s

!

5:72 rad/s J

13.69

13.70


13.71

13.72 v

=

an

=

_ R = 1:2(2:2) = 2:640 m/s v2 2:6402 2 = = 3:168 m/s R 2:2

Assume impending sliding when

35o

at = 0

= 35 .

9.81m

man

µs N n

t

N FBD Fn N Ft s

MAD

= man N 9:81m cos 35 = m(3:168) = m(9:81 cos 35 + 3:168) = 11:204m = mat 9:81m sin 35 = 0 sN 9:81 sin 35 9:81 sin 35 = = = 0:502 J N 11:204


13.73

13.74


13.75

T maθ

θ

maR

mg FBD

MAD

• = 0, so that the acceleration components are aR = Note that R_ = R • a =R . mg sin = mR•

F = ma

g sin R

)•=

d_ d d_ _ = d dt d

But • = d_ _ d

2

R _ and

g sin R

=

_ d_ =

_2

g sin d R

2

=

g cos + C R

= 0 then _ = _ A

Initial condition: when _2

A

2

= )

_2 g g +C C= A R 2 R _2 _ 2 = 2g (1 cos ) A R

(a) Rigid rod: pendulum will just reach position B if _ = 0 at r 4g ) _A = J R

= 180

(b) Flexible string: FR = maR

T + mg cos =

Pendulum will just reach position B if T = 0 at mg _A

2

mR _ r 5g = J R

=

_2 = g R

mR _

2

= 180

2 2 2g 5g ) _ A = _ + (2) = R R


13.76 aR a

θ

aθ

R_ R_ cos

R _ sin

R θ 2m O

=

4 m/s (given)

=

0

a

= =

a =

R cos = 2 m

1 cos = R 2

m

1

R_ = R _ tan = 4 tan 1 d(4R) 1 1 d(R2 _ ) = = 4 R_ R dt R dt R cos 2 4 (4 tan ) = 8 sin m/s 2 a 8 sin 2 = = 8 tan m/s cos cos

F = ma = 0:6(8 tan ) = 4:8 tan N J

13.77 We must show that a = 0. R2 _

= constant ) R2 • + 2RR_ _ = 0 ) a = R• + 2R_ _ = 0 Q.E.D

R• + 2R_ _ = 0

*13.78 (a)

θ

maR mg

N FR = maR

maθ

FBD

MAD

• mg sin = m(R

R_ )

d d d d = = !0 dt d dt d

2

• g sin = R

• = !2 )R 0

! 20 R

d2 R d 2


= ! 20

g sin d2 R d 2

R

=

d2 R d 2

g sin ! 20

R J

(a)

(b) Solution of the homogeneous equation d2 R d 2 A particular integral is

R = 0 is R = C1 sinh + C2 cosh

R=

g sin 2! 20

Therefore, the solution of the di¤erential equation in Eq. (a) is R = C1 sinh + C2 cosh

g sin 2! 20

Initial conditions: R R_

=

0 when

=0

=

0 when

=0 R=

) C2 = 0 C1

g (sinh 2! 20

g =0 2! 20

C1 =

g 2! 20

sin ) J

13.79


*13.80

13.81


13.82

13.83


13.84


13.85

Initial conditions: Letting x1 = are

g sin = L

•=

Equation of motion:

= 60 =

3

9:81 sin = sin 9:81 rad _ = 0 at t = 0

and x2 = _ , the …rst-order equations and the initial conditions x_ 1 x1 (0)

= x2

x_ 2 =

=

x2 (0) = 0

3

sin x1

The MATLAB program for solving the equations at intervals of 0.1 s is: function problem13_85 [t,x] = ode45(@f,(0:0.1:2),[pi/3,0]); printSol(t,x) function dxdt = f(t,x) dxdt = [x(2) -sin(x(1))]; end end The two lines of printout that span x1 = 0 are: t x1 x2 1.6000e+000 8.5645e-002 -9.9633e-001 1.7000e+000 -1.4249e-002 -9.9990e-001 Linear interpolation: t 1:6 = 1:7 1:6

0 0:085 645 0:014 249 0:085 645

t = 1:686 s J

which agrees with the analytical solution

13.86 • = 0, we have aR = (a) Since R_ = R

2

R _ and a = R•

θ mg FBD

=

µN

maR

N FR F

= maR = ma

+. +&

mg sin mg cos

maθ

MAD

2 N = mR _ N = mR•


Elimination of N yields mg cos

mg sin

= mR• + mR_ 2 • = g (cos sin ) R

_2

Q.E.D.

(b) Substituting numerical values, the equation of motion becomes • = 9:81 (cos 2

0:3 _

0:3 sin )

2

Letting x1 = and x2 = _ , the equivalent …rst-order equations and the initial conditions are x_ 1 = x2 x1 (0) = 0

x_ 2 = 4:905(cos x2 (0) = 0

0:3 sin )

0:3x22

The corresponding MATLAB program is function problem13_86 [t,x] = ode45(@f,(0:0.02:1.6),[0,0]); printSol(t,x) function dxdt = f(t,x) dxdt = [x(2) 4.905*(cos(x(1)) - 0.3*sin(x(1))) - 0.3*x(2)^2]; end end The two lines of output spanning x2 = 0 are t 1.5200e+000 1.5400e+000

x1 x2 2.0982e+000 1.1455e-002 2.0977e+000 -6.3349e-002

Linear interpolation: x1 2:0982 2:0977 2:0982 x1

= =

0 ( 0:063349) 0:011455 ( 0:063349) = 2:098 rad = 120:2 J

13.87 (a)

θ mg

maθ

maR

= FBD

N

MAD


= !

0

)_=

cos !t

=

rad/s

FR

= maR

• )R • R R(0)

= R_

0

sin !t

g sin = R( !

= 15 =

0

sin !t)2

12

rad

2

R_ ) g sin(

0

cos !t)

2

2

12

0

• mg sin = m(R

+%

2

= R

!

sin t

32:2 sin

12

cos t

J

_ = 2 ft, R(0) =0 J

_ the equivalent …rst-order equations and the (b) Letting x1 = R and x2 = R, initial conditions are 2

2

x_ 1 x1 (0)

= x2 = 2 ft

x_ 2 = x1

12 x2 (0) = 0

sin t

32:2 sin

12

cos t

The corresponding MATLAB program is function problem13_87 [t,x] = ode45(@f,(0:0.025:3.5),[2,0]); plot(t,x(:,1),’linewidth’,1.5) xlabel(’t (s)’); ylabel(’R (ft)’) grid on printSol(t,x) function dxdt = f(t,x) dxdt = [x(2) x(1)*(pi^2/12*sin(pi*t))^2-32.2*sin(pi/12*cos(pi*t))]; end which produces the following plot:


4.5

4

3.5

R (ft)

3

2.5

2

1.5

1

0.5

0

0.5

1

1.5

2

2.5

3

3.5

t (s)

(c) The two lines lines of output spanning R = 4 ft are: t 3.4000e+000 3.4250e+000

x1 3.9520e+000 4.0718e+000

x2 4.7323e+000 4.8523e+000

Linear interpolation: 4:0718 3:9529 4 3:9529 = 3:4250 3:4 t 3:4 R_ 4:7323 4:8523 4:7323 = 3:410 3:4 3:4250 3:4

t = 3:410 s J R_ = 4:78 ft/s J

13.88 (a) Kinematics: R aR a

• = z• tan = z tan R_ = z_ tan R 2 2 • R _ = (• = R z z _ ) tan = R• + 2R_ _ = (z • + 2z_ _ ) tan az = z•


6 tanβ R 6 ft

β

s z

FBD

β

mg N

=

maz maθ

s maR

β MAD

Kinetics: 0 = m(z • + 2z_ _ ) tan

F

= ma

+

Fs

= mas

+%

g

= az + aR tan

mg cos

= m(az cos

2z_ _ Q.E.D. z + aR sin ) )•= 2

2

z _ ) tan2

= z• + (• z

z• =

z _ tan2 1 + tan2

g

Q.E.D.

Initial conditions: z(0) = 6 ft (b) With

z(0) _ =0

(0) = 0

_ (0) =

v1 6 tan

=

10 6 tan

J

= 20 and g = 32:2 ft/s2 , we have

z• = 0:116978z _

2

28:433 ft/s

2

•=

2z_ _ z

_ (0) = 4:5791 rad/s

Letting x1 = , x2 = _ , x3 = z and x4 = z, _ the equivalent …rst-order equation and the initial conditions are x_ 1 x_ 3

= x2 = x4

x_ 2 = 2x4 x2 =x3 x_ 4 = 0:116978x3 x22 28:433 10 x1 (0) = 0 x2 (0) = = 4:5791 rad/s 6 tan 20 x3 (0) = 6 ft x4 (0) = 0 The MATLAB program for solving the equation of motion is function problem13_88 [t,x] = ode45(@f,(0:0.02:2),[0,4.5791,6,0]); plot(x(:,1),x(:,3),’linewidth’,1.5) xlabel(’theta (rad)’); ylabel(’z (ft)’) grid on printSol(t,x) function dxdt = f(t,x) dxdt = [x(2) -2*x(4)*x(2)/x(3) x(4) 0.116978*x(3)*x(2)^2-28.433]; end end


6.5

6

z (ft)

5.5

5

4.5

4

3.5

0

2

4

6 8 theta (rad)

10

12

14

(c) The two lines of output spanning x4 = 0 are: t 7.6000e-001 7.8000e-001

x1 5.2777e+000 5.4915e+000

x2 1.0695e+001 1.0671e+001

x3 x4 3.9257e+000 -1.3783e-002 3.9302e+000 4.6720e-001

By inspection, minimum value of x3 is 3:93 ft ) hmin = 6

3:93 = 2:07 ft J

13.89 (a)

k(R − L0)

FBD θ

= maθ

mg FR

F

MAD maR

• + & mg sin k(R L0 ) = m(R k • = R_2 ) R (R L0 ) + g sin m = ma + . mg cos = m(R• + 2R_ _ ) 1 ) • = (g cos 2R_ _ ) R = maR

2

R_ )


Substituting given data, the equations of motion become • = R_2 R

40(R

0:5) + 9:81 sin

J

• = 1 (9:81 cos R

2R_ _ ) J

The initial conditions are R(0) = 0:5 m

_ R(0) = (0) = _ (0) = 0 J

_ x3 = and x4 = _ , the equivalent …rst-order (b) Letting x1 = R, x2 = R, equation and the initial conditions are x_ 1 = x2 x_ 2 = x1 x24 40(x1 0:5) + 9:81 sin x3 x_ 3 = x4 x_ 4 = (9:81 cos x3 2x2 x4 )=x1 x1 (0) = 0:5 m x2 (0) = x3 (0) = x4 (0) = 0 The MATLAB program for numerical integration becomes function problem13_89 [t,x] = ode45(@f,(0:0.01:0.75),[0.5,0,0,0]); plot(x(:,3)*180/pi,x(:,1),’linewidth’,1.5) xlabel(’theta (deg)’); ylabel(’R (ft)’) grid on printSol(t,x) function dxdt = f(t,x) dxdt = [x(2) x(1)*x(4)^2-40*(x(1)-0.5)+9.81*sin(x(3)) x(4) (9.81*cos(x(3))-2*x(2)*x(4))/x(1)]; end end


1.3

1.2

1.1

R (ft)

1

0.9

0.8

0.7

0.6

0.5

0

20

40

60 theta (deg)

80

100

120

(c) From the partial printout shown below, we see that Rmax = 1:186 m J t 6.5000e-001 6.6000e-001 6.7000e-001

x1 x2 1.1859e+000 1.4175e-002 1.1854e+000 -1.2450e-001 1.1834e+000 -2.6261e-001

x3 1.5646e+000 1.5823e+000 1.6002e+000

x4 1.7783e+000 1.7798e+000 1.7839e+000

13.90 (a) The equations of motion are identical to those derived in Prob. 13.89: • = R_2 R

40(R

0:5) + 9:81 sin

J

• = 1 (9:81 cos R

2R_ _ ) J

The initial conditions are also the same, except for R(0): R(0) = 0:75 m

_ R(0) = (0) = _ (0) = 0 J

(b) The only changes in the MATLAB program are in the arguments of ode45— changing R(0) to 0:75 and extending the integration period to 1:1 s, as shown below


[t,x] = ode45(@f,(0:0.01:1.1),[0.75,0,0,0]);

1.6

1.4

1.2

R (ft)

1

0.8

0.6

0.4

0.2

0

25

50

75 theta (deg)

100

125

150

(c) Partial printout of the results: t 6.7000e-001 6.8000e-001 6.9000e-001

x1 x2 1.2815e+000 1.4977e-001 1.2820e+000 -5.1364e-002 1.2805e+000 -2.5326e-001

x3 1.8433e+000 1.8549e+000 1.8663e+000

x4 1.1749e+000 1.1529e+000 1.1337e+000

1.0000e+000 1.0100e+000

5.2318e-001 -3.0221e+000 4.9342e-001 -2.9271e+000

2.2686e+000 2.2902e+000

2.1048e+000 2.2322e+000

By inspection, Rmax = 1:282 m J The cord becomes slack when R = 0:5 m. Using linear interpolation: 0:49342 2:2902

0:52318 0:5 = 2:2686

0:52318 2:2686

= 2:285 rad = 130:9

J

13.91 v = 150

mi h

1h 3600 s

5280 ft = 220 ft/s 1 mi

at = 0


F1 F2

mat man MAD

W FBD Ft

= mat

F2 = 0

Fn

= man

F1 + W =

F1

= W

F

=

v2 g 28:1 lb # J 1+

Path

W v2 g

= 180

1+

2202 32:2(1300)

= 28:12 lb

13.92


13.93

13.94 Consider a water particle about to exit the pipe at C. Using polar coordinates, we have R

=

_

=

0:12 m R_ = 0:6 m/s 2 160 = 16:755 rad/s 60 2

•=0 R •=0

2

R _ = 0 0:12(16:755)2 = 33:69 m/s 2 a = R• + 2R_ _ = 0 + 2(0:6)(16:755) = 20:11 m/s q p 2 a = a2R + a2 = ( 33:69)2 + 20:112 = 39:2 m/s J

aR

• = R


13.95

13.96 Kinematics: • = 5 rad/s2 When

_ = 5t rad/s

= 2:5t2 rad

R = 0:18 m (constant)

= 30 = =6 rad: t2 = 2

R_ = 0

2:5

=

=6 = s2 2:5 15

aR

• = R

a

2 = R• + 2R_ _ = 0:18(5) + 0 = 0:900 m/s

0:18(5t)2 =

0:18(25)

15

=

0:9425 m/s

2

Kinetics (assume impending sliding):

30o 9.81m

maθ

maR

µs N FBD F FR

N

MAD

= ma N 9:81m cos 30 = 0:900m N = (0:900 + 9:81 cos 30 ) m = maR 9:81m sin 30 = 0:9425m sN (0:900 + 9:81 cos 30 ) m 9:81m sin 30 = s 9:396 s = 3:963 s = 0:422 J

0:9425m


13.97

13.98 y = sinh x dy dx d2 y dx2

x = sinh

1

y = sinh

1

(3:8) = 2:045 m

=

cosh x = cosh(2:045) = 3:929

=

sinh x = sinh(2:045) = 3:800 m 1 h i3=2 2 1 + (dy=dx) 1 + 3:9292 = d2 y=dx2 3:800

=

an =

v2

=

3=2

= 17:537 m

82 2 = 3:649 m/s 17:537


man mg FBD = tan Fn

Ft

1

N

θ

ma t MAD

(dy=dx) = tan

1

(3:929) = 75:72

= man N mg cos = man N 12(9:81) cos 75:72 = 12(3:649) N = 72:8 N J

= mat mg sin = mat 2 v_ = at = g sin = 9:81 sin 75:72 = 9:51 m/s J

13.99 an =

v2 4:22 2 = = 8:82 ft/s J R 2

3 lb mat o

45 1.8 lb

ma n N FBD

Fn

= man

MAD

3 + 1:8 sin 45

N=

N = 3:451 lb J Ft

= mat

at

=

1:8 cos 45 = 2

13:66 rad/s

3 (8:82) 32:2

3 at 32:2

J

13.100

mg

β man µsN

z

N n FBD

MAD


Fz

=

N

=

0 cos

Fn v

N cos mg s sin

sN

sin

mg = 0 m(32:2) = 37:30m lb = cos 10 0:7 sin 10

v2 = man = s N cos + N sin r N = ( cos + sin ) m s r 300(37:30m) = (0:7 cos 10 + sin 10 ) m 3600 = 98:27 ft/s = 98:27 = 67:0 mi/h J 5280

13.101

13.102

θ mg

maR FBD

N

maθ

MAD


F

= ma

mg sin = m(R• + 2R_ _ ) d_ _ g • = g sin = sin R d R 1 _2 g = cos + C 2 R

+.

= R• + 0

g sin _ d_

=

g sin d R

= 0. ) C = g=R.

Initial condition: _ = 0 when

g = (1 cos ) R p = R _ = 2gR(1

1 2 ) _ 2 v

FR

= maR

When N

=

cos

+-

0:

=

N

g cos = 0

2(1

r

2g (1 R cos ) J _=

cos )

cos )

• mg cos = m(R 2

R_ = = cos

1

2

R_ )

2g(1 cos ) 2 = 48:2 J 3

13.103 T ββT

T FBD

FBD

β mg

mg Before cut

=

n MAD mat

After cut

(a) Due to symmetry, TAB = TCB = T Fy T

=

0

=

mg 2 cos

+#

mg 2T cos = 0 mg = = 0:610mg J 2 cos 35

(b) Immediately after release v = 0; hence an = 0 Fn T

= man = mg cos

+ % T mg cos = 0 = mg cos 35 = 0:819mg J

(c) The results would be equal if mg 2 cos

= mg cos

cos2

= 0:5

= 45

J


13.104

13.105


13.106

R

θ 2a R = 2a cos

R_ = 2a sin

_ = 2a! 0 sin

vR = R_ = 2a! 0 sin v = R _ = 2a! 0 cos q 2 + v 2 = 2a! (constant) J v = vR 0

13.107


13.108

13.109

z A

z A m 0.6

θ

T R

B

FBD

ma n MAD

mg

B

(a) Fz T

= maz T cos mg = 0 mg 1:4(9:81) = = = 79:09 N J cos cos 80

(b) Fn v2 v

v2 R T R sin 79:09(0:6 sin 80 ) sin 80 2 = = = 32:87 (m/s) m 1:4 p = 32:87 = 5:73 m/s J

= man

T sin = m


Chapter 14 14.1

14.2


14.3

14.4

14.5 (a) When L0 = b: A

UA

B

=

0

B

1 k 2

=

=

p

2 B

2b

2 A

b = 0:4142b 1 h 2 = k (0:4142b) 2

i 0 =

0:0858kb2 J

(b) When L0 = 0:8b:

UA

A

= b

B

=

0:8b = 0:2b 1 k 2

2 B

2 A

p = 2b 0:8b = 0:6142b i 1 h 2 k (0:6142b) (0:2b)2 = 2

B

=

0:1686kb2 J


14.6

14.7

180 lb FBD

µkN

30 lb 20o

N Fy = 0 U1

N

180 + 30 sin 20 = 0

N = 169:74 lb

1 mv 2 0 2 2 1 180 [ 0:15 (169:74) + 30 cos 20 ] x = 4:02 2 32:2 x = 16:38 ft J = T2

2

T2

(

sN

+ 30 cos 20 )x =

14.8 UA

B

= TB

d

=

TA

mghA

k mgd

=0

1 mv 2 2 A

2 2ghA + vA 2(32:2)(4) + 102 = = 13:88 ft J 2g k 2(32:2)(0:4)


14.9 U1

2

= T2

T1

(

k mg) x

1 k 2

2

=

0

1 mv 2 2 1

1 1 (150)(x 8)2 = (25)(8)2 2 2 75x2 1151x + 4000 = 0 The larger root is x = 10:03 m

0:2(25)(9:81)x

14.10 v1 =

U1

2

v2

80 km 1h

1000 m 1 km

1h = 22:22 m/s 3600 s

1 = T2 T1 mgh2 = m v22 v12 2 q p 2 2 = v1 2gh2 = 22:22 2(9:81)(10) = 17:249 m/s =

17:249

m s

1 km 1000 m

3600 s = 62:1 km/h J 1h

14.11

R2 = 0.2 m 2

.

0.75 m

A R1

1

.

The tension in the rope is a central force always passing through point A. U1

T1 U1

2

2

=

=

R1 ) mg h p = 10(0:2 0:22 + 0:752 ) = 1:3476 N m 0

= T2

F (R2

T2 = T1 :

1 2 mv 2 = 0:3vB 2 B 2 1:3476 = 0:3vB

0:6(9:81)(0:75)

vB = 2:12 m/s J


14.12

14.13

h

2

1 T1 U1

= T2 = 0

2

= T2

h

=

T1 :

1:223 m J

U1

2

=

1 k( 2

1 (1200)(0 2

2 2

2 1)

0:12 )

mgh = 0 0:5(9:81)h = 0


14.14

14.15

14.16 In the limiting case, package will arrive at D with zero velocity. ) TA = TD = 0.

2P

UA D = area under P -x diagram 10(9:81)(2) = 0 P = 98:1 N J

mghD = 0


14.17

A

2R

25o 20o

B

d

UA

TA UA

B

p

= Fd = F

B

=

2:769 lb ft

0

TB =

=

= TB

TA :

2R cos 25

= 1:2

p

2(1:8) cos 25

1 1 2 2 2 2 mvB = v = 0:03106vB 2 2 32:2 B 2 2:769 = 0:03106vB vB = 9:44 ft/s J

14.18 U1

2

v22

= T2

=

1 k 2

T1

2

1 (2000) (0:122 ) 2 2 89:43 (m/s)

30o

n

mg (1

cos 30 ) =

0:3(9:81)(2:5)(1

1 mv 2 2 2

0

cos 30 ) =

1 (0:3)v22 2

mg man

t FBD

mat MAD

N Fn

= man

N

mg cos 30 = m

N

= m g cos 30 +

v22

v22

= 0:3 9:81 cos 30 +

89:43 2:5

= 13:28 N J


14.19 mg 35o y

µk N

FBD x

N Fy N

= 0 + % N mg cos 35 = 0 = mg cos 35 = 5(9:81) cos 35 = 40:18 N

The distance traveled by the block is 3 + spring. U1

2

= T2

T2

mg(3 + ) sin 35

5(9:81)(3 + ) sin 35

m, where

k N (3

+ )

= deformation of the 1 k 2

2

=

0

1 mv 2 2 1

1 1 (4000) 2 = (5)(6)2 2 2 2000 2 + 18:089 144:27 = 0 The positive root is = 0:2641 m

0:25(40:18)(3 + )

Fmax = k = 4000(0:2641) = 1056 N J

14.20 Release position (position 1) Return position (position 2) T1 = 0

U1

2

U1

T2 =

1 2

= x1 = 1:5 ft = x2 = 0

1 20 2 1 mv 2 = v = 0:3106v22 2 2 2 32:2 2

1 2 k( 22 x1 ) k mg(x2 1) 2 1 = (22:5)(0 1:52 ) 0:25(20)(0 + 1:5) 2 = 17:813 lb ft

=

2

v2

= T2 T1 17:813 = 0:3106v22 = 7:57 ft/s J


14.21

14.22


14.23

14.24 Position 1 = position at entry (x = 0); Position 2 = position at exit (x = 54 in.). U1

2

= = =

T2

T1

(area under the F -x diagram) 10 + 18 18 + 24 24 + 16 (24) + (6) + (24) 2 2 2 942 lb in. = 78:5 lb ft

= =

U1

2

v2

1 1 3:2=16 2 m(v22 v12 ) = (v 2102 ) 2 2 32:2 2 3:106 10 3 v22 136:96 lb ft

= T2 T1 78:5 = 3:106 = 137:2 ft/s J

10

3 2 v2

136:96


14.25

14.26 Position 1 = release position ( = 0); choose as datum for gravitational potential energy. Position 2 = position where spring has maximum deformation ( = max ). V1 = T 1 = 0

T1 + V1 Fmax

V2 =

= T 2 + V2 = k

max

0=

W

max

W

= 2W J

1 + k 2

max

2 max

1 + k 2

T2 = 0

2 max

max

=

2W k

14.27 Let B denote the lowest point of the drop and let A be the datum for gravitational potenetial energy. TA

= TB = VA = 0

VB

= =

1 + k 2B = 2 150 B 24 000

mg(L + 120 VB h

2 B

B)

= 0: = L+

Pmax = k

150(160 +

B)

1 + (240) 2

2 B

= 14:78 ft = 160 + 14:78 = 174:8 ft J

B B B

= 240(14:78) = 3550 lb J

14.28 Position 1: position just before car hits the spring.


Position 2: position where car come to a stop. T 1 + V1 1 18 2

103

10 103 3600

1 1 mv 2 + 0 = 0 + k 2 1 2

= T 2 + V2

2

=

1 k(0:52 ) 2

k = 556

2

103 N/m J

14.29 Position 1: position just before car hits the spring. Position 2: position where car come to a stop.

2 δ2 1

δ1

0.3 m

Spring 1 Spring 2 2

= 0:5 m

T 1 + V1 1 18 2

103

10 103 3600

1

=

= T 2 + V2

2

=

2

0:3 = 0:2 m 1 1 mv12 + 0 = 0 + k 2 2

1 k(0:22 + 0:52 ) 2

k = 479

2 1

+

2 2

103 N/m J

14.30


14.31 Choosing position A as the datum for gravitational potential energy: VA VB TB

1 2 1 k = (2000) 2A = 1000 2A TA = 0 2 A 2 = mgh = 0:5(9:81)(0:8) = 3:924 N m 1 1 = mv 2 = (0:5)(1:52 ) = 0:5625 N m 2 B 2 =

VA + TA A

= VB + TB : 1000 = 0:0670 m J

2 A

= 3:924 + 0:5625

14.32


14.33 A

0.4 m 0.2 m

0.1 m Datum

B

Position A: p = 0:42 + 0:12 0:15 = 0:2623 m A 1 1 VA = mg(2R) + k 2A = 0:6(9:81)(0:4) + (200)(0:2623)2 = 9:235 N m 2 2 1 1 TA = mv 2 = (0:6)(3)2 = 2:70 N m 2 A 2 Position B: B

=

VB

=

TB

=

T A + VA = T B + VB

0:3 0:15 = 0:15 m 1 2 1 k B = (200)(0:15)2 = 2:25 N m 2 2 1 1 2 2 2 mvB = (0:6)vB = 0:3vB 2 2 2 2:70 + 9:235 = 0:3vB + 2:25

vB = 5:68 m/s J

14.34


14.35

14.36 Position 1 = position where = max ; choose as datum for gravitational potential energy. Position 2 = position where = 0. T 1 + V1 v22

= T2 + V2 =

2Lg(1

1 mv 2 mgL(1 cos 50 ) 2 2 2 cos 50 ) = 2(2)(9:81)(1 cos 50 ) = 14:017 (m/s) 0+0=

T

man FBD

MAD

mg Fn T

= man = m g+

+" v22 L

v22 L 14:017 = 0:5 9:81 + 2 T

mg = m

= 8:41 N J

14.37 The potential energy of a spring in terms of the spring force F is Ve =

1 k 2

2

=

1 k 2

F k

2

=

1 F2 2 k


Position 1 = position just before drum stops (choose as datum for gravitational potential). Position 2 = position of maximum tension in cable. In Position 1 the cable force is the weight W of the elevator. ) V1

=

T1

=

1 12002 1 W2 = 1:0909 N m = 2 k 2 660 103 1W 2 1 1200 2 v = (8) = 1192:5 N m 2 g 1 2 32:2

(Ve )1 =

Let F be the cable force in Position 2 V2 = (Ve )2 + (Vg )2 =

=

1 F2 2 660 103

T 1 + V1 = T 2 + V 2 7:576

1 F2 F W 2 k k F F (F 2400) 1200 = 3 660 10 1:32 106

)

10

7

F2

1:8182

1192:5 + 1:0909 10

3

F

=

0+

F (F 2400) 1:32 106

1193:6 = 0 F = 40 900 lb J

14.38 (a) Choose the release position (x = 0) as the datum for gravitational potential energy. Consequently, V1 = T1 = 0 in the release position. Noting that the elongation of the spring is = 2x, we obtain for arbitrary x V2

=

T2

=

1 1 k(2x)2 mgx = (50)(4x2 ) 2 2 1 120 2 1 2 mv = v = 1:8634v 2 2 2 32:2

V2 + T2 v2 v (b)

120x = 100x2

120x

= V1 + T 1 : 100x2 120x + 1:8634v 2 = 0 120x 100x2 = = 64:40x 53:67x2 1:8634 p = 64:40x 53:67x2 ft/s J

xmax occurs when v 64:40xmax 53:67x2max

= =

0: 0

xmax = 1:200 ft J


vmax

14.39

vmax occurs when d(v 2 )=dx = 0: 64:40 2(53:67)x = 0 x = 0:60 ft p = 64:40(0:60) 53:67(0:60)2 = 4:40 ft/s J

1 2.5 in. 2 1 2

V1 V2 T2

10 in. . 78 in 10.30

= 0:15 in. = 0:0125 ft = (10:307 8 10) + 0:15 = 0:4578 in. = 0:0381 5 ft 1 2 = k 21 = 28(0:012 5)2 = 0:004 375 lb ft = 2 k 2 1 1 2 2:5 = Wy + 2 k 2 = 0:8 + 28(0:038 15)2 = 2 12 1 0:8 1 mv22 = v22 = 0:012 422v22 = 2 2 32:2 V1 + T1 v2

= V2 + T2 0:004 375 + 0 = = 3:24 ft/s J

0:125 92 lb ft

0:125 92 + 0:012 422v22

14.40 Datum

5 ft

1.0 ft 6 ft Position 1 V1

=

1:2(1:0)(0:5) =

V2

=

1:2(6)(3) =

T 1 + V1 v2

Position 2 0:6 lb ft

21:6 lb ft

= T2 + V2 0 = 13:71 ft/s J

v2

T1 = 0 1 1:2(6) 2 T2 = v = 0:11180v22 2 32:2 2 0:6 = 0:11180v22

21:6


14.41 Choose line AC as the datum for gravitational potential. V1

V2 T2

1 h (Vg )1 + (Ve )1 = mgR sin 30 + k R(45 2 1 15 = 0:21(9:81)(0:5) sin 30 + (80) 0:5 2 180 = mgR = 0:21(9:81)(0:5) = 1:0301 N m 1 1 mv 2 = (0:21)v22 = 0:105v22 = 2 2 2 =

T 1 + V1 v2

30) 2

180

i2

= 1:2004 N m

= T2 + V2 0 + 1:2004 = 0:105v22 + 1:0301 = 1:274 m/s J

14.42


14.43

14.44


14.45

14.46


14.47 Choose point O as the datum for gravitational potential energy. VA VB TA

1 1 = mgR + k(OA L0 )2 = 2:4(1:5) + (10)(1:5 1:0)2 = 4:850 lb ft 2 2 p 1 1 2 = mgR + k(OB L0 ) = 2:4(1:5) + (10)(1:5 2 1:0)2 2 2 = 2:687 lb ft 1 2:4 2 1 2 2 = v = 0:03727vB = 0 TB = mvB 2 2 32:2 B VA + T A = V B + T B 2 4:850 + 0 = 2:687 + 0:03727vB

vB = 7:62 ft/s J


14.48 On surface of earth (R = Re ): V1

= =

6:673 GMe m = Re 7:50 1010 J

10

11

(5:974 1024 )(1200) 6378 103

In outer space (R ! 1), V2 = 0. ) Energy required = V2

V1 = 7:50

1010 J J

14.49 V1

=

T1

=

V2

=

(3:98 1014 )m GMe m = 6:633 107 m N m = R1 6 106 1 1 mv 2 = m(10 500)2 = 5:513 107 m N m 2 1 2 GMe m (3:98 1014 )m = 1:730 4 107 m N m = R2 23 106 V1 + T1

6:633

107 m + 5:513

= V2 + T 2

107 m = v2

=

1 107 m + mv22 2 3490 m/s J 1:730 4

14.50


14.51

mg ma F N FBD F P

MAD

4200 (12) = 1565:2 lb 32:2 5280 = 137 740 lb ft/s = 250 hp J = F v = 1565:2 60 3600 = ma =

14.52 Let F be the driving force. P

= F v = (ma) v = P dx = v 2 dv m

)

dv dx dv v = mv 2 dx dt dx P 1 3 x= v +C m 3

m

Initial condition: v = 30 mi/h = 44 ft/s when x = 0. 1 2 (44)3 (ft/s) 3

)C=

When v = 60 mi/h = 88 ft/s: x =

x=

m 3 (v 3P

3600=32:2 (883 3(40 550)

443 )

443 ) = 1010 ft J

14.53 Let F be the driving force. P v dv

dv v dt 1 2 P v = t+C 2 m

= F v = (ma) v = m =

P dt m

Initial condition: v = v0 when t = 0. ) C = v02 =2 r 1 2 P 1 2 P ) v = t + v0 v = 2 t + v02 2 m 2 m When t = 60 s: v=

s

2

450 150

103 (60) + 102 = 21:4 m/s J 103


14.54 P =

dW h 15 = 600 = 12 857 lb ft/s = 23:4 hp J dt 0:7

14.55 Let P be the power supplied to driving wheels. From Sample Problem 14.6: Px =

When x = )

1 mv 3 + C 3 5280 3600

0, v = 30 mi/h = 30 0=

When x = ) P

=

Pengine

=

1 3200 (44:0)3 + C 3 32:2

= 44:0 ft/s C=

320 ft, v = 50 mi/h = 50

2:822

5280 3600

= 73:33 ft/s

3200 (73:33)3 2:822 32:2 32 000 32 000 lb ft/s = = 58:18 hp 550 P 58:18 = = 71:0 hp J 0:82

P (320) =

1 3

106 lb ft2 /s

106

14.56 Pin Pout

240 hp = 132 103 lb ft/s dW 62:4 1600 = h= (58) = 96:51 103 lb ft/s dt 60 Pout 96:51 103 = 100% = 100% = 73:1% J Pin 132 103 =

14.57

v

y. x.

100

2.8

From the velocity diagram: y_ = p

2:8 v = 0:02799v 1002 + 2:82


P v

= mg y_ 4800 = 12:49 m/s J

103 = (1400

103 )(9:81)(0:02799v)

14.58

14.59

14.60

320(9.81) N F

20o

y 0.35N

Fy Fx

= =

0 0 F

x N FBD

N 320(9:81) cos 20 = 0 N = 2950 N F 0:35N 320(9:81) sin 20 = 0 0:35(2950) 320(9:81) sin 20 = 0 F = 2106 N

P = F v0

4000 = 2106v0

v0 = 1:899 m/s J


14.61 (a) Pout

= =

(mg)v = 500(9:81)(0:68) = 3335 W 3335 Pout 100% = 100% = 79:4% J Pin 4200

(b) v=

Pout 3335 = = 0:425 m/s J mg 800(9:81)

14.62

14.63

14.64 The propelling force during acceleration is F = ma + FD = ma + 0:14v 2


The power of F is )a=

P = F v = mav + 0:14v 3

P

0:14v 3 mv

(a) When v = 50 km/h = 50(1000=3600) = 13:889 m/s: a=

103 0:14(13:889)3 2 = 5:75 m/s J 2000(13:889)

160

(b) When v = 110 km/h = 110(1000=3600) = 30: 56 m/s: a=

103 0:14(30:56)3 2 = 2:55 m/s J 2000(30:56)

160

14.65 ) P = F v = F0 v + cv 3

F = F0 + cv 2 19:3 = F0 (50) + c(50)3 32:3 = F0 (60) + c(60)3 The solution is: c = 1:384 8 When v = 70 mi/h, we have

10

4

19:3 = 50F0 + 1:25 32:3 = 60F0 + 2:16

105 105

, F0 = 0:03979

P = 0:03979(70) + (1:3848

10

4

)(703 ) = 50:3 hp J

14.66

y

15 lb x

F 35o

0.6N

vx

35o 3 ft/s

Velocity diagram N FBD

Fy Fx

=

0 N cos 35 + 0:6N sin 35 15 cos 35 = 0 N = 10:562 lb = 0 F 15 sin 35 + N sin 35 0:6N cos 35 = 0 F 15 sin 35 + 10:562 sin 35 0:6(10:562) cos 35 = 0 F = 7:737 lb P = F vx = 7:737(3 cos 35 ) = 19:01 lb ft/s J


*14.67

14.68 T0 v1 L0

1

L0

1

14.69 L1

2

=

Z

1 1 T1 v2 mv02 T1 = mv12 ) = 12 = 0:85 2 2 T0 v0 p p = 0:85v0 = 0:85(20) = 18:439 m/s =

= p1

+ " L0 1 = mv1 ( mv0 ) 0:25 (20 + 18:439) = 0:298 lb s J = m(v0 + v1 ) = 32:2

5s

F dt =

0

p0

Z

5s

(2ti

0:6t2 j)dt = t2 i

0:2t3 j

0

mv1 + L1

2

v2

= mv2 = 22:5i

2(10i) + 25(i 12:5j m/s J

5s 0

= 25(i

j) N s

j) = 2v2

14.70 v1 v2 L1

2

= 60(cos 30 i + sin 30 j) = 51:96i + 30j ft/s = 60j ft/s = P t = 2:0P lb s = m(v2 v1 ) 3 2:0P = [(0 51:96)i + (60 32:2 P = 2:42i + 1:398j lb J

L1

2

30)j]


14.71

80 lb

40 lb o

30 y 0.4N N FBD

(L1

2 )x

x

80 + 40 sin 30 = 0

Fy = 0

N

= m(v2

v1 )

[40 cos 30

N = 60 lb

0:4(60)] (4) =

v2 = 17:13 ft/s J

80 (v2 32:2

0)

14.72 W FBD F =µsW

N=W

The limiting force between the tires and the road is F = L0

1

t

sW

= 0:65(2800) = 1820 lb

= p1 p0 + ! F t = mv1 mv0 m 2800=32:2 5280 = (v0 v1 ) = 60 0 = 4:20 s J F 1820 3600


14.73

14.74

14.75

W

mv2 60o

y x

N FBD during contact

68o mv1 Momentum just before impact

Momentum just after impact

(a) (L1

2 )x

v2

= m [(v2 )x (v1 )x ] 0 = m(v2 cos 60 cos 68 cos 68 = v1 = 28 = 20:98 ft/s J cos 60 cos 60

v1 cos 68 )


(b) (L1

2 )y

= m [(v2 )y (v1 )y ] = m(v2 sin 60 + v1 sin 68 ) 2:6=16 = (20:98 sin 60 + 28 sin 68 ) = 0:223 lb s J 32:2

14.76 W

x N FBD during contact L1 2 1:8j

θ2

y 45o mv1 Momentum just before impact

= m(v2 v1) = 0:05 [v2 (cos

2i

+ sin

mv2

Momentum just after impact

20(cos 45 i

2 j)

sin 45 j)]

Equating like components: 0 = v2 cos 1:8 = v2 sin 0:05 The solution is v2 cos ) v2 2

20 cos 45

2 2

+ 20 sin 45

= 14:142, v2 sin 2 = 21:86. p = 14:1422 + 21:862 = 26:0 m/s J 21:86 = tan 1 = 57:1 J 14:142

2

14.77 Use principle of impulse and momentum to compute the velocity v2 immediately after the initial impulse: L1

2

= m (v2

v1 )

10 =

25 (v2 32:2

0)

v2 = 12:880 ft/s

25 lb

0.45(25) = 11.25 lb N = 25 lb FBD


Use principle of work and kinetic energy to …nd the distance x3 traveled: U2

= T3

3

T2

11:25x3 = 0

1 2

25 32:2

(12:8802 )

x3 = 5:72 ft J Use principle of impulse and momentum to compute the time t3 of travel: L2

3

= m(v3 t3 =

v2 )

11:25t3 =

25 (0 32:2

12:880)

12:880 = 0:889 s J 0:45(32:2)

14.78

14.79


14.80 y m

L0 1 (mg sin 15 ) t j ) v1

mg

v0 60o x

15

= m(v1 = m [v1

o

n15 mg si o 15

o

v0 ) v0 (cos 60 i + sin 60 j)]

= v0 (cos 60 i + sin 60 j) g t sin 15 j = 2(cos 60 i + sin 60 j) 9:81(0:5) sin 15 j = 1:0i + 0:463j m/s J

14.81 L1

2

= m(v2 v1 ) = 10 [3:6i = 7:30i + 25j N s J

5(i cos 30

j sin 30 )]

14.82 F ) L1 L1

2

2

(4)2 = pA = 4(1 e 0:5t ) = 50:27(1 e 0:5t ) lb 4 Z 10 Z 10 = F dt = 50:27 (1 e 0:5t )dt = 402:8 lb s

= m(v2

0

v1 )

0

402:8 =

500 (v2 32:2

0)

v2 = 25:9 ft/s J

14.83


14.84

14.85

14.86 L1

2

=

Z

t2

F (t) dt =

0

L1

2

Z

t2

(600

0

= m(v2 t2

v1 )

600t2

240t + 4200 = 0

5t) dt = 600t2

5 2 t 2 2

5 2 t = 3500(3 0) 2 2 t2 = 19:0 s J


14.87 Impulse of F

= = =

(L1

2 )F

=

Z

1:0 + 1:2 1:0 + 1:2 (0:75) + 1:2(1:25) + (1:0) W 2 2 3:425W "

Impulse of W = (L1 (L1

2 )F

(L1

3:425W

F dt = area under F -t diagram

2 )W

3W

2 )W

=W

= m(v2 v1 ) W (v2 0) = 9:81

t = 3W #

v2 = 4:17 m/s J

14.88

14.89

14.90 v

= =

r = hO

= =

0:2i + 0:2j 0:3k 0:22 + 0:22 + 0:32 2:425i + 2:425j 3:638k m/s ! OA = 0:6j m i j k 2:425 2:425 3:638 (0:2) (r v)m = 0 0:6 0 0:437i 0:291k N m s J (5) p


14.91

y A

O

mv sin40o mv mv cos40o x

R 40o

(a) hO = mvR =

2=16 (12)(1:5) = 0:0699 lb ft s 32:2

J

(b) hA

= mv cos 40 (R cos 40 ) mv sin 40 (R 2=16 (12)(1:5)(1 = mvR(1 sin 40 ) = 32:2 = 0:0250 lb ft s J

R sin 40 ) sin 40 )

14.92 2θ

v

y

m r O

R 2θ

θ

r = 2R cos (i cos + j sin )

h = r =

(mv) = 2mvR cos

x

v = v( i sin 2 + j cos 2 ) i cos sin 2

j sin cos 2

k 0 0

2mvR cos (cos cos 2 + sin sin 2 )k = 2mvR cos2

kJ


14.93


14.94

14.95

14.96

. θ

z

R M

=

h1

= Z

t2

M dt

=

0

h2 = h1 +

Z

M m 12 N.m 0

0

t2 t

1.0 s

mv1 Rk = m(R _ 1 )Rk = mR2 _ 1 k 12 (3)2 (5)k = 16:770k lb ft s 32:2 1 (12)(1:0) + 12(t2 1:0) k = (12t2 6) k lb ft s 2

t2

M dt

0 = [ 16:770 + (12t2

0

6)] k

t2 = 1:898 s J


14.97

z L

θ

R

ω

m

L

h = mRvk = m(L sin )(!L sin )k = mL2 ! sin k h1 = 0:4(0:6)2 (12) sin2 60 k = 1:296k N m s h2 = 0:4(0:6)2 ! 2 sin2 25 k = 0:02572! 2 k N m s h1 !2

= h2 1:296k =0:02572! 2 k = 50:4 rad/s J

14.98

x mg

A

FBD

Because there are no horizontal forces acting on the projectile, the horizontal component of velocity is constant. Therefore, x = (v0 cos

(AA )1

2

=

Z

0) t

= (200 cos 30 ) t = 173:21t m

10 s

MA dt =

0

=

2(9:81)(173:21)

Z

10 s

mgx dt =

0

Z

10 s

2(9:81)(173:21t) dt

0

102 2

= 169:9

103 N m s


(hA )2

(hA )1 (hA )2

= (AA )1 2 (hA )2 0 = 169:9 3 = 169:9 10 N m s J

103 N m s

14.99

14.100

14.101 (a) Energy is conserved. Choose Position 2 as the datum for gravitational potential energy

z R N T 1 + V1 v2

mg

FBD

1 1 = T 2 + V2 mv12 + mgh = mv22 + 0 2 2 q p 2 2 = v1 + 2gh = 6 + 2(9:81)(0:5) = 6:768 m/s J

(b) Angular momentum about the z-axis is conserved (hz )1 cos

=

(hz )2 (mv1 ) R = (mv2 cos ) R v1 6 = = = 0:8865 = 27:6 J v2 6:768


14.102

14.103 Angular momentum about the z-axis is conserved: (hz )1 = (hz )2 (m! 1 R1 )R1 = (m! 2 R2 ) R2 !2 = !1

R12 = 10 R22

62 152

= 1:6 rad/s

(v )2 = ! 2 R2 = 1:6(15) = 24:0 in./s J Kinetic energy is conserved: T1 = T2 1 1 m(! 1 R1 )2 = m (v )22 + (vR )22 2 2 q (vR )2

=

14.104

(! 1 R1 )2

=

p

(v )22

24:02 = 55:0 in./s J

(102 )(62 )

z

W T

θ

N=W

R


(a) Angular momentum about z-axis is conserved. (hz )1

=

_2

=

(hz )2 R1 (mv )1 = R2 (mv )2 R1 (mR1 _ 1 ) = R2 (mR2 _ 2 ) R1 R2

2

_1 =

1:5 0:8

2

(6) = 21:09 rad/s J

(b) hz

= R(mv ) = R(mR _ ) = mR2 _

h_ z

= m(2RR_ + R2 •) = 0

)•=

2R_ _ R

When R = 0:8 ft: •2 =

2R_ _ 2 = R2

2( 0:3)(21:09) 2 = 15:82 rad/s J 0:8

14.105

14.106


14.107 h20 GMe (1 + e cos ) RA 10:5 106 1 + e cos 65 1 + e cos 70 ) = = RB 1 + e cos 0 16 106 1+e Since e > 1, the trajectory is a hyperbola J R=

10:5

RA

=

106

=

h0

=

h0 = RA v0

e = 1:4713

h20 GMe (1 + e cos 0) (6:674 10:170 v0 =

h20 10 11 )(5:972 10

10

1024 )(1 + 1:4713)

2

m =s

h0 10:170 1010 = 9690 m/s J = RA 10:5 106

14.108 GMe Re

= (6:674 10 11 )(5:972 = 6378 103 m

v1

1024 ) = 3:986

1014 m3 =s2

h2

h1 O

v2 R1

R2


R1 v1 ) R2

= R2 v2 (angular momentum about O is conserved) v1 6 = R1 = R1 = 0:8R1 v2 7:5

1 2 GMe 1 2 GMe v1 = v (energy is conserved) 2 R1 2 2 R2 1 1 1 2 (v v22 ) = GMe 2 1 R1 R2 1 1 1 (60002 75002 ) = 3:986 1014 2 R1 0:8R1 R1 h1 h2

= 9:842 106 m R2 = 0:8(9:842 106 ) = 7:874 106 m = R1 Re = (9:842 6:378) 106 = 3:464 106 m J = R2 Re = (7:874 6:378) 106 = 1:496 106 m J

14.109


14.110 From Prob.(14.109):

= 2:359

106 s

14.111 From Table 14.2 (for Mars): GM = 6:674

10

11

(0:6417

1024 ) = 4:283

1013 m3 /s

2

14.112 (a) Rmin Rmax GMe

= 6378 + 2500 = 8878 km = 6378 + 4500 = 10 878 km = (6:674 10 11 )(5:972 1024 ) = 3:986 1014 m3 =s2 Rmax Rmin 10 878 8878 e = = 0:101 24 = Rmax + Rmin 10 878 + 8878


vmax vmin

=

=

s s

GMe (1 + e) = Rmin GMe (1 e) = Rmax

r r

(3:986

1014 )(1 + 0:101 24) = 7032 m/s J 8878 103

(3:986

1014 )(1 0:101 24) = 5739 m/s J 10 878 103

(b) h0

103 (5739) = 6:243

= Rmax vmin = 10 878

1010 m2 =s

2 h30 2 (6:243 1010 )3 = (GMe )2 (1 e2 )3=2 (3:986 1014 )2 (1 0:101 242 )3=2 11 290 s J

= =

14.113 Find speed at A just before retrorockets are …red. The orbit is circular (e = 0). R ) h0 h0

h20 GM p e p = RGMe = (6688 103 )(6:674 10 11 )(5:972 = 5:163 1010 m2 =s 5:163 1010 h0 = = 7720 m/s = Rv1 v1 = R 6688 103

=

1024 )

Find speed at A just after the retrorockets are …red. The new orbit is elliptic with Rmax = R and Rmin = Re e=

6378

Rmax Rmin 6688 6378 = 0:02373 = Rmax + Rmin 6688 + 6378

Rmin

=

103

=

h0

=

h0 = Rv2 L1

2

= m(v2

h20 GMe (1 + e) (6:674 5:101 v2 =

h20 10 11 )(5:972 1024 )(1 + 0:02373) 1010 m2 =s h0 5:101 = R 6688

v1 ) = 6000(7627

1010 = 7627 m/s 103

7720) =

5:6

105 N s J

The answer is accurate to only 2 signi…cant …gures because 2 …gures were lost in the subtraction.


14.114 From Table 14.2 (for Venus): GM Rv

1016 ft3 /s

= (3:440 10 8 )(333:5 1021 ) = 1:1472 = 3761 mi = 19:858 106 ft

x1

2

F Rv

y1

R1

vmax

Hmin

v1 (a) q

x21 + y12 =

p

52 + 12

R1

=

h0

= v1 y1 = 8000(1:0

108 = 5:099

108 ) = 800

108 ft

109 ft2 /s

Equation (14.55): E0

=

1 2 v 2 1

=

9:501

GM 1 = (8000)2 R1 2

1016 108

1:1472 5:099

2

106 (ft/s)

Since E > 0, the trajectory is a hyperbola J (b) Equation (14.57): s 2 h0 e = 1 + 2E0 GM s 800 109 = 1 + 2(9:501 106 ) 1:1472 1016

2

= 1:0452

Equation (14.62): 2

800 109 h20 = GM (1 + e) 1:1472 1016 (1 + 1:0452)

Rmin

=

Hmin

= 27:278 106 ft = Rmin Rv = (27:278 19:858) = 7:42 106 ft = 1405 mi J

106


(c) Angular momentum about F is conserved: vmax Rmin = h0

vmax =

h0 800 109 = 29 300 ft/s J = Rmin 27:278 106

14.115 From Table 14.2 (for Venus): GM Rv

(3:440 10 8 )(333:5 1021 ) = 1:1472 3761 mi = 19:858 106 ft

= =

2

Rv

x1 y1

1016 ft3 /s

Rmin

R1 v1

(a) R1 = Equation (14.55):

q p x21 + y12 = 52 + 12 E0 =

1 2 v 2 1

108 = 5:099

108 ft

GM R1

For elliptical orbit E0 < 0. In the limit E0 = 0 (parabola), in which case Eq. (14.55) yields 0 v1

1 2 GM v 2 1 R1 r r 2GM 2(1:1472 1016 ) = = = 6710 ft/s J R1 5:099 108 =

(b) Equation (14.62): Rmin =

h20 GM (1 + e)

Substituting h0 = v1 y1 and e = 1, we get Rmin =

(6710)2 (1:0 108 )2 = 19:623 (1:1472 1016 )(2)

106 ft

Since Rmin < Rv , the spacecraft would crash. J


14.116 (a) At insertion point: = Re + H = (6:378 + 0:240) 106 = 6:618 106 m = vR cos = 9600(6:618 106 ) cos 4 = 63:38 109 m2 /s

R h0

Equation (14.55): E0

1 2 GM v = 2 R 14:150 106

= =

(b) Equation (14.57): s

1 + 2E0

e =

s

1 3:986 1014 (9600)2 2 6:618 106 2 (m/s) < 0 ) Trajectory is an ellipse J

h0 GM

1 + 2( 14:150

=

2

106 )

63:38 3:986

109 1014

2

= 0:5334

Solving Eq. (14.61) for : =

1

cos

1

=

cos

=

11:45

"

1 e

h20 R GM

1 0:5334

1

63:38 109 (6:618 106 )(3:986

2

1014 )

!#

1

J

(c) Equation (14.62): 2

63:38 109 h20 = = 6:572 GM (1 + e) (3:986 1014 )(1 + 0:5334)

Rmin

=

Hmin

= Rmin

Re = (6:572

6:378)

106 = 1:94

106 m

105 m = 194 km J

14.117 (a) At point A: R E0

= Re + H = (6:378 + 0:240) 106 = 6:618 106 m 1 3:986 1014 1 2 GMe v = (11:4 103 )2 = = 4:750 2 R 2 6:618 106

2

106 (m/s)

Since E0 > 0 , the trajectory is a hyperbola. Q.E.D.


(b) When R ! 1, then E0 ! ) v1 =

14.118

p

2E0 =

1 2 v 2 1

p 2 (4:750

106 ) = 3080 m/s J

For the circular orbits: R1 = 3800 mi = 20:06

106 ft

R2 = 25 400 mi = 134:11

106 ft

Equation (14.66): v1 v2

= =

r

r

GMe = R1 GMe = R2

r r

1:4076 20:06

1016 = 26:49 106

1:4076 134:11

1016 = 10:245 106

103 ft/s 103 ft/s

For the transfer orbit: Rmin = R1

Rmax = R2

Equation (14.68): e=

Rmax Rmin 134:11 20:06 = 0:7398 = Rmax + Rmin 134:11 + 20:06

From Table 14.2: vmax

=

s

GMe (1 + e) = Rmin

r

(1:4076

1016 ) (1 + 0:7398) 20:06 106

34:94 103 ft/s s r GMe (1 e) (1:4076 1016 ) (1 0:7398) = = Rmax 134:11 106

= vmin

= At A

: =

At B

: =

5:226

103 ft/s

Lpark-tranfer = m(vmax v1 ) 2000 (34:94 26:49) 103 = 525 103 lb s J 32:2 Ltransfer-geo = m(v2 vmin ) 2000 (10:245 5:225) 103 = 312 103 lb s J 32:2


14.119

14.120 GMe Re

= (6:674 10 = 6378 km

11

)(5:972 1024 ) = 3:986 Rmax = Re + H

1014 m3 =s2

At point A: (h0 )A

=

(vA cos ) Re = (2500 cos 45 ) (6378

(E0 )A

=

1 2 v 2 A

GMe 1 = (2500)2 Re 2

3:986 6378

103 ) = 1:1275 1014 = 103

5:937

1010 m2 =s 107 m2 =s2

At point B: (h0 )B = vB Rmax

(E0 )B =

1 2 v 2 B

GMe 1 2 = vB Rmax 2

3:986 1014 Rmax


Angular momentum and mechanical energy are conserved: (h0 )B

=

(h0 )A :

(E0 )B

=

(E0 )A :

vB Rmax = 1:1275 1010 m2 =s 1 2 3:986 1014 vB = 5:937 2 Rmax

(a) 107 m2 =s2

(b)

2 Multiply Eq. (b) by Rmax :

1 2 2 v R 2 B max

1014 Rmax + 5:937

3:986

2 107 Rmax =0

Substitute for vB Rmax from Eq. (a): 1 2 (1:1275 1010 )2 3:986 1014 Rmax + 5:937 107 Rmax 2 2 5:937 107 Rmax 3:986 1014 Rmax + 6:356 1019 2 5:937Rmax

3:986

10

7

Rmax + 6:356

10

12

=

0

=

0

=

0

Solution is Rmax H

106 m = 6550 km Re = 6550 6378 = 172:0 km J

= 6:550 = Rmax

14.121 (a)

maθ R

FR

= maR

• R

= R_

2

• R

= R_

2

=

F

θ FBD

O

+%

• F = m(R

+-

MAD

O

3:440 2 GMe = R_ R2 1:4076 1016 J R2

F = ma

maR

2

R_ ) 10

8

GmMe • = m(R R2 (409:2 1021 )

2

R_ )

R2

0 = m(R• + 2R_ _ )

•=

2R_ _ J R


= Re + H0 = (3963 + 480) (5280) = 2:346 107 ft J 5280 _ R(0) = v0 sin 5 = 17 500 sin 5 = 2237 ft/s J 3600 (0) = 0 J cos 5 _ (0) = v0 cos 5 = 17 500 5280 R(0) 3600 2:346 107

R(0)

=

1:0899

10

3

rad/s J

_ x3 = and x4 = _ , the equivalent …rst-order (b) Letting x1 = R, x2 = R, equations and the initial conditions are: x_ =

x2

x(0) =

x1 x24 2:346

1:4076 1016 x21 107

2237

0

x4 1:0899

2x2 x4 x1 10

3

T

T

The corresponding MATLAB program is: function problem14_121 [t,x] = ode45(@f,(0:50:7500),[2.346e7,2237,0,1.0899e-3]); printSol(t,x) plot(t,x(:,1),’linewidth’,1.5) xlabel(’t (s)’); ylabel(’R (ft)’) grid on printSol(t,x) function dxdt = f(t,x) dxdt = [x(2) x(1)*x(4)^2-1.4076e16/x(1)^2 x(4) -2*x(2)*x(4)/x(1)]; end end


7

3

x 10

2.9 2.8

R (ft)

2.7 2.6 2.5 2.4 2.3 2.2

0

1000

2000

3000

4000 t (s)

5000

6000

(c) The following lines of printout span the time when t 7.0000e+003 7.0500e+003

7000

8000

=2 :

x1 x2 x3 x4 2.3402e+007 2.1687e+003 6.2548e+000 1.0953e-003 2.3513e+007 2.2835e+003 6.3093e+000 1.0850e-003

Use linear interpolation to …nd t at = 2 : 6:3093 6:2548 2 6:2548 = 7050 7000 t 7000

t = 7026 s J

(d) The partial printouts shown below span Rmax and Rmin : t 2.7500e+003 2.8000e+003 2.8500e+003

x1 x2 x3 x4 2.9411e+007 1.0793e+002 2.2914e+000 6.9351e-004 2.9414e+007 1.5877e+000 2.3261e+000 6.9338e-004 2.9412e+007 -1.0476e+002 2.3608e+000 6.9350e-004

6.2500e+003 6.3000e+003 6.3500e+003

2.2616e+007 -2.3204e+002 2.2609e+007 -5.2497e+001 2.2611e+007 1.2733e+002

5.3933e+000 5.4520e+000 5.5107e+000

1.1728e-003 1.1735e-003 1.1733e-003

By inspection, we …nd that Rmax

= 29:41 106 ft = 5570 mi Rmin = 22:61 106 ft = 4282 mi ) Hmax = Rmax Re = 5570 3963 = 1607 mi J ) Hmin = Rmin Re = 4282 3963 = 319 mi J


14.122 (a) The equations of motion were derived in the solution of Prob. 14.121: 1:4076 1016 J R2

• = R_2 R

•=

2R_ _ J R

The initial conditions are: = Re + H0 = (3963 + 480) (5280) = 2:346 107 ft J 5280 _ sin 5 = 1917:4 ft/s J R(0) = v0 sin 5 = 15 000 3600 (0) = 0 J cos 5 _ (0) = v0 cos 5 = 15 000 5280 R(0) 3600 2:346 107

R(0)

=

0:9342

10

3

rad/s J

_ x3 = and x4 = _ , the equivalent …rst-order (b) Letting x1 = R, x2 = R, equations and the initial conditions are: x_ = x(0) =

x2

x1 x24

2:346

1:4076 1016 x21 107

1917:4

0

x4 0:9342

T

2x2 x4 x1 10

3

T

The corresponding MATLAB program is: function problem14_122 [t,x] = ode45(@f,(0:10:1500),[2.346e7,1917.4,0,0.9342e-3]); printSol(t,x); plot(x(:,3)*180/pi,x(:,1)/5280,’linewidth’,1.5) xlabel(’theta (deg)’); ylabel(’R (mi)’) grid on printSol(t,x) function dxdt = f(t,x) dxdt = [x(2) x(1)*x(4)^2-1.4076e16/x(1)^2 x(4) -2*x(2)*x(4)/x(1)]; end end


4600

4500

4400

R (mi)

4300

4200

4100

4000

3900

0

10

20

30

40 50 theta (deg)

60

(c) Partial printout spanning R = Re = 2:0925 t 1.4400e+003 1.4500e+003

x1 x2 2.0942e+007 -5.0507e+003 2.0892e+007 -5.0835e+003

70

80

90

107 ft:

x3 1.3984e+000 1.4102e+000

x4 1.1723e-003 1.1780e-003

Linear interpolation: 2:0892 1:4102

2:0942 2:0925 2:0942 = 1:3984 1:3984

= 1:4024 rad = 80:4

J

14.123

y 24 lb 10 lb

FBD

o

0.2N Fy L1 2

30 x N

= 0 N 24 + 10 sin 30 = 0 N = 19:0 lb = m(v2 v1 ): Fx t = m(v2 v1 ) 24 [10 cos 30 0:2(19:0)] t = (50 15) 32:2 t = 5:37 s J


14.124 Since there is no impulse in the x-direction, vx does not change: vx = 3 cos 35 = 2:458 m/s (constant) (L1

2 )y

=

(py )2

Z

(py )1 :

10 s

P dt = m(vy )2

m(vy )1

0

(vy )2

v2

14.125

Z 1 10 s 0:3t e dt = 3 sin 35 m 0 = 2:239 m/s p 2:4582 + 2:2392 = 3:32 m/s J = =

1 (3:1674) 0:8

(vy )1

(a) T A + VA

= TB + VB = TC + VC 1 1 2 = mv 2 + 0 = mvC + 8mg 2 B 2

0 + 24mg

p 2(24)(9:81) = 21:7 m/s J p = 2(16)(9:81) = 17:72 m/s J

vB

=

vC (b)

mg MAD

FBD man

N Fn = man

+#

mg

N =m

2 vC

=

2 mvC mg N

Contact is lost when N = 0. )

=

2 vC 17:722 = = 32:0 m J g 9:81

14.126 U1

2

= T2

T1 :

k W (5

1 2 k =0 2 1 35 = (12)2 2 32:2

+ )

1 0:2(35)(5 + ) + (280) 2 = 0:5314 ft

2

1 mv 2 2 1

F = k = 280(0:5314) = 148:8 lb J


14.127 Choose O as the datum for gravitational potential energy. Between positions A and B energy is conserved. TA + VA 2 vB

= T B + VB =

2gR1 (1

0 cos

mgR1 cos

1

1 mv 2 2 B

=

mgR1

1)

Between positions B and C angular momentum about point O is conserved. (h0 )C 2 vC

=

(h0 )B

2 = vB

mvC R2 = mvB R1

R1 R2

2

= 2gR1 (1

cos

R1 R2

1)

2

Between positions C and D energy is conserved. 1 mv 2 2 C 2 vC = 2gR2 (1

mgR2 = 0

TC + VC = TD + VD

2gR1 (1 (1 cos

2

=1

(1

cos

cos cos 1)

1)

1)

R1 R2 2

R1 R2

cos

mgR2 cos

2

2)

2

R1 R2

= 2gR2 (1

cos

2)

3

=1

cos

2

3

=1

(1

= 48:6

J

cos 45 )(1:05)3 = 0:6609

14.128


14.129

y mg

P(N) 60 30 0 0 4

FBD Ps

µsmg

x N = mg

8

12

t(s)

First determine if the crate will move. The smallest force that would move the crate is Ps = s mg = 0:3(16)(9:81) = 47:09 N Since Ps < 60 N, the crate will move. L1

L1

2

2

= (area under P -t diagram) t k mg 1 = 60 8 (30 4) 0:25(16)(9:81)(8) = 106:08 N s 2 = m(v2

v1 ):

106:08 = 16(v2

0)

v2 = 6:63 m/s J

14.130 Energy is conserved. Choose horizontal plane at O as the datum for Vg . p = AB L0 = 52 + 42 + 32 2:5 = 4:571 ft A 1 1 VA = W OA + k 2A = 8(5) + (4:2)(4:571)2 = 83:88 lb ft 2 2 O

VO TO TA + VA vO

p = OB L0 = 42 + 32 2:5 = 2:50 ft 1 2 1 = k O = (4:2)(2:50)2 = 13:125 lb ft 2 2 1 1 8 2 2 2 = mvO = v = 0:12422vO lb ft 2 2 32:2 O 2 = TO + VO 0 + 83:88 = 0:12422vO + 13:125 = 23:9 ft/s J

14.131


14.132

14.133 Angular momentum about AB is conserved: m(R1 ! 1 )R1 = m(R2 ! 2 )R2

T

= T2

T1 =

T T1

=

R2 ! 2 R1 ! 1

% energy lost

=

T T1

2

1 m(R2 ! 2 )2 2 R2 1= R1 "

100% = 1

!2 = !1

R1 R2

2

1 m(R1 ! 1 )2 2 2 4 R1 R1 1= R2 R2 # 2 R1 100% J R2

2

1

14.134 Energy is conserved: TA + VA vB

1W 2 1 1W 2 1 = TB + VB v + k 2 = v + k 2 2 g A 2 A 2 g B 2 B r r gk 2 32:2(2:5) 2 2 = vA + ( A 252 + (0:52 1:02 ) B) = W 1:2 = 23:97 ft/s J


Angular momentum about O is conserved: RA vA = RB (v )B

(v )B =

Rate of elongation

14.135 (px )A vB

=

RA 2 vA = (25) = 20 ft/s RB 2:5

(vR )B = p = 23:972

q v22

2

(v )B

202 = 13:21 ft/s J

= (px )B : mvA cos = mvB = vA cos = 20 cos 35 = 16:383 m/s

(a)

mg FBD (LA

B )y

t

=

(py )B vA sin = g

(py )A : mgt = 0 mvA sin 20 sin 30 = = 1:019 s J 9:81

(b) TA + VA h

= T B + VB : =

2 vB

2 vA

2g

1 1 2 mv 2 + 0 = mvB + mgh 2 A 2 202 16:3832 = = 6:71 m J 2(9:81)

14.136

Dimensions in 108 feet

38.41o

y

3.960

5.0 1.0 A

vB B 3.382 x

O vA = 16 000 ft/s


Angular momentum about point O is conserved. m(16 000)(1:0) = mvB (3:960 cos 38:41 3:382 sin 38:41 ) 16 000 = 1:0018vB vB = 15 970 ft/s J

14.137


14.138

14.139 LA

A

m 0.4

30o 0.2 m

Datum L2A = 0:42 + 0:22 VA

= = =

VB

= =

TB

=

2(0:4)(0:2) cos 30

LA = 0:2479 m

1 k(LA L0 )2 + mgR cos 30 2 1 (110)(0:2479 0:08)2 + 0:6(9:81)(0:4) cos 30 2 3:589 N m 1 k(LB L0 )2 + mgR 2 1 (110)(0:2 0:08)2 + 0:6(9:81)(0:4) = 3:146 N m 2 1 1 2 2 2 mvB = (0:6)vB = 0:3vB 2 2

TA + VA = TB + VB

2 0 + 3:589 = 0:3vB + 3:146

vB = 1:215 m/s J


14.140 Choose the horizontal line through O as the datum for Vg . V1 =

V2

T2

1 k(L1 2

L0 )2 =

1 (200) 2

52

42

1 k(L2 2

=

1 (200) 2

=

1 40 2 1W 2 v = v = 0:6211v22 lb ft 2 g 2 2 32:2 2

T 1 + V1 v2

p

= 69:44 lb ft

12

=

L0 )2

2

Wh

402 + 122 12

42

!2

40

48 12

=

= T2 + V 2 0 + 69:44 = 0:6211v22 = 19:22 ft/s J

159:96 lb ft

159:96


Chapter 15 15.1 vB=A

= vB vA = 260(i cos 30 j sin 30 ) = 445i 601j mi/h J

520( i cos 65 + j sin 65 )

15.2 vA = 40 m/s

vB = 15 + 0:15t m/s

vB=A

= vB

xB=A

=

vA =

25 + 0:15t m/s

25t + 0:075t2 + C

Initial condition: xB=A = 3000 m when t = 0. ) C = 3000 m 25t + 0:075t2 + 3000

xB=A = (xB=A )min occurs when vB=A xB=A

min

= =

0:

25 + 0:15t = 0

t = 166:67 s J

25(166:67) + 0:075(166:672 ) + 3000 = 917 m J

15.3 For aB=A to be zero, aB must be parallel to aA (vertical).

(at )B o

60

(an )B

=

aB

=

aB=A = aB

2 vB

aB

(an)B

162 = 1:280 m/s 200 (an )B 1:280 2 = = 1:478 m/s sin 60 sin 60

aA = 0

=

aA = aB = 1:478 m/s # J 2


15.4 Let P and W refer to the plane and the wind, respectively. = vW + vP=W

vP

y x

vP (sin i + cos j)

=

55(cos 28 i + sin 28 j) + 520j

Equating like components: vP sin vP cos

= 55 cos 28 = 48:56 mi/h = 55 sin 28 + 520 = 545:82 mi/h

48:56 tan 1 = 5:084 J 545:82 p = 48:562 + 545:822 = 548 mi/h J =

vP

15.5

Let W refer to the water and B to the boat

vW

vB

θ

vB/W

vB = vW + vB=W (a) sin =

vW 10 = = 0:4167 vB=W 24

= 24:6

J

(b) vB

=

t

=

q

2 vB=W

2 = vW

p 242

102 = 21:82 km/h

d 4:8 = = 0:220 h = 13:2 min J vB 21:82

15.6 Balls collide if vA=B is directed from A toward B. vA

= vB + vA=B

y x

2j

=

3 (cos 30 i + sin 30 j) + vA=B ( sin i + cos j)


Equating like components: 0 = 3 cos 30 vA=B sin 2 = 3 sin 30 + vA=B cos = tan

1

vA=B sin = 2:598 vA=B cos = 0:5

2:598 = 79:11 0:5

J

15.7

15.8 vA vB

vA=B vA=B

15.9

= 50i mi/h = (60 cos 25 ) i + (60 sin 25 ) j = 54:38i + 25:36j mi/h = vA vB = 4:38i 25:36j mi/h p = 4:382 + 25:362 = 25:74 mi/h J


15.10

15.11


15.12

15.13 vA vB vB=A

= 70j ft/s = 40(i cos 30 + j sin 30 ) = 34:64i + 20:0j ft/s = vB vA = 34:6i 50:0j ft/s J

6j ft/s

2

aA

=

aB

=

(aB )t + (aB )n = v_ B (i cos 30 + j sin 30 ) +

=

4(i cos 30 + j sin 30 ) +

402 (i sin 30 200

2 vB

(i sin 30

j cos 30 )

j cos 30 )

2

aB=A

= 7:464i 4:928j ft/s = aB aA = 7:464i 4:928j =

7:46i + 1:07j ft/s

2

( 6j)

J

15.14 Let W and B refer to the wind and the boat, respectively. Position (a): vW = (vB )a + (vW=B )a = 6j + va ( i cos 5 + j sin 5 )


Position (b): vW = (vB )b + (vW=B )b = 6i + vb ( i cos 28 + j sin 28 ) Equating like components: va cos 5 6 + va sin 5

= 6 vb cos 28 = vb sin 28

Solution is va = 6:349 mi/h

vb = 13:959 mi/h 6:32i + 6:55j mi/h J

vW = 6j + (6:349) ( i cos 5 + j sin 5 ) =

15.15

A

B xA

xB Length of cable: dL dt vB

L = 3xA + 2xB + constant =

0:

=

3vA + 2vB = 0

1:5vA =

1:5(4) =

6 in./s = 6 in./s ! J

15.16

A xA

yB B Length of cable dL dt vA

= L = xA + 3yB + constant = vA + 3vB = 0 =

3vB =

3( 0:4) = 1:2 m/s

! J

15.17

yB

yA B A


Length of cable: dL dt

vA=B vB

= vA =

vB

1 (8) = 2

L = 2yB + yA + constant =

0: 2vB + vA = 0

12 = vA

1 vA 2

vB =

1 vA 2

vA = 8 in./s # J

4 in./s = 4 in./s " J

15.18

15.19


15.20 yA

yB A

B

yC

C

Length of cable

= L = (yC yA ) + 2yC + (yC = 4yC yA yB + constant

dL dt

=

vC

=

vC

=

4vC

vA

yB ) + constant

vB = 0

vA + vB 3 + 1:2 = = 4 4 0:45 ft/s " J

0:45 ft/s

15.21

1.2 m yB

yA B

A

Length of cable = L = 2yB + dL dt

=

vB

=

vB

=

q 2 + 22 + constant yA

yA vA 2vB + p 2 =0 yA + 22 1 y 1 2 p A p vA = (0:8) = 2 2 2 2 yA + 2 2 2 + 1:22

0:343 m/s

0:343 m/s " J

15.22 2.4 m

2.4 m yB

yA A

B


Length of cable dL dt vB

q 2 + 2:42 + constant = L = y A + 2 yB

yB vB = 0 = vA + 2 p 2 yB + 2:42 p 2 + 2:42 yB = vA 2yB p 22 + 2:42 = ( 3:6) = 2:81 m/s " J 2(2)

15.23 20 0

300 B

β

θ

xB

300 cos + 200 cos 300 sin 200 sin

= xB = 0

(a) (b)

200 200 sin = sin 60 = 0:5774 = 35:26 300 300 From (b) : 300 _ cos 200 _ cos = 0 _ _ = 200 cos = 200(2) cos 60 = 0:8165 rad/s 300 cos 300 cos 35:26 From (a) : 300 _ sin 200 _ sin = vB vB = 300(0:8165) sin 35:26 200(2) sin 60 = 488 m/s vB = 488 m/s ! J From (b)

:

sin

=

15.24 x2 = L2 But L_ =

242

) 2xx_ = 2LL_

x_ =

LL_ x

10 in./s ) x_ =

10

L x

Using x = 18 in. L = )

p 182 + 242 = 30 in. 30 x_ = 10 = 16:67 in./s = 16:67 in./s ! J 18


15.25

15.26


15.27

1

yC yA

. D

.

C

yB B

2 A

L1 L2 L_ 1 L_ 2

= yB + yC + (constant) = 2(yA yC ) + yA + (constant) = 3yA = vB + vC = 0 = 3vA 2vC = 0

2yC + (constant)

6 + vC = 0 vC = 6 in./s 3vA 2(6) = 0 vA = 4:00 in./s # J

15.28


15.29

15.30


15.31

y

xB xA

.

GA

B

A

x

(a) The coordinate x of the mass center of the system is given by mA xA + mB xB = (mA + mB )x Since there are no external forces acting on the system in the x-direction, x does not change. Therefore, di¤erentiation with respect to time gives mA vA + mB vB = 0 ) vB =

vA

mA = mB

( 4:4)

18 = 1320 ft/s J 0:06

(b) The muzzle velocity is vB=A = vB

( 4:4) = 1324 ft/s J

vA = 1320

15.32

20 ft

A

B

G d

x 20 ft

A

B x

The mass center G of the system is located at x=

mi xi (10)600 = = 7:692 ft mi 600 + 180


The mass center of the system does not move. Hence the displacement d of the log is given by ) d + 2x = L

d=L

2x = 20

2(7:692) = 4:62 ft J

15.33

15.34 The motion of the mass center is not changed by the explosion. ax x ay y

= 0 vx = 240 cos 60 = 120 m/s = 120t m = g vy = gt + 240 sin 60 = 9:81t + 207:9 m/s = 4:905t2 + 207:9t m


When t = 35 s: x = y =

120(35) = 4200 m 4:905(35)2 + 207:9(35) = 1267:9 m

(mA + mB )x = mA xA + mB xB (20 + 40)(4200) = 20(3820) + 40xB (mA + mB )y = mA yA + mB yB (20 + 40)(1267:9) = 20(1960) + 40yB

xB = 4390 m J yB = 922 m J

15.35 (a)

2000g

1400g B

A

B

A y

0.8(2000)g 1400g

(2000+1400)a

2000g

FBD

x

MAD

Fx

= max 0:8(2000)g = (2000 + 1400)a 0:8(2000)(9:81) 2 a = = 4:616 m/s J 2000 + 1400

(b)

1400g B FBD Fx = max + !

1400a T

B MAD

1400g

T = 1400a = 1400(4:616) = 6460 N = 6:46 kN J


15.36

15.37 System (crate + cart): Fx = max

+ !

Crate:

100 + 300 a 32:2

P =

100 lb FBD P 0.2N

a = 0:0805P

100 a MAD 32.2

= N = 100 lb

Fx P

0:2(100)

= max =

+ !

100 (0:0805P ) 32:2

P

0:2N =

100 a 32:2

P = 26:7 lb J

15.38 Check if the crate will slide on the cart: s NA

= 0:2(100) = 20 lb < 32 lb ) The crate will slide.


100 lb 32 lb A

0.18(100) = 18 lb

MAD

FBD NA= 100 lb (100+300) = 400 lb 18 lb

300 a 32.2 B

B

B

FBD 400 lb Crate A : Cart B

:

100 a 32.2 A

A

MAD

Fx = mA aA + !

32

Fx = mB aB + !

18 =

18 =

100 aA 32:2

300 aB 32:2

aA = 4:51 ft/s

aB = 1:932 ft/s

2

2

J

J

15.39 System:

2850 N A 80g N

A 80a

FBD

MAD B

B

110g N

F

= ma + "

110a

2850

a = 5:190 m/s

2

(80 + 110)(9:81) = (80 + 110)a

J


Block A:

2850/2 N A 80g N

FBD

A 80a MAD

T F

= ma + "

T

=

225 N J

2850 2

80(9:81)

T = 80(5:190)

15.40 Kinematics

xB B

yA

A • = 3aA + aB = 0 L

L = 3yA + xB + (constant)

aB =

3aA

Kinetics

3T A 6aA 6(9.81) N FBD MAD

12(9.81) N T

12aB

B FBD N

MAD


Block A : Block B : Solution is

Fy = mA aA + # Fx = mB aB + !

T = 18:59 N J

6(9:81) 3T = 6aA T = 12aB = 12( 3aA )

aA = 0:516 m/s # J 2

15.41


15.42

15.43 Kinematic constraints: aA

= aB ! = aC # = a


Kinetics:

24 a 32.2

24 lb TAB

A

A MAD

0.4(24) = 9.6 lb 24 lb FBD

24 a 32.2

48 lb TAB

TBC

B 0.4(48) = 19.2 lb

B MAD

48 lb TBC FBD FBD C

C

40 lb

Block A :

Fx = maA +

Block B

:

Fx = maB + !

Block C

:

Fy = maC + #

TAB TBC 40

MAD 40 a 32.2

9:6 = TAB

TBC =

24 a 32:2 19:2 =

24 a 32:2

40 a 32:2

The solution is a = 4:10 ft/s

2

TAB = 12:66 lb J

TBC = 34:9 lb J

15.44 Kinematics: aB =

1 aA = 2 m/s. 2

WA T FBD's

P

A

N y T T

A

mA aA MAD's mBaB

x

B B WB


Kinetics: Block B: Fy = ma 2T 8(9:81) = 8(2)

+ " 2T WB = mB aB T = 47:24 N

Block A: Fx = ma P 47:24 = 3(4)

+ ! P T = mA aA P = 59:2 N J

15.45

0.8 lb P

A

N

x

B

FBD

0.2 lb

0.8 ag

y

T T

40o B

A

0.2 ag MAD

Bob B: Fy Fx

=

0

T cos 40

0:2 = 0 T = 0:2611 lb 0:2 a = ma T sin 40 = g 0:2 2 0:2611 sin 40 = a a = 27:02 ft/s 32:2

Block A: Fx

= ma P

0:8 a 32:2 0:8 0:2611 sin 40 = (27:02) 32:2 P

T sin 40 =

P = 0:839 lb J

15.46 o 15o mg 45o mg 15 o T A 15 B t t T N NA B FBD's

Ft = mat

+&

A man

B

mat man MAD's

mg sin 15 + T cos 15 mg sin 45 T cos 15

= mat = mat

mat

(Block A) (Block B)


Subtracting 2nd equation from 1st: mg(sin 15 12(9:81)(sin 15

sin 45 ) + 2T cos 15 sin 45 ) + 2T cos 15 T

= 0 = 0 = 27:3 N J

15.47 40ο

WA

T

40ο

µANA

A

NA

y

x

µBNB

Α

WB B

mAa

NB

T FBD's

Β

mBa MAD's

Block A: Fy = 0 Fx = 0

+ - NA WA cos 40 NA 40 cos 40 = 0 NA + % T + A NA WA sin 40 T + 0:15(30:64)

= = =

0 30:64 lb mA a 40 40 sin 40 = a 32:2 T 21:12 = 1:2422a

(a)

Block B: Fy = 0 Fx = 0

+ - NB WB cos 40 NB 60 cos 40 = 0 NB +% T + B NB WB sin 40 T + 0:3(45:96)

The solution of (a) and (b) is

= 0 = 45:96 lb = mB a 60 60 sin 40 = a 32:2 T 24:78 = 1:8634a

(b)

a = 14:78 ft/s2 and T = 2:76 N J

15.48 Kinematics: x2 + y 2 = 252

2xx_ + 2y y_ = 0

2

x• x + x_ + y y• + y_ 2 = 0 Immediately after the release we have x_ = y_ = 0. Thus x• x + y y• = 0

y• =

x x •= y

20 x •= 15

4 x • 3

(a)


Kinetics:

4 .. 32.2 y

4 lb NA A

A

P 3

FBD

MAD

4

3 .. x 32.2

3 lb P

B

B

3 NB

Block A:

4

MAD

FBD

Fy = mA aA + "

Substituting for y• from Eq. (a), we get Block B:

Fx = mB aB + !

3 4 P 4= y• 5 32:2 3 4 P 4= 5 32:2 3 4 P = x • 5 32:2

4 x • 3

(b) (c)

The solution of Eqs. (b) and (c) yields x • = 16:99 ft/s

2

P = 1:978 lb J

15.49 yC yA

C

1

2

yB

W B W A Kinematics: Cable 1 : Cable 2 :

yA + (yA yC ) = constant ) 2aA aC = 0 yB + 2yC = constant ) aB + 2aC = 0

aC = 2aA aB = 4aA


Kinetics:

F = ma + #

T1

T1

T2

T2

C A

Pulley C

:

T1

Block A :

W

Block B

W

:

B

T1 FBD's

W

T2

W

2T2 = 0 T1 = 2T2 W W 2T1 = aA W 4T2 = aA g g W W aB W T2 = 4 aA T2 = g g

(a) (b)

Solution of (a) and (b) is aA =

3 g 17

T2 =

5 W J 17

) T1 = 2

5 W 17

=

10 W J 17

*15.50


15.51


15.52 (a)

2

mg k(x1 − x2) N

a1

=

a1

=

a2

=

1

F

=

2

N

FBD's

Block 1:

Block 2:

mg

ma2

1

ma1

MAD's

Fx = max + ! F k(x1 x2 ) = ma1 1 32:2 [F k(x1 x2 )] = [1:2 (2 12) (x1 x2 )] m 5 2 7:728 154:56(x1 x2 ) ft/s J

Fx = max + ! k(x1 x2 ) = ma2 2 12 k (x1 x2 ) = (x1 x2 ) = 154:56(x1 m 5=32:2

x2 ) ft/s

2

J

Initial conditions: x1 (0) = x_ 1 (0) = x2 (0) = x_ 2 (0) = 0 J (b) Letting x = x1 x2 x_ 1 and the initial conditions are x_

=

x3

x(0)

=

0

x4 0

7:728 0

0

x_ 2

T

, the equivalent …rst-order equations

154:56(x1

x2 )

154:56(x1

x2 )

T

T

The MATLAB program for integrating these equations is function problem15_52 [t,x] = ode45(@f,[0,0.1],[0,0,0,0]); printSol(t,x); function dxdt = f(t,x) dxdt = [x(3) x(4) 7.728-154.56*(x(1)-x(2)) 154.56*(x(1)-x(2))]; end end From the last line of output t 1.0000e-001

x1 3.4149e-002

x2 4.4914e-003

x3 6.0233e-001

x4 1.7047e-001

we deduce that v1 P

= 0:602 33 ft/s = 7:23 in./s J = k(x1 x2 ) = (2 12)(34:149

v2 = 0:170 47 ft/s = 2:05 in./s J 4:491) 10 3 = 0:712 lb J


15.53 (a)

xB xA

d F F

A

FBD's

Particle A

:

aA

=

B

mAaA

=

mBaB

MAD's

Fx = max + ! F = mA aA 2 F c=d 0:005=(xB xA )2 = = mA mA 0:015 1 2 m/s J 3(xB xA )2

=

Particle B

:

aB

= =

Fx = max + ! F 0:005=(xB xA )2 = mB 0:01 1 2 m/s J 2(xB xA )2

F = mB aB

Initial conditions: xA (0) = 0

xB (0) = 0:5 m

(b) Letting x = xA xB x_ A and the initial conditions are x_

=

x3

x(0)

=

0

x4 0:5 m

x_ A (0) = 0 x_ B

T

x_ B (0) =

, the equivalent …rst-order equations 1

1 3(x2 0

2 m/s J

x1 )2

2(x2

x1 )2

2 m/s

The MATLAB program that solves the equations numerically is function problem15_53 [t,x] = ode45(@f,[0:0.005:0.25],[0,0.5,0,-2]); printSol(t,x); function dxdt = f(t,x) xx =1/(x(2)-x(1))^2; dxdt = [x(3) x(4) -xx/3 xx/2]; end end


The condition for minimizing d is d_ = x_ B x_ A = 0. Hence x_ B = x_ A when d is minimized. The following two lines of the output span the instance where x_ B = x_ A : t x1 2.1500e-001 -6.2911e-002 2.2000e-001 -6.6964e-002

When t When t

x2 x3 x4 1.6437e-001 -7.9442e-001 -8.0837e-001 1.6045e-001 -8.2667e-001 -7.5999e-001

= 0:215 s, d = 0:16437 = 0:220 s, d = 0:16045 ) dmin = 0:227 m J

( 0:06291) = 0:2273 m ( 0:06696) = 0:2274 m

We …nd x_ A = x_ B = v by linear interpolation. 0:75999 0:82667

( 0:80837) v = ( 0:79442) v

( 0:80837) ( 0:79442)

Therefore, x_ A = x_ B = 0:800 m/s

v=

0:800 m/s

J

15.54 (a)

F F A

B

=

mBaB

mAaA

FBD's MAD's (Only horizontal forces are shown)

Block: aA

Bullet: aB

=

=

Fx = max + ! F = mA aA 10 F 50(vB vA ) 2 = (vB vA ) m/s J = mA 15 3 Fx = max + ! F = mB aB 50(vB vA ) F 2 = = 2000(vB vA ) m/s J mB 0:025

Initial conditions: xA (0) = xB (0) = vA (0) = 0

vB (0) = 600 m/s J


(b) Letting x = xA xB vA and the initial conditions are x_

=

x3

x(0)

=

0

vB

T


10 (x4 x3 ) 3 0 600 m/s

x4 0

2000(x4

x3 )

The corresponding MATLAB program is function problem15_54 [t,x] = ode45(@f,[0:0.005e-3:1e-3],[0,0,0,600]); printSol(t,x); function dxdt = f(t,x) dxdt = [x(3) x(4) (x(4)-x(3))*10/3 -(x(4)-x(3))*2000]; end end The last line of output is shown below t 1.0000e-003

x1 5.6722e-004

x2 2.5967e-001

x3 8.6368e-001

x4 8.1795e+001

We see that at t = 0:001s we have xA = 5:67

4

10

m = 0:567 mm J

vB = 81:8 m/s J

15.55 (a) Let xA and xB be the displacements of the cars, measured from the position where the bumpers just touch.

xA FF

A

xB B

=

mAaA

mBaB

MAD's FBD's (Only horizontal forces are shown)

Car A

:

aA

=

Fx = max + ! F = mA aA F 20 000(xA xB ) = = 53:67(xA mA 12 000=32:2

Car B

:

aB

=

xB ) ft/s

2

J

Fx = max + ! F = mB aB F 20 000(xA xB ) 2 = = 35:78(xA xB ) ft/s J mB 18 000=32:2


Initial conditions: xA (0) = xB (0) = 0


=

x3

x(0)

=

0

x4 0

vB (0) = 5 ft/s J

vA (0) = 8 ft/s vB

T


53:67(x1 8 ft/s

x2 )

35:78(x1

x2 )

5 ft/s

The MATLAB program shown below was used for integration. Note that only t and the contact force F were printed out (x1 on the printout represents F ) function problem15_55 [t,x] = ode45(@f,[0:0.005:0.4],[0,0,8,5]); F = 20000*(x(:,1)-x(:,2)); printSol(t,F) function dxdt = f(t,x) dxdt = [x(3) x(4) -53.67*(x(1)-x(2)) 35.78*(x(1)-x(2))]; end end (c) The following partial printout spans Fmax : t 1.6000e-001 1.6500e-001 1.7000e-001

x1 6.3335e+003 6.3436e+003 6.3396e+003

By inspection we determine that Fmax = 6340 lb J The two lines of output spanning F = 0 are: 3.3000e-001 1.3013e+002 3.3500e-001 -1.6984e+002 We use linear interpolation to …nd t when contact is lost: 169:84 130:13 0 = 0:335 0:330 t

130:13 0:330

t = 0:332 s J


15.56 (a) Let xA and xB be the displacements of the blocks from the equilibrium position (where the spring is unstretched)

mAg

FBD's

xB B

mAaA

P

A

F

N N F mBg

=

MAD's mBaB

NB F =

kN

Block A

:

aA

=

k mA g

sgn (vA

vB )

P = mA aA k vB ) xA mA 2000 0:3(9:81) sgn (vA vB ) xA 2 2 2:943 sgn (vA vB ) 1000xA m/s J

=

aB

vB ) =

Fx = max + ! F P = k g sgn (vA mA

=

Block B

sgn (vA

F

:

Fx = max + ! F = mB aB mA F = k g sgn (vA vB ) = mB mB 2 = 0:3 (9:81) sgn (vA vB ) 4 =

1:4715 sgn (vA

vB ) m/s

2

J

Initial conditions: xA (0) = xB (0) =


=

x(0)

=

x3

x4

0:02 m

vB

2:943 sgn (x3 0:02 m

vA (0) = vB (0) = 0 J

0:02 m

0

T


x4 )

1000x1

1:4715 sgn (x3

x4 )

0

The ‡ollowing MATLAB program was used for the plot:


function problem15_56 [t,x] = ode45(@f,[0:0.0025:0.2],[-0.02,-0.02,0,0]); axes(’fontsize’,13) plot(t,x(:,3),t,x(:,4),’linewidth’,1.5) xlabel(’t (s)’); ylabel(’v (m/s’) gtext(’A’);gtext(’B’) % Creates mouse-movable text grid on function dxdt = f(t,x) dxdt = [x(3) x(4) -2.943*sign(x(3)-x(4))-1000*x(1) 1.4715*sign(x(3)-x(4))]; end end

0.6

A

0.4

v (m/s

0.2 B

0

-0.2

-0.4 0

0.05

0.1 t (s)

0.15

0.2


15.57 (a) Let v be the velocity of A relative to the disk q ) vR = R_ v = R( _ !) v = R_ 2 + R2 ( _ !

=

2 60

45

The friction force F = ) FR = F FR • R

F

= 4:7124 rad/s

O

FR

FBD F

maθ

R

= θ

MAD maR

Fθ

k mg

vR = v

opposes the relative velocity vector v

k mg

R_ v

= maR = R_

2

= R_

2

where v= Initial conditions:

x2

x(0)

=

1:0 ft 0

9:66 0

where v=

9:66

!) v

2

R_ )

R_ J v

4:7124)2

_ R(0) = (0) = _ (0) = 0 J _

x1 x24

=

• FR = m(R R_ kg v

R( _

F = m(R• + 2R_ _ ) _ ! F 2R_ _ = kg mR R v _ 4:7124 9:66 J v

R(0) = 1:0 ft

x_

k mg

+%

q R_ 2 + R2 ( _

(b) Letting x = R R_ the initial conditions are

v = v

2 FR = R_ m 2 R_ 0:3(32:2) = R _ v

2R_ _ R 2R_ _ R

=

F =F +&

= ma

• =

!)2

T

x2 v

, the equivalent …rst-order equations and x4

2x2 x4 x1

9:66

x4

4:7124 v

0

q x22 + x21 (x4

4:7124)2

The corresponding MATLAB program is


function problem15_57 [t,x] = ode45(@f,[0:0.01:1],[1,0,0,0]); printSol(t,x) function dxdt = f(t,x) v = sqrt(x(2)^2 + x(1)^2*(x(4) - 4.7124 )^2); dxdt = [x(2) x(1)*x(4)^2 - 9.66*x(2)/v x(4) -2*x(2)*x(4)/x(1) - 9.66*(x(4) - 4.7124)/v]; end end The two lines of output spanning R = 2:5 ft are: t 9.4000e-001 9.5000e-001

x1 2.4831e+000 2.5203e+000

x2 3.6938e+000 3.7325e+000

x3 2.0583e+000 2.0765e+000

x4 1.8289e+000 1.8092e+000

We …nd R_ and _ at R = 2:5 ft by linear interpolation: 3:7325 2:5203

R_ 3:6938 3:6938 = 2:4831 2:5 2:4831

_ 1:8289 1:8092 1:8289 = 2:5203 2:4831 2:5 2:4831 q q ) vA = R_ 2 + (R _ )2 = 3:7112 + (2:5

R_ = 3:711 ft/s _ = 1:8200 rad/s 1:8200) = 5:87 ft/s J 2

15.58

O

θ 400

A

Spring

B

L

1

400(1 - cosθ) 400 sinθ

350 = sin

0 40

θ

0 60

Dimensions in mm

400 = 41:81 600

q 2 2 L = (0:35 + 0:4 sin 41:81 ) + [0:4(1 cos 41:81 )] = 0:6250 m = L L0 = 0:6250 0:35 = 0:2750 m


Because the system rotates about O, we have !=

vA vB = 0:6 0:4

vB =

2 vA 3

2

T2

= =

1 2 2 1 1 2 2 = mA vA vA mA + mB vA + mB 2 2 3 2 9 1 2 2 2 (4) + (2) vA = 2:444vA 2 9

Choose the initial position as the datum for gravitational potential energy. yA

=

V2

= = =

cos 41:81 ) = 0:101 86 m " 1 2 mA g yA + mA g yB + k 2 1 4(9:81)(0:4) + 2(9:81)(0:101 86) + (240)(0:2750)2 2 4:623 N m

0:4 m #

T1 + V 1 = T 2 + V 2

yB = 0:4(1

2 0 + 0 = 2:444vA

vA = 1:375 m/s J

4:623

15.59 vA = 22 ft/s U1

2

= T2

vB = 11 ft/s "

)

xA = 2 yB

T1 :

1 2 1 1 2 2 k xA WB yB = 0 mA vA + mB vB k WA 2 2 2 1 1 5 9 1 k(5:42 ) 0:2(5)(5:4) 9(2:7) = (222 ) 2 2 32:2 2 32:2 k = 1:700 lb/ft J

(112 )


15.60

15.61


15.62 Horizontal momentum is conserved: p1 = p2 + ! 0 = mb oat vb oat + mb oy vb oy cos 35 0 = 60vb oat + 40(10 cos 35 ) vb oat = 5:46 m/s vb oat = 5:46 m/s J

15.63

y

δ x The mass center of the system does not move: mA xA + mB xB = mA + mB (xB + ) where xB is the centroidal coordinate of the cart in the initial position. mA xA = (mA + mB )

mA xA 15(2:5) = 0:798 m J = mA + mB 15 + 32

=

15.64 Horizontal momentum is conserved: p1

= p2

+ ! mA vA = mB

vB

=

0 = mA vA + mB vB 12 vA = 1:5vA 8

Energy is conserved: T 1 + V1 1 (2 2

12)

1:8 2 12

1 1 1 2 2 0 + k 2 = mA vA + mB vB +0 2 2 2 1 0:75 2 1 0:5 v + ( 1:5vA )2 2 32:2 A 2 32:2

= T2 + V2

2

=

0:0675 = vB = vA = 1:523 ft/s

2 0:029 11vA vA = 1:5228 ft/s 1:5( 1:5228) = 2:28 ft/s

J

vB = 2:28 ft/s ! J


15.65

15.66 ) 2vA + vB = 0

Constraint: 2yA + xB = constant

T

vB =

2vA

T T

B

A 7 lb Block A: + # L1 (7

2

2T ) t

Block B: + ! L1

2

Tt

= p2 =

(7

2T ) t =

7 (6) 32:2

(7

2T ) t = 1:3043

= p2 =

7 vA 32:2

p1 :

p1 :

Tt =

12 ( 12) = 4:472 32:2

12 vB 32:2

0 (a)

0 (b)


Substituting (b) into (a): 7t

t = 1:464 s J

2(4:472) = 1:3043

15.67 Let vB = …nal velocity of block and vC = …nal velocity of cart, both positive to the right. Mometum is conserved: p1

= p2 + ! mB = vB = mC

vC

0 = mC vC + mB vB 8 vB = 0:4vB 20

Energy is conserved: T 2 + V2 = T 1 + V1

vC vB=C vB=C

1 1 2 2 mC vC + mB vB +0 = 2 2 1 1 2 (20)( 0:4vB )2 + 8vB = 2 2 vB =

= 0:4( 3:273) = 1:309 m/s = vB vC = 3:273 1:309 = = 4:58 m/s J

1 0+ k 2 2 1 (12 000) (0:1)2 2 3:273 m/s

4:582 m/s

15.68 Constraint: 2yA + yB = constant

T

) 2vA + vB = 0

T

2

= p2

2vA

T B

A 8 lb + " L1

vB =

6 lb

p1 :

Block A :

(2T

WA )t = mA vA

0

(2T

Block B

(T

WB ) t = mB vB

0

(T

:

8 vA 32:2 6 6)(7) = ( 2vA ) 32:2 8)(7) =

The solution is T = 4:50 lb and vA = 28:2 ft/s " J


15.69

15.70 Since there are no external forces acting on the system in the horizontal direction, we have p 1 = p2

0=

400 800 vA + vB g g

vA =

2vB

400 lb 180 lb (0.35)(400) lb N =400 lb FBD (U1 0 + [180

2 )ext

+ (U1

2 )int

= T2 T1 1 800 2 1 400 2 vA + v (0:35)(400)] (8) = 2 32:2 2 32:2 B 1 400 1 800 2 2 2 320 = ( 2vB ) + vB = 37:27vB 2 32:2 2 32:2 r 320 vB = = 2:93 ft/s J 37:27


15.71

15.72


15.73


15.74

15.75


15.76 L

mωL

L O

Momentum diagram

L/3

mωL

G

2L/3

mωL

hG = 2(m!L)L + (m!L)

2L 8 = m!L2 3 3

J

15.77 z 8 m/s

2m

6 m/s

O

m 2 12 m 3 /s y x 1 4 3m m m 4 A i 1 2 3

ri 4i 4i + 4j + 3k 3i + 2j

ri=A 4j 3k 7i 2j

=

X

=

340i

vi 8k 12j 6i

3

ri

vi 32j 36i + 48k 12k 36i 32j + 60k

(a) hO

= m

(b) hA

180i 160j + 300k N m s J X = m ri=A vi = 5( 68i 12k)

ri

vi = 5( 36i

ri=A vi 32i 36i 12k 68i 12k

32j + 60k)

60k N m s J


15.78 (a) (AO )1

2

=

(hO )2

8 ( t)

=

140 4(0:62 ) + 7(0:32 )

(b) +

!=

(hO )1

Z

+

C0 t = 0

_ ( mA r2 A

t = 36:23 s J

dt + C1 = t + C1

( is constant)

When t = 0, ! = 140 rad/s. ) C1 = 140 rad/s When t = 36:23 s, ! = 0. ) 0 = (36:23) 140 ) + d When

= 0, ! =

When ! = 0,

=

2 mB r B )

= 3:864 rad/s2

d! d d! d! = = ! dt d dt d 1 = ! d! 3:864 = ! 2 + C2 2 =

1 ( 140)2 = 9800 (rad/s)2 140 rad/s. ) C2 = 2 2536 9800 = 2536 rad. ) = = 404 rev J 3:864 2

15.79 (a)

.

L

O

mv 2 L

mv1

. O

Just after impact

L

Momentum diagrams

Just before impact

2mv2

Angular momentum of the system about O is conserved: (hO )1

=

(hO )2

mv1 L = 3mv2 L

!2

=

v2 v1 = J L 3L

v2 =

1 v1 3


(b) 2

1 1 1 1 1 mv 2 T2 = (3m)v22 = (3m) v1 = mv12 2 1 2 2 3 6 (1=6) (1=2) T2 T1 % energy change = 100% = 100% = 66:7% T1 (1=2) % energy lost = 66:7% J T1

=

15.80


15.81 A

mv0

3L/4 m(vA)y 3L/4 L/4 3m(vB)y B A mv L/4 3mvx G x 3mv0

G

B Initial momenta (p1 )x

=

(p2 )x

+ !

(p1 )y

=

(p2 )y

+"

(hG )1 3 v0 2

Final momenta 3mv0

mv0 = 3mvx + mvx

0 = m(vA )y + 3m(vB )y

1 v0 2 3(vB )y

vx =

(vA )y =

3L L 3L L = (hG )2 + mv0 + 3mv0 = m(vA )y + 3m(vB )y 4 4 4 4 3 3 1 = [(vB )y (vA )y ] = [(vB )y + 3(vB )y ] (vB )y = v0 4 4 2 1 3 v0 = v0 ) (vA )y = 3 2 2 ) vA =

1 i 2

3 j v0 J 2

1 1 i + j v0 J 2 2

vB =

15.82

0.3 m yA

0.3 m

.

yB A B

q 2 + 0:32 y A + 2 yB

= L

) vA

=

When yB

=

yA = L 2y v p B B 2 + 0:32 yB

0:4 m: vA =

q 2 + 0:32 2 yB

p

2(0:4)vB = 0:42 + 0:32

1:60vB


0.3 m

∆yA A

m

A

m

0.5 m

= 0

.

m 0.5

B 2m Position 2

Position 1 (Datum)

yA = yA jyB

0.3 m

0.4 m

B 2m

.

yA jyB =0:25 m = [L

2(0:3)]

[L

2(0:5)] = 0:4 m

T1 + V1 = T2 + V2 : 1 1 2 mv 2 + (2m) vB + mg yA 2 A 2 1 2 0 = ( 1:60vB )2 + vB + 9:81(0:4) 2 vB = 1:312 m/s # J

0+0

=

2mg(0:4) 2(9:81)(0:4)

15.83


15.84

2m(vB)y B A mω0 L

L/2

.G L/2

B ω 0

2m (vB)x

L/2

G.

Position 2

L/2

Position 1

A m(vA)y

m(vA)x

Because there are no external forces in the xy-plane, linear and angular momenta about the mass center G are conserved. Because the bar is rigid (vA )y = (vB )y (px )1

= )

(py )1

= )

(px )2 : 0 = m(vA )x + 2m(vB )x 1 (vB )x = (vA )x 2

(py )2 :

m! 0 L = m(vA )y + 2m(vB )y !0 L (vA )y = (vB )y = 3

!0 L =

3(vA )y


(hG )1

=

!0 L = )

(hG )2 :

(m! 0 L)

(vA )x

2(vB )x

(vA )x =

!0 L 2

vA = ! 0 L

1 i 2

1 j 3

L L = m(vA )x 2 2 ! 0 L = (vA )x (vB )x = J

2m(vB )x 2

L 2

1 (vA )x = 2(vA )x 2

!0 L 4

vB = ! 0 L

1 i 4

1 j 3

J

15.85

15.86


#15.87


15.88

15.89 +! (px )1 = (px )2 : 0 = 125 000v

106(1640) cos 35

v = 1:139 m/s J


15.90 p1 v…nal

= p2 + ! mA vA + mB vB = (mA + mB )v…nal mA vA + mB vB 40(5) + 55(3) = = = 3:842 m/s ! J mA + mB 40 + 55

% energy lost

=

T2 T1

1

=

1

100% = 1

95(3:842)2 40(5)2 + 55(3)2

2 (mA + mB ) v…nal 2 + m v2 mA vA B B

100%

100% = 6:20% J

15.91 Position 1 = just before impact Position 2 = just after impact Position 3 = max. compression of spring + ! p1 = p 2 : mA vA = (mA + mB ) v…nal mA 0:008 v…nal = vA = vA = 0:004 975vA mA + mB 1:6 + 0:008 1 1 2 (mA + mB )v…nal +0 = 0+ k 2 2 2 1 1 2 (1:608) (0:004975vA ) = (6500)(0:052)2 2 2 vA = 665 m/s J

T2 + V2 = T3 + V3 :

15.92 Position 1 = release position; Position 2 = just before impact; Position 3 = just after impact; Position 4 = …nal (rest) position U1

2

v2

U3

1 WA 2 = T2 T1 WA h = v 2 g 2 p p = 2gh = 2(32:2)(8) = 22:70 ft/s

p3

= p3

v3

=

4

= T4

d

=

+ mA v2 = (mA + mB )v3 mA 8 v2 = (22:70) = 6:985 ft/s mA + mB 26

T3

k (WA

+ WB )d = 0

1 (mA + mB )v32 2

(mA + mB )v32 v2 6:9852 = 3 = = 3:79 ft J 2 k (WA + WB ) 2 kg 2(0:2)(32:2)


15.93 (a) Position 1 = just before impact; Position 2 = just after impact; Position 3 = …nal (rest) position p1 = p2 + ! 0:02(600) = (10 + 0:02)v2 U2 k (WA

3

+ WB )d

= T3 =

0:25(10 + 0:02)(9:81)d =

0

mB v1 = (mA + mB )v2 v2 = 1:1976 m/s

T2 1 (mA + mB )v22 2

1 (10 + 0:02)(1:1976)2 2

d = 0:292 m J

(b) T1

=

T2

=

% energy loss

=

1 1 mB v12 = (0:02)(600)2 = 3600 J 2 2 1 1 (mA + mB )v22 = (10 + 0:02)(1:1976)2 = 7:186 J 2 2 T1 T2 3600 7:186 100% = 100% = 99:8% J T1 3600

15.94 (a) Horizontal momentum is conserved: p1 v…nal

= p2

+ ! mB vB cos 30 = (mA + mB )v…nal mB 60 (8) cos 30 = 1:4846 ft/s J vB cos 30 = mA + mB 280

=

(b) % energy lost =

1

=

1

T2 T1

100% = 1 ! 2 280 (1:4846) 60 (8)

2

2 (mA + mB ) v…nal 2 mB vB

100%

100% = 83:9% J

15.95 Position 1 = just before collision; Position 2 = just after collision; Position 3 = …nal (rest) position p1 v…nal

= p2 =

+ ! mA vA = (mA + mB )v…nal mA 5000 vA = vA = 0:5814vA mA + mB 8600


U2

3

= T3

0:8(3600)(22) =

T2

k WB d

1 8600 2 (0:5814vA ) 2 32:2

=0

1 2 (mA + mB )v…nal 2

vA = 37:5 ft/s J

15.96 Choose Position 2 as the datum for Vg . Energy beween Positions 1 and 2 is conserved: 1 T 1 + V 1 = T 2 + V2 : 0 + mgh1 = mv22 + 0 2 1 (8)v22 v2 = 3:388 m/s 8(9:81)(2:5)(1 cos 40 ) = 2 Horizontal momentum is conserved during impact: + ! p2 = p02 : mv2 = mtot v20 8(3:388) = 11v20 v20 = 2:464 m/s Energy beween Positions 2 and 3 is conserved: 1 2 mtot (v20 ) + 0 = 0 + mtot gh3 T 2 + V2 = T 3 + V3 : 2 1 (11)(2:464)2 + 0 = 0 + (11)(9:81)(2:5)(1 cos ) 2 cos = 0:8762 = 28:8 J

15.97 (a) Momentum is conserved during impact between carts (note that the parcel keeps going at v = 15 ft/s): p1

= p01

+ ! (mA + mC )v = (mA + mB )v 0 + mC v mA 50 15 = 7:5 ft/s J v= mA + mB 100

v0

=

(b) Impact between the front of cart A and the parcel conserves the momentum: p01 v…nal

= p…nal + ! (mA + mB ) v 0 + mC v = (mA + mB + mC )v…nal 0 100(7:5) + 35(15) (mA + mB ) v + mC v = = 9:44 ft/s J = mA + mB + mC 135

15.98 (a) p1 = p2 : [mA (vA )1 [4200(44)

mB (vB )1 sin 30 ] i + mB (vB )1 (cos 30 )j = (mA + mB )v2 3500(60) sin 30 ] i + 3500(60)(cos 30 )j = (4200 + 3500)v2 v2 = 10:36i + 23:6j ft/s J


(b) T1

= = T2

1 1 1 4200 1 3500 mA (vA )21 + mB (vB )21 = (442 ) + (602 ) 2 2 2 32:2 2 32:2 321:9 103 lb ft = =

1 4200 + 3500 1 (mA + mB )v22 = (10:362 + 23:62 ) 2 2 32:2 79:4 103 lb ft

Energy absorbed = T1

T2 = (321:9

79:4)

103 = 243

103 lb ft J

15.99 p1

p2

= mA vA (i cos 25 + j sin 25 ) mB vB j 8 2:8=16 = (130)(i cos 25 + j sin 25 ) (14)j 32:2 32:2 = 0:6403i 3:180j slug ft/s (2:8=16) + 8 = (mA + mB )v2 = v2 = 0:2539v2 32:2 p1 v2

= p2 0:6403i 3:180j = 0:2539v2 = 2:52i 12:52j ft/s J

15.100

A Datum

4 ft

δ

B

A B Position 4

Position 1

Position 1: Release position. Note that the springs are compressed by 0 = WB =k = 48=1200 = 0:04 ft Position 2: Just before the impact. Position 3: Just after the impact. Position 4: Position of maximum displacement. Springs are compressed by additional . T1 + V1 = T2 + V2 : p p 1 v2 = 2gh = 2(32:2)(4) = 16:050 ft/s 0 + mA gh = mA v22 + 0 2


p 2 = p3 : mA v2 = (mA + mB )v3

v3 =

mA 60 v2 = (16:050) = 8:917 ft/s mA + mB 60 + 48

T3 + V3 = T4 + V4 : 1 1 (mA + mB )v32 + 0 = 0 (mA + mB )g + k( 2 2 1 60 + 48 1 (8:9172 ) = (60 + 48) + (1200) 2 32:2 2 133:34 = 108 + 600 0 = 600 2 108 = 0:572 ft J 4

2

2

2 0)

2

0:042

0:96 134:30

15.101


15.102


15.103

15.104

15.105


15.106


*15.107

15.108


15.109 p 1 = p2 : m(vA )1 + m(vB )1 (vA )1 + (vB )1

= m(vA )2 + m(vB )2 = (vA )2 + (vB )2

(a)

e = vsep =vapp : (vB )2 (vA )2 (vA )1 (vB )1 e(vA )1 e(vB )1 = (vB )2

e =

(vA )2

(b)

Subtract Eq. (a) from Eq. (b): (vA )1 (1

e) + (vB )1 (1 + e) = 2(vA )2

Q.E.D.

Add Eqs. (a) and (b): (vA )1 (1 + e) + (vB )1 (1

e) = 2(vB )2

Q.E.D.

15.110 With e = 1=2 the formulas in Prob. 15.109 become (vA )2 =

3 1 (vA )1 + (vB )1 4 4

(vB )2 =

3 1 (vA )1 + (vB )1 4 4

(vB )2 =

3 v0 4

After impact of A and B: (vA )2 =

1 v0 4


After impact of B and C: (vB )3

=

(vC )3

=

1 1 (vB )2 = 4 4 3 3 (vB )2 = 4 4

3 v0 4 3 v0 4

3 v0 16 9 = v0 16 =

After second impact of A and B: (vA )4

=

(vB )4

=

1 (vA )2 + 4 3 (vA )2 + 4

3 1 (vB )3 = 4 4 1 3 (vB )3 = 4 4

1 v0 4 1 v0 4

3 4 1 + 4 +

3 v0 16 3 v0 16

13 v0 64 15 = v0 64 =

Final velocities are: vA =

13 v0 J 64

vB =

15 v0 J 64

vC =

9 v0 J 16

15.111 p 1 = p2 : (vB )2 1:2 = = 0:667 (vA )1 1:8

1:2 (vA )1 = 1:8 (vB )2 e = vsep =vapp : e=

(vB )2 = 0:667 J (vA )1

15.112


15.113

+ ! p1 = p2 : mA (vA )1 = mA (vA )2 + mB (vB )2 0:3(6) = 0:3( 2:5) + 1:5(vB )2 (vB )2 = 1:70 m/s ! e=

vsep (vB )2 (vA )2 1:70 ( 2:5) = = = 0:70 J vapp (vA )1 6

15.114


15.115

15.116

θ2

θ2'

θ3

θ1' θ1

Rebound formula

:

Geometry

:

Rebound formula Geometry

0 1

tan tan

2

= e tan 1 = tan 01

1

e tan 01 1 tan 01 : tan 3 = = = tan e tan 02 ) Q.E.D. 1 = 3 :

tan

0 2

= e tan

2

=

1


15.117 The x-component of the momentum is conserved for each disk: (vAx )2 = 12 cos 70 = 4:104 m/s

(vBx )2 = 0

Momentum of the system is conserved in the y-direction: + " mA (vAy )1 = mA (vAy )2 + mB (vBy )2 m(12 sin 70 ) = m(vAy )2 + (2m)(vBy )2 (vAy )2 + 2(vBy )2 = 11:276 m/s

e=

(vBy )2 (vAy )2 vsep = vapp (vAy )1 (vBy )2 (vAy )2

(vBy )2 (vAy )2 12 sin 70 9:585 m/s

(a)

0:85 = =

(b)

Solution of (a) and (b) is (vAy )2 = ) (vA )2 = 4:10i

2:631 m/s

(vBy )2 = 6:954 m/s

2:63j m/s J

(vB )2 = 6:95j m/s J

15.118 Let v2 = velocity of pellet and ! 2 = angular velocity of the bar after impact. (hO )1 = (hO )2 + mp ellet v0 L / = mp ellet v2 L / + 2m(L! 2 )L / 0:16 0:16 (300) = v2 + 2 (0:5) (0:75)! 2 3 = 0:01v2 + 0:75! 2 (a) 16 16 e=

L! 2 v2 v0

Solution of (a) and (b) is v2 =

0:75 =

0:75! 2 v2 300

(b)

220 ft/s and ! 2 = 6:93 rad/s J

15.119 Since the release positions are the same, the velocities of the pendulums are equal just before the impact: (vA )1 = (vB )1 = v Because A returns to its release position, its velocity just after the impact equals the velocity just before the impact: (vA )2 = (vA )1 = v


mv 3mv

A

B

mv

3mv'

mv + 3mv = mv + 3mv 0

+

B

Momenta after impact

Momenta before impact p 1 = p2

A

v0 =

1 v 3

v v0 v v=3 1 = = J v+v 2v 3

e=

15.120 Position Position Position Position Position

1 2 3 4 5

= = = = =

release position just before A hits the ground (choose as the datum position) just after A hits the ground but before impacting with B just after A and B impact B is at its maximum elevation

Position 1 to 2:

T1 + V1 (vA )2

=

Position 2 to 3: (vA )3 (vB )3

1 0 + mgh = mv22 (valid for each ball) p p2 (vB )2 = v2 = 2gh = 2(32:2)(5) = 17:944 ft/s

= T 2 + V2

= e(vA )2 (vA )3 = 0:85(17:944) = 15:252 ft/s = (vB )2 = 17:944 ft/s

Position 3 to 4:

B Before impact

m(vB)3 10m(vA)3

A

p3 = p4

e=

+"

(vB )4 (vA )4 (vA )3 + (vB )3

B

m(vB)4 After 10m(vA)4 impact

A

10m(vA )3 m(vB )3 = 10m(vA )4 + m(vB )4 10(15:252) 17:944 = 10(vA )4 + (vB )4 134:58 = 10(vA )4 + (vB )4 0:85 =

(vB )4 (vA )4 15:252 + 17:944

28:22 = (vB )4

(a)

(vA )4 (b)


Solution of (a) and (b) is (vA )4 = 9:669 ft/s, (vB )4 = 37:89 ft/s. Position 4 to 5: Ball B :

1 mB (vB )24 + 0 = 0 + mB g (h 2 37:892 (vB )24 =5+ = 27:3 ft J = dA + 2g 2(32:2) T 4 + V 4 = T 5 + V5

h

dA )

*15.121


15.122

A 18θ0 90

.

0.5(vA)y

y

.

6N s

4.8 N s

x

30o B Just before impact

0.6(vB)y

0.5(vA)x 0.6(vB)x

Just after impact

90 = 30 180 The x-component of the impulse of each disk is unchanged. = sin1

Disk A : Disk B :

6 sin 30 = 0:5(vA )x 4:8 sin 30 = 0:6(vB )x

(vA )x = 6:0 m/s (vB )x = 4:0 m/s

Momentum of the system is unchanged in the y-direction. 4:8 cos 30 6 cos 30 0:5(vA )y + 0:6(vB )y e=

vsep vapp

0:6 =

= =

0:6(vB )y + 0:5(vA )y 1:0392

(vA )y (vB )y 12 cos 30 + 8 cos 30

(vA )y

(vB )y = 10:392

(a) (b)

Solution of Eqs. (a) and (b) is (vA )y = 4:724 m/s

(vB )y =

5:668 m/s

The …nal speeds are vA vB

15.123

p 6:02 + 4:7242 = 7:64 m/s J p = 4:02 + 5:6682 = 6:94 m/s J =


15.124

15.125


15.126

Mg T/2

T/2 FBD

From FBD: T = M g = 250 lb Equation (15.49): T = mu _ P

250 = m(1100) _

m _ = 0:2273 slugs/s

1 1 mu _ 2 = (0:2273)(11002 ) = 1:3752 2 2 1:3752 105 = 250:0 hp J 550

= =

105 lb ft/s

15.127 Choose the water jet outside the hose as the control volume vin = 50 m/s

A=

1 2 1 d = (0:04)2 = 1:2566 4 4

10

3

m2

(a) Stationary plate: vout = 0 m _ = Av = 1000(1:2566

10

3

)(50) = 62:83 kg/s

The force acting on control volume: F = m(v _ out

vin ) = 62:83(0

Force acting on plate: P =

50) =

3142 N

F = 3140 N J

(b) Moving plate: vout = 4 m/s m _ = A(vin F P

vout ) = 1000(1:2566

10

= m(v _ out vin ) = 57:80(4 = F = 2660 N J

3

)(50 50) =

4) = 57:80 kg/s 2659 N


15.128

y vin

x

vout 28o

Control volume m _ = vin

=

180(8:34) = 0:7770 slugs/s 32:2(60) 45i ft/s vout = 45(i cos 28 + j sin 28 ) = 39:73i + 21:13j ft/s

Force acting on control volume: F = m(v _ out

vin ) = 0:7770(39:73i + 21:13j 45i) =

4:095i + 16:418j lb

Force acting on vane: P =

F = 4:10i

16:42j lb J

15.129


15.130 Choosing the fan as the control volume, we have vin

= v = 20 ft/s

m _ =

Au = 2:33

vout = u = 50 ft/s (6:42 ) 10 3 (50) = 3:748 slugs/s 4

Equation (15.47): T = F = m(v _ out

vin ) = 3:784(50

20) = 113:5 lb J

15.131


15.132

Pin R y x FBD of control vol. Pout Let R = force exerted by the bend on water. Equation (15.47): F = m(v _ out vin ) From FBD: F = Pin Pout + R Pin i + Pout j + R = m( _ vout j vin i) Substitute m _ = vin

=

Ain vin = 1000

0:42 (4:2) = 527:8 kg/s

4:2 m/s (given) Ain = 4:2 Aout

vout

= vin

Pin

= pin Ain = 12

Pout

4

= pout Aout = (18

0:4 0:6

2

= 1:8667 m/s

(0:42 ) = 1508 N 4 (0:62 ) 103 ) = 5089 N 4

103

) 1508i + 5089j + R = 527:8( 1:8667j R = 3720i 6070j N

4:2i)

Force exerted by the water on the bend is F=

R = 3720i + 6070j N J


15.133

T A

ρgL

L

FBD

B Let be the mass of the chain per unit length. Choose the portion AB of the chain as the control volume (the bottom link lies on the ground). We have steady ‡ow, where the force acting on the control volume is F = m(v _ out

vin )

where vin = m _ =

0 (bottom link has no velocity) vout = 3:2(1:5) = 4:8 kg/s ) F = 4:8((1:5

vout = 1:5 m/s

0) = 7:2 N

From FBD: F =T

) T = F + gL = 7:2 + 3:2(9:81)(4) = 132:8 N J

gL

15.134

vout 40

y vin

Control volume

Mg

o

P x N

Let the xy axes be attached to the control volume. vin = 3i ft/s

vout = 30( i cos 40 + j sin 40 ) =

20 ( 22:98i + 19:284j 3i) 32:2 16:137i + 11:978j lb ( F)x = 16:14 lb J

F = m(v _ out P

= =

22:98i + 19:284j ft/s

vin ) =


15.135 d

a 5 in.

P

25.4 ft/s a

d2 d2 62:4 vout = (25:4) = 38:66d2 slugs/s 4 4 32:2 (5=12)2 D2 p= (624) = 85:09 lb = Aa a p = 4 4 = P = m(v _ out vin ) 85:09 = 38:66d2 (25:4 0) = 0:2944 ft = 3:53 in. J

m _ = P F d

15.136 W

y

FBD

x

θ

2T

Thrust of one engine = T = m _ out u = (80 + 1:6)(660) = 53 860 N Fy

= =

0 +" 2T sin W =0 8000(9:81) W = sin 1 = 46:77 sin 1 2T 2(53 860)

J

Fx

= max + ! 2T cos = ma 2T cos 2(53 860) cos 46:77 2 a = = = 9:22 m/s J m 8000

15.137 y x

FBD T T =m _ out u = Fy

Mg

θ

250 (1500) = 11 646 lb 32:2 = =

M=

5000 = 155:28 slugs 32:2

0 +" T sin Mg = 0 Mg 155:28(28:8) sin 1 = sin 1 = 22:6 T 11 646

J


15.138

Mg

FBD 4.5 km A T Control volume = rocket + wire above A.

M = 70 + 0:004(4500) = 88:0 kg p_V

d(M v) = M_ v + M a = ( v) v + M a dt 0:004(220)2 + 88:0a = 193:6 + 88:0a N

= =

+" F = p_V

=

a =

F

+"

Mg

T =

88:0(9:81)

193:6 + 88:0a =

18 =

881:3

881:3 N a=

12:21 m/s

2

12:21 m/s # J 2

15.139 Equation (15.48): mu _ Substituting m _ =

M g = M v_

dM=dt and v_ = a (constant), we get dM u dt

1 dM = M

Mg = Ma

Integration yields ln M =

a+g dt u

a+g t+C u

Initial condition: M

= M0 when t = 0 ) C = ln M0 M a+g ) ln = t M0 u

When M = 0:25M0 : ln (0:25) =

34:4 + 9:81 t 1300

t = 40:8 s J


15.140

Mg FBD

30 in. P m _ =

" 0:075 vin Ain = (80) 32:2 4

10 12

2

#

= 0:101 63 slugs/s

2

20

30 = 4:909p 4 12 vin ) + # M g P = m(0 _

P

= pAout = p

F

= m(v _ out

4:909p =

0:101 63(80)

p = 5:73 lb/ft

2

vin )

J

15.141


15.142

15.143

30o N2

A

y 60 a A g A

x

40 lb

60 lb FBD N1

N2

MAD

a B B 40 g A

30o 40 a g B/A

30o FBD

MAD

Block A: + +

"

Fx = mA aA

N2 sin 30 =

Fy = 0

N2 cos 30

N1

60 aA 32:2 60 = 0

Block B: 40 aB=A cos 30 aA 32:2 40 aB=A sin 30 N2 cos 30 = 32:2

+ !

Fx = mB (aB )x

N2 sin 30 =

+

Fy = mB (aB )y

40

#


After simplifying, the equations become 0:5N2 1:8634aA N1 0:8660N2 0:5N2 + 1:2422aA 1:0758aB=A 0:8660N2 + 0:6211aB=A

= 0 = 60 = 0 = 40

The solution is N1 = 85:7 lb aA aB

N2 = 29:7 lb

aA = 7:97 ft/s

2

= 7:97i ft/s J = aA + aB=A = ( 7:97 + 23:0 cos 30 ) i

aB=A = 23:0 ft/s

2

2

=

11:95i

11:50j ft/s

2

(23:0 sin 30 ) j

J

15.144

15.145 p1 = p 2 + 4000(25) = 36 000v2

mcar v1 = mtot v2 v2 = 2:78 mi/h J


15.146 p1 = p 2

+ !

0 0 mA vA + mB vB = mA vA + mB vB 0 8000(6) + 5000(2) = 8000vA + 5000(5:2) 0 vA = 4:0 mi/h J

15.147 yA

yB A B

L_ = 4vA + vB = 0 vB = 8 m/s " J

L = 4yA + yB + constant 4(2) + vB = 0 vB = 8 m/s

15.148 Position 1 = just before impact Position 2 = just after impact (datum position for Vg ) Position 3 = maximum displacement of block Momentum parallel to the incline is conserved: + % p1 = p2 : mbullet v0 cos 20 = mtot v1 0:8 0:8 v0 cos 20 = + 5 v1 v1 = 9:304 10 3 v0 16 16 Energy is conserved after the impact : T 2 + V2 = T 3 + V3 : 1 (9:304 2

1 mtot v12 + 0 2 10

3

)v0

2

v0

=

0 + mtot gd sin 20

=

32:2(1:6) sin 20

=

638 ft/s J


15.149

15.150 y

µsW A

250

ω T

FBD T

B

400

µsW

x

Block B tends to slide to the right and block A tends to slide downward. The friction forces shown oppose impending sliding. Block B: Fx 0:25(2 9:81) T Block A: Fy 0:25(2 9:81) T

= max + ! T = mrB ! 2 sW = 2(0:4)! 2 4:905 + T = +0:8! 2 = may + " = 2(0:25)! 2

T = mrA ! 2 4:905 T = 0:5! 2

(a)

sW

(b)

Solution of (a) and (b) is T = 21:3 N and ! = 5:72 rad/s J

15.151


15.152


15.153


15.154

15.155

L Datum

L

L L/2

y2 s1 y1

WA Position 1

L L

WA

WB

WB s2

Position 2


Length of rope is constant: 2s1 + y1 p 2 L2 + (0:5L)2 + y1 y1 y2

= = =

2s2 + y2 p 2 2L2 + y2 0:5924L

Energy is conserved: V1 = V2 WA (0:5L)

W B y1 0:5WA L 0:5WA L WA WB

= WA L W B y 2 = WB (y1 y2 ) = 0:5924WB L 0:5924 = = 1:185 J 0:5

15.156


15.157 Position 1: release position. Position 2: just before impact. Position 3: just after impact. T1 + V1 = T2 + V2 : 1 1 mA (vA )22 + k 22 2 2 1 1 1 (300) (0:32 ) = (0:5)(vA )22 + (300)(0:052 ) 2 2 2 (vA )2 = 7:246 m/s 1 0+ k 2

2 1

=

e = vsep =vapp : e=

(vB )3 (vA )3 (vA )2

0:8 =

(vB )3 (vA )3 7:246

(a)

p 2 = p3 : mA (vA )2 = mA (vA )3 + mB (vB )3 0:5(7:246) = 0:5(vA )3 + 0:2(vB )3

(b)

Solution of Eqs. (a) and (b) is (vB )3 = 9:32 m/s J

(vA )3 = 3:52 m/s

15.158 Position 1 = just before shell is …red Position 2 = just after the shell is …red Position 3 = maximum compression of spring System: + ! p1 = p2 : 0 = mA (vA )2 0 = 24(1800) 650 (vB )2

Barrel only: T2 + V2 1 2

650 32:2

(66:46)2

= T3 =

1 1 2 mB (vB )2 + 0 = 0 + k 2 2

V3 :

1 (26 2

mB (vB )2 (vB )2 = 66:46 ft/s

103 )

2 3

3

2 3

= 1:852 ft J

15.159 Block A:

3 lb

y x

o

25

FBD NA

= 0.2NA

3 a 32.2 MAD


Fy

=

Fx

= max

a =

0

+"

NA cos 25

+ !

0:2NA sin 25

3=0

NA sin 25 + 0:2NA cos 25 =

NA = 3:651 lb 3 a 32:2

32:2 2 (3:651) (sin 25 + 0:2 cos 25 ) = 23:66 ft/s 3

System (block A and wedge B):

9 lb

P

9 a 32.2

=

FBD 0.4NB NB Fy

=

Fx

= max

+ !

=

9 (23:66) = 10:21 lb J 32:2

P

0

MAD

+"

0:4(9) +

NB

9=0 P

NB = 9 lb 9 a 0:4NB = 32:2

15.160

(mA+ mB)v2

mB(vB )1

mA (vA )1

B A Before impact

A

B

After impact

(vA )1

=

40 mi/h = 40

(vB )1

=

26 mi/h = 26

5280 3600 5280 3600

= 58:67 ft/s = 38:13 ft/s

(a) Momentum is conserved: mA (vA )1 + mB (vB )1 = (mA + mB )v2 3800 3200 + 3800 3200 (58:67) + (38:13) = v2 g g g v2 = 47:52 ft/s J (b) T1

= =

1 1 mA (vA )21 + mB (vB )21 2 2 1 3200 1 3800 (58:67)2 + 2 32:2 2 32:2

(38:13)2 = 256:8

103 lb ft


T2 =

1 1 (mA + mB )v22 = 2 2

% energy loss =

T1

T2 T1

3200 + 3800 32:2 100% =

(47:52)2 = 245:5

256:8 245:5 256:8

103 lb ft

100% = 4:40% J

15.161


Chapter 16 16.1

16.2 = 10e

0:5t

!=

20e

0:5t

+ C1

= 40e

0:5t

+ C1 t + C2

= 0 when t = 0. ) C1 = 20 rad/s, C2 =

Initial conditions: ! = )!=

20e

0:5t

+ 20 rad/s

= 40e

0:5t

+ 20t

40 rad

40 rad

When t ! 1, ! ! 20 rad/s. ) ! 1 = 20 rad/s J When ! = 0:5! 1 : 10

= =

20e 0:5t + 20 e 40(0:5) + 20(1:3863)

0:5t

= 0:5 t = 1:3863 s 40 = 7:726 rad = 1:230 rev J

16.3 = ) ! d!

=

d! d d! d! = = ! dt d dt d 1 2 d ! = + C1 2

Initial condition: ! = 6000 rev/min when ) When

1 2 (! 2

( is constant)

= 0. ) C1 = 60002 =2 (rev/min)

2

60002 ) =

= 3600 rev, ! = 1200 rev/min. )

1 (12002 2

60002 ) = (3600)

=

4800 rev/min

2

d! dt = d! t = ! + C2 dt Initial condition: ! = 6000 rev/min when t = 0. ) C2 = 6000 rev/min When ! = 0: =

t = C2

4800t =

6000

t = 1:250 min = 75:0 s J


16.4 =

4t2 + 24t

10 rad

!=

8t + 24 rad/s

=

8 rad/s

2

(a) When t = 4 s: !=

8(4) + 24 =

8 rad/s J

=

8 rad/s

2

J

(b) Note that the rotation reverses direction when t = 3 s (obtained by setting ! = 0). When t = 0, = 10 rad When t = 3 s, = 4(3)2 + 24(3) 10 = 26 rad When t = 4 s, = 4(4)2 + 24(4) 10 = 22 rad The total angle turned through is tot = (10 + 26) + (26 22) = 40 rad J

16.5 Z 4 + 6t != dt = 4t + 3t2 + C1 Z = ! dt = 2t2 + t3 + C1 t + C2 =

Initial conditions: ! = 0 and

= 0 when t = 0: ) C1 = C2 = 0

) ! = 4t + 3t2

= 2t2 + t3

When ! = 24 rad/s: 24 = 4t + 3t2

t = 2:239 s

)

= 2(2:239)2 + 2:2393 = 21:3 rad J

16.6


16.7

16.8 2

= 12 rad/s Initial conditions: C2 = 0.

= 6t2 + C1 t + C2 rad

! = 12t + C1 rad/s = 0, ! =

) ! = 12t

24 rad/s when t = 0. ) C1 = 24 rad/s

= 6t2

24 rad/s and

24t rad

Note that the rotation reverses direction when t = 2 s (obtained by setting ! = 0). When t = 0, = 0: When t = 2 s, = 6(2)2 24(2) = 24:0 rad When t = 4 s, = 6(4)2 24(4) = 0 The total angle turned through is tot = 48 rad J


16.9 !

=

4t1=2

=

=

jt=6 s

Z

! dt =

8 3=2 t +C 3

8 3=2 (6) + C 3

jt=0 s =

J

C = 39:2 rad

16.10 _=

= sin t

•=

cos t

2

sin t

= R• = R 2 sin t 2 an = R _ = R 2 cos2 t q q a = a2t + a2n = R 2 sin2 t + cos4 t at

The acceleration is maximized when

d (sin2 t + cos4 t) d( t)

=

0

2 sin t cos t + 4 cos3 t( sin t)

=

0

There are three solutions: t

=

t

=

t

=

2

0 yielding a = R 2 4

yielding a = R p

3 R 2

yielding a =

) amax = R

2

2

2

J

16.11 Pulley B: v

=

a =

(RB )o ! B (RB )o

B

40

! B = 24 rad/s J

12 = 20! B 28

12 = 20

B

B

=

16:8 rad/s

2

J

Belt between B and C: v0

=

(RB )i ! B = 8(24) = 192 in./s

0

=

(RB )i

a

B

= 8( 16:8) =

134:4 in./s

2

Pulley C: v0 0

a

= RC ! C = RC

C

! C = 8:0 rad/s J

192 = 24! C 134:4 = 24

C

C

=

2

5:60 rad/s

J


16.12 Left pulley: vA

=

(aA )n

=

a2A

=

(RA )i ! A

12 = 0:75! A ! A = 16 rad/s 2 2 2 (RA )o ! A = 2(16) = 512 ft/s (aA )t = (RA )o A = 2 2 2 2 (aA )n + (aA )t 600 = 512 + 4 2A A = 156:41

2

A 2

rad/s

Right pulley: vB

= RB ! B

(aB )n

=

RB ! 2B

(aB )t

= ab elt = (RA )i A = 0:75(156:41) = 117:31 ft/s q p 2 (aB )2n + (aB )2t = 115:22 + 117:312 = 164:4 ft/s J =

aB

16.13

12 = 1:25! B 2

! B = 9:6 rad/s

= 1:25(9:6) = 115:2 ft/s

2 2

!B !C !D

rA 50 (320) = 457:1 rad/s !A = rB 35 = ! B = 457:1 rad/s rC 20 = !C = (457:1) = 203 rad/s J rD 45 =

16.14


16.15

16.16 AC

vB

= !

vB

=

0:36i + 0:54j 0:3k =p = 0:5035i + 0:7553j 0:362 + 0:542 + 0:32

rB=C = 4 0:9063i

AC

i ( 0:54j) = 4 0:5035 0

0:4196k

j 0:7553 0:54

k 0:4196 0

1:0876k m/s J


B

= = 6

rB=C + ! (! rB=C ) = rB=C + ! vB ( 0:54j) + 2 AC vB AC i j i j k 0:4196 + 4 0:5035 0:7553 6 0:5035 0:7553 0:9063 0 0 0:54 0

= =

4:65i + 3:71j

1:107k m/s

2

k 0:4196 1:0876

J

16.17 Let w be the width of the tape. dV1 dV2 dV1

!

= Vol. of tape leaving the reel during time dt is v0 hw dt = Vol. change of tape on the reel (approx.) is 2 Rw dR dR vo h = dV2 : v0 h dt = 2 R dR = dt 2 R = =

v0 R d! d! dR = = dt dR dt

v0 dR = R2 dt

v0 R2

v0 h 2 R

=

v02 h J 2 R3

16.18 ! rB=A vB

= ! =

12i

= !

15j + 9k = 25 p = 152 + 92 9k in.

21:44j + 12:862k rad/s

CA

rB=A =

i 0 12

j 21:44 0

k 12:862 9

= 192:96i + 154:34j + 257:3k in./s = 16:08i + 12:86j + 21:4k ft/s J

aB

= !

vB =

i 0 192:96

j 21:44 154:34

=

7502i + 2482j + 4137k in./s

=

625i + 207j + 345k ft/s

2

k 12:862 257:3 2

J

16.19 vB = vA + vB=A


v)

v)

B

+

!

=

A

(vB )x = 8 cos 60 = 4 ft/s q p 2 (vB )y = vB (vB )2x = 62

) + !

( B x

" =

2ω

8 ft/s 60o + ω A

( B y

2 ft

B

42 = 4:472 ft/s

(vB )y = 8 sin 60 + 2! 4:472 = 8 sin 60 + 2! 1:228 rad/s ! = 1:228 rad/s J

16.20 Wheel rolls without slipping: vC = R! = 1:75! vB = vC + vB=C

8 ft/s B +"

0.75ω = 1.75ω + B 0.75 ft (vB)x C

ω

C

= 0:75! ! = 10:667 rad/s J = 1:75! = 1:75(10:667) = 18:67 ft/s

8 vC

! J

16.21 Wheel rolls without slipping: vC = R! vA = vC + vB=C + vA=B

vA

A =

Rω

ω

+ C C

Rω ω AB

R

B +

B 2

2R

45o

A

2 2RωAB

+

+"

0 vA vA

p = R! 2 2R! AB sin 45 ! AB = 0:5! J p p = R! + 2 2R! AB cos 45 = R! + 2 2R(0:5!) cos 45 = 2R! J


16.22 vO = vA + vO=A

6 ft/s

=

O

4 ft/s

2.5ω

+

A

O 2.5 ft

ω +!

6=4

2:5!

.A

! = 4:0 rad/s

J

16.23

A 0.5 m/s A

=

0.6 m/s B

+

0.16 m

vA = vB + vA=B

B + ! 0:5 =

0:6 + 0:16!

0.16ω

ω

! = 6:875 rad/s

J

vC = vB + vC=B

C

=

0.6 m/s B

+

0.6875 m/s 0.1 m

vC C

6.875 rad/s B + ! vC =

0:6 + 0:6875 = 0:0875 m/s ! J


16.24


16.25

16.26 vA = vB + vA=B

ω A

=

12 in./s B

in. 10

vA

.

B

+

A 10ω

4

3


+ +

"

3 (10!) ! = 2:0 rad/s 5 4 4 vA = (10!) = (10 2) = 16:0 in./s 5 5

0 = 12

J

16.27

16.28

D B y A 2 ft/s

ω

θ

d

x

vB 2.5 ft

d = 2:5 csc


Solution I (scalar notation) vB = vA + vB=A

ωd vB

θ

B + -

=A

vA

d

+ ωA

0 = vA sin + !d ! = 0:8 sin2 rad/s

0= J

θ

B

2 sin + !(2:5 csc )

Solution II (vector notation) vB vB i vB i

= vA + ! = 2 cos i = 2 cos i

rB=A 2 sin j + !k 2:5 csc i 2 sin j + 2:5! csc j

Equating j-components: !=

! = 0:8 sin2

2 sin + 2:5! csc

J

rad/s

16.29 = vB + vC=B = ! AB rB=A + ! BC rC=B = 2:8k ( 30i) + ! BC k (30i + 60j) = 84j + ! BC (30j 60i)

vC ! CD rC=D ! CD k 60i 60! CD j

Equating like components: 60! BC 60! CD

= =

0 84

) ! BC = 0 J ) ! CD = 1:40 rad/s

J

16.30

y

B

D 6

15

A

8 E

x


= rB=A = ( 6i) = 36j = vB

! AB 6k

vD + vB=D ! DE rD=E + ! BD rB=D ! DE k ( 6i + 8j) + ! BD k ( 15i ! DE ( 6j 8i) + ! BD ( 15j + 8i)

8j)

Equating like components: 0 ! BD

= 8! DE + 8! BD = ! DE = 1:714 rad/s

36 = J

6! DE

15! BD

16.31


16.32

B

A

Geometry: rB=A rC=B

18 30o x

27

y

30

φ

C

18 cos 30 + 27 cos = 30 = 57:74 = 18(i cos 30 + j sin 30 ) = 15:588i + 9j in. = 27(i cos 57:74 j sin 57:74 ) = 14:412i 22:83j in.

= vB + vC=B = ! AB rB=A + ! BC rC=B = 20k (15:588i + 9j) + ! BC k (14:412i 22:83j) = 311:8j 180i + ! BC (14:412i 22:83j) in./s

vC vC j

Equating like components: 0 vC

= 180 + 22:83! BC ! BC = 7:884 rad/s = 311:8 + 7:884(14:412) = 425 in./s " J

16.33 vB = vA + vB=A

30 o

vB

Β + ! vB sin 30 + " vB cos 30 (0:3420!) cos 30

= = =

ω

Α

20o 0.5ω + 2 m/s Α 0.5 m Β

=

0:5! sin 20 vB = 0:3420! 2 + 0:5! cos 20 2 + 0:5! cos 20 ! = 11:516 rad/s

J

vC = vA + vC=A

(vC)y

Α (vC)x =

C + +

! "

vC

=

2 m/s

11.516 rad/s +

20o

Α 1.0 m

11.516 m/s C

(vC )x = 11:516 sin 20 = 3:939 m/s (vC )y = 2 + 11:516 cos 20 = 8:822 m/s p 3:9392 + 8:8222 = 9:66 m/s J


16.34

A

m B 4 . 0 θ 0.25 m

0:25 = 38:68 0:4 vC = vB + vC=B

= sin

1

1.0 m/s vC C

= 2.5 rad/s

+ +

"

.

m 0.4

C 38.68o o 0.4 38.68 m B 0.4ωBC ωBC B

+

.

A

0 = (1:0 0:4! BC ) cos 38:68 ! BC = 2:50 rad/s vC = (1:0 + 0:4! BC ) sin 38:68 = [1:0 + (0:4 2:5)] sin 38:68 = 1:250 m/s J

16.35


16.36

16.37

vB

A vB 30j 30j

80

B 160 o y 30 vA

D

x

= vA + vB=A 30j = vA i + ! BD rB=A = vA i + ! BD k 160(i cos 30 + j sin 30 ) = vA i + ! BD (138:56j 80i)

Equating y-components: 30 = 138:56! BD vD

vD

! BD = 0:2165 rad/s

= vB + vD=B = 30j + ! BD rD=B = 30j + 0:2165k 80(i cos 30 + j sin 30 ) = 30j + 15:0j 8:660i = 45:0j 8:660i mm/s p = 45:02 + ( 8:660)2 = 45:8 mm/s J


16.38


16.39

16.40

13 i n.

12 in.

O 5 in. B ω 20 in./s

.

A

vA

Point B is the I.C. of the disk. vO vA

= ! OB 20 = 5! ! = 4 rad/s = ! AB = 4(13) = 52 in./s J


16.41

12 in./s 6 in.

B

ω

O

in. 10

vA

8 in.

A

Point O is the I.C. of the disk. vB vA

= OB ! 12 = 6! ! = 2 rad/s = OA ! = 8(2) = 16 in./s J

16.42

ω = 2.5 rad/s

300 vG

O vO= 500 mm/s

G 3 2

d = 200

e C

Dimensions in mm vO 500 = = 200 mm ! 2:5 determine the location of the instant center C of

d=

The directions of vO and vG the wheel. p e = 3002 + 2002 = 360:6 mm vG = e! = 360:6(2:5) = 902 mm/s %56:3

J


16.43

16.44


16.45

16.46 vC 24 rad/s

80 A

vB vB

vB 120 C B

= vA + vA=B = 0 + ! AB AB = 4:8(200) = 960 mm/s vC vC

= vA + vC=A = 0 + ! A AC = 24(80) = 1920 mm/s

ωB I.C. of B

(A and B on the arm AB)

(A and C on gear A)

The velocities vC and vB establish the instant center for velocities of gear B. ) !B =

vC 1920 = = 8 rad/s 240 240

J


16.47

0.5 m/s

0.5 m/s A

a I.C.

60 mm

C 160 − a

100 mm B

0.6 m/s 0.6 m/s Velocities of A and B establish the I.C. of the gear. a 0:5

=

!

=

vC

=

160 a a = 72:73 mm 0:6 vA 500 = = 6:875 rad/s J a 72:73 (a AC)! = (72:73 60) (6:875) = 87:5 mm/s ! J

16.48

vB

B

30

6 in .

o

6 rad/s A Geometry: BE = vB ! BC vC ! CD

vC 8 in.

60o ωCD

8 = 16:0 in. cos 60

ωBC 13.856 in.

16 in.

E

C 4 in. D CE = 8 tan 60 = 13:856 in.

= ! AB AB = 6(6) = 36 in./s vB 36 = = = 2:25 rad/s 16 BE = ! BC CE = 2:25(13:856) = 31:18 in./s vC 31:18 = = = 7:80 rad/s J 4 CD


16.49

16.50

16.51


16.52

16.53

D

vC 60

Dimensions in mm

C

60

ωCD

vB B

30

A

2.8 rad/s

I.C. of link AB is point A. This determines the direction of vB .


I.C. of link CD is point D. This determines the direction of vC . Since vB and vC are parallel, link BC is translating. Therefore, = 0 J = vC

! BC vB From motion of link AB:

vB = AB ! AB = 30(2:8) = 84:0 mm/s From motion of link CD: vC = CD ! CD

84:0 = 60! CD

! CD = 1:40 rad/s

J

16.54 D

C 12 i n.

ωBC

B vB

Geometry:

AC BD

! BC vB ! AB

γ

30 in./s

β

6 i 45o n.

ωAB A

6 12 = = 20:70 sin sin 45 = 180 45 20:70 = 114:3 p 2 = CD = 12 + 62 2(12)(6) cos 114:3 = 15:468 in. AC sin 45

=

6=

15:468 sin 45

6 = 15:875 in.

vC 30 = = 1:940 rad/s 15:468 CD = ! BC BD = 1:940(15:875) = 30:80 in./s vB 30:80 = = = 5:13 rad/s J 6 AB =


16.55

A

27 in. 60o

12 rad/s

E

in. 15

vD

D

B

vB

ωBC F Geometry: BF = vB ! BC

AE cos 60

AB =

27 cos 60

15 = 39 in.

= ! AB AB = 12(15) = 180 in./s vB 180 = = = 4:62 rad/s J 39 BF

16.56

13 in.

12 rad/s B A 15 in. vB

F

BE

D

BF = tan DF = BF + DF cot = (27 = tan

vB ! BC

β

E

β

27 in.

Geometry:

ωBC

1

vD 1

27

15 = 42:71 13 15) + 13 cot 42:71 = 26:08 in.

= ! AB AB = 12(15) = 180 in./s vB 180 = = = 6:90 rad/s J 26:08 BE


16.57

vC

C

ωBCD

24

E 18 F

B 24

A

20 o 35 vB 48 rad/s

18 D

BF

=

BE

=

DE

q 2 BC

2

CF =

Dimensions in inches

p 242

vD 182 = 15:875 in.

BF 15:875 = = 19:380 in. cos 35 cos 35 = DF + BF tan 35 = 18 + 15:875 tan 35 = 29:12 in. vB

! BCD vD

= ! AB AB = 48(20) = 960 in./s 960 vB = = 49:54 rad/s J = 19:380 BE = ! BCD DE = 49:54(29:12) = 1443 in./s ! J


16.58

ωAD

E 30o

22 .27 in.

in. 78 20.

24 in.

D C

. 8 in

.

in. 12 18 in./s

.

vD

vC 60o

30o

A

.B

Point E is the I.C. of link AD. Its location is determined by the known directions of the velocities at A and C. AE CE DE vA vD

12 = 24 in. sin 30 12 = = 20:78 in. tan p 30 = 20:782 + 82 = 22:27 in. =

18 = 24! AD ! AD = 0:75 rad/s = AE ! AD = DE ! AD = 22:27(0:75) = 16:70 in./s J


16.59

16.60

ωAB 8

E

vB B

vA O 16 in./s

A 3 5

4 C ωdisk

Geometry: OB BE AE

! disk ! AB vB

q p 2 2 = AB AO = 82 32 = 7:416 in. 3 3 = AO + OB = 3 + (7:416) = 8:562 in. 4 4 5 5 = OB = (7:416) = 9:270 in. 4 4

vO 16 = = 4 rad/s vA = ! disk AC = 4(5) = 20 in./s 4 OC vA 20 = = = 2:1575 rad/s 9:270 AE = ! AB BE = 2:1575(8:562) = 18:47 in./s " J =


16.61 y

B o

30

B

8 m/s

2

=

x

+

o

6 m/s2

30

A

A

0.2ωAB2

m 0.2 o 30

0.2αAB α ωABAB(sense indeterminate)

aB = aA + aB=A x+ %

y+ -

6 cos 30 = 8 sin 30

0:2! 2AB

6 sin 30 = 8 cos 30

0:2

AB

! AB = 6:78 rad/s J AB

= 49:6 rad/s

2

J

16.62


16.63

B

ω = 5 rad/s

rB/A

10 "

α = 12 rad/s2

.

A 4" r D/A

y

v0 a0

D

x No-slip condition gives

aA = R = 10(12) = 120 in./s

rD=A

= =

2

2

12k rad/s ! = 5k rad/s 4j in. rB=A = 10j in. R = 10 in.

(a) aD

= aA + aD=A = R i + rD=A + ! = 10(12)i + ( 12k) ( 4j) + ( 5k) =

120i

48i + 100j =72i + 100j in./s

(! rD=A ) [( 5k) ( 4j)] 2

J

(b) aB

= aA + aB=A = R i + rB=A + ! (! rB=A ) = 10(12)i + ( 12k) 10j + ( 5k) [( 5k) 10j] =

120i + 120i

250j = 240i

250j in./s

2

J

(c) The acceleration of the end of the string is the same as the horizintal component of the acceleration of point D on the spool: a0 = 72i in./s

2

J


16.64 B

ω = 5 rad/s α = 12 rad/s2

10 "

.

A 4"

y

v0 a0

D

x

Note that the acceleration of point D on the spool is aD = (a0 !) + (aD )y " aD = aA + aD=A


(aD)y D

= 16 in./s2 A

aA

A

+

4"

4(12) = 48 in./s2

D

4(5)2 =100 in./s2 Equating like components: + !

+

"

16 = aA

48

aA = 64 in./s

(aD )y = 100 in./s

) aD = 16i + 100j in./s

J

2

2

2

aA = 64i in./s

2

J

aB = aA + aB=A

10(5)2 = 250 in./s2 aB

=

64 in./s2 A

+

10(12) = 120 in./s2

B 10"

A aB = (64 + 120)i


250j = 184i

2

250j in./s

J


16.65 = B

A

1.2(2)2 m/s2 30o α B ω = 2 rad/s

8 m/s2 +

m 1.2

aA

A

1.2α

aA = aB + aA=B + + The solution is

aA = 1:2 sin 30 1:2(2)2 cos 30 0 = 8 1:2 cos 30 + 1:2(2)2 sin 30

" !

= 10:007 rad/s2 and aA =

10:161 m/s2

) aA = 10:16 m/s # J 2

16.66 Let A and B be points on the disk. vA + vB=A = vB

3 m/s A

0.4 m

+ω

A

+" 3

B

=

0.4ω

0:4! = 1:0

1.0 m/s B J

! = 5 rad/s

Note that the belt accelerations shown in Fig. P16.66 are the tangential components of the accelerations of points A and B on the disk. aA + aB=A = aB

0.5 m/s2 A

0.2ω2

+

+ " 0:5

A

0.4 m

α, ω 0:4 =

1:2

0.4ω2 B 0.4α

=

= 4:25 rad/s

0.2ω2

B

1.2 m/s2 2

J


16.67

vB B

16 in.

vC C

β 15 rad/s A

8 in.

! BC = 0 (vB and vC are parallel)

= sin

1

6 = 30 12

aC = aB + aC=B

=

16αBC C 30o

+α

BC

16 i n.

aC

8(15)2 in./s2 8 in. 15 rad/s B A

B 0 = 8(15)2

+ !

+

"

aC = 16

16

BC

2

cos 30

BC

= 129:9 rad/s

sin 30 = 16(129:9) sin 30 = 1039 in./s " J 2

BC

16.68

3 rad/s

A

vC ! BC sin

2

180

β

t 6f

6 ft

γ ft 4 C

vB

B

= ! AB = 3 rad/s (A is the I.C. for BC) 2 1 = = 2 sin 1 = 38:94 6 3 38:94 = +2 = 90 = 90 = 70:53 2 2


t 6f

aC

3 rad/s A 12 rad/s2

β

C =

B

0 = 0 = + = BC aC = = +

+

γ

6(12) ft/s2 +

4(32) ft/s2

6(32) ft/s2

B 4 ft

C

αBC

3 rad/s

= aB + aC=B

aC +"

4αBC

aC

6(32 ) cos + 6(12) sin + 4(32 ) cos + 4 BC sin 54 cos 38:94 + 72 sin 38:94 + 36 cos 70:53 4 BC sin 70:53 2 26:32 rad/s 2 6(3 ) sin 6(12) cos 4(32 ) sin + 4 BC cos 54 sin 38:94 72 cos 38:94 36 sin 70:53 4( 26:32) cos 70:53 2

=

91:1 ft/s = 91:1 ft/s

2

! J

16.69 ) ! BC =

By inspection: b! = 2b! BC

aC

ω, α C =

A

bα b aC

+

aC +"

0

bω +

ω/2, αBC

2

B

= aB + aC=B ! = b! 2 + 2b 2 2b

2b(ω/2)2 2b

B

= b

1 2!

2

BC

=

3 2 b! 2 BC

=

C

2bαBC

J 1 2

J

16.70

B 30

D ! BC =

4 m/s 6m 0.1

o

ωBC

C vC

vB 4 = = 28:87 rad/s (D is the I.C. for BC) 0:16 cos 30 BD aC = aB + aC=B (note that aB = 0)


C

28.87 rad/s

B

6m 0.1

αBC

aC =

30o

C 0.16(28.872) m/s2

0.16αBC +

0

=

0:16

BC

cos 30 + 0:16(28:872 ) sin 30 2

aC

= 481:2 rad/s = 0:16 BC sin 30 0:16(28:872 ) cos 30 = 0:16( 481:2) sin 30 0:16(28:872 ) cos 30

aC

=

BC

+#

154:0 m/s = 154:0 m/s " J 2

2

16.71

A

vA = 2 m/s

0.5

0.4

E 0.3

ωBC

vB

ωAB B

C

0.6

Point E is the I.C. of bar AB ! AB vB ! BC

vA 2 = = 5:0 rad/s 0:4 EA = EB ! AB = 0:3(5) = 1:5 m/s vB 1:5 = 2:5 rad/s = = 0:6 BC =

aB = aA + aB=A

0.6αBC

ωBC = 2.5 rad/s 1.2 m/s2 =

C

αBC

A + 0.4

ωAB = 5 rad/s αAB 0.5

0.6(2.5)2 m/s2 B 0.6

A

0.3 0.5αAB

+

!

+

"

0.5(5)2 m/s2 B

4 3 1:2 + (0:5 AB ) (0:5)(5)2 5 5 4 3 = (0:5)(5)2 + (0:5 AB ) 5 5

0:6(2:5)2 = 0:6

BC


The solution is )

AB

= 31:125 rad/s2 and J

2

AB

= 31:1 rad/s

BC

=

32:23 rad/s2 J

2

BC

= 32:2 rad/s

16.72

ωAB

vB

2 ft

C

B

1.5 ft A

6 ft/s

Point C is the I.C. of bar AB. 1:5! AB vB 2! BC

= =

6 ! AB = 4 rad/s 2! AB = 2(4) = 8 ft/s 8 vB = = 4 rad/s = vB ! BC = 2 2 aA = aB + aA=B

B 2αBC 2 ft/s2 A

=

αBC ωBC= 4 rad/s

+ !

2= AB

+

BC

3

+A

B 2(4)2 ft/s2

2 ft

C

ft 2.5

2.5(4)2 ft/s2

αAB ωAB= 4 rad/s

4

2.5αAB

4 3 2(4)2 + (2:5)(4)2 + (2:5) 5 5 2 = 1:3333 rad/s J

AB

3 4 + (2:5)(4)2 (2:5) AB 5 5 3 4 0 = 2 BC + (2:5)(4)2 (2:5)(1:3333) 5 5 2 2 = 10:67 rad/s = 10:67 rad/s J "

0=2

BC


16.73 vB = AB! AB = 1:2(10) = 12:0 m/s " Note that vB and vD are parallel. Therefore, bar BD is translating in the position shown. ) ! BD = 0

vD = vB = 12:0 m/s "

! DE =

vD DE

12 = 5:0 rad/s 2:4

aD = aB + aD=B 3

ωDE= 5.0 rad/s 2.4α DE αDE 2.4 m

E

1.5αBD

D 2.4(5)2 m/s2

=

10.0 rad/s 1.2(10)2 m/s2 A B 1.2 m

+

4

D

m

.5 αBD 1 3 4

B + ! +

"

2:4(5)2 = 1:2(10)2 2:4 DE

4 (1:5 BD ) 5 2 = 100 rad/s J

DE

=

3 (1:5 5 2:4

BD )

DE

=

BD

= 200 rad/s

2

J

4 (1:5)(200) 5

16.74

Acceleration Analysis

Assume

AB

to be counterclockwise.


16.75

B

vB

0.8 m

ωBC A 0.6 m

0.4 m C vC

.

D

8 rad/s

Point A is the I.C. of bar BC. vC 0:6! BC vB 0:8! AB

=

0:4(8) = 3:2 m/s 3:2 = vC ! BC = = 5:333 rad/s 0:6 = 0:8! BC = 0:8(5:333) = 4:266 m/s 4:266 = vB ! AB = = 5:333 rad/s 0:8 aC = aB + aC=B


0.8(5.333)2 m/s2 0.8αAB

5.333 rad/s

B

=

+ !

AB

=

3

αBC

5.333 rad/s

A

4 (5:333)2 5

0 = 0:8(5:333)2 BC

4

C

αAB + #

+

0.8 m

αBC

m 1.0

0.4(8)2 m/s2 0.4 m D C 8 rad/s

B

=0 J

3 5

(5.333)2 m/s2

BC

3 4 (5:333)2 + BC 5 5 4 3 (5:333)2 + (0) 0:4(8)2 = 0:8 AB 5 5 2 2 53:3 rad/s = 53:3 rad/s J 0:4(8)2 =

0:8

AB

16.76

n. i 25 15 in.

B

C vC ! AB

A

ωAB

D vB 6 rad/s . in 5 1 vC

= vB = ! CD CD = 6(15) = 90 in./s vB 90 = = = 3:6 rad/s ! BC = 0 (vB and vC are parallel) 25 AB aB + aC=B = aC

αAB

20 rad/s2 D A αBC 45o 3.6 rad/s 45o 6 rad/s + 15 = 5 C 15 2 B 15(20) in./s2 C 2 2 15(6 ) in./s α 25 15αBC AB 25(3.62) in./s2 B


25(3:62 ) + 15

+%

BC

sin 45

=

= 20:36 rad/s = 15(20) = 15(20)

BC

+&

25 25

+ 15 BC cos 45 + 15(20:36) cos 45

AB

AB

15(62 )

=

AB

J

2

J

2

3:36 rad/s

16.77 D

B vB

15 in.

ωBD

8 in.

in. 17 vD

6 in.

E

Point E is the I.C. of bar BD vB ! BD

= AB ! AB = 6(3) = 18 in./s # vB 18 = = = 0:8571 rad/s 21 BE

! DE = ! BD = 0:8571 rad/s

aB = aD + aB=D

+ !

+"

A

10αDE =

8

D 10 ω

2 DE

10

2 6ωAB B 6

ωAB

D

6

ωDE αDE

E

+

17 2 17ωBD B 15 17αBD

ωBD αBD

8

6 8 15 8 (10! 2DE ) (10 DE ) + (17! 2BD ) + (17 10 10 17 17 6(3)2 = 6(0:8571)2 8 DE + 15(0:8571)2 + 8 BD 38:57 = 8 BD 8 DE

6! 2AB

0

=

0 = 0 =

=

8 6 8 (10! 2DE ) (10 DE ) + (17! 2BD ) 10 10 17 8(0:8571)2 6 DE + 8(0:8571)2 15 15 BD 6 DE

Solution of (a) and (b) is )

BD

= 1:3775 rad/s2 and 2

BD

= 1:378 rad/s

J

DE

DE

=

15 (17 17

BD )

(a)

BD )

(b) 3:444 rad/s2

= 3:44 rad/s

J


16.78

16.79


16.80


16.81


16.82

16.83

rA/O y ω x O 0.6 ft A, A' ! = 30

2 60

=

rad/s


Let A0 be a point on OB that is coincident with A at the instant under consideration aA

= aA0 + aA=OB + aC = ! (! rA=O ) + 0 + 2! vA=OB = k ( k 0:6i) + 2( k 3i) = k (0:6 j) + 6 j = 0:6 2 i + 6 j

aA

=

5:92i + 18:85j ft/s

2

J

16.84

16.85

P

30o A

16 .97 1i n.

in. 24

12 in.

45o

B

Let P 0 be …xed to bar BC. vP = vP 0 + vP=BC


16.971ωBC

24(5) in./s in. 24 30o

5 rad/s

16 .97 1i n.

P

=

45o

ωBC

A

P'

vP/BC

+

45o

B

+ - 45

24(5) cos 15 = 16:971! BC

2

! BC = 6:83 rad/s

J

16.86 6 rad/s2

y

3 rad/s

rP/Ao 30 A B

x ! !

rP=A rP=A vP

aP

36 in./s P

15 in. 2

= 3k rad/s = 6k rad/s vP=B = 36j in./s = 15(i + j tan 30 ) = 15i + 8:660j in. = 3k (15i + 8:660j) = 45j 25:98i in./s = vA + vP 0 =A + vP=B = 0 + ! rP=A + vP=B = 45j 25:98i + 36j = 51:96i + 126:0j = 26:0i + 81:0j in./s J

= aA + aP 0 =A + aP=B + aC = 0+ rP=A + ! (! rP=A ) + 0 + 2! vP=B = 6k (15i + 8:660j) + 3k (45j 25:98i) + 2(3k) = ( 90j + 51:96i) + ( 135i 77:94j) 216i =

299i

167:9j in./s

2

36j

J


16.87 aP = aA + aP 0 =A + aP=disk + aC Noting that A is a …xed point and that P 0 coincides with A, we have aA = aP 0 =A = 0. Hence aP = aP=disk + aC

82 = 8 in/s2 8 vP/disk = 8 in./s P aP

=

2(1.2)(8) = 19.2 in./s2

+

8 in.

vP/disk = 8 in./s ω = 1.2 rad/s

=

11.2 in./s2

B

aP = 11:2 in./s " J

16.88 vP = vA + vP 0 =A + vP=AB = 0 + vP 0 =A + vP=AB

P

ft/s 45o 4(1.6971) P' 45o = + 1 7 vP/AB 4 rad/s .69 1 A

0

=

4(1:6971)

vP vP

= =

vP

+

+"

vP=AB sin 45

vP=AB = 6:788 ft/s

4(1:6971) + vP=AB cos 45 [4(1:6971) + 6:788] cos 45 = 9:60 ft/s J

vP2 9:602 2 = = 76:8 ft/s R 1:2 aP = aA + aP 0 =A + aP=AB + aC = 0 + aP 0 =A + aP=AB + aC (aP )y =

1.6971(12) ft/s2 45o P

(aP)x 76.8 ft/s2

=

P' 1 7 4 rad/s 69 1. A 12 rad/s2

1.6971(42) ft/s2 +

ω 45o 45o + vP/AB aP/AB 2(4)(6.788) ft/s2


+" + !

76:8

42 )

=

1:6971(12

aP=AB

=

2

(aP )x

=

1:6971(12 + 42 )

(aP )x

=

28:8 ft/s

2

) aP

=

28:8i

47:52 ft/s

76:8j ft/s

aP=AB

2(4)(6:788) sin 45

47:52 + 2(4)(6:788) cos 45 2

J

16.89 vP = vA + vP 0 =A + vP=AB = 0 + vP 0 =A + vP=AB

ωAB

0.2 309 m

0.2309ωAB P' vP/AB x 30o 1.2 m/s + = 60 o P y A +x +y .

1:2 cos 30 1:2 sin 30

J

= 0:2309! AB ! AB = 4:50 rad/s = vP=AB vP=AB = 0:6 m/s

aP = aA + aP 0 =A + aP=AB + aC 0 = 0 + aP 0 =A + aP=AB + aC

0.2309αΑΒ

0.2 309 m

30o

0 = 4.50 rad/s A +x -

0.2309 (4.52) m/s2 P'

ωAB

+ 60 o + v P/AB aP/AB

30o 2(4.50)(0.6) m/s2

αΑΒ

0 = 0:2309

AB

2(4:50)(0:6)

2

AB

= 23:4 rad/s

J

16.90 aO vP=disk

20 2 (2) = 3:333 ft/s 12 2 6 ft/s # aP=disk = 28 ft/s #

= R = =

aP = aO + aP 0 =O + aP=disk + aC


aP = 3.333 ft/s O

2

+ O

+ +

!

"

)

1.0(52) ft/s2 P'

1.0 ft

1.0(2) ft/s

2

+ 5 rad/s 2 rad/s2

(aP )x =

3:333

(aP )y =

1:0(52 )

aP =

65:3i

2(5)(6) ft/s2 ω + 28 ft/s2 vP/disk

1:0(2)

2(5)(6) =

28 =

53:0j ft/s

2

53:0 ft/s

65:3 ft/s

2

2

J

*16.91


16.92 ! = 5k rad/s

=0

rP 0 =A = 0:4(i cot 60 + j) = 0:2309i + 0:4j) m

(a) vP 0 =A vP=AB

= ! rP 0 =A = 5k (0:2309i + 0:4j) = 1:1547j 2i m/s = vP=AB (i cos 60 + j sin 60 ) = vP=AB (0:5i + 0:8660j) m/s vP vP i

= vA + vP 0 =A + vP=AB = 0 + 1:1547j 2i + vP=AB (0:5i + 0:8660j)

Equating y-components: +

" )

0 = 1:1547 + 0:8660vP=AB vP=AB = 1:333 m/s vP=AB = 1:333(i cos 60 + j sin 60 ) = 0:6665i 1:1544j m/s J

(b) = !

aP 0 =A

(!

rP 0 =A ) = 5k

(1:1547j

2i) =

5:776i

10j m/s

2

2

= aP=AB (0:5i + 0:8660j) m/s = 2! vP=AB = 2(5k) ( 0:6665i

aP=AB aC

=

6:665j + 11:544i m/s

1:1544j)

2

= aA + aP 0 =A + aP=AB + aC = 0 + ( 5:776i 10j) + aP=AB (0:5i + 0:8660j) + ( 6:665j + 11:544i)

aP aP i

Equating y-components: +

"

)

0=

10 + 0:8660aP=AB

6:665

aP=AB = 19:244 m/s

aP=AB = 19:244(0:5i + 0:8660j) = 9:62i + 16:67j m/s

2

2

J

16.93 vP = vP 0 =A + vP=AB

P

=

0.8 m

0.9 m

A 0.8(5) m/s 45o

5 rad/s C

P'

ωAB

+

vP/AB

0.9ωAB


+ ! + " )

0:8(5) cos 45 = 0:9! AB 0:8(5) sin 45 = vP=AB

J

! AB = 3:143 rad/s vP=AB = 2:828 m/s

aC = 2! AB vP=AB = 2(3:143)(2:828) = 17:777 m/s

2

Let P 0 be the point on AB that is coincident with P at this instant. Coordinates are embedded in bar AB: aP = aP 0 =A + aP=AB + aC

45o P 0.8 m


=

A 0.9 m

0.8(5)2 m/s2 0.8(3) m/s2

αAB ωAB = 3.143 rad/s

+

aP/AB

vP/AB = 2.828 m/s

+ 17.777 m/s2 ωAB = 3.143 rad/s

0.9αAB

C

0.9(3.143)2 + !

0:8(52 ) sin 45 + 0:8(3) cos 45 = 0:9 AB

= 37:4 rad/s

2

AB

17:777

J

16.94


16.95

16.96


16.97 B P 390

A

150

2 rad/s

22.62o 360

D

vP = vA + vP 0 =A + vP=AB = 0 + vP 0 =A + vP=AB


P 0.3 m

0.3(2) m/s 2 rad/s

+x +y .

m 0.39 o 22.62

= ωAB A D

0:3(2) sin 22:62 = 0:39! AB 0:3(2) cos 22:62 = vP=AB

0.39ωAB 22.62o P' + v P/AB

x y

! AB = 0:5917 rad/s vP=AB = 0:5538 m/s

J

aP = aA + aP 0 =A + aP=AB + aC = 0 + aP 0 =A + aP=AB + aC

P

0.15 m

0.15(22) m/s2

A

0.39(0.59172) m/s2 +

22.62

ωAB

22.62o 22.62 + vP/AB vP/AB 2(0.5917)(0.5538) m/s2 o

0.5917 rad/s

+x AB

αAB 0.39 m P' o

=

2 rad/s O

0.39αAB

=

0:15(22 ) cos 22:62 = 0:39 0:260 rad/s

2

AB

2(0:5917)(0:5538)

J

16.98


16.99

A

9i n.

θ

xD xD vD aD

= =

9 cos in. 9 _ sin =

B

9 (8) sin 40 = 2

=

9• sin

9 _ cos in./s

=

9(140) sin 40

46:3 in./s

2

9(8)2 cos 40 =

) vD = 46:3 in./s ! J

1251 in./s

aD = 1251 in./s

2

2

! J

16.100

A yA

θ L B xB

2 Constraint: x2B + yA = L2

(a) Di¤erentiating with respect to time: 2xB vB + 2yA vA xB ) vA = vB yA ) vA

= = =

0

(a) vB tan =

1:4 tan 20 =

0:5096 m/s

0:510 m/s # J

(b) Di¤erentiating (a) with respect to time: 2 2 vB + xB aB + vA + yA aA 2 2 vB + vA + xB aB ) aA = yA

aA

= = =

0 1:42 + ( 0:5096)2 + 0 = 1:8 cos 20

1:312 m/s

2

1:312 m/s # J 2


16.101

C

.

β 3 ft

.B

2.5 ft t 2f

θ

0.8 rad/s A Geometry: 2 sin + 3 sin

= 2:5

(a)

Di¤erentiation yields: _ + 3 cos

2 cos When

_ =0

(b)

= 45 : Eq. (a): 2 sin 45 + 3 sin = 2:5 = 21:22 Eq. (b): (2 cos 45 ) (0:8) + (3 cos 21:22 ) _ = 0

_ =

0:4046 rad/s

Di¤erentiate Eq. (b): 2 sin When

_ 2 + 2 cos •

3 sin

_ 2 + 3 cos

•=0

= 45 : (2 sin 45 ) (0:8)2 + 0

(3 sin 21:22 ) ( 0:4046)2 + (3 cos 21:22 ) • = 0 • = 0:3872 rad/s2

Therefore, J

! BC = 0:405 rad/s

2

BC

= 0:387 rad/s

J

16.102

B 0.25 m

φ

5m 0.7

0.5 m

C

x θ

D


Geometry: x = 0:75 cos + 0:5 cos 0:25 = 0:75 sin 0:5 sin

(a)

Di¤erentiation with respect to time: 0:75 _ sin 0:75 _ cos

x_ = 0 = Substitute

0:5 _ sin 0:5 _ cos

(b) (c)

= 60 and x_ = 1:2 m/s:

From (a): From (b): From (c):

0:25 = 0:75 sin 60 0:5 sin = 53:04 _ _ 1:2 = 0:75 sin 60 0:5 sin 53:04 _ _ 0 = 0:75 cos 60 0:5 cos 53:04

Solving (d) and (e) simultaneously, we get _ = rad/s. Therefore, ! CD = 1:045 rad/s

(d) (e)

1:302 rad/s and _ =

1:045

J

16.103 Geometry: 0:4 sin 1 + 0:3 sin 2 0:4 cos 1 _ 1 + 0:3 cos 2 _ 2 0:4 sin

1

_ 2 + 0:4 cos 1

1

•1

0:3 sin

2

_ 2 + 0:3 cos 2

2

•2

= 0:4 = 0

(a) (b)

=

(c)

0

When 1 = 30 and _ 1 = 0:6 rad/s (note that •1 = 0): Eqs. (a) and (b): 0:4 sin 30 + 0:3 sin 2 (0:4 cos 30 ) (0:6) + (0:3 cos 41:81 ) _ 2

= =

0:4 0

2 = 41:81 _ 2 = 0:9295 rad/s J

Eq. (c): ( 0:4 sin 30 ) (0:6)2

(0:3 sin 41:81 )( 0:9295)2 + (0:3 cos 41:81 ) •2 = 0 •2 = 1:095 rad/s2 J


16.104

16.105


16.106

16.107

A R y

θ

e

O

_ = 800

2 60

= 83:78 rad/s

Law of cosines: R2 802

= y 2 + e2 2ye cos = y 2 + 402 2y(40) cos 50

(a) y = 99:61 mm


Di¤erentiating Eq. (a): 0

=

y_

= =

2y y_

_

2ye _ cos + 2ye sin

ye _ sin 99:61(40) (83:78) (sin 50 ) = e cos y 40 cos 50 99:61 3460 mm/s = 3:47 m/s # J

*16.108

A

θ

9 ft

C

φ

6 ft

L

B

Geometry: 6 sin L sin 6 cos + L cos

= =

0 9 ft

(a) (b)

Di¤erentiating with respect to time: 6 _ cos 6 _ sin Substituting

L _ cos L_ sin L _ sin + L_ cos

= 20 and L_ =

0 0

(c) (d)

2 ft/s, we get

From (a): From (b):

6 sin 20 L sin = 0 6 cos 20 + L cos = 9 ft

The solution is L = 3:939 ft and From (c): 6 _ cos 20 From (d): 6 _ sin 20 The solution is _ =

= =

= 31:40 .

3:939 _ cos 31:40 ( 2) sin 31:40 _ 3:939 sin 31:40 + ( 2) cos 31:40

0:4053 rad/s and _ =

= 0 = 0

0:4265 rad/s. Therefore, J

! AB = 0:427 rad/s

16.109 D y A

0.5 m

θ θ

C

.5 m 0 m 0.5 B

θ


y y_

= 3(0:5) sin = 1:5 sin m = 1:5 _ cos m/s

y• = Substitute

2

1:5 _ sin + 1:5• cos m/s

(a) 2

(b)

= 30 , y_ = 3 m/s and y• = 0:

From (a): From (b):

3 = 1:5 _ cos 30 0=

_ = 2:309 rad/s 1:5(2:309) sin 30 + 1:5• cos 30

! ABC = 2:31 rad/s

• = 3:078 rad/s2

2

J

2

ABC

= 3:08 rad/s

J

16.110


16.111

16.112 x

G

2R θ A

x = 2R tan x_ = 2R _ sec2 2 x• = 2R _ (2 sec )(sec tan ) + 2R• sec2 = 4R! 2 sec2 tan (note that _ = ! and • = 0) = 4R! 2 sec2 50 tan 50 = 11:537R! 2 )

gear

=

x• = 11:54! 2 R

J


16.113

D B 2 ft

θ

vA A

x

Geometry: 2 cot

x • =

_2

2 csc2 (2 cot

= When

) x_ =

2 csc2 _ 2 2 [2 csc ( csc cot )] _ 2 csc2

x =

•

•)

= 30 : x_ = x • =

2 csc2 30 _ 2 •=0 (2 cot 30 ) (0:25)

2 ft/s 0

! AD (aB )t (aB )n

=

2=

_ = 0:25 rad/s

_ = 0:25 rad/s • = 0:2165 rad/s2

J

2

2 = AB _ = 4(0:25)2 = 0:25 ft/s 30 . J 2 = AB • = 4(0:2165) = 0:866 ft/s 60 - J


16.114


16.115

16.116


16.117

1.5 ft

vO

A

O

25o 8 ft/s

25o C

ω

The directions of the velocties of points O and A locate the instant center C of the disk. AC = 1:5 csc 25 = 3:549 ft !=

OC = 1:5 cot 25 = 3:217 ft

vA 8 = 2:254 rad/s = 3:549 AC

J

vO = OC ! = 3:217(2:254) = 7:25 ft/s ! J

16.118 Because the two arms form a parallelogram linkage, member BC translates (BC remains horizontal). Hence the velocity and acceleration vectors of B and C are identical.

vC aC

rB=A

=

2(i cos 30 + j sin 30 ) = 1:7321i + 1:0j m

! AB

=

2k rad/s

AB

=

1:5k rad/s

2

= vB = ! AB rB=A = 2k (1:7321i + 1:0j) = 2:0i + 3:464j m/s J = aB = ! AB vB + AB rB=A = 2k ( 2:0i + 3:464j) + ( 1:5k) (1:7321i + 1:0j) = 4:0j 6:928i 2:598j + 1:5i 2 = 5:43i 6:60j m/s J


16.119 vC = vB + vC=B

60 in./s

= ωAB A

+ !

0

=

+"

60

=

in. 10

C

17ωAB Bω BC B . n i 17 8 in. + 8 in. 10ωBC 15 in. 6 in.C

8 8 + 10! BC = 8! AB + 8! BC 17 10 15 6 17! AB + 10! BC = 15! AB + 6! BC 17 10 17! AB

The solution is ! AB = ! BC = 2:86 rad/s

J

16.120 !

=

AB

=

rad/s = 20 AB rad/s rC=B = 0:6j m 0:6i + 0:8j + 0:4k p = 0:5571i + 0:7428j + 0:3714k 0:62 + 0:82 + 0:42

=

AB

= !

vC

aC

2

4

=

0:8914i

rC=B + !

(!

i 0:5571 0

=

20

=

0:484i

i 0:5571 0

rC=B = 4

j 0:7428 0:6

k 0:3714 0

1:337k m/s J rC=B ) =

j 0:7428 0:6

rC=B + !

k 0:3714 0

4:30j + 9:33k m/s

2

+4

vC

i 0:5571 0:8914

j 0:7428 0

k 0:3714 1:337

J

16.121 AB

=

rBC

=

vC

aC

= ! =

3i + 5j + 4k = 32 + 52 + 42 5j ft

p

0:4243i + 0:7071j + 0:5657k

= ! rBC = 5 ( 0:4243i + 0:7071j + 0:5657k) = 14:143i + 10:608k ft/s J vC = 5 ( 0:4243i + 0:7071j + 0:5657k)

37:5i + 62:5j

50:0k ft/s

2

( 5j)

(14:143i + 10:608k)

J


16.122

16.123


16.124

A

805 .0

E

4 3

15 0

A

201 .2

2

60

1

Geometry: AD AE

vB ! BC

1

B

360 vA

D 7 rad/s

D 90

! AB

720

2

120

O

vA

ωAB

vB

180

C

ωBC

p

902 + 1802 = 201:2 mm p = 3602 + 7202 = 805:0 mm

=

= AD! OA = 201:2(7) = 1408:4 mm/s 1408:4 vA = = 1:7496 rad/s = 805:0 AE = BE! AB = 720(1:7496) = 1259:7 mm/s 1259:7 vB = = = 7:00 rad/s J 180 BC

16.125 (a)

vB

0 25 4 rad/s

5 37 O

300

A 200

B 150 225 200 C vC

ωCD D


Point O is the I.C. for bar BC. vB ! BC vC ! CD

= AB! AB = 250(4) = 1000 mm/s vB 1000 = = = 1:60 rad/s Q.E.D 625 OB = OC! BC = 500(1:60) = 800 mm/s vC 800 = = 4 rad/s Q.E.D = 200 CD

(b) aB = aC + aB=C

375(1.6)2 mm/s2 375αBC B

250(4)2 mm/s 2

0 25 4 rad/s

B

=

3

200(4)2 mm/s2 200 C

4 rad/s D

αCD

200αCD

4

+

A

375

αBC 1.6 rad/s C

+ !

+ #

4 (250)(4)2 = 200(4)2 375 5 2 J BC = 17:07 rad/s 3 (250)(4)2 = 200 5 2 CD = 7:20 rad/s

CD

BC

+ 375(1:6)2 J

vC

C

60 0

16.126 F

B

E

vB

0 60

00 72 rad/s 5 o 35

ωBCD

450

A D

vD


Geometry: BE BF DF vB

p 6002 4502 = 396:9 mm BE 396:9 = = = 484:5 mm cos 35 cos 35 = DE + BE tan 35 = 450 + 396:9 tan 35 = 727:9 mm =

= ! AB AB = 72(500) = 36 000 mm/s vB 36 000 = = 74:30 rad/s J = 484:5 BF = ! BCD DF = 74:30(727:9) = 54 080 mm/s = 54:1 m/s ! J

! BCD vD

16.127 Let D0 be a point on rod AB that is coincident with D at the instant under consideration aD

= aD0 + aD=AB + aC = rD0 =A + ! (! rD0 =A ) + aD=AB + 2! vD=AB = 18k (12i 5j) + 6k [6k (12i 5j)] + 48j + 2(6k) = (216j + 90i) + 6k (72j + 30i) + 48j + 432i =

216j + 90i

( 36j)

432i + 180j + 48j + 432i = 90i + 444j in./s

2

J

16.128 vC = vB + vC=B

C vC

0.1 2m

B

= .

4

3

3

0.12(1.5) m/s

+

C

4

0.09ωBC

1.5 rad/s

0.0 9m

ωBC

. B

A + ! +

#

4 3 (0:12) (1:5) (0:09)! BC ! BC = 2:667 rad/s 5 5 3 4 vC = (0:12)(1:5) + (0:09)! BC 5 5 3 4 = (0:12)(1:5) + (0:09)(2:667) = 0:300 m/s # J 5 5 0=

aC = aB + aC=B


C

0.1 2m

0.12(1.5)2 m/s2

=

aC

.

4

B

+

1.5 rad/s

A

.

3 4 3 (0:12) (1:5)2 + (0:09)(2:667)2 (0:09) BC 5 5 5 2 BC = 6:484 rad/s 3 4 4 aC = (0:12)(1:5)2 + (0:09)(2:667)2 + (0:09) BC 5 5 5 4 3 4 2 2 = (0:12)(1:5) + (0:09)(2:667) + (0:09)(6:484) 5 5 5 2 = 1:067 m/s # J

+ ! +

C

0.09(2.667)2 m/s2 4 3 0.0 4 9m 0.09αBC 2.667 rad/s αBC B 3

3

0=

#

16.129 (a) Because bars AB and CD are parallel, the velocities of points B and C are also parallel. Therefore, the bar BC translates in the position shown. ) ! BC = 0 vC = vB = AB! AB = 75(12) = 900 mm/s 900 vC = ! CD = = 6 rad/s J 150 CD (b) aC = aB + aC=B

3

150(6)2 mm/s2

B

C 4

75(12)2 mm/s2

=

75 mm

αCD

15 0m m

150αCD

3

12 rad/s

6 rad/s

4

+

αBC B

165αBC 165 mm C

A

D +

4 3 3 (150 CD ) + (150)(6)2 = (75)(12)2 5 5 5 2 = 27:0 rad/s J CD


+ "

3 4 4 (150 CD ) (150)(6)2 = (75)(12)2 + 165 5 5 5 3 4 4 (150)(27:0) (150)(6)2 = (75)(12)2 + 165 5 5 5 2 J BC = 40:9 rad/s

BC

BC

16.130 (a)

O 30o B o vB 40 60 o 800

Geometry: BC OC OA

ωAB = 6 rad/s y

C

x A

vA

= 800 cos 40 = 612:8 mm = BC tan 30 = 612:8 tan 30 = 353:8 mm = OC + AB sin 40 = 353:8 + 800 sin 40 = 868:0 mm

vA = ! AB OA = 6(868:0) = 5208 mm/s = 5:21 m/s ! J (b)

! AB aA aA i aA i

! AB aA rA=B rA=B

2

= 6k rad/s AB = 8k rad/s = aA i aB = aB (i cos 60 j sin 60 ) = aB (0:5i 0:8660j) = BC i AC j = 612:8i 800j sin 40 = 612:8i 514:2j mm = 6k (612:8i 514:2j) = 3677j + 3085i

= aB + AB rA=B + ! AB (! AB rA=B ) = aB (0:5i 0:8660j) + 8k (612:8i 514:2j) + 6k (3677j + 3085i) = aB (0:5i 0:8660j) + (4902j + 4114i) + ( 22 060i + 18 510j)

Equating like components: aA = 0:5aB + 4114 22 060 0 = 0:8660aB + 4902 + 18 510 The solution is

aB = 27 050 mm/s2 and aA = aA = 4:42 m/s

2

4423 mm/s2 . Therefore, J


16.131


17.1

17.2

. 2

R R

2R

1

.

R 1 3 1 2 R (R) = R 3 3 1 5 3 R R3 = 3 3 V2 1 m2 = m= m V 5

R2 (2R) = 2 R3

V1

=

V

= V1

V2 =

2

m1

=

V1 12 m= m V 5

Iz

=

(Iz )1

=

1 2

(Iz )2 =

12 m R2 5

V2 =

1 3 m1 R2 m2 R2 2 10 3 1 57 m R2 = mR2 J 10 5 50


17.3

17.4


17.5

y 90 mm

90 mm

8 mm

.

2 144 mm

y- = 104.0 mm

1 x

10 mm m1

=

4

d21 L1 =

4

(7850)(0:010)2 (0:144) = 0:08878 kg

d2 L2 = (7850)(0:008)2 (0:180) = 0:07103 kg 4 2 4 m = m1 + m2 = 0:08878 + 0:07103 = 0:159 81 kg

m2

y

=

0:08878(0:072) + 0:07103(0:144) mi y i = = 0:1040 m = 104:0 mm m 0:159 81 Coordinates of mass center are (0, 104.0 mm, 0) J

= ) Iz Iz

1 1 m1 L21 + m2 y22 = (0:08878)(0:144)2 + 0:07103(0:148)2 3 3 = 2:170 10 3 kg m2 = Iz my 2 = 2:170 10 3 0:159 81(0:1040)2 = 4:42 10 4 kg m2 J =

17.6 "

Iz

=

1 3 mro d b2 + mro d 12

) Iz

=

1 1 mrod b2 = mb2 J 2 6

b p 2 3

2

#

where mro d =

m 3

17.7 "Thin" implies that t

b, so that t can be neglected in comparison to b.

One side (mass = m=4): (Iz )1 =

1 m 2 m b + 12 4 4

b 2

2

=

1 mb2 12


Bottom (mass = m=2): (Iz )2 = Iz = 2(Iz )1 + (Iz )2 = 2

1 m 2 m (b + b2 ) + 12 2 2 1 mb2 12

p b 2 2

2

=

1 2 mb 3

1 1 + mb2 = mb2 J 3 2

17.8 2

75 y

1

180

150 x

(Ix )1

=

(Ix )2

=

(Ix )3

=

Ix

=

(Iz )1

=

(Iz )2

=

(Iz )3

=

Iz

=

m1

=

m2

=

m3

=

60 y

3 x

Thickness = 80

0:15(0:18)(0:08)(2650) = 5:724 kg 2

(0:075)2 (0:08)(2650) = 1:8732 kg (0:06)2 (0:08)(2650) =

2:398 kg

1 (5:724)(0:152 + 0:082 ) = 0:013 785 kg m2 12 1 (1:8732) 3(0:075)2 + 0:082 = 0:003 63 kg m2 12 1 (2:398) 3(0:06)2 + 0:082 = 0:003 44 kg m2 12 0:003 44 = 0:013 98 kg m2 J i (Ix )i = 0:013 785 + 0:003 63 1 (5:724)(0:152 + 0:182 ) + 5:724(0:09)2 = 0:072 55 kg m2 12 1 (1:8732)(0:075)2 = 0:002 63 kg m2 4 1 (2:398)(0:06)2 = 0:004 32 kg m2 2 0:004 32 = 0:0709 kg m2 J i (Iz )i = 0:072 55 + 0:002 63


17.9 y

y

50

100

50

z

x

200 50 100

150 100 Block: Cylinder: Hole:

(Iz )1

=

(Iz )2

=

(Iz )3

=

Iz =

200 m1 = 0:2(0:4)(0:1)(7850) = 62:80 kg m2 = (0:05)2 (0:15)(7850) = 9:248 kg m3 = (0:05)2 (0:1)(7850) = 6:165 kg

62:80 (0:22 + 0:42 ) + 62:80(0:1)2 = 1:6747 kg m2 12 9:248 (0:05)2 + 9:248(0:2)2 = 0:3815 kg m2 2 6:165 (0:05)2 = 0:0077 kg m2 2

i (Iz )i

= 1:6747 + 0:3815

0:0077 = 2:05 kg m2 J

17.10

y x

1

_ r

2 ft G

2

1.5 ft m1 =

1:2 2 = 21:30 32:2 3:5

10

3

slugs

m2 =

1:2 1:5 = 15:972 32:2 3:5

10

3

slugs


(Iz )1

=

(Iz )2

= =

Iz

=

1 1 m1 L21 = 21:30 10 3 (2)2 = 28:40 10 3 slug ft2 3 3 " # 2 1 L2 1 2 2 m2 L2 + m2 L1 + = m2 (3L21 + L22 ) 12 2 3 1 (15:972 10 3 ) 3(2)2 + 1:52 = 75:87 10 3 slug ft2 3 (Iz )1 + (Iz )2 = (28:40 + 75:87) 10 3 = 104:27 10 3 slug ft2 J 2(0) + 1:5(0:75) Li xi = = 0:3214 ft Li 2 + 1:5 Li yi 2( 1:0) + 1:5( 2) = = 1:4286 ft Li 2 + 1:5

x = y

Iz

= Iz =

24:4

=

mr2 = 104:27 10

3

(21:30 + 15:972)(0:32142 + 1:42862 )

10

3

slug ft2 J

17.11


17.12 I

1 msp oke R2 3 8 8 + 32:2 3

= Irim + 8(Isp oke ) = mrim R2 + 8 8 = R2 mrim + msp oke = 1:252 3 =

0:5 32:2

0:453 slug ft2 J

17.13 (a) Sphere: Ix

= =

2 mR2 + m(L + R)2 5 2 2 130 5 130 + 5 32:2 12 32:2

60 + 5 12

2

= 118:73 slug ft2

Rod: Ix

Pendulum:

=

1 mL2 + m 12

=

1 3

25 32:2

2

L 2 60 12

=

1 mL2 3

2

= 6:47 slug ft2

Ix = 118:73 + 6:47 = 125:20 slug ft2 J

(b) Using the speci…ed approximation: 2

Ix

130 60 + 5 = 118:45 slug ft2 32:2 12 118:45 125:20 100% = 5:39% J 125:20

= m(L + R)2 =

% error =

17.14 I = IO mx2 408:5 120x2 )I

= IC m(2 x)2 = 145:5 120(2 x)2 x = 1:5479 m J = 408:5 120(1:5479)2 = 121:0 kg m2 J


17.15

*17.16

17.17


17.18

17.19 B 20 lb O

( MO )FBD P

=

ma

O

2'

P A N

B

2'

A

MAD

FBD

= ( MO )M AD = 23:1 lb J

60o

+

P (2 sin 60 )

20(4 cos 60 ) = 0


17.20

B

P

R mg

FBD

R

µs mg

A

ma

= A

MAD

N = mg

Assume impending sliding. MA

= maR +

2RP = maR

Fx

= ma + !

s mg

a =

2

sg

+ P = ma

J

P =

ma 2

s mg

+

ma = ma 2

17.21


17.22

NA

y

4.5'

.2.25' = B. 55 lb N G

20 lb

B

FBD Fx = max

O 20 =

.

G

55 a g

MAD O 55 a 32:2

( MO )FBD NA (4:5)

x

5.25'

A

a = 11:709 ft/s

2

J

=

( MO )M AD + 55 (11:709)(5:25) 55(2:25) = 32:2 NA = 50:8 lb J

17.23


17.24 The bar AB is translating. Therfore, all points on the bar have the same velocity and acceleration.

TA

35o

A

.

2 ft

y

TB

.G

.B =

2 ft mg FBD

x

ma n

A

G

ma t

.

B

MAD

an = 3 _

2

at = 3•

( MG )FBD = ( MG )M AD + : (TA cos 35 ) (2)

(TB cos 35 )(2) = 0

TA = TB

( Fx )FBD = ( Fx )M AD + .: mg sin 35 =

mat

48 sin 35 =

48 3• 32:2

•=

6:16 rad/s

2

J

( Fy )FBD = ( Fy )M AD + -: TA + TB

mg cos 35

2TA

48 cos 35

= man =

48 (3 32:2

2TA 32 )

48 cos 35 =

2 48 (3 _ ) 32:2

TA = TB = 39:8 lb J

17.25 y

80 lb

FBD F

80 a 32.2 n o = 30

MAD

x N

The box translates along a circular path of 5-ft radius ) an = R! 2 = 5(2)2 = 20 ft/s Fy

= may

+"

Fx

= max

+ !

N

2

80 (20) sin 30 32:2 80 (20) cos 30 F = 32:2 80 =

N = 104:8 lb J F = 43:0 lb J


17.26 L/2

A

B = A mg FBD

N ( MA )FBD L + mg 2 aB

L/2

mL2α 12

B MAD

m Lα 2

=

( MA )M AD L L mL2 = m + 2 2 12 3 = L = g #J 2

=

3g 2L

17.27

A N2 N1

=

7a y

x

MAD

FBD

A

Ax

= 35 3a 3(9.81) N

2m

35 o

35 o T B

o

B

FBD

1.0 m

Ay

7(9.81) N

MAD

System: Fx = max

+ . 7(9:81) sin 35 = 7a

a = 5:627 m/s

2

Bar: ( MA )FBD T

= ( MA )M AD + (T sin 35 )(2) = (3a cos 35 )(1:0) = 2:142a = 2:142 (5:627) = 12:05 N J

17.28 1 2 B 12 mLαAC

mg B

C C = A L/2 L/2 L/2 R L/2 T FBD's 1 mL2α MAD's mg DE T 12 F D = D E E L/2 L/2 L/2 L/2 F Lα m 2 ( AC + αDE) A


Kinematics: # aE = aC =

L 2

# aF = aE +

AC

L 2

DE

=

L ( 2

AC

+

DE )

Kinetics— bar AC: ( MB )FBD = ( MB )M AD

1 L T = mL2 2 12

+

AC

AC

6T mL

=

Kinetics— barDE: ( MF )FBD Fy T

=

( MF )M AD

= may

1 L T = mL2 2 12 L T = m ( AC + 2

+

+#

mg

DE DE )

DE

=m

6T mL 6T 2 mL

=

L 2

1 mg J 7

=

17.29

.G

NB

=

mg o

60

. A

.A

Ax

Ay

60o MAD

FBD

( MA )FBD = ( MA )M AD +

mg/2

8f t

8f t

10 ft

.G

B

:

mg (8 cos 60 )

10NB

=

60(8 cos 60 )

10NB

=

mg (8 sin 60 ) 2 60 (8 sin 60 ) 2

NB = 3:215 lb J

( Fx )FBD = ( Fx )M AD +!: Ax

NB sin 60 Ax

= =

mg Ax 2 32:78 lb

3:215 sin 60 =

60 2


( Fy )FBD = ( Fy )M AD + ": Ay + NB cos 60

17.30

mg = 0 Ay + 3:215 cos 60 Ay = 58:39 lb p A = 32:782 + 58:392 = 67:0 lb J

60 = 0


17.31

17.32

40(0.252)α

40(9.81) N

G 0.6 m

0.6 m 0.6 m = G T 800 N R T FBD's A 60(9.81) N

60(0.6α) A

MAD's

System:

800(0:6)

( MG )FBD = ( MG )M AD + : 60(9:81)(0:6) = 40(0:25)2 + 60(0:6 )(0:6) =

2

5:263 rad/s

J


Block A: Fy T

= may + " T 60(9:81) = 60(0:6) = 60(9:81) + 60(0:6)(5:263) = 778 N J

17.33 FBD

mg 2 ft

=

A

B

2 ft

µs NA

( MB )FBD ma 4:5mg 7:5

= max

4.5 ft 3 ft

+ !

=

( MB )M AD

=

2ma

ma A

B

NB 4.5 ft 3 ft NA Fx

MAD

G

s NA

= ma

NA =

ma s

+

4:5mg 7:5NA = 2ma a 2 7:5 = 2a a = 12:74 ft/s J 0:8

4:5(32:2)

s

17.34

Ay

.A

.G

R

Ax

I

2R

2R b

= IC

b

2R

π

MAD

= R2 + =

.

mbα

FBD

sin

A

θ

.

R

=

mg

b2

C I-α G

2

=

cos = m

1+

4 2

R2

R b

2

2R

= mR2

m

2R

2

=

4

1

2

mR2

( MA )FBD = ( MA )M AD + : mgR g

= I + (mb )b = =

2R

=

g 2R

1

4 2

mR2 + 1 +

4 2

mR2

J


( Fx )FBD = (Fx )M AD +!: Ax = mb sin = mb

g 2R

2R b

=

mg

! J

( Fy )FBD = (Fy )M AD +": Ay

mg =

mb cos =

mb

g 2R

R b

Ay =

mg " J 2

17.35

FA

FA 3' C 2' FBD

FB

2(aC + 3α) A

A 1180 lb.ft =

5aC

20α

FB B

4(-aC +2α)

B MAD

Only horizontal forces are shown Kinematics: ! aA = aC + 3

aB =

aC + 2

Kinetics— racks: FA = 2(aC + 3 )

FB = 4( aC + 2 )

Kinetics— gear: 2 IC = mkC = 5(2)2 = 20 slug ft2

( MC )FBD = ( MC )M AD 1180

+ 1180 3FA 2FB = 20 6(aC + 3 ) 8( aC + 2 ) = 20 2aC 54 = 1180 (a)

Fx = max + ! FB FA = 5aC 4( aC + 2 ) 2(aC + 3 ) = 5aC 2 11aC = 0 Solution of (a) and (b) is

= 22:0 rad/s2

(b)

and aC = 4:0 ft/s2 ! J


17.36

x aA

A

6' T

B aB

y

α 8' G

B

B

G

8'

mg

=

A

6'

1 m(102)α 12 3mα

4mα A

MAD

N FBD

Kinematics:

aG

= k rG=A = rG=B = 3i + 4j ft rG=A = ( 3j + 4i) = aA + aG=A = aA i + rG=A = aA i + ( 3j + 4i) ) (aG )y = 3 aG

= aB + aG=B = ) (aG )x = 4

aB j +

rG=B =

aB j + (3j

4i)

Kinetics: Fx = max + ( MA )FBD

T = 4m

=

( MA )M AD + 1 m(102 ) + 3m (3) 3mg 8T = 12 3mg 8(4m ) = 1:3333m = T =4

4m (4) 2

0:090g = 0:090(32:2) = 2:898 rad/s 50 (2:898) = 18:0 lb 32:2

J

J


17.37


17.38


17.39

17.40 R2 h = (0:12)2 (0:36)(7850) = 42:62 kg 3 3 = mg = 42:62(9:81) = 418:1 N 3 3 = m(4R2 + h2 ) = (42:62) 4(0:12)2 + 0:362 = 0:2992 kg m2 80 80

m = W I

22o 0.2 7m

G =

418.1 N F A N FBD

0.2 7m

0.2992α

A

=

Fx

11.507α

y x

MAD

( MA )FBD = ( MA )M AD 418:1(0:27 sin 22 ) = 0:2992 + 11:507 (0:27)

+

Fy

G

= ) = )

12:416 rad/s

2

J

may + " N 418:1 = 11:507 sin 22 N = 418:1 11:507(12:416) sin 22 = 365 N " J max + ! F = 11:507 cos 22 F = 11:507(12:416) cos 22 = 132:5 N ! J


17.41 4 = 0:124 22 slugs 32:2 1 1 mL2 = (0:124 22)(32 ) = 0:093 17 slug ft2 = 12 12 = mr = 0:124 22(0:75) = 0:093 17 " = mr! 2 = 0:124 22(0:75)(42 ) = 1:4906 lb

m = I may max

y

Cy Cx G 1.5 lb ft C

.

..

0.75'

=

FBD

x MAD

0.09317α G 1.4906 lb

.C . 0.09317α

Only the horizontal forces are shown on the FBD. ( MC )FBD

= =

Fx Fy

( MC )M AD + 2

9:20 rad/s

= max + = may + " C=

17.42

1:5 = 0:093 17 + 0:093 17 (0:75) J

Cx = 1:4906 lb Cy = 0:093 17 = (0:09317) (9:20) = 0:8572 lb p

1:49062 + 0:85722 = 1:720 lb J


17.43

Ay A

d

W

L

=

.C

P

A

Ax = 0

Iα

G C

mat d − L/2

.

FBD

MAD

B

B

( MC )FBD = ( MC )M AD + : 0

= I

d

=

mat d

L 2

=

mL2 12

m

L 2

d

L 2

2 L J 3


17.44

17.45


17.46 30 = 0:9317 slugs 32:2 2 1 1 16 = mR2 = (0:9317) = 0:8282 slug ft2 2 2 12 ! p 302 + 162 = 2:640 = mL = 0:9317 12

m = I mat System:

16"

A

.

By 34"

30 lb

. B

Bx y

= x

FBD

0.8282α 34" " 16 A 2.640α

.

.

B

MAD


( MB )FBD

=

( MB )M AD +

=

10:231 rad/s

30

34 12

= 0:8282 + 2:640

34 12

J

2

Disk:

30 lb Ax

.A 16"

Ay

FBD ( MA )FBD

=

T

=

.

A

= T

0.8282α

2.640α

C

MAD

16 T = 0:8282 12 0:6212 = 0:6212(10:231) = 6:36 lb J

( MA )M AD +

17.47

A

.G

Iα

4 ft

= θ

40 lb O

C

θ

4 ft

θ

A

C

NA

B FBD NB

.G

ma n

ma t B

O MAD

The path of the mass center G is a circle of radius r = 4 ft centered at O. Let the angular acceleration of line OG be = • clockwise. Then the angular


acceleration of the bar is also , but it is directed counterclockwise (see MAD). m = man mat

= =

I

= =

W 40 = = 1:2422 slugs g 32:2 mr! 2 = 1:2422(4)(2)2 = 19:875 lb mr = 1:2422(4) = 4:969 1:2422(8)2 mL2 = = 6:625 slug ft2 12 12 30

( MC )FBD = ( MC )M AD +

:

W (4 sin 30 )

= I + mat (4)

40(4 sin 30 )

=

6:625 + 4(4:969 )

= 3:019 rad/s

2

J

( Fx )FBD = ( Fx )M AD + !: NA

= =

man sin 30 + mat cos 30 19:875 sin 30 + 4:969(3:019) cos 30 = 3:05 lb ! J

( Fy )FBD = ( Fy )M AD +": NB NB

W = man cos 30 mat sin 30 40 = 19:875 cos 30 4:969(3:019) sin 30 NB = 15:29 lb " J

17.48


17.49 20o FBD

y

Mg

Iα

x G

R

=

Ma-

0.075N C N Fy N Fx = max

G

MAD C

= 0 + - N M g cos 20 = 0 = M (9:81) cos 20 = 9:218M

+ % 0:075N M g sin 20 0:075(9:218M ) M (9:81) sin 20

= =

a =

Ma Ma 2:66 m/s

2

J

17.50


17.51


17.52


17.53


17.54

17.55 Kinematics:

y

1.2 m

ω, α O G x 0.4 m


a = aO + aG=O = R i + rG=O + ! ! rG=O = 1:2(3)i + ( 3k) 0:4i + ( !k) ( !k 0:4i) = 3:6i 1:2j + ( !k) ( 0:4!j) =

0:4! 2 i

3:6

1:2j m/s

Kinetics:

2

Iα O G 0.4 m Mg FBD N

C F

Ma-y O G 0.4 m

=

MAD

Ma-x 1.2 m

C

I = M k 2 = M (0:4)2 = 0:16M ( MC )FBD = ( MC )M AD + 0:4M g = 1:2M ax 0:4M ay + I 0:4M (9:81) = 1:2M 3:6 0:4! 2 0:4M ( 1:2) + 0:16M (3) 3:924

=

5:280

0:48! 2

! = 1:681 rad/s J

17.56


17.57


17.58 aC = aO + aC=O Kinematics (C is the midpoint of bar AB):

aC

aO

=O

+

0.15 α 0.15 C 2 0.15ω

ω, α O

Kinetics:

NB

0.3 0.2

4

0.15

3

C mg

0.2

Ay

Iα

=

C

m(a0 + 0.15ω2)

A

Ax

A

B

m(0.15α)

B

FBD

MAD 2

! = 5 rad/s = 8 rad/s a0 = R = 0:45(8) = 3:60 m/s m = 25 kg mg = 25(9:81) = 245:3 N mL2 25(0:5)2 I = = = 0:5208 kg m2 12 12

2

( MA )FBD = ( MA )M AD : 0:15mg

0:5NB NB

= I + m(a0 + 0:15! 2 )(0:2) (0:15m ) (0:15) = 0:5208(8) + 25 3:60 + 0:15(5)2 (0:2) 0:15(25)(8)(0:15) = 0:757 N & J

17.59 1 1 mL2 = (4:2) (1:22 ) = 0:5040 kg m2 12 12 Kinematics (the system has 2 DOF): Bar AB:

I=

aA + aG=A = aG


aA A

α

+

A

=

aA− 0.6α G

0.6 m 0.6α

G

Kinetics:

WA A

. NA

G

A

.

14 N 0.6 m

.

=

G 0.5040α

.

2.4aA 4.2(aA − 0.6α)

WAB

y MAD

FBD Fx = max + !

x

14 = 2:4aA + 4:2(aA

( MA )FBD = ( MA )M AD + 0 = 0:5040 4:2(aA Solution is

=

5:07 rad/s

2

J

0:6 )

0:6 )(0:6) aA = 4:06 m/s ! J 2


17.60


17.61

17.62 For the assembly: MB

= IB =

40 = 2

6:822 rad/s

1 2

40 32:2

(1:2)2 +

40 (2)2 32:2

J


For the disk:

Ay

. A

NC 0.8 ft

IAα

=

Ax C

.A man mat

FBD

MAD

MA

= IA

NC

=

0:8NC =

1 2

40 32:2

(1:2)2 (6:822)

7:63 lb " J

17.63 W

= mg = 150(9:81) = 1471:5 N = m(AG! 2 ) = 150(0:6)

man

v0 1:8

!= 2

= 27:78v02

y

Ax

y

0.6 G

θ

A 1.8 Ay 1471.5 N

G x

=

27.78v02

P

θ

x

MAD

FBD Fy Ay

v0 v0 = R 1:8

= may + " Ay 1471:5 = = 1471:5 27:78v02 sin

27:78v02 sin


Disk stays in bearing if Ay this condition is

0 for all . The largest v0 that does not violate r 1471:5 v0 = = 7:28 m/s J 27:78

17.64


17.65

*17.66


17.67


17.68 (a) MA = IA

k =

mL2 3

3k 3k = 2 mL mL2

=

J

(b) 3k d! ! ! d! = d mL2 Initial condition: ! = 0 when = 0 =

0

3k 2 2mL r

= )

!=

2 0

+C

3k ( mL2

(c) ! max = !j

1 2 ! = 2

d

=0

=

C= 2

2 0

r

3k 2mL2

3k 2mL2

2

+C

2 0

) J

3k mL2

0

J

17.69 (a)

1.2 m mg FBD

360

Fx 30v

FD

N d

P 0.45 m = C y

ma 0.6 m C x

MAD

= max + ! P FD = ma 2 = 60a a = 6:0 0:5v m/s Q.E.D

(a)


(b) dv Zdt

a = t

=

dv = dt a dv = 6:0 0:5v

dv = dt 6:0 0:5v 2:0 ln(6:0

0:5v) + C

Initial condition: v = 0 when t = 0. ) C = 2 ln 6:0 )t=

2:0 ln

6:0

0:5v = 6:0

v 12

2:0 ln 1

(b)

When tipping impends, we have d = 0 on the FBD. ( MC )FBD = ( MC )M AD + 1:2P 0:45mg = 0:6ma 2 1:2(360) 0:45(60)(9:81) = 0:6(60)a a = 4:643 m/s Eq. (a):

4:643 = 6:0

0:5v v = 2:714 m/s 2:714 2:0 ln 1 = 0:513 s J 12

Eq. (b):

t=

17.70 (a) W I

= mg = 2:4(9:81) = 23:54 N 1 1 = mR2 = (2:4)(0:18)2 = 0:03888 kg m2 2 2

23.54 N O 45

o

0.18 m

0.03888α

=

Fy NC

= may +% = 20:47 N

C

y

NO C 0.15NC FBD NC

O x

MAD

NC cos 45 + 0:15NC sin 45

23:54 cos 45 = 0

( MO )FBD

=

( MO )M AD

+

0:15NC (0:18) = 0:03888

0:15(20:47)(0:18)

=

0:03888

= 14:215 rad/s

2

J


(b) =

d! ! d

)

d = ! d!

14:215 d = ! d!

14:215 =

1 2 ! +C 2

= 0: ) C = 0 1 ) 14:215 = ! 2 ) = 0:03517! 2 2

Initial condition: ! = 0 when

Final angular speed of the disk is ! 1 = v=R = 6=0:18 = 33:33 rad/s = 0:03517(33:33)2 = 39:07 rad = 6:22 rev J

When ! = ! 1 :

17.71 (a) Wheel: I = mR2 =

Ay

A

1.0' 6 lb 1.0' 20o

18 (2)2 = 2:236 slug ft2 32:2

N Ax

0.75N

40o

0.75N B FBD N Bar AB : + N = Wheel +

: =

2.236α

20o 2' 18 lb C Cx Cy

=

C MAD

FBD

MA = 0 6(1:0 sin 20 ) + 0:75N (2 cos 40 ) 15:033 lb

N (2 sin 40 ) = 0

( MC )FBD = ( MC )M AD 0:75N (2) = 2:236 0:75(15:033)(2) = 2

10:085 rad/s = 10:085 rad/s

2

2:236

J

(b) =

d! dt

) d! =

dt =

10:085 dt

)!=

10:085t + C

Initial condition: ! = 400 rev/min = 41:89 rad/s when t = 0 ) C = 41:89 rad/s

)!=

When ! = 0: t =

10:085t + 41:89

41:89 = 4:15 s J 10:085


17.72 (a) Wheel: I = mR2 =

Ay

N

A

Ax

1.0' 6 lb 40

1.0' 20o

0.75N

o

0.75N B FBD N Bar AB : + N = Wheel +

2.236α

20o 2' 18 lb C Cx Cy

=

C MAD

FBD

MA = 0 6(1:0 sin 20 ) 0:8429 lb

:

18 (2)2 = 2:236 32:2

0:75N (2 cos 40 )

N (2 sin 40 ) = 0

( MC )FBD = ( MC )M AD 0:75N (2) = 2:236 0:75(0:8429)(2) = 2

=

0:5654 rad/s = 0:5654 rad/s

2

2:236

J

(b) =

d! dt

) d! =

dt =

0:5654 dt

)!=

0:5654t + C

Initial condition: ! = 400 rev/min = 41:89 rad/s when t = 0 ) C = 41:89 rad/s

)!=

When ! = 0: t =

0:5654t + 41:89

41:89 = 74:1 s J 0:5654

17.73 (a) IA

=

IB

=

1 2 1 2

18 32:2 30 32:2

(1:52 ) = 0:6289 slug ft2 (2:02 ) = 1:8634 slug ft2


18 lb

Ax

' 1.5

A

=

0.6289αA

0.3N = 5.4 lb N = 18 lb N = 18 lb 0.3N = 5.4 lb 30 lb

2'

B

=

Bx

1.8634αB

By FBD's

( MA )FBD A

( MB )FBD B

MAD's

=

( MA )M AD +

= =

12:880 rad/s = 12:880 rad/s J ( MB )M AD + 5:4(2:0) = 1:8634

=

5:4(1:5) = 2

5:796 rad/s

2

0:6289

A

2

B

J

(b) d! != dt During sliding: =

!A

=

!B

=

dt

! = t + ! 0 ( constant, ! 0 = initial speed)

2 = 12:880t + 18:850 rad/s 60 5:796t + 0 = 5:796t rad/s 12:880t + 180

Sliding stops when RA ! A t

= RB ! B = 0:9147 s

1:5( 12:880t + 18:850) = 2:0(5:796t)

Final speeds: !A !B

= =

12:880(0:9147) + 18:850 = 7:07 rad/s J 5:796(0:9147) = 5:30 rad/s J


17.74 (a) Let _ be the angular velocity of the disk. Noting that A is the I.C. of the disk, we have vA = vG + vA=G

0

. G

.

O

θ

β

=

G 0 = R_

r

Rθ

.A

R )•= • r

mg

ma n

G

G .. Iβ r

at = R•

ma t

A

A

F an = R _

MAD 2

(point G rotates about O)

:

= I • + (mat ) r 3 • R 2

=

N

( MA )FBD = ( MA )M AD +

=

. rβ

.

R ) _ = _ r

FBD

g sin

.

r_

θ

mg(r sin )

r

+

R

•=

mgr sin = 2g sin 3R

mr2 2

R• + mR• r r

J

(b) The di¤erential equation is equivalent to d_ _ d

= )

2g _ d_ = sin 3R 1 _2 2g = cos + C 2 3R

2g sin d 3R


Initial condition: _ = 0 when 0

=

1 _2 2

=

= 60

2g cos 60 + C 3R 2g 1g cos 3R 3R

C= _=

1g 3R r 2g (2 cos 3R

1) J

17.75


17.76


17.77


17.78 Assume impending sliding.

B

Iα

G

30o

L/2

B

30o

mg F = µsN

A

A MAD

N FBD ( MA )FBD = ( MA )M AD + L mg sin 30 2

=

:

L 2 3g 3g sin 30 = 2L 4L

= I +

mL α 2

mL 2

L mg sin 30 = 2

mL2 mL2 + 12 4

( Fx )FBD = ( Fx )M AD + !: sN

=

mL mL cos 30 = 2 2

3g sin 30 2L

cos 30 = 0:3248mg

( Fy )FBD = ( Fy )M AD + ": N

mg N

mL sin 30 2 mL 3g = mg sin 30 2 2L =

s

=

sN

N

=

sin 30 = 0:8125mg

0:3248 = 0:400 J 0:8125


17.79 A

θ

L/2

θ L/2

mg 2

=

( MA )FBD

=

m L g( + L) cos 2 2

d! ! d

m L 2 2 18 g cos 17 L

=

) ! d! =


mLα 2

( MA )M AD

=

=

m Lω2 22 mLα 22

mLω2 2

1m 2 12 2 L α MAD

mg 2

FBD

+

1m 2 12 2 L α

A

= 0: r

)!=

2

+

m 2 L +2 2

1 m 2 L 12 2

J

d

=

1 2 ! 2

=

18 g cos d 17 L 18 g sin + C 17 L

)C=0 36 g sin 17 L

J

17.80

A

An

6 ft

At t

Bt n 750 lb . ft B FBD Bn mat man

100 lb 2 ft 2 ft θ A C Cn An At

ma t = A

θ

C

man MAD

FBD 100 (6) = 18:634 lb 32:2 = mL! 2 = 18:634! 2 lb

= mL =

(a) Bar AB: MB = 0 +

6At

750 = 0

At = 125 lb


Bar AC: 125

Ft 100 cos

= mat + & = 18:634

At 100 cos = 18:634 2 = 6:708 5:367 cos rad/s J

(b) =

d! ! d

! d! =

d = (6:708

5:367 cos ) d

1 2 ! = 6:708 5:367 sin + C 2 Initial condition: ! = 0 when = 0: ) C = 0 p ) ! = 13:412 10:734 sin rad/s J

17.81


17.82

17.83


*17.84


=

d! ! d

! d! =

1 2 ! = 2

d = 4:905 sin

4:905 cos + C

Initial condition: ! = 0 when = 30 ) C = 4:905 cos 30 p 2(4:905)(cos 30 cos ) ! = p = 3:132 0:8660 cos rad/s J (b) Substituting the expressions for ! and Eq. (a), we get NA

= =

, and the values of M and L into

25(3) [(4:905 sin ) cos 2(4:905)(0:8660 2 183:94(3 cos 1:7320) sin N J

(c) NA = 0 when 3 cos

1:7320 = 0

)

= 54:7

cos ) sin ]

J

17.85 (a) Kinematics of bar AB:

aA

L/2

A L + 2α

G

ω, α

Lω2 2

B a = aA + aG=A


Kinetics:

m2g Ax

P0

FBD's

Collar: Bar:

MAD's

Ay A

G θ m1g

Fx = max

Ax m Lα = 12

L/2

L/2

B

m2aA

=

N Ay

B

+ !

P0

A m1 Lω2 2 m1aA

I-α

Ax = m2 aA

Ax = P0

L Ax = m1 aA + m1 ! 2 sin 2 L 2 L ) P0 m2 aA = m1 aA + m1 ! sin m1 cos 2 2 1 m1 L(! 2 sin + cos ) P0 2 ) aA = m1 + m2 Fx = max

+ !

Bar: ( MA )FBD L m1 g sin 2

L m1 g sin 2 g sin

=

= =

1 m1 L2 3

=

( MA )M AD L = I + m1 2

P0 m1

+ L 2

(m1 aA )

1 m1 L(! 2 sin + 2 m1 + m2

m2 aA L cos 2

m1

L cos 2

cos ) L cos 2

1 P0 m1 L(! 2 sin + cos ) 2 2 L cos 3 m1 + m2 2P0 cos 2g(m1 + m2 ) sin L! 2 m1 cos sin (4L=3) (m1 + m2 ) + m1 L cos2

Q.E.D.

(b) Substituting the given data, the equation of motion becomes = =

2(12) cos 24 cos

2(9:81)(3:6 + 2:0) sin 0:8! 2 (3:6) cos sin (4 0:8=3)(3:6 + 2:0) + 3:6(0:8) cos2 109:87 sin 2:88! 2 sin cos 5:973 + 2:88 cos2


The initial conditions are = ! = 0 at t = 0. Letting x1 = and x2 = !, the equivalent …rst-order equations and the initial conditions are x_ 1 x1 (0)

= x2

x_ 2 =

24 cos x1

109:87 sin x1 2:88x22 sin x1 cos x1 5:973 + 2:88 cos2 (x1 )

= x2 (0) = 0

The corresponding MATLAB program is: function problem17_85 [t,x] = ode45(@f,[0:0.02:2],[0,0]); printSol(t,x*180/pi) axes(’fontsize’,14) plot(t,x(:,1)*180/pi,’linewidth’,1.5) xlabel(’t (s)’); ylabel(’theta (deg)’) grid on function dxdt = f(t,x) s = sin(x(1)); c = cos(x(1)); num = 24*c - 109.87*s - 2.88*x(2)^2*s*c; den = 5.973 + 2.88*c^2; dxdt = [x(2); num/den]; end end Below is a partial printout spanning the location where that is in degrees and ! is in deg/s t 8.6000e-001 8.8000e-001

is maximized. Note

x1 x2 2.4173e+001 1.8418e+000 2.4179e+001 -1.3238e+000

By inspection we see that

max

= 24:2 J


(c) 25

theta (deg)

20

15

10

5

0 0

0.5

1 t (s)

1.5

2

17.86 (a) In Prob. 17.85 the di¤erential equation of motion was = Setting sin = =

24

24 cos

109:87 sin 2:88! 2 sin cos 5:973 + 2:88 cos2

and cos = 1, we obtain the "linearized" form 109:87 2:88! 2 = 2:711 5:973 + 2:88

(0:3253! 2 + 12:411)

(b) Using x1 = and x2 = !, the equivalent …rst-order equations and the initial conditions are x_ 1 = x2

x_ 2 = 2:711 (0:3253x22 + 12:411)x1 x1 (0) = x2 (0) = 0

The corresponding MATLAB program is: function problem17_86 [t,x] = ode45(@f,[0:0.02:2],[0,0]); printSol(t,x*180/pi) axes(’fontsize’,14) plot(t,x(:,1)*180/pi,’linewidth’,1.5) xlabel(’t (s)’); ylabel(’theta (deg)’)


grid on function dxdt = f(t,x) dxdt = [x(2) 2.711 - (0.3253*x(2)^2 + 12.411)*x(1)]; end end Below is a partial printout spanning the location where degrees and ! is in deg/s) t 8.8000e-001 9.0000e-001

is maximized ( is in

x1 x2 2.4771e+001 1.5291e+000 2.4771e+001 -1.5149e+000

By inspection, we have max = 24:8 J The error caused by linearization is % error =

24:2 24:8 24:6

100% =

2:4%

(c) 25

theta (deg)

20

15

10

5

0 0

0.5

1 t (s)

1.5

2

17.87 (a) Geometry:

B A

φ

L

R

θ C


L sin

) sin

= R sin

tan

sin

p

=

1

Kinetics:

φ

θ

R sin L sin

=q (L=R)2

sin2

sin2

Iα

P B

P tanφ

=

R Cx

=

C

C

Cy MAD

FBD

( MC )FBD = ( MC )M AD + P (R sin ) + (P tan )(R cos ) = I 0 1 sin A (R cos ) = W k 2 (P0 sin ) (R sin ) + @P0 q g (L=R)2 sin2 0 cos gP0 R sin @ sin + q = 2 Wk (L=R)2

2

sin

1

A Q.E.D.

(b) Substituting the given data, the equation of motion becomes 0 1 cos 32:2(24)(0:75) sin @ A sin + q = 180(0:6)2 (1:5=0:75)2 sin2 ! cos = 8:944 sin sin + p 4 sin2

The initial conditions are = =2 rad and ! = 0 at t = 0. Letting x1 = and x2 = !, the equivalent …rst-order equations and the initial conditions are ! cos x1 x_ 1 = x2 x_ 2 = 8:944 sin x1 sin x1 + p 4 sin2 x1 x1 (0)

=

2

rad

x2 (0) = 0

The corresponding MATLAB program is function problem17_87 [t,x] = ode45(@f,[0:0.02:2.4],[pi/2,0]);


printSol(t,x) axes(’fontsize’,14) plot(t,x(:,2),’linewidth’,1.5) xlabel(’t (s)’); ylabel(’omega (rad/s)’) grid on function dxdt = f(t,x) s = sin(x(1)); c = cos(x(1)); dxdt = [x(2) 8.944*s*(s + c/sqrt(4-s^2))]; end end The two lines of output that span ! = 10 rad/s are: t 2.2400e+000 2.2600e+000

x1 1.3274e+001 1.3473e+001

x2 9.8875e+000 1.0031e+001

Using linear interpolation to …nd t when ! = 10 rad/s: 2:26 10:031

t 2:24 = 9:8875 10

2:24 9:8975

t = 2:25 s J

(c) 12

omega (rad/s)

10 8 6 4 2 0 0

0.5

1

1.5

2

2.5

t (s)


17.88 (a) Kinematics (note that AB has two DOF: angle

L/2

A

β +

aA

θ

G Lα 2

and displacement of A):

A

ω, α

L ω2 2

a = aA + aG=A Kinetics (note that

FBD

=

):

A

y

B

= B

Fy

= may

+.

aA

= g sin

L ( sin 2

+

( MA )FBD L mg cos 2 L mg cos 2 1 g cos 2 1 g cos 2 )

=

= = = = =

MAD I-α

N

L/2 θ β x mg

G

mg sin

φ θ

y

A m L ω2 φ 2

maA m Lα x 2

L = maA + m ( sin 2

! 2 cos )

! 2 cos )

( MA )M AD L L L I + m + (maA ) sin 2 2 2 L 1 L L mL2 + m + (maA ) sin 12 2 2 2 1 1 L + aA sin 3 2 1 1 L L + sin g sin ( sin ! 2 cos ) 3 2 2

(2g=L)(cos

sin sin ) (4=3) sin2

! 2 sin cos

Q.E.D.

(b) Using the given data, we have 2g L

= )

2(32:2) = 8:050 = 60 = rad 8 3 8:050(cos sin sin ) ! 2 sin cos a= (4=3) sin2

=

=

3


The initial conditions are = ! = 0 when t = 0. Letting x1 = and x2 = ! 2 , the equivalent …rst-order equations and the initial conditions are: x_ 1 x_ 2 x1 (0)

= x2 0:8050 [cos x1 =

sin sin( (4=3)

x22 sin( x2 )

x1 )] sin2 (

x1 ) cos(

x1 )

= x2 (0) = 0

The MATLAB program is: function problem17_88 [t,x] = ode45(@f,[0:0.01:1.0],[0,0]); printSol(t,x) function dxdt = f(t,x) s = sin(pi/3-x(1)); c = cos(pi/3-x(1)); num = 8.050*(cos(x(1))-sin(pi/3)*s)- x(2)^2*c*s; den =4/3-s^2; dxdt = [x(2); num/den]; end end The two lines of output spanning t 8.4000e-001 8.5000e-001

x1 1.0249e+000 1.0477e+000

We compute ! at 1:0477 2:2872

= =3 = 1:0472 rad are: x2 2.2575e+000 2.2872e+000

= =3 by linear interpolation: 1:0249 =3 1:0249 = 2:2575 ! 2:2575

! = 2:29 rad/s J

17.89 (a)

.. m1 Lθ B 2 r - .. L/2 G N L/2 L .I21θ mg m1 θ A θ Ax 1 2 .. . . = A m2(rθ + 2rθ) θ m2g Ay .. . m2(r − rθ 2) MAD's FBD's N B

+

Bar: ( MA )FBD L m1 g cos + N r 2 L m1 g cos + N r 2

= = =

( MA )M AD L L m1 • 2 2 1 1 m1 L2 • m1 L2 • = 12 4

I1 •

1 m1 L2 • 3

(a)


Collar:

F = ma +- N • = m2 (g cos + r + 2r_ _ )

N

m2 g cos = m2 (r• + 2r_ _ ) (b)

Substituting (b) into (a): L cos + m2 (g cos + r• + 2r_ _ )r 2 3 m1 gL cos + 2r(g cos + 2r_ _ ) 2 m2 3 m1 g cos L + 2r + 4rr_ _ 2 m2

m1 g

3 2

)•=

Collar:

g cos

= =

m1 L + 2r + 4rr_ _ m2 Q.E.D. m1 2 L + 3r2 m2

Fr = mar r• =

1 m1 L2 • 3 m1 2 L + 3r2 • m2 m1 2 L + 3r2 • m2

=

2

+%

m2 g sin = m2 (• r

2

r_ )

g sin + r _ Q.E.D.

(b) Substituting the given data, we get • =

3 32:2 cos (2(1:5) + 2r) + 4rr_ _ = 2 2(1:5)2 + 3r2

r• =

32:2 sin + r _

The initial conditions are: _ r r_ Letting x = are

32:2 cos (3 + 2r) + 4rr_ _ 3 + 2r2

2

T

= =3, _ = 0, r = 1:5 ft and r_ = 0 when t = 0 , the …rst-order equations and the initial conditions

x_ 1

= x2

x_ 2 =

32:2 cos x1 (3 + 2x3 ) + 4x3 x4 x2 3 + 2x23

x_ 3

= x4

x_ 4 =

32:2 sin x1 + x3 x22

x(0)

=

=3 rad

0

1:5 ft

0

T

The MATLAB program that produced the plot is function problem17_89 [t,x] = ode45(@f,[0:0.01:0.5],[pi/3,0,1.5,0]); axes(’fontsize’,14) plot(x(:,1),x(:,3),’linewidth’,1.5) xlabel(’theta (rad)’); ylabel(’r (ft)’) grid on function dxdt = f(t,x)


num = 32.2*cos(x(1))*(3 + 2*x(3)) + 4*x(3)*x(4)*x(2); den =3 + 2*x(3)^2; dxdt = [x(2) -num/den x(4) -32.2*sin(x(1)) + x(3)*x(2)^2]; end end

1.5

r (ft)

1.25

1

0.75

0.5 -2

-1

0 theta (rad)

1

2

17.90 (a) Kinematics:

aθ

Top view .. y

r

ω, α

ar D

C + C

aD = aC + aD=C


Kinetics:

B D L/2 C Cx N r

mABy.. .. Iθ C

Cy

A

θ

A

maθ

D

mar D

N FBD's

A

θ MAD's

y = a sin pt

) y• =

ap2 sin pt

:

( MC )FBD = ( MC )M AD 1 • I ) N= r

Collar D N

.. my

=

k(r − r0)

Rod AB

=

B

+

N r = I• (a)

: F = ma + - N = ma + m• y cos 2 • _ = m(r + 2r_ ) + m( ap sin pt) cos

(b)

Equating (a) and (b), we obtain 1 • I = m(r• + 2r_ _ ) + m( ap2 sin pt) cos r 1 I + mr2 • = m 2r_ _ ap2 sin pt cos r r • = ap2 sin pt cos 2r_ _ Q.E.D. I=m + r2 Rod AB k(r

:

Fr = mar 2

+%

k(r

r0 ) = mar + m• y sin

= m(• r r _ ) + m( ap2 sin pt) sin 2 k (r r0 ) + r _ + ap2 sin pt sin Q.E.D r• = m

r0 )

(b) Substituting the given data, we get • =

r (312:5

10

r sin 10t cos = r• = =

6 ) =0:125

2r_ _

+ r2

h

0:01(10)2 sin 10t cos

2r_ _

i

0:0025 + r2 2 3:125 (r 0:05) + r _ + 0:01(10)2 sin 10t sin 0:125 2 25(r 0:05) + r _ + sin 10t sin


The initial conditions are: = _ = 0, r = 0:05 m and r_ = 0 when t = 0 Letting _ r r_ T , the …rst-order equations and the initial conditions are x= x_ 1

= x2

x_ 2 =

x_ 3

= x4

x_ 4 =

=

x(0)

0

0

x3 (sin 10t cos x1 2x4 x2 ) 0:0025 + x23 0:05) + x3 x22 + sin 10t sin x1

25(x3

0:05 m

0

T

The following MATLAB program was used to produce the plot: function problem17_90 [t,x] = ode45(@f,[0:0.01:3],[0,0,0.05,0]); axes(’fontsize’,14) plot(t,x(:,1),’linewidth’,1.5) xlabel(’time (s)’); ylabel(’theta (rad)’) grid on function dxdt = f(t,x) num = x(3)*(sin(10*t)*cos(x(1)) - 2*x(4)*x(2)); den = 0.0025 + x(3)^2; dxdt = [x(2) num/den x(4) -25*(x(3) - 0.05) + x(3)*x(2)^2 + sin(10*t)*sin(x(1))]; end end

2

theta (rad)

1.5

1

0.5

0 0

Since

0.5

1

1.5 time (s)

2

2.5

3

is positive, the rotation is counter-clockwise (as viewed from above)


17.91

17.92 Consider the disk as a composite of the uniform disk 1 and the half-disk 2.

m

2m

+

O

O x- G 2 2 2

1

(a) x2

=

x =

4 R 3 m1 x1 + m2 x2 0 + mx2 4 = = R J m1 + m2 3m 9


(b) IO = (IO )1 + (IO )2 = I = IO

3mx2 =

m1 R 2 m2 R 2 2mR2 mR2 5 + = + = mR2 2 4 2 4 4 5 mR2 4

3m

4 R 9

2

= 1:190mR2 J

17.93 30o

mg

x

G A = 0.4 0.2N N

0.4 0.4

P

y

FBD

ma G MAD

Assume impending tipping Fy N

= 0 + - N mg cos 30 = 0 = mg cos 30 = 50(9:81) cos 30 = 424:8 N

( MG )FBD = ( MG )M AD + 0:4N 0:4(0:2N ) 0:4P = 0 P = 0:8N = 0:8(424:8) = 340 N J

17.94


17.95 Assume impending loss of contact at B.

N

0.9 '

0.9 '

A

B

Ay

A

Ax

0.9 '

0.2N FBD's

W + 12 a g

18 lb =

W + 12 lb

= W

MAD's Wa g

60o

60o

Bar AB: ( MA )FBD

=

a =

( MA )M AD +

W (0:9 cos 60 ) =

W a(0:9 sin 60 ) g

0:5774g

System: Fy Fx

=

0+"

N

(W + 12) = 0 N = W + 12 lb W + 12 a = max + ! 18 0:2N = g W + 12 18 0:2(W + 12) = (0:5774g) W = 11:15 lb J g


17.96 D

T

G

A

B FBD y

L/2 L/2 o 45 mg 1 mL2α aG = 45o B + G L/2 B D 12 α Lα aB L/2 L/2 2 A G aG = aB + aG/B maB o 45 1mLα 2 ( MD )FBD = ( MD )M AD

+

mg

L 4

1 1 mL2 + mL 12 2 6g 5L

= =

Fx T + mg sin 45

= max =

1 mL 2

+& 6g 5L

x B MAD

T + mg sin 45 =

L 4

1 mL sin 45 2

T = 0:283mg J

sin 45

17.97

0.24 m

45o

B

=

( MA )FBD = ( MA )M AD

45o

_ ma 45o

G m

B MAD

A

Ax

ma = mCG! 2 = 6

C

0.2 2 4

A

0.2 4

FBD Ay

2

ω T m

0:24 p 2 +

(10)2 = 101:82 N

0:24T

=

T

=

0:24 p 2 72:0 N J 101:82

17.98 y 279.5α 1800 lb o 62.11aA MAD α 419.3 40 A B = x A 4.5 ft 4.5 ft B 4.5 ft 4.5 ft 0.8 N N 2000 lb

FBD


I

=

max

=

1 2000 2 1 mL2 = 9 = 419:3 slug ft2 12 12 32:2 2000 L 2000 aA = 62:11aA may = m = 32:2 2 32:2

9 2

= 279:5

( MA )FBD = ( MA )M AD (1800 sin 40 )(9) 2000(4:5) = 419:3 + 279:5 (4:5) = Fy = may

Fx = max

+"

N N

2000 + 1800 sin 40 2000 + 1800 sin 40 N

+ ! 0:8N 0:8(1078:5)

J

2

0:8427 rad/s

1800 cos 40 1800 cos 40

= =

aA

=

aA

=

= = =

279:5 279:5(0:8427) 1078:5 lb

62:11aA 62:11aA 8:309 ft/s 8:31 ft/s

2

2

J

17.99


17.100 aG I

= aA + aG=A = 50 -aA + # 2:25 =

mL2 80(4:5)2 = = 135:0 kg m2 12 12

y'

40o

T A

40o

x'

G

80(9.81) N ( MG )FBD T Fy 0

B

2.25 m

2.25 m

80aA

=

2.25 m

135.0α 2.25 m

o

40 80(2.25)α

FBD

= ( MG )M AD + = 93:34

MAD

(T sin 40 ) (2:25) = 135:0

T 80(9:81) sin 40 = 80(2:25) sin 40 = may0 + % 93:34 80(9:81) sin 40 = 80(2:25) sin 40 = 2:413 rad/s J T = 93:34(2:413) = 225 N J 2

17.101


17.102 Assume impending sliding

B 60 lb

Iα G

=

120 lb

8 ft

b

F = µsN A

ma t

FBD

A

MAD

N : 120 162 mL2 = = 79:50 slug ft2 12 32:2 12 L 120 16 = m = = 29:81 lb 2 32:2 2

I

=

mat

Fy = 0: N = 120 lb ( Fx )FBD = ( Fx )M AD + !: 60

sN

= mat

60

( MA )FBD = ( MA )M AD + 60b b

0:3(120) = 29:81

2

= 0:8051 rad/s

:

= 79:50 + 29:81 (8) = 5:230 = 5:230(0:8051) = 4:21 ft J


17.103

17.104

G

110 lb y 200 lb 2 ft 1.0 ft

D

T FBD

N

x =

0.3N

G 20.12α D 12.422α MAD


The spool "rolls" on point D. ) a = DG = 2 ft/s2 I

mk

2

200 (1:8)2 = 20:12 slug ft2 32:2 200 (2 ) = 12:422 lb 32:2

=

ma = Fy = 0 + "

N

200 = 0

N = 200 lb

( MD )FBD = ( MD )M AD + 110(5) 0:3N (1:0) = 20:12 + 12:422 (2) 110(5) 0:3(200)(1:0) = 20:12 + 12:422 (2) =

10:898 rad/s

a = Fx

2

2(10:898) = 21:8 ft/s

2

J

= ma + ! 110 T + 0:3N = 12:422 110 T + 0:3(200) = 12:422(10:898) T = 34:6 lb J

17.105 n

A An FBD

R

At

=

θ mg

t mRα

A mRω2 MAD mR2α

(a) ( MA )FBD

= =

( MA )M AD g sin J 2R

+

mg(R sin ) =

mR2

mR (R)

(b) = )

d! ! ) ! d! = d = d 1 2 g ! = cos + C 2 2R

g sin d 2R

= =2. ) C = 0 r g )!= cos J R



(c) Fn An

mg cos Ft

&

At + mg sin

A=

s

= man % An g = mR cos R At + mg sin = g mR sin 2R

=

2

1 mg sin 2

2

(2mg cos ) +

= mg

mg cos = mR! 2 An = 2mg cos mR 1 mg sin 2

At = r

4 cos2 +

1 sin2 4

J

17.106 B y 3m

20(9.81) N P A 3m FBD NA Fx ( MA )FBD 20(9:81)(3) + 1:5P 20(9:81)(3) + 1:5P

+

B x =

1.5 m

A

20a 2.25 m MAD

= max + ! P = 20a ) a = 0:05P = ( MA )M AD = 20a(2:25) = 20(0:05P )(2:25) P = 785 N J

17.107 (a)

θC

1.0' G

. .

F N

mg FBD

( MC )FBD = ( MC )M AD + mg(1:0) cos 40 cos

=

Iα

. .

C

man

1.0' G mat MAD

:

= I + mat (1:0) 40 82 40 = + (1:02 ) 32:2 12 32:2

= 5:084 cos rad/s

2

J


(b) d! ! = 5:084 cos d

1 2 ! = 2

! d! = 5:084 cos d


5:084 sin + C

= 0: ) C = 0

p ! = 3:189 sin rad/s

) ! 2 = 10:168 sin

J

(c) ( Ft )FBD = ( Ft )M AD + %: N

mg cos N

=

mat

=

40 cos

N

40 (1:0) 32:2

40 cos =

40 (5:084 cos ) = 33:68 cos lb J 32:2

( Fn )FBD = ( Fn )M AD + -: F

mg sin

= man

F

=

F

40 sin +

40 sin =

40 (1:0)! 2 32:2

40 (10:168 sin ) = 52:63 sin lb J 32:2

(d) Sliding impends when F=N =

s:

52:63 tan = 0:6 33:68

= 21:0

J

17.108 (a)

P A

y

L/2

mg θ FBD

+

P (L cos )

= )

L/2 By

B

x = Bx

( MB )FBD L mg cos 2

α mL 2 ω2 mL 1 mL2α 2 12 θ B MAD

A

=

( MB )M AD 1 L L = mL2 + m 12 2 2 3 2P = g cos J 2L m

3 d! ! ) ! d! = d = d 2L 1 2 3 2P ! = g sin + C 2 2L m

2P m

g cos d


Initial condition: ! = 0 when = 0: ) C = 0 s 3 2P )!= g sin L m

J

(b) Fx Bx

L L Bx = m ! 2 cos + m sin 2 2 L 3 2P L 3 2P = m g sin cos + m g cos 2 L m 2 2L m 9 9 = (2P mg) sin cos = (2P mg) sin 2 ! J 4 8 = max

+ !

sin

17.109 R FBD O x

O

=

mg

O

1 mR2α 2

MAD

µkmg

N = mg Final angular speed of disk: ! 0 =

( MO )FBD

= )

v 5 = = 20 rad/s R 0:25

( MO )M AD + =2

k mg(R)

=

1 mR2 2

0:25(9:81) 2 kg =2 = 19:62 rad/s R 0:25

d! ) d! = dt = 19:62dt dt Initial condition: ! = 0 when t = 0. ) C = 0 =

! = ! 0 when 19:62t = 20

) ! = 19:62t + C

) t = 1:019 s J


Chapter 18 18.1

18.2

18.3


18.4

B 2.4 ft 4 . 10 ft 2 lb lb 10 k = 12 lb/ft θ L0 =2.5 ft

40 lb.ft A Work of gravity:

(U1

Work of couple:

(U1

2 )c

=

Work of spring:

(U1

2 )s

=

= = Total work: U1

2

=

C

2 )g

=

= 40

C

2W

h=

2(10)(1:2 sin 35 ) =

35 180

= 24:43 lb ft

1 k (L2 L0 )2 (L1 L0 )2 2 1 (12) (4:8 cos 35 2:5)2 (4:8 2 19:437 lb ft

13:766 lb

ft

2:5)2

13:766 + 24:43 + 19:473 = 30:1 lb ft J

18.5 U1

2

=

12(9:81)(0:375)

=

78:9 J J

8(9:81)

p 0:752 + 0:32

0:45 + 40

2

18.6 6 rad/s 250 mm

6 rad/s

200 N

B

250 mm 150 mm

150 mm

A

A 200 N

(a) Both cases: P Case (a): P Case (b): P

B (b)

= F vB = F (AB !) = 200(6)AB = 1200 AB = 1200(0:4) = 480 W J = 1200(0:1) = 120 W J


18.7 De…nition of power: De…nition of e¢ ciency:

(a) C

=

(b) C

=

Pout = C! Pout = Pin

)C=

Pin !

0:78(12 103 ) = 49:7 N m J 1800(2 =60) 0:78(12 103 ) = 24:8 N m J 3600(2 =60)

18.8 M =I =I

d! dt

But P = M !, so that P d! =I ! dt

P dt = I! d!

Pt =

1 2 I! + C 2

Initial condition: ! = 0 when t = 0: ) C = 0 r 2P t != J I

18.9


18.10

18.11


18.12

18.13


18.14

18.15 I IC T

= mk 2 = 60(0:24)2 = 3:456 kg m2 2

= I + m CG = 3:456 + 60(0:252 + 0:122 ) = 8:070 kg m2 1 1 = IC ! 2 = (8:070)(2)2 = 16:14 J J 2 2


18.16

C ω

ft 10

d

8 ft

3 ft

A vA = 12 ft/s

3 ft G

B 3

vB

4

Instant center of zero velocity is at C. From geometry AC

=

d2

=

IC = I + md2 = T =

vA 12 = = 1:5 rad/s 8 AC 82 + 32 = 73 ft2

8 ft

!=

62 + 73 12

24 1 W 2 W 2 L + d = 12 g g 32:2

= 56:65 slug ft2

1 1 IC ! 2 = (56:65)(1:5)2 = 63:7 lb ft J 2 2

18.17

0.3 rad/s C d 30o

G

4 ft

6 ft 1.8 ft/s

A

Point C is the I.C. of the bar. From geometry: d2 = (4 sin 30 )2 + (6

4 cos 30 )2 = 10:431 ft2


I IC T

1 1 mL2 = 12 12

=

20 32:2

= I + md2 = 3:313 +

(8)2 = 3:313 slug ft2

20 (10:431) = 9:792 slug ft2 32:2

1 1 IC ! 2 = (9:792)(0:3)2 = 0:441 lb ft J 2 2

=

18.18

4 rad/s

A

0.8 ft

3.2 ft/s

C

2.4 f t 10 lb 3.2 ft/s

B 2 lb

15 lb Bar AB is translating with the velocity v = R! = 0:8(4) = 3:2 ft/s Disk: IC =

T

= =

1 1 mR2 = 2 2

15 32:2

(0:8)2 = 0:14907 slug ft2

1 1 1 IC ! 2 + mAB v 2 + mB v 2 2 2 2 1 2 10 2 (3:2)2 + (3:2)2 0:14907(4) + 2 32:2 32:2

= 3:10 lb ft J

18.19

1.5 m/s

1.4 kg

110 mm

B

220 mm 2.8 kg

1.5 m/s A

13.64 rad/s

C


Bar BC is translating with the velocity v = 1:5 m/s. ! AB

=

TAB

= =

TBC

=

1:5 = 13:64 rad/s 0:110 1 1 1 2 IA ! 2AB = mAB AB ! 2AB 2 2 3 1 (1:4)(0:110)2 (13:64)2 = 0:5253 J 6 1 1 mBC v 2 = (2:8)(1:5)2 = 3:150 J 2 2

T = 0:5253 + 3:150 = 3:68 J J

18.20 5 = 0:155 28 slugs 32:2 2 = mR = 0:155 28(1:5)2 = 0:3494 slug 1 1 1 = mv 2 + I! 2 = (0:155 28)(30)2 + 2 2 2

m = I T

ft2 1 (0:3494)(102 ) = 87:3 lb ft J 2

18.21 vB

B

3m

2 m 30o vC C 7.5 rad/s A D 30o

ωBC

y x

E Point E is the I.C. of bar BC vB ! BC vC

vBC

= =

= ! AB AB = 7:5(2) = 15 m/s 15 vB = = = 2:50 rad/s 3 csc 30 EB = ! BC EC = 2:50(3 cot 30 ) = 12:990 m/s

1 1 (vB + vC ) = [ 15i + 12:990( i cos 30 + j sin 30 )] 2 2 13:125i + 3:248j m/s


TBC

1 1 2 IBC ! 2BC + mBC vBC 2 2 1 (12)(3)2 1 (2:5)2 + (12)(13:1252 + 3:2482 ) = 1125 J J 2 12 2

= =

18.22

A

= =

+# 3=8

+# v=8 I=

T

B

G v

vB = vA + vB=A v = vA + vG=A

ω 1.2 m

1.2 m

2:4!

1:2! = 8

! = 2:083 rad/s 1:2(2:083) = 5:500 m/s

2

18(2:4) = 8:640 kg m2 12

1 2 1 1 1 I! + mv 2 = (8:640)(2:083)2 + (18)(5:500)2 2 2 2 2 291 J J

18.23

ωDE E

vD

.D

1.6 m m 1.0

vB B

.

0.8 m

A

ωAB

Kinematics vB vD ! ED

= ! AB LAB = 12(0:8) = 9:6 m/s = vB = 9:6 m/s ! BD = 0 v BD = 9:6 m/s vD 9:6 = = = 6:0 rad/s LED 1:6

Mass properties (IAB )A

=

(IED )E

=

mBD

=

1 1 mAB L2AB = (2 3 3 1 1 2 mED LED = (2 3 3 2 1:0 = 2:0 kg

0:8)(0:8)2 = 0:3413 kg m2 1:6)(1:6)2 = 2:731 kg m2


Kinetic energy T

1 1 1 2 (IAB )A ! 2AB + (IED )E ! 2ED + mBD vBD 2 2 2 1 1 1 (0:3413)(12)2 + (2:731)(6:0)2 + (2:0)(9:6)2 2 2 2 165:9 J J

= = =

18.24

.

B

5 m/s

25 rad/s

.A

.C8.333 rad/s

5 m/s

Disk A: IA

=

TA

=

mA R2 8(0:2)2 = = 0:160 kg m2 2 2 1 1 1 1 2 IA ! 2A + mA vA = (0:160)(25)2 + (8)(5)2 2 2 2 2 150:0 J

= Bar AB (translates):

TAB =

1 1 2 mAB vAB = (6)(5)2 = 75:0 J 2 2

Bar BC (rotates about C): TBC =

1 1 mBC L2BC 2 1 4(0:6)2 (IBC )C ! 2BC = ! BC = (8:333)2 = 16:67 J 2 2 3 2 3

System: T = TA + TAB + TBC = 150:0 + 75:0 + 16:67 = 242 J J

18.25 Choose the horizontal plane through point A as the datum for gravitational potential energy ) V ( ) = mg

L cos 2

T =

1 1 IA ! 2 = 2 2

1 mL2 ! 2 3


1 1 mgL cos + mL2 ! 2 = C 2 6 1 Initial condition: ! = 0 when = 0: ) C = mgL 2 r 1 g 1 1 2 2 ) mgL cos + mL ! = mgL ! = 3 (1 cos ) J 2 6 2 L V + T = C (constant)

18.26

ω m = 4 kg k- = 0.18 m

6 kg A

0.25 m 0.15 m C

vA

12 kg

B

vB

The displacements of the blocks during a 90 clockwise turn of the pulley are

U1

2

T2

U1

2

yA

= RA

= 0:25

yB

= RB

= 0:15

2

= 0:3927 m "

2

= 0:2356 m #

= =

mA g yA + mB g yB [ 6(0:3927) + 12(0:2356)] 9:81 = 4:621 J 1 1 1 2 2 mA vA + mB vB + mC k 2 ! 2 = 2 2 2 1 2 = 6(0:25!) + 12(0:15!)2 + 4(0:18)2 ! 2 = 0:3873! 2 2

= T2

T1

4:621 = 0:3873! 2

0

! = 3:45 rad/s J


18.27

.

.

C

C

3 ft

B.

6 ft

.A (IA )AB = V1 T1 V2 T2

5 rad/s

9 ft

2

.Aω

t 7f .81 10

1

Datum 6 ft

B

mAB L2AB (20=32:2) (6)2 = = 7:453 slug ft2 3 3

= WAB (3) = 20(3) = 60 lb ft 1 1 = (IA )AB ! 20 = (7:453) (5)2 = 93:16 lb ft 2 2 1 2 1 = k = (5)(10:817 3)2 = 152:76 lb ft 2 2 1 1 = (IA )AB ! 2 = (7:453) ! 2 = 3:727! 2 2 2 V 1 + T 1 = V 2 + T2 60 + 93:16 = 152:76 + 3:727! 2 ! = 0:328 rad/s J

18.28


18.29 M (lb.ft) 0.88 0.50 0 0

π

θ (rad)

Open

θ

M

Closed

Mass moment of inertia of the jaw about axis AD: I

= IAB + IBC + IDC = 2IAB + IBC = 2 =

2 2:22 10 3 32:2

=

1:7236

U1

U1

10

3

5

(3)

3 12

2

+

(2:22

=

area under M - diagram =

T2

=

1 2 1 I! = (1:7236 2 2

= T2

!

=

10 32:2

3

3 12

)(2)

+ mBC L2AB 2

slug ft2

2

2

1 mAB L2AB 3

T1

501:6 rad/s

2:168 =

10

5

0:88 + 0:50 2

= 2:168 lb ft

)! 2

1 (1:7236 2

10

) vBC = LAB ! =

5

)! 2

0

3 (501:6) = 125:4 ft/s J 12

*18.30


18.31


18.32

18.33


18.34

vB v

1.5 ft

0.5 ft

ω

Kinematics: Let vB be the velocity of the block. From the velocity diagram of the spool we see that !=

vB = 0:5vB 2:0

v = 1:5! = 0:75vB

It follows that sA = 0:75sB Conservation of energy: Let the release position (position 1) be the datum for potential energy. V1 = 0 T1 = 0


V2 T2

= =

WB sB + WA sA sin 30 = 38(6) + 64(0:75 6)(0:5) 84:0 lb ft 1 1 1 2 = mA v 2 + mA k 2 ! 2 + mB vB 2 2 2 1 64 1 38 2 1 64 (0:75vB )2 + (1:25)2 (0:5vB )2 + v = 2 32:2 2 32:2 2 32:2 B 2 = 1:5373vB V1 + T1 0+0

= V2 + T2 2 = 84:0 + 1:5373vB

vB = 7:39 ft/s J

18.35

M = 0.15P

G C

P 0.4 m

Replace P by the equivalent force-couple system shown. Displacement of G: U1

2

T2

s=R

= 0:4(2 ) = 0:8 m

=

M + P s = 0:15P (2 ) + P (0:8 ) = 0:5 P N m 1 1 1 2 = IC ! = I + mR2 ! 2 = 1:6 + 40(0:4)2 (6)2 = 144:0 N m 2 2 2 U1 2 = T2 T1 0:5 P = 144:0 0 P = 91:7 N J

18.36


18.37

18.38 (a)

A

0.3 m

0.075 m

o

30 Datum

B

A

vA

0.3 m

B

ωAB

Position 2

Position 1

In position 2 point B is stationary. ) Point B is the I.C. of bar AB. V1 T2

= mAB g(0:15 sin 30 ) = 3:2(9:81)(0:15 sin 30 ) = 2:354 J 1 1 1 1 = (IAB )B ! 2AB = mAB L2AB ! 2AB = (3:2)(0:3)2 ! 2AB 2 2 3 6 =

0:0480! 2AB

T 1 + V1 = T 2 + V2

0 + 2:354 = 0:0480! 2AB + 0

! AB = 7:003 rad/s


vA = LAB ! AB = 0:3(7:003) = 2:10 m/s #J

(b)

Datum

δ

B

0.3 m

A

Position 3 V3

1 + k 2 2 1 3:2(9:81) + (1200) 2 2

=

mAB g

= T1 + V1

2

2

= 600

2

= T 3 + V3 0 + 2:354 = 0 + 600 = 0:0771 m J

15:696 2

15:696

18.39 3 rad/s

G 2.7 ft/s 0.25' 0.65'

L0 C Position 1 I v1 V1 T1 V2 T2

Datum

ω2

0.25' 0.4ω2 0.4' G C L0 + 0.65π Position 2

16 (0:25)2 = 0:198 94 slug ft2 32:2 = CG ! 1 = 0:9(3) = 2:7 ft/s v2 = CG ! 2 = 0:4! 2 = IO

md2 = 0:23

= mge = 16(0:25) = 4:0 lb ft 1 2 1 1 1 16 = I! 1 + mv12 = (0:198 94)(3)2 + (2:7)2 = 2:706 lb ft 2 2 2 2 32:2 1 1 = mge + k 2 = 4:0 + (5)(0:65 )2 = 6:425 lb ft 2 2 1 2 1 1 1 16 = I! 2 + mv2 = (0:198 94)! 22 + (0:4! 2 )2 = 0:139 22! 22 2 2 2 2 32:2 T1 + V1 !2

= T 2 + V2 2:706 + 4:0 = 0:139 22! 22 + 6:425 = 1:421 rad/s J


18.40

L/2

30o

L/2 B

L/2

B

Position 1

ω2

A

A

P

C

L/2

C

Position 2

P

In Position2, point A is the I.C. of the bar U1 U1

2

2

= P (L sin 30 ) = T2

T2 =

1 1 IA ! 22 = 2 2

1 P L sin 30 = mL2 ! 22 6

T1

1 mL2 ! 22 3 0

!2 =

r

3P J mL

18.41 IC T

2

2

2

= I + mCG = m(k 2 + CG )! 2 = 320(0:42 + CG ) = 51:20 + 320CG 1 1 2 2 IC ! 2 = 51:20 + 320CG ! 2 = 25:60 + 160CG ! 2 = 2 2

2

Dimensions in meters 0.3 Datum 1.4

G

G

0.3

ω3 ω2

G

0.3 1.7

1.1 C Position 2

C Position 1

C Position 3

(a) Position 1 (initial position): V1 = 0 2

CG T1

= =

1:42 + 0:32 = 2:050 m2 [25:60 + 160(2:050)] (2:8)2 = 2772 J

Position 2 (position of maximum T ): 2

CG T2 V2

= = =

1:12 = 1:21 m2 [(25:60 + 160(1:21)] ! 22 = 219:2! 22 mg(0:3) = 320(9:81)(0:3) = 941:8 J


T1 + V1 2772 + 0

= T 2 + V2 = 219:2! 22

941:8

! 2 = 4:12 rad/s J

(b) Position 3 (position of minimum T ): 2

CG T3 V3 T1 + V 1 2772 + 0

= 1:72 = 2:89 m2 = [25:60 + 160(2:89)] ! 23 = 488:0! 23 = mg(0:3) = 320(9:81)(0:3) = 941:8 J = T3 + V3 = 488:0! 23 + 941:8

! 3 = 1:937 rad/s J


18.42

G G a

b 1 Datum

ω

A

2

In position 2: IA

= I + mb2 = m = m

V1 V2 T2

2

+ b2 a2 + 5b2 + mb2 = m 4 4

1:02 + 5(0:75)2 = 0:9531m 4

= mga = m(32:2)(1:0) = 32:2m T1 = 0 = mgb = m(32:2)(0:75) = 24:15m 1 1 IA ! 2 = (0:9531m)! 2 = 0:4766m! 2 = 2 2 V 1 + T 1 = V 2 + T2 32:2m + 0 = 24:15m + 0:4766m! 2 ! = 4:11 rad/s J

18.43


18.44

18.45


18.46

18.47

.

A

.B

2 ft

ft 2.5

.

ω2 A

1.5 ft

. D

. C Position 1

2 ft

Datum vB

.

2 ft 0.5 ft

ωCD

.

D

B

2 ft

2.5 ft

Position 2

vC

.C

Position 1 (release position): T1 = 0

V1 = 2w0 (1:5) + 2:5w0 (0:75) = 4:875w0


Position 2: Note that BC is translating (! BC = 0) V2 = 2w0 (0:5) vB ! CD mBC (IAB )A

T2

= = =

2:5w0 (1:25)

2w0 (1:0) =

4:125w0

= vC = vBC = LAB ! 2 = 2! 2 vC 2! 2 = !2 = LCD 2 2:5w0 = 0:07764w0 = 32:2 1 2w0 = (ICD )D = (2)2 = 0:08282w0 3 32:2

1 1 1 2 (IAB )A ! 22 + (ICD )D ! 2CD + mBC vBC 2 2 2 1 1 1 (0:08282w0 )! 22 + (0:08282w0 )! 22 + (0:07764w0 )(2! 2 )2 2 2 2 0:2381w0 ! 22 T 1 + V1 0 + 4:875w0 !2

= T2 + V2 = 0:2381w0 ! 22 4:125w0 = 6:15 rad/s J

18.48 Kinematics ( = 45 ):

2.121 ft

C

1.5 ft

A 45o

.

vA

G

ωAB

vG B

vB

Point C is the I.C. of bar AB. vG

=

!B

=

1:5! AB vB = 2:121! AB vB 2:121! AB = = 1:414! AB R 1:5


Conservation of energy:

. 1.5'

IAB

=

IB

=

1

45o Datum

. 1.5'

1 1 12 mAB L2AB = (3)2 = 0:2795 slug ft2 12 12 32:2 1 15 1 mB R2 = (1:5)2 = 0:5241 slug ft2 2 2 32:2

V1 = WAB (1:5) = 12(1:5) = 18 lb ft V2 (T2 )AB

(T2 )B

2

T1 = 0

= WAB (1:5 sin 45 ) = 12(1:5 sin 45 ) = 12:728 lb ft 1 1 2 = IAB ! 2AB + mAB vG 2 2 1 12 1 (0:2795)! 2AB + (1:5! AB )2 = 0:5590! 2AB = 2 2 32:2 1 1 2 IB ! 2B + mB vB = 2 2 1 1 15 = (0:5241)(1:414! AB )2 + (2:121! AB )2 2 2 32:2 = 1:5718! 2AB V1 + T 1 18 + 0 ! AB

= V2 + T 2 = 12:728 + (0:5590 + 1:5718)! 2AB = 1:5730 rad/s

vB = 2:121! AB = 2:121(1:5730) = 3:34 ft/s J

18.49


18.50 60 lb A v2

60 lb C

C 3 ft

3 ft

3 ft

3 ft

A

o

240 lb.ft

60 240 lb.ft

Position 1 D

B

Position 2 B

D

Bar AC undergoes curvilinear translation U1

2

T2

= C

h = 240

30 180

60(3

3 sin 60 ) = 101:55 lb ft

1 60 2 1 mv 2 = v 2 2 2 32:2 2

= U1

mg

2

= T2

T1

101:55 =

1 60 2 v +0 2 32:2 2

v2 = 10:44 ft/s J


18.51 30o 80(9.81) N µkN =

FBD T

Fy kN U1

x

y MAD

80a

N

= 0 + - N 80(9:81) cos 30 = 0 = 0:4(679:7) = 271:9 N

2

=

2mgy

T2

=

2

= U1

2

kN x

= 2(80)(9:81)(x sin 30 )

N = 679:7 N

271:9x = 512:9x

1 1 1 1 2 mv22 + IA ! 2A = mv22 + mRA 2 2 2 2 1 1 mv22 = 1 + 80(5)2 = 2500 J 1+ 4 4

= T2

T1

512:9x = 2500

v2 RA

2

x = 4:87 m J

0

18.52

C A

0.5 m

0.3 m

P 0.4 m B Position 1

ωC

C 0.08 m 0.5 m ωAB B Position 2

Datum A 2 m/s

P

Position 1: V1

1 = mC g(0:3) + mAB g(0:15) + k 2 = =

2 1

1 2(9:81)(0:3) + 3(9:81)(0:15) + (72)(0:4 2 13:541 J

0:1)2


Position 2 (B is the I.C. of bar AB): IC

=

(IAB )B

=

! AB

=

!C

=

V2

T2

1 1 2 mC RC = (2)(0:08)2 = 0:0064 kg m2 2 2 1 1 2 mAB LAB = (3)(0:5)2 = 0:25 kg m2 3 3 vA 2 = = 4 rad/s LAB 0:5 vA 2 = = 25 rad/s RC 0:08

1 1 2 k 2 P x = (72)(0:7 0:1)2 P (0:5 0:4) 2 2 = 12:96 0:1P 1 1 1 2 IC ! 2C + mC vA + (IAB )B ! 2AB = 2 2 2 1 1 1 (0:0064)(25)2 + (2)(2)2 + (0:25)(4)2 = 8:0 J = 2 2 2

=

T1 + V 1 = T 2 + V 2 0 + 13:541 = 8:0 + 12:96

0:1P

P = 74:2 N J

18.53 Kinematics: ! A = vB =0:25 = 4vB Choose the ground as the datum for potential energy. V1 = mB gh = 5(9:81)(6) = 294:3 J V2 T2

0 1 1 2 = mA R! 2A + mB vB 2 2 1 2 = (8)(0:25)2 (16vB )+ 2

T1 = 0

=

=

1 1 2 mA R2 (4vB )2 + mB vB 2 2

1 2 2 (5)vB = 6:50vB 2

V1 + T1 = V2 + T2 2 294:3 + 0 = 0 + 6:50vB

vB = 6:73 m/s J


*18.54


18.55

vC = 2vA

15"

vA

ωA vA !A

= vB = 0:5vC ) sA = sB = 0:5 vC vC = = 0:4vC = !B = 2R 30=12 250 32:2

IA = IB = mR2 =

15 12

sC

2

= 12:131 slug ft2

Cylinder A reaches corner of the slab when sC T2

sA = 20 in. = =

U1

2

U1

1 2 mC vC +2 2 1 1500 2 vC 2 32:2

= P 2

sC

= T2

sC = 180 T1

0:5 sC = 20 in.

sC = 40 in.

1 1 2 IA ! 2A + mA vA 2 2 + 12:131(0:4vC )2 + 40 12

250 2 (0:5vC )2 = 27:174vC 32:2

= 600 lb ft

2 600 = 27:174vC

0

vC = 4:70 ft/s J

18.56


18.57

D

180

180 A

B

G

120

x- = 380 100 x= IAB

=

ID

=

8(180) + 16(480) mi xi = = 380 mm mi 8 + 16 1 1 mAB L2AB = (8)(0:36)2 = 0:0864 kg m2 12 12 1 1 mD R2 = (16)(0:120)2 = 0:1152 kg m2 2 2

(a) I h

= IAB + mAB (0:2)2 + ID + mD (0:1)2 = 0:0864 + 8(0:2)2 + 0:1152 + 16(0:1)2 = 0:6816 kg m2 = I! = 0:6816(5) = 3:408 N m s J

(b) IA hA

= I + (mAB + mD )x2 = 0:6816 + (8 + 16)(0:38)2 = 4:147 kg m2 = IA ! = 4:147(5) = 20:7 N m s J

18.58


18.59

mABv-AB

IABω

m v- BC BC IBCω B

A 1.2'

d C

3.2' (a)

mv-

= B

A

C

(b)

Momentum diagram (a): IAB !

=

IBC !

=

mAB vAB mBC vBC

hA

= =

1 mAB L2AB ! = 12 1 mBC L2BC ! = 12

1 3 (2:4)2 ! = 0:04472! 12 32:2 1 5:5 (1:6)2 ! = 0:03644! 12 32:2

3 (1:2!) = 0:11180! 32:2 5:5 = mBC (rBC !) = (3:2!) = 0:5466! 32:2 = mAB (rAB !) =

[0:04472 + 0:03644 + 0:11180(1:2) + 0:5466(3:2)] ! 1:9644!

Momentum diagram (b): mv =

8:5 (!d) = 0:2640!d 32:2

hA = mvd = 0:2640!d2

Diagrams (a) and (b) are equivalent: 1:9644! = 0:2640!d2

d = 2:73 ft J


18.60

18.61

18.62 (AO )1

2

=

Z

60 s

C(t)dt = 12

0

= (hO )1 (AO )1

2

=

= (hO )2

60 s

1

2t

e

0

714:0 lb ft s

0

Z

(hO )2 = mk 2 ! 2 = (hO )1 :

40 32:2

9 12

714:0 = 0:6988! 2

1 dt = 12 t + e 2

60 s 2t 0

2

! 2 = 0:6988! 2 0

! 2 = 1022 rad/s J


18.63

20 in. Ay Ax A 120 lb

20 in.

.

.

Iω1

A mBω1R B 50 lb

FBD

IA =

Momentum diagram

1 1 120 mA R2 = 2 2 32:2

2

20 12

= 5:176 slug ft2

Z

(hA )1 IA ! 1 + (mB ! 1 R) R

Z

t

MA dt

=

(hA )2

0 t

mB gR dt =

0

0

5:176(12) +

20 12

50 (12) 32:2

2

20 12

50

t

=

0

t = 1:367 s J

18.64 h1 A1

2

= =

0 Z

0

A1

2

!1

= h2 =

h1

t

h2 = I! Z t M (t)dt = M0 e

t=t0

dt = M0 t0 1

e

t=t0

0

M 0 t0 1

e

t=t0

= I!

0

!=

M0 t 0 1 I

e

t=t0

M 0 t0 4:6(3:8) = = 24:3 rad/s J 0:72 I


18.65


18.66 400

P W 3P 3P 200 Ox O Oy 3Pµk

200

FBD's 2P +

(AO )1

2

=

Z

12 s

(3P

k) R

dt = 3

k R(area

under P -t diagram)

0

=

3(0:3)(0:2)(6P0 ) = 1:08P0

+

mk 2 ! 1 =

(hO )1 =

(AO )1

2

= (hO )2

(hO )1

20(0:16)2 400 1:08P0 = 0

2 60

=

21:45 N m s P0 = 19:86 N J

( 21:45)

18.67

mrodvG C Irodω G

.

A

mrodvG C Irodω G

mdiskvB

.

.B R

L/2 L/2 Momentum diagram (a)

A

mdiskvB

.B R

Idiskω

L/2 L/2 Momentum diagram (b)

(a) Angular speed of disk A will remain zero. hC

=

(hC )disk + (hC )ro d

=

(mdisk vB )L + Iro d ! + (mro d vG )

=

(mdisk !L)L +

=

(AC )1

2

1 L mro d L2 ! + mro d ! 12 2

1 mdisk + mro d 3 (AC )1 = (hC )2

2

L 2

L2 ! =

44 + (18=3) 32:2

L 2 16 12

2

! = 2:761!

= M0 t = 6(3) = 18 lb ft s

(hC )1

18 = 2:761!

0

! = 6:52 rad/s J


(b) Disk and rod have the same angular speed. We must add the following term to hC : Idisk !

1 1 44 mdisk R2 ! = (1:0)2 ! = 0:6832! 2 2 32:2 hC = 2:761! + 0:6832! = 3:444!

= )

(AC )1

2

= (hC )2

(hC )1

18 = 3:444!

0

! = 5:23 rad/s J

18.68 I-ω

mg 0.3 m FBD

O

O

µkmg

Initial mv- momentum diagram

mg Let t be the time when cylinder stops +

!

t

=

+

L1 2 = p2 p1 k mgt = 0 v 2 = = 0:3398 s g 0:6(9:81) k

(AO )1

k mgRt

=

!

=

2

= (hO )2

(hO )1

mv

k mgRt

1 mR2 ! 2 2 k gt 2(0:6)(9:81)(0:3398) = = 13:33 rad/s R 0:3

=0

I!

J

18.69


18.70

P R

Iω

R

mωR mg F

C FBD

C Momentum diagram

N = mg

(a) + Z

(AC )1

2

= (hC )2

(hC )1

t

P (2R) dt

= I! + m!R(R)

0

0

2P Rt = t

=

mR2 ! + m!R2 2 3 m!R 3 60(10)(0:4) = = 0:900 s J 4 P 4 200

(b) Z

+ ! L1

2

= p2

p1

t

(P + F )dt

= m!R

0

(P + F )t = m!R

0

0

F

=

Pt

m!R = t

200(0:9)

(60)(10)(0:4) = 66:67 N 0:9


Smallest

s

that would prevent sliding is s

=

F 66:67 = = 0:1133 J N 60(9:81)

18.71 B A

IAω1

mB Lω2 2 I-

IAω2

ω

B 2

10" 10" Final momenta

Initial momenta

IA

=

1 1 6 mA R2 = 2 2 32:2

IB

=

1 4 1 mB L2 = 12 12 32:2

(hO )1

=

0:052 41(16)

=

0:8386

=

+

mALω2

9 12

2

= 0:052 41 slug ft2

20 12

2

= 0:028 76 slug ft2

IA ! 1 = IA ! 2 + IB ! 2 + mA L2 ! 2 + mB

(hO )2 "

0:052 41 + 0:028 76 +

6 32:2

20 12

2

+

4 32:2

10 12

2

L 2 #

2

!2

!2

! 2 = 1:224 rad/s J

0:6850! 2

18.72

L/2 x C

A

1 m L2ω 12 AB B

mC xω mAB(L/2)ω Momentum diagram (a) Angular momentum of the system about A is conserved. hA

= =

1 L mAB L2 ! + mAB ! 12 2 1 mAB L2 + mC x2 ! 3

L + (mC x!) x 2


When x = When x =

1 2 (1:6)2 + 3 32:2 1 2 1:6 ft: (hA )2 = (1:6)2 + 3 32:2

0:5 ft: (hA )1 =

(hA )1 = (hA )2

0:2213 = 0:07685!

0:3 (0:5)2 (4) = 0:2213 lb ft s 32:2 0:3 (1:6)2 ! = 0:07685! 32:2 ! = 2:88 rad/s J

(b) The act of the mass leaving the rod does not change the angular momentum of the rod. ) ! = 2:88 rad/s J

*18.73

18.74


18.75

18.76

18.77

O

0.125mω1 O

Iω1 0.125 m Initial momenta I=

0.375 m

0.375mω2 Iω2

Final momenta

1 1 mL2 = (18)(0:75)2 = 0:8438 kg m2 12 12 (hO )1

=

(hO )2

2

I! 1 + (0:125) m! 1 = I! 2 + (0:375)2 m! 2 0:8438 + (0:125)2 (18) (16) = 0:8438 + (0:375)2 (18) ! 2 ! 2 = 5:33 rad/s J


18.78 With rods vertical: (Iz )1

=

2 "

1 mR (a2 + b2 ) + mR d21 + mB k 2 12

9 12

2

1 mR (a2 + L2 ) + mR d22 + mB k 2 12 " # 2 1 18 1:52 + 182 18 15 24 2 + + 12 32:2 122 32:2 12 32:2

9 12

2

=

0:7043 slug ft2

6 12

2

#

24 32:2

=

18 1 18 1:52 + 2:52 + 2 2 12 32:2 12 32:2

+

With rods horizontal: (Iz )2

= = =

2

2:377 slug ft2

Angular momentum about z-axis is conserved: (Iz )1 ! 1 = (Iz )2 ! 2 0:7043(40) = 2:377! 2

! 2 = 11:85 rad/s J


*18.79


18.80

18.81


18.82

Ι-ABω1 I-1ω1 m(2.5ω1)

I1ω1 m(2.5ω1)

2.5' O 2.5' Initial momenta

m(2.5ω2)

I2ω2

m(2.5ω2) ΙABω2

I2ω2

A

O B Final momenta

One plate: 1 1 12 mb2 = (2)2 = 0:124 22 slug ft2 12 12 32:2 1 m(2b2 ) = 2I1 = 0:2484 slug ft2 I2 = 12

In position 1:

I1 =

In position 2: IAB =

1 1 8 mAB L2 = (3)2 = 0:186 34 slug ft2 12 12 32:2

+ (hO )1 2 I1 ! 1 + m(2:5) ! 1 + IAB ! 1

= (hO )2 = 2 I2 ! 2 + m(2:5)2 ! 2 + IAB ! 2

2

12 (2:5)2 + 0:186 34 (12) 32:2 12 (2:5)2 + 0:186 34 ! 2 2 0:2484 + 32:2 5:342! 2 ! 2 = 11:44 rad/s J 2 0:124 22 +

= 61:12

=

18.83


18.84

A

Iω1

.

=

+ C

C Momenta before impact

0.25mL ω2 F dt

Impulse during impact

G C

. .

Iω2 0.25L


Angular momentum about C is conserved: I! 1 2

mL !1 12 !2

= I! 2 + 0:25mL! 2 (0:25L) mL2 = ! 2 + 0:25mL! 2 (0:25L) 12 = 0:5714! 1 = 0:5714(25) = 14:29 rad/s J

18.85


18.86 (a)

ω2 Datum ω2d

Velocities after impact d

θ C G

Position of maximum displacement

0:12 (1600)(0:60) = 3:578 lb ft s 32:2 + (hC )2 = (mAB + mD )! 2 d2 + I! 2 1 24 24 + 0:12 ! 2 (0:6)2 + (6)2 ! 2 = 2:506! 2 = 32:2 12 32:2 (hC )1 = (hC )2 3:578 = 2:506! 2 ! 2 = 1:4278 rad/s J +

(hC )1 = mD v1 d =

(b)

A C D mv D1

^ F

B Momenta before impact

T2

V2 V3

L/2

^ R ^ F

d

Iω2 (mAB + mD)ω2d

L/2

FBD during impact


1 2 1 I! + (mD + mAB )(! 2 d)2 2 2 2 1 1 24 1 0:12 + 24 = (6)2 (1:4278)2 + (1:4278 0:6)2 2 12 32:2 2 32:2 = 2:554 lb ft = (WD + WAB )d = (0:12 + 24)(0:6) = 14:472 lb ft = (WD + WAB )d cos = (0:12 + 24)(0:6 cos ) = 14:472 cos =

2:554

T2 + V2 = T3 + V3 14:472 = 0 14:472 cos

= 34:6

J


18.87 P^

A L/2

A Iω mv-

A

ω L/2

FBD during impact +

(hA )1 = (h2 )2 : vC = v

Momenta after impact 0 = I!

L 2

!

y

=v

mv

L 2

6 v L

C

y

vvC

Velocities after impact 1 L mL2 ! = mv 12 2

L 2

y

=0

y=

!=6

v L

L J 3

18.88

mv1

0.25mω2 G 0.25 m

5m 0.2

G 0.2 m

Iω2

A^x A A A ^ Ay 1 Momentum FBD during 2 Momenta before impact impact after impact

A Datum 3 Imminent tipping

(a) I=

1 m(0:32 + 0:42 ) = 0:02083m 12

+ (hA )1 = (hA )2 : mv1 (0:2) = I! 2 + (0:25)2 m! 2 0:2mv1 = 0:02083m! 2 + (0:25)2 m! 2 ! 2 = 2:40v1 J (b) 1 2 1 I! + m(0:25! 2 )2 + 0:2mg = 0 + 0:25mg 2 2 2 1 1 (0:02083m)(2:40v1 )2 + m(0:25 2:40v1 )2 + 0:2m(9:81) 2 2 0:25m(9:81) 0:4905 v1 = 1:430 m/s J

T2 + V2

= T 3 + V3 :

0:240v12

= =


18.89 (a) + !2

(hA )1 = (hA )2 =

mv1

L = IA ! 2 2

mv1

3 3 v1 = (5:5) = 9:167 rad/s 2L 2(0:9)

L 1 = mL2 ! 2 2 3

J

(b) Choose the horizontal plane through A as the datum for potential energy. V2 + T 2 L 1 mg + IA ! 22 2 2 L 1 1 mL2 ! 22 mg + 2 2 3 s r g ! 3 = ! 22 6 = 9:1672 L

= V3 + T3 L = mg + 2 L = mg + 2 6

9:81 0:9

1 IA ! 23 2 1 1 mL2 ! 23 2 3 J

= 4:32 rad/s

18.90 1 mR2ω 1 2

1 mR2ω2 2

mRω1

20o R A Momenta before impact

mRω2 R

A ^ ^ N F FBD during impact

A Momenta after impact

+ (hA )1 = (hA )2 : 1 1 mR2 ! 1 + (m! 1 R) R cos 20 = mR2 ! 2 + (m! 2 R) R 2 2 ! 1 (1 + 2 cos 20 ) = 3! 2 !2 =

1 + 2 cos 20 1 + 2 cos 20 !1 = (4) = 3:84 rad/s 3 3

J


18.91

18.92


18.93

2

1 G mv1

.

+

.

G I-ω 2

F dt

Momenta A after impact

ft 2.5

=

2 ft Momenta before impact

A

A

mv2

Impulse during impact

Angular momentum about A is conserved during impact: mv1 (2) 2mv1 !2

= I! 2 + mv2 (2:5) m 2 = (3 + 42 )! 2 + m! 2 (2:5)2 12 = 0:240v1

3

2

.G

.G 2 ft

2.5 ft

ω2

Datum A

A Energy is conserved after the impact: For the box to ‡ip, it has to reach position 3. V2

=

T2

= =

2mg = 64:4m V3 = 2:5mg = 2:5(32:2m) = 80:50m 1 2 1 1m 2 1 I! + mv 2 (2:5) = 3 + 42 ! 22 + m! 22 (2:5)2 2 2 2 2 2 12 2 2 4:167m! 2 V2 + T2 64:4m + 4:167m! 22 !2 v1 =

T3 = 0

= V3 + T 3 = 80:50m + 0 = 1:9656 rad/s

1:9656 = 8:19 ft/s J 0:240


18.94

18.95 (a)

R/2

R G Momenta before impact

FBD during impact

F^

mv-2 G

Iω2



Z

F^ dt = mv2 = 2:8m Z R F^ dt = I! 2 0 (AG )1 2 = (hG )2 (hG )1 + 2 R 2 3:5 3:5 (2:8m) = mR2 ! 2 !2 = = = 33:60 rad/s J 2 5 R 1:25=12 L1

= p2

2

p1

+

(b)

mg mv-2

Iω2

G R C Momenta after impact (hC )1

mv2 R

=

G R

m(Rω3)

F N FBD during slipping C

G R C

Iω3

Momenta during rolling

+ mv2 R I! 2 = (mR! 3 ) R + I! 3 2 = mR2 ! 3 + mR2 ! 3 5 5v2 2R! 2 5(2:8) 2(1:25=12)(33:60) = = 7R 7(1:25=12) = 9:60 rad/s J

2 mR2 ! 2 5 !3

(hC )1

18.96 Angular momentum about the z-axis is conserved: mA (0:8)2 !2 4 12(0:8)2 !2 0:009(850)(0:6) = 0:009(310)(0:6) + 4 ! 2 = 1:519 rad/s J mB v1 (0:6)

= mB v2 (0:6) +

18.97 IA

= = =

1 2 2 2 2 mAB AB + mC RC + mc AB + RC 3 5 2 2 20 2 5 3 5 23 1 2 + + 3 32:2 12 5 32:2 12 32:2 12

2

0:6318 slug ft2

Let v2 be the velocity of ball D after the impact + 0:6318! 1

=

(hA )1 = (hA )2 : IA ! 1 = mD (AB + RC )v2 W 23 v2 ! 1 = 0:094 21W v2 32:2 12

(a)


Since impact is elastic, energy is conserved T1

1 1 IA ! 21 = mD v22 2 2

= T2 :

1 (0:6318) ! 21 2

=

1 2

W 32:2

p ! 1 = 0:2217 W v2

v22

(b)

Equating (a) and (b): p 0:094 21W = 0:2217 W

W = 5:54 lb J

18.98 (a)

mCv1 30o

mCω2b

b

b IBω2

C B

A

A

Momentum before impact Bar AB: IB =

L/2 mAB L ω 2 2 Momenta after impact

B

1 60 1 mAB L2 = (8)2 = 39:75 slug ft2 3 3 32:2

(hB )1 = (hB )2 (mC v1 sin 30 ) b = IB ! 2 + (mC ! 2 b)b 28 28 (15 sin 30 )(6) = 39:75! 2 + ! 2 (6)2 32:2 32:2 ! 2 = 0:5507 rad/s J (b) Initial deformation of the spring is 1

=

0:5WAB 0:5(60) = = 1:5 in. k 20

L b L/2

Lθ A

Datum bθ Lθ/2 θ G C

B


T2

= =

V3

= = =

1 IB ! 22 + mC (! 2 b)2 2 1 28 39:75(0:5507)2 + (0:5507 6)2 = 10:774 lb ft 2 32:2 1 2 k(L (mAB g) (b ) 1) 2 2 1:5 1 (20 12) 8 60(4 ) 28(6 ) 2 12 7680

2

648 + 1:875

V 2 + T 2 = V 3 + T3 0 + 10:774 = 7680 2 648 + 1:875 + 0 0 = 7680:0 2 648:0 8:90 = 0:09639 rad = 5:52 J

18.99


18.100

18.101 (a)

B

B L/2

L mb v1

A Momentum before impact

+ ! mb v1

=

+ (mb v1 ) L

=

mb v1

=

!2

=

G mv2 1 2 12 mL ω2

A Momenta after impact

p 1 = p2 mv 2

v2 =

mb 0:04 v1 = (2800) = 6:222 ft/s m 18

(hB )1 = (hB )2 1 L mL2 ! 2 + (mv 2 ) 12 2 1 1 mb mL! 2 + m v1 12 2 m 6mb v1 6(0:04)(2800) = = 7:467 rad/s J mL 18(5)


(b) Due to absence of horizontal forces, v remains constant after impact (v3 = v2 ).

B

B

Datum

θ

L/2

v-2

G

G ω3 = 0

ω2

L cosθ 2 v-3 = v-2 A

A Immediately after impact

V2

=

T2

= =

Position of maximum θ

L = 18(2:5) = 45:0 lb ft 2 1 mL2 ! 22 + mv 22 12 1 18 18 (5)2 (7:467)2 + (6:222)2 12 32:2 32:2

mg 1 2 1 2

V3

=

T3

=

= 43:29 lb ft

L cos = 18(2:5 cos ) = 45:0 cos lb ft 2 1 1 18 mv 22 = (6:222)2 = 10:820 lb ft 2 2 32:2 mg

V2 + T 2 = V 3 + T 3 45:0 + 43:29 = 45:0 cos + 10:820 45:0 + 43:29 10:820 = cos 1 45:0

= 73:8

J

18.102 Kinematics When = 0, the rods are vertical and the collars at B and D are at the limit of their travel. Therefore, vB = vD = 0


Also note that due to symmetry the angular velocities of the rods are equal in magnitude but opposite sense: ! AC =

! CE = !

vA and ! are related by the relative velocity equation vB = vA + vB=A

.

A ω 0

vA

=

+

L/2

ωL/2

B 0=

vA +

!L 2

!=

2 vA L

Conservation of energy Choose the horizontal rod that guides collar A as the datum for potential energy. V1

=

V2

=

T2

=

0

2

T1 = 0 L 3L mg + 2 2 1 2 I! 2

=

V1 + T 1 0+0 vA

18.103

=

2mgL

mL2 2 mL2 ! = 12 12

2vA L

2

=

1 mv 2 3 A

= V2 + T2 1 2 2mgL + mvA 3 p = 6Lg J

=


18.104

18.105

Iω G

G

R

mg F = µkmg FBD

mv-

C Momentum diagram

N = mg

Sliding stops when vC = v0

v + !R = v0

(a)

(a) During sliding: p1 +

Z

t

F dt = p2

0+

k mgt

= mv

v=

k gt

(b)

0


(hC )1 +

Z

t

F R dt =

0 k mgRt

=

(hC )2

0+

mR2 ! 2

k mgRt

!=

2

= I!

k gt R

(c)

J

(d)

Substituting Eqs. (b) and (c) into Eq. (a) yields k gt

+

2

k gt R = v0 R

t=

v0 3

kg

(b) Subsitution of Eq. (d) into Eqs (b) and (c) results in v=

kg

v0 3

kg

=

v0 J 3

!=

2

kg

R

v0 3

kg

=

2 v0 J 3R

18.106


18.107

18.108

3 ft

3 ft

L − 6.708' 3 ft C

8' 6.70

3 ft

B

A

B

A

Position 1

3 ft L − 3'

ωAB Position 2

C vC

vC = AB! AB = 3! AB Let L be the length of the rope. U1

2

= =

T2

= = =

AB + WC [(L 3) (L 6:708)] 2 W (1:5) + W (3:708) = 2:208W 1 1 2 (IAB )A ! 2AB + mC vC 2 2 11 W 1 W (3)2 ! 2AB + (3! AB )2 2 3 32:2 2 32:2 0:18634W ! 2AB WAB


U1 2 2:208W vC

= T2 T1 = 0:18634W ! 2AB 0 ! AB = 3:442 rad/s = 3(3:442) = 10:33 ft/s # J

18.109 ^ N

T^A

C A

G

^ N o

45o 45

FBD during impact

45o 45o

Momentum before impact

2.4 m 2.68 3m

G

O

T^B

O 2.4m ω 1.2 m C ΑΒ C 30 N.s Iω 2.683mCω B A B G

1.2 m A

B

O


Point O is the I.C. of bar AB. I=

1 1 mAB L2 = (15)(4:8)2 = 28:80 kg m2 12 12 +

(30)(1:2)

(hO )1 = (hO )2 2

= I! + (2:4)2 mAB ! + (2:863) mC ! 2

(30)(1:2) = 28:80! + (2:4)2 (15)! + (2:683) (5)! ! = 0:238 rad/s J

18.110


18.111

18.112 V1 T2

= mgh 1 2 = I! + 2

T 1 + V1 !

T1 = 0 V2 = 0 1 1 2 mv 2 = mR2 ! 2 2 2 5

1 9 + m(!R)2 = mR2 ! 2 2 10

9 = T 2 + V2 0 + mgh = mR2 ! 2 10 p p p 10 gh gh = = 1:054 J 3 R R


18.113

18.114

^y O

L

L

mv1

F^

A Momentum before impact

O Datum Lm ω 2 OA 2

Iω2 F^

FBD during impact

I=

O

^ O x

L/2

O

θ

L/2 L/2

A

mv2

A

Momenta after Position of max. displacement impact

1 1 30 mOA L2 = (3)2 = 0:6988 slug ft2 12 12 32:2

System +

(hO )1 = (hO )1

mv1 L = I! 2 + mOA

L 2

2

! 2 + mv2 L 2

5 30 3 5 (1800)(3) = 0:6988! 2 + !2 + (1400)(3) 16 32:2 32:2 2 16 32:2 52:41 = 2: 795 ! 2 + 40: 76 ! 2 = 4:1667 rad/s


Bar only T 2 + V2

= T3 + V3

2

1 2 1 L L ! 22 WOA I! + mOA 2 2 2 2 2 " # 2 1 L I + mOA ! 22 2 2 1 2

0:6988 +

30 2 (1:5)2 (4:167) 32:2

=

0

WOA

L cos 2

L = WOA (1 2

cos )

=

30(1:5)(1

cos )

=

62:6

J

18.115 IAω1

Α^y Α^x Α 1.5' IAωA2

IA

=

IB

=

Momenta before impact

F^

Β^y Β^ x 1.2' Β IBωB2

F^

1 mA L2A = 12 1 mB LB = 12

FBD's during impact Momenta after impact

1 8 (3)2 = 0:18634 slug ft2 12 32:2 1 6 (2:4)2 = 0:08944 slug ft2 12 32:2

Kinematics (plastic impact): 1:5! A2 = 1:2! B2

! B2 = 1:25! A2

Bar A: + 1:5

Z

Z

A1

=

2

(hA )2

(hA )1

F^ dt = IA (! A2 F^ dt =

! A1 ) = 0:18634 (! A2

0:12423 (! A2

5)

5)

Bar B: + 1:2

Z

Z

A1

2

=

(hA )2

F^ dt = IB (! B2 F^ dt =

(hA )1 ! B1 ) = 0:08944 (1:25! A2 )

0

0:09317! A2


0:12423 (! A2 J

! A2 = 2:857 rad/s

5) = 0:09317! A2 J

! B2 = 1:25(2:857) = 3:57 rad/s

18.116 0.08 m

mg

FBD

C

0.4N2 N2 0.4N1

N1

Fx Fy

= 0: = 0:

+ +"

N2 0:4N1 = 0 N1 + 0:4N2 = mg = 12(9:81)

The solution is N1 = 101:48 N + =

A1 2 = MA t = [C 0:4(N1 + N2 )R] t [5 0:4(101:48 + 40:59)(0:08)] 3 = 1:3613 N m s

(hA )2 = I! 2 = A1

2

= (hA )2

N2 = 40:59 N

(hA )1 :

2 2 mR2 ! 2 = (12)(0:08)2 ! 2 = 0:030 73! 2 5 5 1:3613 = 0:030 72! 2 0 ! 2 = 44:3

rad/s J

18.117 θ

Position 1

1.8 m

1.2 m

Position 2

1.5 m/s

!1 = V2 T1

0.6 m Datum 2.5 rad/s

v1 1:5 = = 2:5 rad/s R 0:6

= W h = (9:81m) [1:2(1 cos )] = 11:772(1 1 1 1 1 2 = mv 2 + I! 21 = mv12 + mr ! 21 2 1 2 2 2 1 1 2 2 = m (1:5) + (0:6) (2:5)2 = 1:6875m 2 2

cos )m


V1 + T1 cos

= V2 + T 2 : 0 + 1:6875m = 11:772(1 cos )m + 0 1:6875 = 1 = 0:8567 = 31:1 J 11:772

18.118 Ox

Oy

ωA

A C

O

2 ft

C C

O

B

(mΒ + mC)vI- ω

IΒωΒ Final momenta

FBD Kinematics:

! C = ! A (C and A are rigidly connected) ! B = ! A + ! B=A = ! A 25 rad/s Final momentum diagram: +

"

20 + 12 (2! A ) = 1:987 6! A lb s 32:2 2 20 10 2 = mB kB !B = (! A 25) = 0:4313! A 32:2 12

(mB + mC )v =

+

IB ! B

+

2 IC ! C = mC kC !A =

12 32:2

3 12

10:783

2

! A = 0:023 29! A

Conservation of angular momentum:

1:987 6! A (2) + (0:4313! A

+ (hO )2 10:783) + 0:023 29! A !A

= (hO )1 = 0 = 2:43 rad/s J


Chapter 19 19.1

19.2 From Sample Problem 19.4:

aC=Q

= ( 623:6 + 141:42 y + 200:0 z )i + (720:2 141:42 x + 173:21 z )j + ( 509:0 200:0 z 173:21 y )k


Equating like components of aC = aC k = aQ + aC=Q , we get 0 = 1621:4 + 141:42 y + 200:0 z 0 = 720:2 141:42 x + 173:21 z aC = 787:0 200:0 x 173:21 y

x z

2

=

1:020 rad/s

=

2

4:99 rad/s

J J

2

=

(a) (b) (c)

4:41 rad/s

aC = 1755 mm/s

2

2

J

J

19.3

19.4 !

20i + 30j + 50k 202 + 302 + 502 = ! ( 0:3244i + 0:4867j + 0:8111k)

= !

OC

rB =

= !p

i 0:3244 0

j 0:4867 30

k 0:8111 50

vB

= !

vB

= !(0:002i + 16:220j 9:732k) p = ! 0:0022 + 16:2202 + 9:7322 = 18:916! vB 160 ) != = = 8:46 rad/s J 18:916 18:916


19.5

19.6


19.7 Let the y-axis be embedded in cone A, so that the coordinate system rotates about the z-axis with the angular velocity (a) From the velocity diagram of cone A:

z

!A

ω1 = 4 rad/s y 40o Ω ωA

= ! 1 tan 40 = 2 tan 40 = 1:6782 rad/s = !1 j k = 2j 1:6782k rad/s J

(b) = =

d! A dt

+

!A = ! _1+

!A

=xyz

7j + ( 1:6782k)

(2j

1:6782k) = 3:36i

2

7j rad/s

J

19.8


19.9

19.10

z

ω1 = 36 rad/s x

ω2 = 14 rad/s

23.2o

(a) Let the xyz-axes be attached to the frame B. !1 !2 !

= 36(i cos 23:2 + k sin 23:2 ) = 33:09i + 14:182k rad/s = 14k rad/s = ! 1 + ! 2 = 33:09i + 28:18k rad/s (valid only at the instant) J


(b) Since the xyz-axes are attached to the frame, we have ) =

=

d! dt

320k + 14k

+ !2

!=! _ 2 + !2

= !2 .

!

=B 2

(33:09i + 28:18k) = 463j + 320k rad/s

J

19.11

19.12


19.13 rA=B = 0:18i

vA=B

= ! =

rA=B =

i !x 0:18

0:2j + 0:1k m

j !y 0:2

k !z 0:1

(0:1! y + 0:2! z )i + ( 0:1! x + 0:18! z ) j + ( 0:2! x vA = vA i

0:18! y )k

vB = 2k m/s

vA = vB + vA=B Equating like components, we get vA = 2 + 0:1! y + 0:2! z 0 = 0:1! x + 0:18! z 0 = 2 0:2! x 0:18! y

(a) (b) (c)

Assuming the spin velocity to be zero, we have ! rA=B = 0: 0 = 0:18! x

0:2! y + 0:1! z

(d)

Solution of Eqs. (a)-(d) is !x vA

= 4:85 rad/s = 3:11 m/s J

! y = 5:72 rad/s

! z = 2:70 rad/s

) ! = 4:85i + 5:72j + 2:70k rad/s


19.14 (a) Let the xyz-coordinate system be attached to the arm AB. !1

=

8k rad/s

!2

=

2i rad/s

6 ! 3 = p (j + k) = 4:243(j + k) rad/s 2

Angular velocity of the disk is !

= ! 1 + ! 2 + ! 3 = 8k + 2i + 4:243(j + k) = 2i + 4:243j + 12:243k rad/s J

(b) = ! 2 + ! 3 = 2i + 4:243(j + k) rad/s Angular acceleration of the disk is = ! _ = (!) _ =xyz + =

33:9i

! =0+

i 2 2

j 4:243 4:243

k 4:243 12:243

J

2

16:0j rad/s

19.15

z P

2.4 ft

5

4

4 ft

3.2 ft

3

A

ωPQ/OA

O

Q

y

(a) Let the xyz-axes be attached to OA. ! OA ! P Q=OA !P Q rP=O

= !2 =

16k rad/s 3j + 4k = ! 1 = 25 = 15j + 20k rad/s 5 = ! OA + ! P Q=OA = 15j + 4k rad/s = 4k ft


Noting that point O is …xed, we have vP = ! P Q (b) Note that PQ

=

aP

rP=O = (15j + 4k)

4k = 60i ft/s J

= ! OA .

d! P Q dt

+ ! OA

! P Q = 0 + ( 16k)

=OA

= rP=O + ! P Q (! P Q rP=O ) = P Q rP=O + ! P Q PQ = 240i 4k+ (15j + 4k) 60i = 960j + ( 900k) + 240j =

720j

2

(15j + 4k) = 240i rad/s

900k ft/s

2

vP

J

19.16

19.17 rB=A vA

= =

0:75i + 0:9j + 0:27k m 2:1i m/s vB = vB j


= !

vB=A

=

rB=A =

(0:27! y

The equations vB 0 vB 0 0

i !x 0:75

j !y 0:9

k !z 0:27

0:9! z )i + ( 0:27! x

0:75! z )j + (0:9! x + 0:75! y )k

= vA + vB=A and ! rB=A = 0 (no spin velocity) yield = 2:1 + 0:27! y 0:9! z = 0:27! x 0:75! z = 0:9! x + 0:75! y = 0:75! x + 0:9! y + 0:27! z

The solution is !x vB

= =

0:327 rad/s 1:750 m/s J

! y = 0:392 rad/s

!z =

2:22 rad/s

19.18 z A

0.8 m

θ

B

y cos

= )

When

=

y 0:8

y

_ sin = y_ = 0:2 = 0:25 0:8 0:8 0:25 cos _= ) • = 0:25 sin sin2 • = 0:8660 rad/s2 30 : _ = 0:5 rad/s )

Let the xyz-coordinates be attached to the supporting frame. ) !1

= !1

= 1:5k rad/s ! 2 = _ i = 0:5i rad/s = ! _ =! _1+! _ 2 = (! _ 1 )=xyz + ! 1 ! 1 + (! _ 2 )=xyz + ! 1 = 0+

•i + 1:5k

0:5i =

0:866i + 0:75j rad/s

2

!2

J


19.19

19.20


19.21


19.22 (a) Let the xyz-axes be attached to arm AB, so that ! 1 = 3k rad/s

! 2 = 10i rad/s

= ! _ =! _1+! _ 2 = (! _ 1 )=xyz + ! 1 =

(0 + 0) + (0 + ! 1

= !1

! 2 ) = 3k

! 1 + (! _ 2 )=xyz + ! 1 10i = 30j rad/s

!2

J

2

(b) !

= (10i + 3k) ( 10k) = 100j in./s = ! (! rQ=B ) + rQ=B = (10i + 3k) 100j + 30j ( 10k)

rQ=B aQ=B

2

= (1000k 300i) 300i = 1000k 600i in./s = aB + aQ=B = AB! 21 j + aQ=B = 30(3)2 j + 1000k

aQ

=

600i

270j + 1000k in./s

2

600i

J

19.23 (a) Let the xyz-axes be attached to arm OA, so that !1 !

= !1 .

= 3k rad/s ! 2 = 5(j cos 35 + k sin 35 ) = 4:096j + 2:868k rad/s = ! 1 + ! 2 = 4:096j + 5:868k rad/s = ! _ = (!) _ =xyz + =

12k

!=

12:288i rad/s

2

12k + 3k

(4:096j + 5:868k)

J

(b) rA=O vA

= 0:480((j cos 35 + k sin 35 ) = 0:3932j + 0:2753k m = ! 1 rA=O = 3k (0:3932j + 0:2753k) = 1:1796i m/s

rP=A vP=A

= = = =

vP

0:2( j sin 35 + k cos 35 ) = 0:11472j + 0:16383k m ! rP=A = (4:096j + 5:868k) ( 0:11472j + 0:16383k) 1:3442i m/s vA + vP=A = ( 1:1796 + 1:3442)i = 0:1646i m/s J


19.24 (a) Let the xyz axes be embedded in the arm AB, so that ! 1 = 12k rad/s

= !1

! 2 = 9i rad/s

For arm AB: !

= ! 1 + ! 2 = 9i + 12k rad/s = ! _1+! _ 2 = (! _ 1 )=xyz + ! 1 = 0 + (0 + 12k

! 1 + (! _ 2 )=xyz + ! 1

9i) = 108j rad/s

2

!2

J

(b) = 1:2 (j cos 38 + k sin 38 ) = 0:9456j + 0:7388k m ! = (9i + 12k) (0:9456j + 0:7388k) = 11:347i 6:649j + 8:510k m/s (! rB=A ) = (9i + 12k) ( 11:347i 6:649j + 8:510k) rB=A rB=A

!

rB=A aB

=

79:79i

=

108j

212:75j

59:84k m/s

2

(0:9456j + 0:7388k) =79:79i m/s

2

= aB=A = ! (! rB=A ) + rB=A = (79:79i 212:75j 59:84k) + 79:79i =

159:6i

212:8j

59:8k m/s

2

J

19.25

z z' O

ω

β y y'

(a) Let the x0 y 0 z 0 axes be the principal axes of the rod ) Ix0

= Iz0 =

Iy0 = 0

! x0

=

hO

= Ix0 ! x0 i0 + Iy0 ! y0 j0 + Iz0 ! z0 k0 = =

0

mL2 12 ! y0 =

mL2 ! (k cos 12

! sin

! z0 = ! cos

+ j sin ) cos

mL2 0 !k cos 12

J


(b) T

= =

1 1 mL2 ! h0 = (!k) ! (k cos 2 2 12 1 mL2 ! 2 cos2 J 24

+ j sin ) cos

19.26 z

y

5m 0.1

0.1 5m

O

G

O

x

G

^ P

Let ! 2 = ! x i + ! y j + ! z k From solution of Sample Problem 19.7: (hO )2 = 0:12! x i + (0:06! y

(AO )1

2

= rG=O =

0:48i

(AO )1

2

0:045! z )j + ( 0:045! y + 0:06! z )k Z

i j 0 0:15 2:2 1:4 0:33j + 0:33k N m s

k 0:15 1:8

^ dt = P

= (hO )2

(hO )1 = (hO )2

0

Equating like components: 0:48 = 0:12! x 0:33 = 0:06! y 0:045! z 0:33 = 0:045! y + 0:06! z The solution is !x

T2

= 4:0 rad/s ) ! 2 = 4:0i = =

! y = 3:143 rad/s ! z = 3:143 rad/s 3:143j + 3:143k rad/s J

1 1 ! 2 (hO )2 = ! 2 (AO )1 2 2 2 1 [4(0:48) + ( 3:143)( 0:33) + 3:143(0:33)] = 1:997 J J 2


19.27

z

m 0.4 kg 2 0. 4 k8 m g

O g 4k

0.8 4k m g

x

m 0.4 kg 2

y (a) Iy

=

Ixy

=

4(0:8)2 (2)(0:4)2 + + 2(4)(0:4)2 = 1:7067 kg m2 3 12 2 [2(0:2)( 0:8)] + 2 [4(0:4)( 0:4)] = 1:920 kg m2

2

!= (hO )x (hO )y (hO )z

18j rad/s

= Ixy ! y = ( 1:920)( 18) = 34:56 N m s = Iy ! y = 1:7067( 18) = 30:72 N m s = 0 hO =

34:56i

30:72j N m s J

(b) T =

1 1 ! hO = ( 18)( 30:72) = 276 J J 2 2

19.28 (a) The xyz-axes are the principal axes of the disk ! = ! 1 + ! 2 = 60j + 20k rad/s

hO

1 1 mR2 = (12)(0:15)2 = 0:0675 kg m2 4 2 2Ix = 0:1350 kg m2

Ix

= Iz =

Iy

=

= Ix ! x i + Iy ! y j + Iz ! z k = 0i + 0:1350(60)j + 0:0675(20)k = 8:10j + 1:35k N m s J


(b) T

= =

1 1 ! hO + mv2 2 2 1 1 [60(8:10) + 20(1:35)] + (12)(20 2 2

0:4)2 = 641 J J

19.29

z

1

. . . b/2 O b b 3 b/2

y

2

With ! x = ! y = 0, ! z = ! Eqs. (19.17a) are equivalent to hO = Ixz

=

Iyz

= m( b)

Iz

Ixz !i

Iyz !j + Iz !k

0 b 2

+ m(b)

2

= m ( b) + mb2 +

b 2

+0=

mb2

8 1 (2m)(2b)2 = mb2 12 3

8 hO = mb2 ! j + k 3

J

From Eq. (19.25) we get T =

4 1 Iz ! 2 = mb2 ! 2 J 2 3

19.30

0.3 m

Position 2

A

x

0.4 m

ω y m 0.5 Position 1

B

δ

ωrod


In Position 2, the spring is undeformed and the rod is rotating about B with the angular velocity vB 0:15! ! ro d = = = 0:3! 0:5 AB . T2

= = =

1 1 (Idisk )O ! 2 + (Iro d )B ! 2ro d 2 2 1 1 1 1 mdisk R2 ! 2 + mro d L2 ! 2ro d 2 2 2 3 1 1 1 1 (6)(0:15)2 ! 2 + (2)(0:5)2 (0:3!)2 2 2 2 3

0:04125! 2 1 2 1 = k = (2000)(0:1)2 = 10 N m 2 2

= V1

T1 + V1 !

= T 2 + V2 0 + 10 = 0:04125! 2 + 0 = 15:57 rad/s J

19.31


19.32


19.33

19.34

O

z 20o L=1 .5'

Instant axis

ω1

3

1

y

R=

10

0.5'

ω0

C !0 !1 !

ω

= 4( j sin 20 + k cos 20 ) = 1:3681j + 3:759k rad/s = !1 j 3j + k p = ! = ( 0:9487j + 0:3162k) ! 10

! ( 0:9487j + 0:3162k) !

= !0 + !1 = ( 1:3681

! 1 ) j + 3:759k


Equating like terms: 0:9487! 0:3162!

= =

1:3681 3:759

!1

The solution is ! = 11: 888 rad/s

! 1 = 9: 910 rad/s

) ! = 11:888 ( 0:9487j + 0:3162k) = Iy Iz

11:278j + 3:759k rad/s

1W 2 1 6 R = (0:5)2 = 0:02329 slug ft2 2 g 2 32:2 W 2 W 1 2 6 0:52 = Iz + L = R + L2 = + 1:52 g g 4 32:2 4 =

= 0:4309 slug ft2

Since the xyz-axes are the principal axes at O, we have hO

= Ix ! x i + Iy ! y j + Iz ! z k = 0:02329( 11:278)j + 0:4309(3:759)k = 0:263j + 1:620k lb ft s J

19.35 (a) OB rB=A vA

vB

= vA + !

) vB = 0 = 0 =

= = =

p 1:82 1:22 0:92 = 0:9950 m 0:9950i 1:2j 0:9k m 1:4j m/s vB = vB i

rB=A

vB i =

1:4j +

i !x 0:9950

j !y 1:2

0:9! y + 1:2! z 0:9! x + 0:9950! z 1:4 1:2! x 0:9950! y

k !z 0:9 (a) (b) (c)

Since the moment of inertia of AB about its axis is negligible, the spin velocity of AB is irrelevant. We assume the spin velocity to be zero, i.e., ! rB=A = 0

0:9950! x

1:2! y

0:9! z = 0

(d)

The solution of (a)-(d) is vB !y

= 1:6884 m/s ! x = 0:3889 rad/s = 0:4690 rad/s ! z = 1: 0553 rad/s 12(1:8)2 (0:3889i 0:4690j + 1: 0553k) 12 1:260i 1:520j + 3:419k N m s J

h = I! = =


(b) v

=

T

= = =

vA + vB 1:4j + 1:6884i = = 0:8442i 0:7j m/s 2 2 1 (I! 2 + mv 2 ) 2 1 12(1:8)2 (0:38892 + 0:46902 + 1: 05532 ) + 12(0:84422 + 0:72 ) 2 12 9:62 N m J

19.36

19.37 z

ω1

A R O

2R

ω

ω2

B

ω2 = 2ω1

y

ω

2

1

ω1

! = ( 2j + k)! 1 For the disk: v Iy Ix

=

2R! 1 = 2(0:5)! 1 = ! 1 1 1 14 = mR2 = (0:5)2 = 0:05435 slug ft2 2 2 32:2 1 = Iz = Iy = 0:02717 slug ft2 2


1 1 Iy ! 2y + Iz ! 2z + mv 2 2 2 1 1 14 2 2 = 0:05435( 2! 1 ) + 0:02717! 21 + ! = 0:3397! 21 2 2 32:2 1 = C0 (4 ) = 0:35(4 ) = 4:398 lb ft

T2

U1 U1

=

2 2

= T2

4:398 = 0:3397! 21

T1

! 1 = 3:60 rad/s J

0

19.38

z A x

4.8 m

vA

4m Position 2

vB = 0 B m 53 2.6 y

Kinematics (position 2) Note that in position 2 collar B has reached the limit of its travel. Therefore, vB = 0. rA=B = 2:653i 4j m

vA = vB + !

rA=B

vA k = 0 +

i !x 2:653

j k !y !z 4 0

Expanding the determinant and equating like terms:

4! x

4! z 2:653! z 2:653! y

= 0 = 0 = vA

Setting the spin velocity of the bar to zero gives the fourth equation: ! rA=B = 0

2:653! x

4! y = 0

The above equations yield !x

= 0:173 62vA ! y = 0:115 16vA !z = 0 2 2 ) ! 2 = 0:173 622 + 0:115 162 vA = 0:043 406vA

Conservation of energy


Using the xy-plane as the datum for potential energy: V1

=

V2

=

T2

= =

1 (1:2mg) = 0:6(9:81)m = 5:886m T1 = 0 2 0 1 1 mv 2 + I! 2 2 2 1 vA 2 1 m(4:8)2 2 2 + m (0:043 406vA ) = 0:166 67mvA 2 2 2 12 V1 + T1 = V2 + T2 2 5:886 + 0 = 0 + 0:166 67vA vA = 5:94 m/s J

19.39


19.40 Ix

=

Iy

=

Iz

=

1 1 300 mb2 = (4)2 = 12:422 slug ft2 12 12 32:2 1 300 1 ma2 = (3)2 = 6:988 slug ft2 12 12 32:2 1 300 2 1 m(a2 + b2 ) = (3 + 42 ) = 19:410 slug ft2 12 12 32:2 (hO )2

(AO )1

2

= rAO

= Ix ! x i + Iy ! y j + Iz ! z k = 12:422! x i + 6:988! y j + 19:410! z k Z k P^ dt = ( 1:5i + 2j) ( 20k) = 40i

(AO )1

2

= (hO )2

(hO )1 = (hO )2

30j lb ft s

0

Equating like components, we get 40 = 12:422! x 30 = 6:988! y 0 = 19:410! z !=

3:22i

) ! x = 3:22rad/s ) !y 4:29 rad/s ) !z = 0 4:29j rad/s J

19.41 (a) L1 2 = mv2 mv1 0:18i = 2v2 0 v2 =

0:09i m/s J

(b) 1 1 mR2 = (2)(0:09)2 = 0:0081 kg m2 2 2 1 = Iz = mR2 = 0:00405 kg m2 4 = Ix ! x i + Iy ! y j + Iz ! z k = 0:0045(2! x i + ! y j + ! z k)

Ix

=

Iy h2 A1

2

= rA=G L1 2 = 0:09(j cos 30 = 0:0081j + 0:01403k N m s

k sin 30 )

( 0:18i)

A1 2 = h2 h1 = h2 0 0:0081j + 0:01403k = 0:0045(2! x i + ! y j + ! z k) !x = 0 ! y = 1:8 rad/s ! z = 3:118 rad/s


T

= = =

1 1 1 1 h2 ! 2 + mv 22 = A1 2 ! 2 + mv 22 2 2 2 2 1 1 [0:0081(1:8) + 0:01403(3:118)] + (2)(0:09)2 2 2 0:0373 N m J

19.42

19.43


19.44


19.45


19.46


0.7 2 ft

19.47

C

1.6 ft

RC

B

ωx

60o

0.7 2 ft

y

x A D

1.6 ft

60o

−ωy

ω

RD

FBD

Let the xyz-axes be attached to bar AB. !x !y !z

= ! cos 60 = 5 cos 60 = 2:5 rad/s = ! sin 60 = 5 sin 60 = 4:330 rad/s = 0

For bar AB (CD does not contribute to dynamic reactions): Ix

=

0

Iy

= Iz =

1 1 mL2 = 12 12

14 32:2

(1:44)2 = 0:07513 slug ft2

Euler’s equation (the other two equations are trivially satis…ed): Mz = ! x ! y (Iy Ix ) RC (1:6) RD (1:6) = 2:5( 4:330)(0:07513) RC + RD = 0:5083 F = 0: RC RD = 0 ) RC = 0:254 lb " J

RD = 0:254 lb # J

19.48 Let the xyz-axes be embedded in the plate. !x = 0 Ix =

mb2 12

!y = Iy =

! 0 sin ma2 12

! z = ! 0 cos Iz =

m(a2 + b2 ) 12


C O R FBD Since the xyz-axes are the principal axes of inertia, we apply Eqs. (19.33). With !_ x = !_ y = !_ z = ! x = 0, these equations become Mx

= ! y ! z (Iz

My Mz

= 0 = 0

Iy )

) Cx = ( ! 0 sin )(! 0 cos )

) Cy = 0 ) Cz = 0

mb2 12

1 mb2 ! 20 sin cos i J 12 F = ma )R=0 J

C=

19.49 Let the xyz-axes be embedded in the plate. !x = 0

!y =

! 0 sin

! z = ! 0 cos

2

Iz

Ix

=

Iy

=

Iz

=

Iy

=

mb2 m b 5 + mb2 = 12 12 2 48 ma2 12 2 m(a2 + b2 ) m b ma2 5 + = + mb2 12 12 2 12 48 5 mb2 48


C O R FBD Since the xyz-axes are the principal axes of inertia, we can apply Eqs. (19.33). With !_ x = !_ y = !_ z = ! x = 0, these become Mx

= ! y ! z (Iz

My Mz

= =

) Cx = ( ! 0 sin )(! 0 cos )

Iy )

5 mb2 48

) Cy = 0 ) Cz = 0

0 0

5 mb2 ! 20 sin cos 48

C=

i J

ω0 z

βm

aA

A

y

b cosβ 2 aA =

b cos ! 20 (j cos 2

+ k sin )

F = mA aA R

= mA aA = =

m 12

b cos ! 20 (j cos 2

1 mb! 20 cos (j cos 24

+ k sin )

+ k sin ) J


19.50 B FBD

L mg R Ay z ω

A

C Az = mg

x

O

y

Rotation about …xed axis: Mx C

= Iyz ! 2z =

C = myz! 2 = m( R)

L 2

!2

1 mRL! 2 J 2

19.51

y

x

RA A

0.012 m G

0.4 8m

B

0.3 2m

RB

z

FBD (only horizontal forces are shown) We have rotation about a …xed axis. Equations (19.39) apply. Mx = Iyz ! 2z

My =

Ixz ! 2z

With origin of xyz-axes at A: Iyz Ixz My

= myz = 20(0)(0:48) = 0 = mxz = 20(0:012)(0:48) = 0:1152 kg m2 = Ixz ! 2z 0:8RB = 0:1152(18)2 RB = 46:7 N J

With origin of xyz-axes at B: Iyz Ixz My

= myz = 20(0)( 0:32) = 0 = mxz = 20(0:012)( 0:32) = 0:0768 kg m2 = Ixz ! 2z 0:8RA = ( 0:0768)(18)2 RA = 31:1 N J


19.52 The xyz-axes are attached to the rod. Because we have rotation about a …xed axis, Eqs. (19.39) apply. Iyz = 0

Ixz = m

y

A

Ax

ma2 a (a) = 2 2

a O 1

a a 2

x

B

Bx

a

z

FBD (only the dynamic reactions are shown) Ixz ! 20

My = max =

mx1 ! 20

Ax a

mx2 ! 20 =

a m ! 20 2

ma! 20 J

ma2 2 ! 2 0

ma! 20 =

3 ma! 20 2

3 ma! 20 2

Ax + Bx =

Fx = max Ax =

Bx a =

1 ma! 20 J 2

Bx =

19.53 5m 0.0 W1 y

z

x 0 .1 m R B W2 0 RA B .1 m

A 0.8 N. m FBD

m 0.1 W1

5m 0.0


m1

=

m2

=

W1

=

Ixz

=

Iyz

=

0 X

mi yi zi = 0:3(0:1)( 0:025) + 0:3( 0:1)( 0:075)

0:0015 kg m2 X = (Iz )i + mi d2i

= Iz

m 1:8 = = 0:3 kg (0.1 m 0.05 m plate) 6 6 4m 4(1:8) = = 1:2 kg (0.2 m 0.1 m plate) 6 6 0:3(9:81) = 2:943 N W2 = 1:2(9:81) = 11:772 N

=

1 1 (1:2)(0:2)2 + 2 (0:3)(0:1)2 + 0:3(0:12 + 0:052 ) 12 12

=

0:012 kg m2

Rotation about …xed axis: Mz = Iz !_ z My =

0:8 = 0:012!_ z

Iyz !_ z (2W1 + W2 )(0:05) (2 2:943 + 11:772) (0:05)

Fx = 0 (2

2

!_ z = 0:1RB 0:1RB RB

66:67 rad/s = = =

(2W1 + W2 ) + RA + RB = 0 2:943 + 11:772) + RA + 7:83 = 0 ) RA = 9:83i N J

J

Iyz !_ z 0:0015( 66:67) 7:83 N

RA = 9:83 N

RB = 7:83i N J

19.54 x y RA . z

RB B m 0.1

A

FBD (only horizontal forces are shown) From solution of Problem 19.53: Iyz = 0:0015 kg m2

Izx = 0


Rotation about a …xed axis: Mx Fy

= Iyz ! 2z 0:1RB = 0:0015(10)2 RB = 1:5 N = 0 RA + R B = 0 RA = RB = 1:5 N ) RA =

1:5j N J

RB = 1:5j N J

19.55 x A y

RA

1.2 5'

B FBD 2 lb D

0.7 5' RB

C ' . 5 0.7 ω z ' 5 0.7 3 lb

Use Eqs. (19.37) with ! z = 0 , !_ z = !_ Ixz

=

0

Iz

=

1 1 mL2 = 3 3

Mz = Iz !_

Iyz = myz =

3(0:75)

My = 1:25RB 2 =

RA + 2:80 ) RA =

3 32:2

(1:5)2 = 0:06988 slug ft2

2(1:5) = 0:06988!_

!_ =

2

10:733 rad/s

Iyz !_ 1:25RB 3(2) + 2(2) = 0:139 75!_ 0:139 75( 10:733) RB = 2:80 lb

Fx

= max

1:0

=

1:050i lb J

3 (0:75)(2) = 0:13975 slug ft2 32:2

RA + R B

3 L !_ 32:2 2

1:0 =

3 (0:75)( 10:733) 32:2 RB = 2:80i lb J

RA = ! _ =

1:050 lb

10:733k rad/s

2

J

19.56 C

z R

FBD

O

x

y mg


Use modi…ed Euler equations with the xyz-axes be attached to the fork !1 ! Ix Iy

= 180i rad/s ! 2 = 40k rad/s = ! 2 = 40k rad/s = ! 1 + ! 2 = 180i + 40k rad/s = mR2 = 0:15(0:08)2 = 9:6 10 4 kg m2 1 = Ix = 4:8 10 4 kg m2 2

Mx = Iz My = Ix Mz = Iy

y !z z !x

Iy Iz

z !y x !z

Cx Cy

x !y

Ix

y !x

Cz

)C=

= 0 0=0 = (9:6 10 4 )(40)( 180) = 6:91 N m = 0 0=0

0

6:91k N m J

19.57

Az Ay A

Ax

0.7 5 ft

z

G

_ an = 0.75ω12 FBD

25 lb

x

y

Let the xyz-axes be attached to the shaft AG. ! = 30j + ! 1 k Iy Ix

= !1 k

1W 2 1 25 R = (1)2 = 0:3882 slug ft2 2 g 2 32:2 1 = Iz = Iy = 0:1941 slug ft2 2 =

First equation of Eqs. (19.34) with G as the moment center: Mx 0:75Az

= Iz y ! z Iy z ! y = 0 (0:3882) ! 1 (30)

(a)


Fz = maz

Az

25 = 0

Az = 25 lb

Equation (a) becomes ! 1 = 1:610 rad/s J

0:75(25) = (0:3882) ! 1 (30)

19.58 z O

ω0

m 0.2

ωy 25o A

Oy y O

m 0.1

ωz

z Oz

y

9.81m

G FBD 25o A

Let the xyz-axes be attached to the bar. These axes are the principal axes of the bar at O: ! Ix Iy

= ! 0 (j sin 25 + k cos 25 ) = ! 0 (0:4226j + 0:9063k) 1 1 = mL2 = m(0:2)2 = 0:013 333m 3 3 = Ix = 0:013 333m Iz = 0

Euler equation: Mx = ! y ! z (Iz Iy ) (9:81m) (0:1 sin 25 ) = (0:4226! 0 ) (0:9063! 0 )(0 ! 0 = 9:01 rad/s J

0:013 333m)

19.59


19.60 ωy y A

x 50o

T

B FBD Ay

Ax A

0.9 m

ωx

0.9 m

ω

20(9.81) N 50o

Let the xyz-axes be attached to the clevis so that the z-axis coincides with the pin !x !y Ix

= ! sin 50 = 6 sin 50 = 4:596 rad/s = ! cos 50 = 6 cos 50 = 3:857 rad/s !z = 0 1 1 = 0 Iy = Iz = mL2 = (20)(1:8)2 = 21:60 kg m2 3 3

Euler equation:

+

T (1:8 sin 50 )

Mz = ! x ! y (Iy Ix ) 20(9:81)(0:9 cos 50 ) = (4:596)(3:857)(21:60 T = 360 N J

0)


19.61 A

L/4 H

ω

A

=

G

mrω2

L/ 2

V

L/ 2

45o G mg y'

x'

FBD

B

From FBD-MAD: r= Fx0 Fy 0

y ω x o 45 G

r MAD

B

L sin 45 2

A

B

L = 0:103 55L 4

H = mr! 2 = 0:103 55mL! 2 mg = 0 V = mg

= max0 = 0 V

Euler equation (xyz-axes embedded in bar AB): Mz

= ! x ! y (Iy Ix ) L sin 45 + V 2 =

(! sin 45 ) (! cos 45 ) mg

=

L cos 45 2 1 mL2 0 12

H

L sin 45 2

0:103 55mL! 2

L cos 45 2

1 mL2 ! 2 sin 45 cos 45 12

The solution is ! = 2:126

r

g = 2:126 L

r

32:2 = 8:53 rad/s J 2

19.62 CA

RA

A x

b/2

b FBD b

G

mg

z

y


Mass center G of the disk travels a circular path of radius rG=A about A. Hence the (normal) acceleration of G is a = !2

(! 2

= !0 i

rG=A ) = ! 0 i

(! 0 bk

! 0 bj) =

F = ma:

RA =

[! 0 i ! 20 bj

(bj + bk)] ! 20 bk

m! 20 b(j + k) J

Note that RA is directed along the line AG; thus its moment about G is zero. Use modi…ed Euler equations with xyz-axes attached to arm AG )

= !2 i = !0 i = ! 2 i + ! 1 k = ! 0 (i + 2k) (angular velocity of disk)

!

Modi…ed Euler equations (note that ! _ = 0): Mx

= Iz

My

=

Mz

= Iy )

y !z

Iz

Iy

x !z x !y

CA =

z !y :

+ Ix

(CA )x = 0

z !x :

Ix

y !x : 2 2 m! 0 b

8

m(b=2)2 (! 0 )(2! 0 ) = 2

(CA )y =

mb2 ! 20 4

(CA )z = 0

j J

19.63 4 rad/s Q

ωx

θ ωz

x

12 rad/s

P

z Q

R

z

CA θ x

C

D

P

FBD

CD

Let the xyz-axes be attached to P Q (the y-axis coinciding with the shaft of the motor at B) ! x = 4 sin rad/s

!y =

) !_ x = (4 cos ) _

12 rad/s !_ y = 0

! z = 4 cos rad/s !_ z = ( 4 sin ) _

Substituting _ = 12 rad/s, we get !_ x = 48 cos rad/s

2

!_ y = 0

!_ z =

1 1 mL2 = (16)(2:4)2 = 7:680 kg m2 12 12 Euler equations with D as the moment center: Ix = Iy =

Mx Cx

2

48 sin rad/s

= Ix !_ x + ! y ! z (Iz Iy ) = 7:680(48 cos ) + ( 12)( 4 cos )(0

Iz = 0

7:680) = 0


My Cy

= Iy !_ y + ! z ! x (Ix Iz ) = 7:680(0) + (4 cos )(4 sin )(7:680 Mz Cz ) CD CA

= Iz ! z + ! x ! y (Iy Ix ) = 0 + (4 cos )( 12)(7:680

0) = 122:88 sin cos

7:680) = 0

= Cy = 122:88 sin cos = 61:4 sin 2 N m J = Cx sin + Cz cos = 0 J

19.64

19.65 The no precession condition is ) Iz = I

= Iz =I = 1

1 1 mR2 = m(3R2 + h2 ) 2 12

p h = 3 J R


19.66

Z,y

.

φ

z R

6 ft

O

x

ψ.

mg A FBD _ =

F

_ = _R = L

16 rad/s

0.8 ft

0:8 6

16

=

2:133 rad/s

Since the precession axis is perpendicular to the spin axis, Eq. (19.48) applies: = Iz _ _

Mx (F

4)(6)

=

1 2

(F

4 32:2

mg)L =

1 mR2 _ _ 2 F = 4:23 lb J

(0:8)2 ( 16)( 2:133)

19.67 z x

Z,y .

φ

mg B

ψ.

O 0.5 m

0.5 m

F A mAg

FBD's

O

0.5 m

R

F A

Since the precession axis is perpendicular to the spin axis, Eq. (19.48) applies to the disk: 1 Mx = Iz _ _ (F mA g) 0:5 = mA R2 _ _ 2 where F is the force applied by the shaft on the disk. Substituting _ = 2 rad/s, and _ = 40 rad/s, we get [F

25(9:81)] 0:5 =

1 (25)(0:3)2 (2)( 40) 2

F = 65:25 N

From the FBD of the shaft: Mx = 0

0:5mg

0:5F = 0

m=

F 65:25 = = 6:65 kg J g 9:81


19.68

z Z 5o

β ω

y

1 Iz 1 =2 tan = tan = tan 2 I = tan 1 (2 tan ) = tan 1 (2 tan 5 ) = 9:925

(19.52):

= )

_ =1

(10.53c): )

_ =

_ cos = 1

2 2

(10) cos 5 =

4:98 rad/s

4:98k rad/s J _

( 4:98) = = 5:056 rad/s (1 ) cos (1 2) cos 9:925 = !(j sin + k cos ) = 5:056(j sin 9:925 + k cos 9:925 ) = 0:872j + 4:980k rad/s J

(19.53a):

!=

! )!

19.69 x

_

=

_

=

Z,y C φ.

z

ψ.

15 000(2 ) = 1570:8 rad/s 60 v 600(5280) 1 = = 0:083 33 rad/s R 3600 2(5280)

Since the precession axis is perpendicular to the spin axis, Eq. (19.48) applies to the rotor: Mx C

= Iz _ _ 500 = mk 2 _ _ = (1:2)2 (0:083 33)(1570:8) = 2930 lb ft J 32:2


19.70 z

mg Z

FBD

L/2 L/2

RO = 2mg y O mg

θ Iz

Iz

=

2 "

1 mR2 2

R

= mR2

I

=

1 2 mR2 + m 4

I

=

1 mR2 1 2

L 2

2

#

=

1 L2 mR2 1 + 2 2 R

L2 R2

We have steady precession with _ = 2 rad/s, _ = 3 rad/s and (19.46): +

Mx = (Iz 0

=

1 mR2 1 2

0

=

0:8988 1

= 32

2 I) _ sin cos + Iz _ _ sin L2 (2)2 sin 32 cos 32 + mR2 (2)(3) sin 32 R2 L L2 + 3:180 = 2:13 J R2 R

19.71


19.72

Z

y

L

θ

B1 B B2

β

FBD

R A z mg C

N

Let the xyz-axes be attached to disk. The motion is steady precession with _ = ! = 3 rad/s = tan

1

R = tan L

1

3 = 26:57 6

Iz

=

1 1 mR2 = 2 2

I

=

1 1 mR2 + mL2 = 4 4

=

49:50

6 32:2

3

10

3 12

mg(L sin )

6

N

= =

6 sin 26:57 12

5: 823 10 2:12 lb J

3

= 153:4

2

= 5: 823 3 12

2

+

10

3

slug ft2

6 32:2

6 12

2

slug ft2

L cos

N

= 180

6 32:2

Mx +

_ = _ =0

N 49:50

=

(Iz

2 I) _ sin cos + Iz _ _ sin

=

(Iz


6=12 cos 26:57 10

3

(3)2 sin 153:4 cos 153:4 + 0

19.73 The xyz coordinate system is embedded in the sphere-rod assembly with the origin at the …xed point O. Iz =

2 mR2 5

I=

2 47 mR2 + m(3R)2 = mR2 5 5


Oz

Oy O

2R

θ R

mg FBD Equation (19.46): 2 (Iz I) _ sin cos + Iz _ _ sin 45 2 mR2 _ sin cos + mR2 _ _ sin mg(3R sin ) = 5 5 2 3g = 9R _ cos + 0:4R _ _ 3(9:81) + 0:4(0:08)(60)(960) 3g + 0:4R _ _ = cos = = 0:7225 2 9(0:08)(60)2 9R _ = 43:7 J

Mx

=


#19.74


19.75

Z, y 1.25 ft

1.25 ft

A

Bz

G

RA

RB

380 lb

'_

=

1:2 rad/s

_ = 160 rad/s

Iz

=

1 1 mR2 = 2 2

380 32:2

Fy = 0

1:25RA

+"

Mx 1:25RB R A RB

5 12

2

RA + RB

= 1:0244 slug ft2 380 = 0

(a)

= Iz '_ _ = 1:0244(1:2)(160) = 157:35

(b)

Solution of (a) and (b) is RA = 269 lb J

RB = 111:3 lb J


19.76

19.77

Z

z

mm 48

G 30o

y

0.5(9.81) N

Oy O Oz

FBD

We have steady precession. Equation (19.46) is

(0:5

Mx

=

9:81) (0:048 sin 30 )

=

0:11772

=

(Iz


2 (5 10 4 20 10 4 ) _ sin 30 cos 30 +(5 10 4 )(120) _ sin 30

6:495

10

4 _2

+ 0:03 _

The two solutions are _ = 4:33 and 41:9 rad/s J


19.78 (a) The third equation of Eqs. (19.45) is Mz = Iz !_ z where Iz = 5

10

4

! z = Iz !_ z

!_ z =

Iz

!z

kg m2 (see Prob. 19.77). The solution is ! z = Ce

( =Iz )t

The initial condition ! z = (! z )0 at t = 0 yields C = (! z )0 . ) ! z = (! z )0 e

( =Iz )t

J

(b) When ! z = 0:5(! z )0 , then 0:5 t

= e

( =Iz )t

=

(ln 0:5)

Iz Iz

=

t = ln 0:5

(ln 0:5)

5 10 4 = 2:77 s J 1:25 10 4

19.79

19.80


#19.81


19.82


19.83

19.84


19.85

z

Z 25 o

80 rad/s

β

.

(a) We have torque-free motion with tan

= )

ω

y

= Iz =I = 2:105. From Eq. (19.52):

tan = 2:105 tan 25 = 0:9816 = 44:47 J

Equation (19.53c): _

=

_

=

(1 (1

) _ cos _ ) cos

=

2:105( 80) = 168:2 rad/s J (1 2:105) cos 25

=

( 80) cos 25 = 65:6 rad/s J (1 2:105)

(b) Equation (19.53a): _

= !(1

!

=

) cos _

(1

) cos

19.86


19.87


Chapter 20 20.1 (a) x(t) E (b)

r

v(t)

r

250 = E sin(pt + ) p= = 5 rad/s 10 q p x20 + (v0 =p)2 = 0:042 + ( 0:08=5)2 = 0:0431 m J =

=

tan

1

x0 p v0

= tan

1

= x(t) _ = pE cos(pt + ) 1 t= p

)

2

1 = 5

k = m

0:04(5) = 0:08

1:1903 rad

v(t) = 0 when pt + 2

=

2

+ 1:1903 = 0:552 s J

20.2 (a) x = E sin(pt + ) where ) f

=

E=

s

x20 +

v0 p

2

1:5 v0 =p = 153:90 rad/s 0:0122 0:0072 E 2 x20 153:90 p = = 24:5 Hz J 2 2

p= p

(b) tan

= x =

x0 p 0:007(153:90) = = 0:7182 v0 1:5 0:012 sin (153:90t + 0:6228) m J

= 0:6228 rad

20.3


20.4

20.5


20.6

20.7


20.8


20.9

θ

ft 1.5

1.8 lb 8(1.0)θ

Oy Ox

.

ft 1.0

ft 1.5 FBD

2.4 lb

= IO • 2:4 1:8 (2:5)2 + (1:5)2 • 8(1:0 )(1:0) = 32:2 32:2 7:10 = 0:5171• MO

1:8(2:5 )

2:4(1:5 )

This equation has the form m• + k = 0. Therfore, r r k 7:10 p= = = 3:71 rad/s J m 0:5171

20.10

Lθ T x FBD

Lθ T

Fx = max For small , we have sin

mx.. MAD

+ !

2T sin = m• x

tan = x=L, so that the equation of motion becomes 2T

x = m• x L

x •+

2T x=0 J mL

Since this equation has the formpx • + p2 x = 0, the motion is simple harmonic with the circular frequency p = 2T =mL.


20.11

.. m Lθ m g kbθ FBD L

.2

m Lθ

=

MAD

θ

bOx

Oy

Use small angle approximations: sin

O

and cos

( MO )FBD = ( MO )M AD

+

mg(L ) • + kb

If

1 kb (b)

O = mL•(L)

2

mgL mL2

=

0

= 1:2 s, then p = 2 = = 2 =1:2 = 5: 236 rad/s kb2 mgL mL2 2 2 p mL + mgL kb2 = 0 2 (5: 236) (0:3)L2 + 0:3(9:81)L 300(0:08)2 = 0 8:225L2 + 2:943L 1:92 = 0 ) p2 =

The positive root is L = 0:336 m J

20.12

a mg

θ

O b R

k(bθ +∆) =

FBD

Use small angle approximations: sin ( MO )FBD = ( MO )M AD where

and cos +

O a .2 maθ .. maθ MAD 1

mga + k(b +

)b =

ma2 •

is the static elongation of the spring. Substituting the static equilibrium


equation

mga + k kb2 )p

b = 0, we get 2 • + kb ma2 • =0 ma2 s r k b 20 12 = = = 7:583 rad/s ma 2:8=32:2 24

=

=

2 2 = = 0:829 s J p 7:583

20.13

#20.14


#20.15 W Water level after displacement x Water level at equilibrium

∆

ρ ρw

Area = A h =

Ww FBD

.. (W/g)x MAD


20.16

20.17

12(2)(θ + θs) Ax

A Ay

2 ft

θ

3 − 8(4)(θ + θs) B

2 ft

C

FBD

= IA • 3 12(2)( + s )(2) + [3 8(4)( + s )] 4 = (4)2 • 32:2 12 176( + s ) = 1:4907• MA

The static displacement s is obtained by setting ft . Therefore, the equation of motion becomes

= • = 0 yielding

s

= 12=176

1:4907• + 176 = 0


r

p 1 ) f= = 2 2

k 1 = m 2

r

176 = 1:729 cps J 1:4907

20.18 (a) s

p

=

r

ccr

=

2mp = 2

=

6 c = = 0:3643 < 1 ccr 16:472

k = m

14 12 = 20:40 rad/s 13=32:2 13 32:2

(20:40) = 16:472 lb s/ft ) Underdamped.

(b) ln

xn+1 xn xn+1 xn

= = e

2 p 1

2:458

2

=

2 (0:3643) p = 1 0:36432

2:458

= 0:0856 J

20.19 Damping is critical if c =1 = p 2 km

s p c = 2 km = 2 (16

20.20 c = p =1 2 km

12)

20 32:2

= 21:8 lb s/ft J

p p ) c = 2 km = 2 80(3) = 31:0 N s/m J

20.21 = )

20.22

q 2 2 2 = 0:5 p ) 1 = 0:5 0:5 d 2 p p 1 p p = 0:8660 c = 2 km = 2(0:8660) 80(3) = 26:8 N s/m J

p

=

x(t)

=

r

r k 80 = = 5:164 rad/s m 3 (A1 + A2 t)e pt x(t) _ = p(A1 + A2 t)e

pt

+ A2 e

pt


Initial conditions: = 0:01 m ) A1 = 0:01 m = 0:2 m/s ) pA1 + A2 = 0:2 5:164(0:01) + A2 = 0:2 A2 = 0:14836 m/s

x(0) x(0) _

) x(0:1) = [0:01

5:164(0:1)

0:14836(0:1)] e

=

2:89

10

3

m J

20.23 (a) Equivalent spring constant is k=

1 1 = 90 lb/ft = 1=k1 + 1=k2 1=225 + 1=150 m=

p

=

1:2 = 0:03727 slugs 32:2

r

r k 90 = = 49:14 rad/s m 0:03727 2(2:8) c = = 1:5288 overdamped J 2mp 2(0:03727)(49:14)

= (b) +

q

q

2

1 p

=

1:5288 +

2

1 p

=

1:5288

) x(t) = A1 e

p 1:52882 p 1:52882

18:30t

+ A2 e

1 49:14 =

18:30 s

1

1 49:14 =

132:0 s

1

132:0t

Initial conditions: 1:5 = 0:125 ft A1 + A2 = 0:125 ft 12 x(0) _ =0 18:30A1 + 132:0A2 = 0

x(0) =

The solution is A1 = 0:14511 ft, A2 = ) x(t) = 0:14511e

0:02011 ft

18:30t

0:02011e

132:0t

ft J


20.24

2x

. 2cx

bx FBD

Ox

kx

b Oy O b x

MAD

=

O b

.. mx

mg

(MO )FBD = (MO )M AD (kx)b + 2cx(2b) _ + mgx = mxb • mg m• x + 4cx_ + k + x = 0 b

p

= =

d

20.25

=

r k + mg=b 1800 + 3(9:81=0:2) = = 25:48 rad/s m 3 4c 4(30) = = 0:7849 2mp 2(3)(25:48) 2 2 p p = 0:398 s J = 2 25:48 1 0:78492 p 1

r


20.26


20.27

20.28


20.29

h L From Eq. (20.31) the amplitude the relative motion (displacement of trailer relative to the wheels) is (!=p)2 Z=Y 1 (!=p)2 where Y

=

p2

=

1:5=12 2 v h = = 0:0625 ft != 2 2 L 2k 2(240 12) 2 = = 231:8 (rad/s) m 800=32:2

The amplitude of the absolute motion is X =Z +Y =Y

(!=p)2 +Y = 1 (!=p)2 1

Y (!=p)2

The displacement of the trailer is x(t) = X sin !t =

1

Y sin !t (!=p)2

x •(t) =

Y !2 sin !t 1 (!=p)2

The wheels are about to leave the road if j• xjmax = g, i.e., Y !2 =g 1 (!=p)2 If !=p < 1: Y !2 1 (!=p)2

= g s

!

=

r

v

=

!L 12:664(6) = = 12:07 ft/s 2 2

g = Y + g=p2

32:2 = 12:644 rad/s 0:0625 + 32:2=231:8


If !=p > 1: Y !2 (!=p)2 1

= g s

!

=

r

v

=

!L 20:53(6) = = 19:60 ft/s 2 2

g g=p2

Y

=

32:2 = 20:53 rad/s 32:2=231:8 0:0625

Wheels lose contact with the road if 12:07 ft/s < v < 19:60 ft/s J

20.30

h L From Eq. (20.31) the maximum relative displacement (displacement of trailer relative to the wheels) is zmax = jZj = Y

(!=p)2 1 (!=p)2

where Y

=

!

=

p2

=

h 1:5=12 = = 0:0625 ft 2 2 v 12(5280=3600) 2 =2 = 18:431 rad/s L 6 2k 2(240 12) 2 = = 231:8 (rad/s) m 800=32:2 18:4312 = 1:4655 231:8 1:4655 = 0:0625 = 0:196 76 ft 1 1:4655 ) (!=p)2 =

zmax

But jZj is also the deformation of each spring, measured from the static equilibrium position. Thus the maximum force in each spring is Fmax

= =

W + kzmax 2 800 + (240 2

12)(0:196 76) = 967 lb J


#20.31


20.32


20.33

20.34 r

p

=

Z

= Y

k = m

s

36 = 43:95 rad/s 0:6=32:2

(!=p)2 = 0:6 1 (!=p)2

0:8283 1 0:8283

! p

2

40 43:95

=

2

= 0:8283

= 2:89 in. J

20.35 With

= 0, Eq. (20.26) is X=

1

P0 =k = (!=p)2 1

P0 =k ! 2 =(k=m)

)k=

P0 + m! 2 X

If !=p < 1 (X = +7:5 mm): 0:25 + 4(62 ) = 177:3 N/m J 0:0075 7:5 mm): k=

If !=p > 1 (X =

k=

0:25 + 4(62 ) = 110:7 N/m J 0:0075

20.36 (a) Using Eq. (20.26) with X1 X2

=

5

=

= 0:

1 (! 2 =p)2 20 1 (10=p)2 = 2 1 (! 1 =p) 4 1 (5=p)2 p2 100 p = 2:5 rad/s J p2 25


(b) X1

=

P0 k

P0 =k j1 (! 1 =p)2 j

=

0:02 1

0:02 =

j1

P0 =k (5=2:5)2 j

(5=2:5)2 = 0:06 m = 60 mm J

#20.37


20.38

20.39 k m

m

4k Equivalent systems

It was shown in the solution of Prob. 20.9 that the e¤ective spring constant of the system is k 0 = 4k. Hence we can replace the mass-spring-pulley system with the equivalent mass-spring system shown. With = 0 and Z = 5Y; Eq. (20.31) yields r (!=p)2 ! 5 5Y = Y ) = 1 (!=p)2 p 6 r ! r r ! r r r 5 5 k0 5 4k k ) != p= = = 1:826 J 6 6 m 6 m m


20.40

20.41


#20.42

20.43

m

k Equivalent system

The system is equivalent to the mass-spring system shown, where k

=

mg

=

m(32:2) = 483:0m 0:8=12 2

k ! 182 2 = 483:0 (rad/s) = = 0:6708 m p 483:0 (!=p)2 0:6708 = Y = 0:2 = 0:4075 ft = 4:89 in. J 1 (!=p)2 1 0:6708 )

Z

p2 =


20.44

20.45

20.46


20.47 X

=

! = 0:5 p c 8 c c p = p = p = 0:5350 = 2mp 2 km 2m k=m) 2 60(30=32:2)

X

=

=

0:25 P0

3 in. = 0:25 ft

= =

q [1

q (1

P0 =k 2

(!=p)2 ] + (2 !=p)2 P0 =60 2 0:52 )

0:25 = 2

+ (2(0:5350)(0:5))

P0 =60 0:9213

13:82 lb J

#20.48



20.49

20.50


20.51 0:8 = 0:024 84 slugs k = 12(12) = 144 lb/ft 32:2 r r k 144 = = = 76:14 rad/s m 0:024 84 c 0:4 ! 50 = = = 0:105 75 = = 0:6567 2mp 2(0:024 84)(76:14) p 76:14

m = p

X

=

Y

q

[1

=

tan

= x(t)

1

2

0:2=144

q

(1

=

2

(!=p)2 ] + (2 !=p)

2: 372 3

2 0:65672 )

10

3

2

+ [2(0:105 75)(0:6567)]

ft = 0:02847 in.

2 !=p 2(0:105 75)(0:6567) = = 0:2442 (!=p)2 1 0:65672 = X sin(!t ) = 0:02847 sin(50t x(t) lags P (t) by 13:72 J

= 13:72

0:2442) in. J


20.52

.

cx

x

FBD

mg

k(y − x − ∆) F = m• x

k(y

x

)

cx_ + mg = m• x

Substituting the static equilibrium condition k

= mg, we get

k(y x) cx_ = m• x m• x + cx_ + kx = kY sin !t Letting P0 =k = Y , the last equation can be written as m• x + cx_ + kx = P0 sin !t

p

= =

X

=

P0 =k

q

[1

=

=

1

x(t)

2

(!=p)2 ] + (2 !=p)2 12

q

(1

tan

r k 90 103 = = 600 rad/s m 0:25 c 60 ! 900 = = 0:2 = = 1:5 2mp 2(0:25)(600) p 600

r

= 8:65 mm

2

2

1:52 ) + [2(0:2)(1:5)]

2(0:2)(1:5) 2 !=p = = (!=p)2 1 1:52

0:48

=

0:448 rad = 25:6

= X sin(!t ) = 8:65 sin(900t + 0:448) mm J x(t) leads y(t) by 25:6 J


20.53

20.54


20.55

20.56


20.57

P(t)

R

2 ft

θ

. 1.5 ft O.

4 lb 2 ft

.

c(2θ)

k(1.5θ + ∆) FBD Assume that the angular displacement

P (t)(3:5)

k(1:5 +

)(1:5)

of the bar is small.

c(2 _ )(2)

MO = IO • 4(2) = m(2)2 •

(a)

where is the deformation of the spring at static equilibrium. Its value is obtained by setting P (t) = = • = 0 which yields 1:5k 8 = 0: Substituting this into Eq. (a) gives us 3:5P (t)

2:25k

4c _

3:5(0:4 sin 4t) 2:25(6) 4(2:2) _ 0:4969• + 8:80 _ + 13:50

4 (4)• 32:2 = 0:4969• = 1:40 sin 4t =

which has the form M • + C _ + K = P0 sin !t


p

r

r K 13:50 = = 5:212 rad/s M 0:4969 C 8:8 = = 1:6989 2M p 2(0:4969)(5:212) 4=5:212 = 0:7675

= =

!=p

P0 =k

=q

=

max

=

[(1

=

(

q (1

2

2

(!=p)2 ] + (2 !=p) 1:40=13:5

= 0:03928 rad 2

0:76752 )2 + [2(1:6989)(0:7675)]

block )max

= 24(

max )

= 24(0:03928) = 0:943 in.

20.58 (a)

mg

6x.

80(0.004 sinωt − x)

FBD 60x N + !

Fx = max

80(0:004 sin !t

x)

6x_

60x = 12• x

12• x + 6x_ + 140x = 0:32 sin !t J (b) Comparing with M x + Cx + Kx = P0 sin !t, we obtain r r K 140 p = = = 3:416 rad/s M 12 C 6 = = = 0:07319 2M p 2(12)(3:416) With ! = p Eqs. (20.26) and (20.27) yield X

=

0:32=140 P0 =K = = 0:015 615 m = 15:62 mm 2 2(0:07319)

=

tan

1

1 = =2 rad

x(t) = 15:62 sin(3:42t + =2) mm J


20.59 (a)

P0 sinωt

Oy Ox O

θ

b

θ

b

.

mg − cbθ

FBD

kb(θ + θs)

(mg

cb _ )b

kb(

MO s )b + (P0 sin !t)b

= IO • = mb2 •

Substituting the static equilibrium equation mgb + kb2 the equation of motion becomes mb2 • + cb2 _ + kb2

p

= =

r

r k 30 ! 6 = = 8:660 rad/s = = 0:6928 m 0:4 p 8:660 c 3 = = 0:4330 2mp 2(0:4)(8:660) =

(P0 =b)=k

q

[1

= = = =

2

(!=p)2 ] + (2 !=p)2 (1:5=0:4)=30

q

(1

tan

=0

= P0 b sin !t P0 sin !t J = b

m• + c _ + k (b)

s

2

2

0:69282 ) + (2(0:4330)(0:6928))

0:157 44 rad = 9:02

J

2 !=p 2(0:4330)(0:6928) = = 1:1537 2 1 (!=p) 1 0:69282 49:1 J


#20.60

#20.61



20.62

mg kx

R

FBD F

C N Fx Fy

= m• x = 0

MC

= IC •

F N

kx = m• x F = kx + m• x mg = 0 N = mg 1 x • (kx) R = mR2 + mR2 2 R

) F = kx

m

2 k x 3m

=

1 kx 3

Fmax =

x •=

2 k x 3m

1 kx0 3

At impending sliding we have Fmax = N

s

kx0 = 3mg

s

k=

3mg x0

s

=

3(8)(9:81)(0:2) = 314 N/m J 0:15

20.63


20.64

20.65 Use formula given in Prob. 20.64: p

2

= )

r

gy k2 + y2 gy k2 = 2 y2 p

(a) (b)

Pin at A: p=

2

=

y

=

0:28 m

=2s

From Eq. (b) k 2

=

9:81(0:28) 3:14162

0:282 = 0:199 91 m2

2 = 3:1416 rad/s 2

Pin at B: y

=

From Eq. (a) p2

= )

0:21 m 9:81(0:21) 2 = 24:51 (rad/s) 0:199 912 + 0:212 2 2 =p = = 1:269 s J p 24:51


20.66 (a) Let

s

be the static angular displacement

mg Ox O L/2

L/2 . c2Lθ/2

Oy

+

c2 L _ 2

mg

Upon substituting becomes

s

L 2

h

θ

FBD . kL(θs + θ) + c1Lθ

MO = IO • i 1 kL( s + ) + c1 L _ L = mL2 • 3

= mg=(2kL) and cancelling L2 , the equation of motion c2 1 • m + c1 + 3 4

_

k =0 J

(b) Comparing with M • + C _ + K = 0, we obtain s s r K k 80 = = = 3:464 rad/s p = M m=3 20=3 Ccr

= =

m 20 p=2 3:464 = 46:19 N s/m 3 3 C c1 + c2 =4 25 + 16=4 = 0:628 J = = Ccr Ccr 46:19

2M p = 2

20.67

Oy O O R x FBD 3mg θ mg MO = IO •

+

(mg) R

=

2:833R• + g

=

1 1 (3m) R2 + m(2R)2 • 2 3 0

Comparing with M • + K = 0, we obtain r r r K g g p = = = 0:5941 M 2:833R R r r p 0:5941 g g f = = = 0:0946 Hz J 2 2 R R


20.68 F2

2

= I + m OG = =

Letting

s

8(9.81) N

G 5m 2 1 0. O 0.15 m Oy

Ox

IO

θ

0.2 m

F1

FBD

1 m(a2 + b2 ) + m 12

b2 a2 + 4 4

=

1 m(a2 + b2 ) 3

1 (8)(0:152 + 0:22 ) = 0:166 67 kg m2 3

be the static angular displacement, the spring forces are F1 F2

= k1 = k2

1 2

= 300 [0:2( s + )] = 60( s + ) = 600 [0:15( s + )] = 90( s + )

MO = IO • + 8(9:81)(0:075) F1 (0:2) F2 (0:15) = 0:166 67• 8(9:81)(0:075) 60( s + )(0:2) 90( s + )(0:15) = 0:166 67• Substituting the static equilibrium equation 8(9:81)(0:075)

60 s (0:2)

90 s (0:15) = 0

we get for the equation of motion 0:166 67• + 25:5 = 0 Comparing with M • + K = 0, we obtain r r p 1 K 1 25:5 )f = = = = 1:969 Hz J 2 2 M 2 0:166 67


20.69

x A

L/4

kx By B

Bx

L/4

θ G

L/2

mg

FBD

k(3x − x0 sinωt)

C

3x

m= IB =

mL2 +m 12

12x

12 = 0:3727 slugs 32:2 L 4

2

=

=

4 2 x= x L 3

7 7 mL2 = (0:3727)(6)2 = 1:9567 slug ft2 48 48

MB L 3L mgx kx k (3x x0 sin !t) 4 4 3 6 150 (3x x0 sin 8t) (6) 150x 4 4 2262x + 675x0 sin 8t 1:3045• x + 2262x K p

= IB • = IB

2 3

= 1:9567 = =

x • 2 x • 3

1:3045• x 675x0 sin 8t

=

2262 lb M = 1:3045 slug ft r r K 2262 = = = 41:64 rad/s M 1:3045 2

(!=p)

2

xmax

8 = 0:03691 41:64 (!=p)2 0:03691 = x0 = 0:0796 in. J 2 = 2 (1 2 0:03691)2 [1 (!=p) ] =


20.70

20.71

x mg kx

R F

FBD C N

(a) m=

8 = 0:2484 slugs 32:2

MC = IC • p=

(kx) R = r

2 k = 3m

r

1 3 mR2 + mR2 = mR2 2 2 3 x • 2 k mR2 x •+ x=0 2 R 3m

IC =

2 110 = 17:182 rad/s J 3 0:2484

(b) Fx = m• x

F

kx = m• x

F = m• x + kx


p2 X sin pt:

Substitute x = X sin pt , x •= F Fmax

= X( mp2 + k) sin pt 1:2 = X( mp2 + k) = 12 s

=

(0:2484)(17:182)2 + 110 = 3:667 lb

Fmax 3:667 = = 0:458 J mg 8

20.72

kLθ

. cLθ L

O L

L/2

θ

L/2

. mg

FBD (only forces contributing to MOare shown) MO L (kL )L (cL _ )L mg 2 mg m• + c _ + k + 2L

= IO • mL2 = 3 3 =

•

0

Comparing with M • + C _ + K = 0, we obtain M K

p !d

1:8 = 0:05590 slugs C=c 32:2 mg 1:8 = k+ = 20 + = 20:60 lb/ft 2L 2(1:5)

= m=

r

K = = M q = p 1

r 2

20:60 = 19:197 rad/s 0:05590 p = 19:1971 1 0:52 = 16:625 rad/s J


20.73 (a)

Oy O

Ox b

2b

θ FBD . 2cbθ

b/2

kb(θs + θ)

mg IO =

1 m(3b)2 + m 12

b 2

2

= mb2

MO +

mg

b 2

2cb _ (2b)

The static angular displacement

s

(mg)

kb(

s

+ )b

= IO • = mb2 •

is given by b 2

kb2

s

=0

(a)

The equation of motion becomes, after utilizing Eq. (a) and cancelling b2 , m• + 4c _ + k = 0 J (b) Comparing with M • + C _ + K = 0, we obtain r r k 3200 p = = = 24:12 rad/s m 5:5 4c mp 5:5(24:12) = ccr = = = 66:3 N s/m J 2mp 2 2(1)

20.74 . chθ

A mg x-

h Bx

MB = IB

B +

By

b mgx

FBD

θ C kb(θs + θ) ch2 _

kb2 (

s

+ ) = IB •

Substituting the static angular displacement s = mgx=(kb2 ), the equation of motion becomes IB • + ch2 _ + kb2 = 0


Comparison with M • + C _ + K = 0 yields p2

= =

(20 12)(2:5)2 kb2 2 = 2 = 7:709 slug ft 2 p (2 2:22) C ch2 c(1:0)2 = = = 0:004 650c 2M p 2IB p 2(7:709)(2 2:22) K kb2 = M IB

Damping is critical when

) IB =

=1

) ccr =

1 = 215 lb s/ft J 0:004 650

20.75


#20.76


20.77


20.78

*20.79


*20.80

20.81

k1Rθ Oy

θ

R

k2Rθ Vg

=

Ve

=

V

W

O

Ox R/2 G

FBD

W R cos 2

W

R 2

1

1 2

2

1 1 k1 (R )2 + k2 (R )2 2 2 WR 1 WR = Vg + Ve = + + (k1 + k2 )R2 2 2 2

2


)K=

WR 44(2:4) + (k1 + k2 )R2 = + (25 + 35)(2:4)2 = 398:4 lb ft 2 2 T )M

1 _2 I 2

=

= I = mk 2 = )p=

r

K = M

r

44 (1:8)2 = 4:427 lb ft s2 32:2 398:4 = 9:49 rad/s J 4:427

20.82 Datum A L V

=

T

= =

θ

B Lθ

C L

L 1 1 k(L )2 2mg = kL2 2 mgL ) K = kL2 2 2 2 2 2 1 1 2 1 (IA + IC ) • = 2 mL2 • ) M = mL2 2 2 3 3 r r r 2 M 2mL2 =3 m = 5:13 =2 =2 J p K kL2 k

20.83 (a)

m

θa Datum

V

T

b

θ

1 1 2 = mga cos + k(b )2 mga 1 + 2 2 1 = mga + kb2 mga 2 ) K = kb2 2 1 = m(a _ )2 ) M = ma2 2 r r K kb2 mga p= = J M ma2

1 2 kb 2

2

mga


(b) kb2

mga > 0

k>

mga J b2

20.84

.

G

h

.G θ

Datum

max

R

θmax

Position of Vmax

Position of Tmax

Use Rayleigh’s principle: h

=

R cos

R=R

max

Vmax

= mgh = mgR

Tmax

=

1

cos max cos max

R

1

2 max =2) 2 max =2

R

2 max

2

2 max

2 1 _2 1 mL2 _ 2 1 IG max = = mL2 p2 2 2 12 max 24 Tmax 1 mL2 p2 24

(1 1

2 max

p2

= Vmax = mgR =

12gR L2

V0 2 max 2

2 max

0 p=

2p 3gR J L

20.85


20.86

20.87


20.88

Datum L

θ m V

=

T

=

mg(L cos )

R

1 2 ) K = mgL 2 2 _2 _ 2 = 1 m L2 + R 2 2

mgL 1

1 1 m(L _ )2 + 2 2

1 mR2 2 R2 ) M = m L2 + 2 s s r m L2 + R2 =2 2 M L2 + R2 =2 = =2 =2 =2 p K mgL gL

The condition

= 1:2 s yields s L2 + R2 =2 2 = gL

1:2

1:22 g R2 L+ = 0 2 4 2 1:1745L + 0:03125 = 0

L2 L2

4 L2

2 2L

+ R2 =2 = 1:22 gL

1:22 (32:2) 1 2 L + (0:25) = 0 4 2 2

The two solutions are L1 = 0:02724 ft = 0:327 in. J

L2 = 1:1473 ft = 13:76 in. J


20.89 Use Rayleigh’s method.

A.

.E

Datum L/2

b

3mg Position of maximum V

Tmax

2(mg

=

4mgL cos

=

4mgL

= =

Tmax p2

= Vmax =

12 g 11 L

.

b

D

Position of maximum T

L cos 2

=

Lθmax

.

C

V0

.

.

.

B

Vmax

.

θmax

L

L/2

mg

.

.

E

L

L/2

mg

.

θmax

L/2

L/2

θmax

A

max )

max

Vmax

3mL cos

max 2 max

4mgL 1 V0 = 2mgL

2 2 max

1 mL2 _ 2 2 + 3m L _ max 2 3 max 2 11 11 mL2 _ max = mL2 p2 2max 6 6 11 mL2 p2 2max = 2mgL 6 r 3g p=2 J 11L

V0

2

2 max

0


20.90

20.91 Datum position (spring undeformed)

β R x+ Let

∆

be the static elongation of the spring V

= = )

T

= ) f=

1 k(x + )2 mg(x + ) sin 2 1 1 k 2 + mg sin + (k mg sin )x + kx2 2 2 K=k 2 1 1 x_ 1 1 3 mR2 + mx_ 2 = m x_ 2 2 2 R 2 2 2 3 M= m 2 r r r p 1 K 1 2k k = = = 0:1299 J 2 2 M 2 3m m


20.92

Area = A

x Water level

h F

The force F of buoyancy is equal to the weight of water displaced: F =A

water x

Buoyancy acts like a spring of sti¤ness k = A V

=

T

=

p

= =

water

1 A x2 mwo o d gx ) K = A water 2 water 1 1 mwood x_ 2 = Ah wo o d x_ 2 ) M = Ah wo o d 2 2 g g r

K = M

r

g h

water wo o d

=

s

32:2(0:036) = 10:266 rad/s 0:5(0:022)

2 2 = = 0:612 s J p 10:266

20.93


20.94


#20.95


20.96 Datum

ω v

θ

R−r

R

r r) _

v = (R

V

=

T

=

mg(R

r) cos

!=

mg(R

v R r_ = r r

r) 1

1 2

) K = mg(R

2

r)

2

=

12 2 R r 1 2 1 _ 2 + 1 m(R r)2 _ 2 I! + mv 2 = mr 2 2 25 r 2 2 1 7 7 m(R r)2 _ ) M = m(R r)2 2 5 5 s r r 1 1 p K 5g g f= = = = 0:1345 J 2 2 M 2 7(R r) R r

20.97 Use Rayleigh’s method

.

θmax

Datum θmax h Position of Vmax

R−r

r R

C R

Position of Tmax


h IC Vmax

=

mgh =

= r + (R r) cos max = I + mR2 = mR2 + mR2 = 2mR2 mg [r + (R

mg r + (R V0

=

Tmax

=

r) 1

r) cos 1 2 2 max

max ]

1 mgR + mg(R 2

=

mgR 2 1 _2 1 2mR2 _ max = mR2 p2 IC max = 2 2

Tmax p

= Vmax V0 mR2 p2 r g(R r) J = 2R2

2 max

=

r)

2 max

2 max

1 mg(R 2

r)

2 max

20.98 (a)

x

Fx = max

(b)

kx

mg

FBD

N

+ !

2kx = m• x

kx

2(150)x = 6• x x • + 50x = 0 J

p 50 p = = 1:125 Hz J f= 2 2

(c) x = E sin(pt + ) s s 2 2 v ( 120) E = x20 + 02 = 252 + = 30:22 mm p 50 p ! x p 25 50 0 = tan 1 = tan 1 = 0:9744 rad v0 120 )

x = 30:22 sin(7:071t

0:9744) mm J


20.99

20.100


20.101

L/2

C(t)

θ

L/2

O

k1 L θ 2

k2 L θ 2

FBD (only forces contributing to MO are shown)

C0 sin !t

(k1 + k2 )

L 2

MO L 2

m• + 3(k1 + k2 )

= IO mL2 • = 12 C0 = 12 2 sin !t L

This equation has the form M • + K = P0 sin !t where

p

= =

X

=

18 = 0:5590 slugs 32:2 3(k1 + k2 ) = 3(20 + 30) = 150:0 lb=ft C0 0:5 12 2 = 12 = 0:16667 lb/ft L 62

M

= m=

K

=

P0

=

r

r K 150 = = 16:381 rad/s M 0:5590 P0 =K 0:16667=150:0 = 2 2 = 0:02582 rad [1 (!=p)2 ] [1 (18=16:381)2 ] L 6 = 0:02582 = 0:07746 ft = 0:930 in. J 2 2

20.102 Momentum is conserved during impact. Letting v0 be the velocity of the block (and the embedded bullet) immediately after the impact, we have 0:012(1200) = (5:012)v0

v0 = 2:873 m/s


Now consider the ensuing damped free vibration with the inital conditions x = 0 and x_ = 2:873 m/s at t = 0. r r k 1800 p = = = 18:951 rad/s m 5:012 ccr = 2mp = 2(5:012)(18:951) = 189:96 N s/m c 50 = = = 0:2632 Since < 1, the system is underdamped ccr 189:96 q p 2 = 18:951 1 0:26322 = 18:283 rad/s !d = p 1 x(t) = Ee pt sin(! d t + ) x(t) _ = E pe pt sin(! d t + ) + E! d e p = 0:2632(18:951) = 4:988 x = x_ = )

pt

cos(! d + )

0 at t = 0 E sin = 0 2:873 m/s at t = 0 E p sin + E! d cos 2:873 2:873 = 0:157 14 m = =0 E= !d 18:283 ) x (t) = 0:1571e

0:499t

= 2:873

sin 18:28t m J

20.103


20.104 Since

= 0:6 < 1, the system is underdamped. s r k 300 p = = = 34:75 rad/s m 8=32:2 p !d

=

0:6(34:75) = 20:85 rad/s q p 2 = p 1 = 34:75 1 0:62 = 27:80 rad/s

x(t) = Ee

pt

sin(! d t + ) = Ee

20:85t

sin(27:80t + )

Initial conditions: x = 3 in. when t = 0. ) E sin = 3 in. x_ = 0 when t = 0. ) 20:85E sin + 27:80E cos = 0 27:80 = 1:3333 = 53:13 (0.9273 rad) ) tan = 20:85 3 ) E= = 3:750 in. sin 53:13 x(t) = 3:750e

20:85t

sin(27:80t + 0:9273) in. J

20.105


20.106

20.107


20.108

20.109

x

mg k1x cx.

k2(y − x) N

Fx = max

FBD

+ ! k2 (y x) k1 x cx_ = m• x k2 Y sin !t (k1 + k2 ) x cx_ = m• x m• x + cx_ + (k1 + k2 ) x = k2 Y sin !t

This has the form m• x + cx_ + kx = P0 sin !t, where k P0

p

= k1 + k2 = 8 + 14 = 22 kN/m = k2 Y = 14(0:024) = 0:336 m

= =

! p

=

r

r k 22 000 = = 52:44 rad/s m 8 c 50 = = 0:05959 2mp 2(8)(52:44) 18 = 0:3432 52:44


The steady-state displacement is x(t) = X sin(!t X

=

=

= )

q q

), where

P0 =k 2

2

(!=p)2 ] + (2 !=p)

[1

0:336=22 2

= 0:017 293 m 2

0:34322 ) + [2(0:05959)(0:3432)]

(1

2 !=p 2(0:05959)(0:3432) = tan 1 = 0:04633 rad 2 1 (!=p) 1 0:34322 x(t) = 0:01729(sin 18t 0:0463) m J x(t) lags y(t) by 0:0463 rad (2:65 ) J tan

1

20.110


20.111

b L/2

Ox O

. Oy

Substituting

s

θ kb(θ + θs)

s

L 1 kb2 ( + s ) = mL2 • 2 3 of the bar is oblained by setting

s

=0

mg

The static angular displacement L 2

.. mg

MO = I0 •

mg

FBD

kb2

)

s

=

(a) = 0:

mgL 2kb2

into Eq. (a) yields mg

L 2

mgL 1 = mL2 • 2kb2 3 2 • + 3kb = 0 mL2 r b 3k ) p= J L m

kb2

+

20.112


20.113 b az

F

θ

z

a

y,z

b

FBD

Oy

x

O

Ox

= may

mg F =k =k

kb But kb is

b + z a

ax = a _

( MO )FBD + F b (mg) a b k + z b mga a b2 mga + k z + ma• z a

2

MAD a max

O

ay = a• + y• = z• + y•

= =

( MO )M AD (may ) a

=

m(• z + y•)a

= maY ! 2 sin !t

mga = 0 (static equilibrium equation), so that the equation of motion z• +

k b2 z = Y ! 2 sin !t J m a2


20.114 From Eq. (20.31):

Y

= Z

! p

=

)Y

= =

rh

1

(!=p)

2

i2

+ (2 !=p)2

(!=p)2

18 = 0:9549 2 (3) q 2 2 (1 0:95492 ) + [2(0:25)(0:9549)] 10 0:95492 5:32 mm J



An Instructor’s Solutions Manual to Accompany

ENGINEERING MECHANICS: DYNAMICS, 4TH EDITION ANDREW PYTEL JAAN KIUSALAAS

ISBN: 978-1-305-88500-4

© 2017, 2010 Cengage Learning WCN: 01-100-101 ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be reproduced, transmitted, stored, or used in any form or by any means graphic, electronic, or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval systems, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without the prior written permission of the publisher except as may be permitted by the license terms below.

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Instructor's Solutions Manual to Accompany

Engineering Mechanics: Dynamics 4th EDITION

ANDREW PYTEL JAAN KIUSALAAS

Contents Chapter 11:.......................................................................................................................... 1 Chapter 12:........................................................................................................................ 83 Chapter 13:...................................................................................................................... 140 Chapter 14:...................................................................................................................... 204 Chapter 15:...................................................................................................................... 294 Chapter 16:..................................................................................................................... .370 Chapter 17:...................................................................................................................... 452 Chapter 18:...................................................................................................................... 521 Chapter 19:...................................................................................................................... 577 Chapter 20:...................................................................................................................... 647

Appendix F F.1

F.2


F.3


F.4

F.5


F.6

F.7


F.8

F.9


F.10

F.11


F.12


F.13

F.14


F.15

F.16


F.17


F.18

F.19


F.20

F.21


F.22


F.23


F.24

F.25


F.26


F.27


F.28


F.29


F.30

Iy0 Iz0

Iy0 z0

m b2 3 m (2b)2 + = mb2 2 3 4 3 4 = (Iz0 )1 + (Iz0 )2 + (Iz0 )3 m (2b)2 m m m b2 3 = + (2b2 ) + b2 + = mb2 2 12 2 4 4 3 2 b m = = (Iy0 z0 )2 = (Iyz )2 + m2 y 0 z 0 = 0 + ( b) 4 2 =

(Iy )1 + (Iy )2 =

IAB = mb2

3 4

1 2

+

3 2

1 2

2

1 8

1 2

1 2 mb 8

= mb2 J

F.31


F.32


F.33

F.34


F.35


F.36

F.37


F.38


F.39


F.40

F.41


F.42



Dynamics Pytel Kiusalaas 4th Solutions

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