300 Creative Physics Problems with Solutions (Anthem Learning).pdf

Second solution of Prohlem 6. In casc b) thc probl em requests an acce lerati on-free state in a pos iti on whcre the normal accelerat io n and the tangenti al accelera ti on cnc loses the same (45° ) angle with the verti ca l. Thc ir rcs ultant should be the oppos ite of the (hori zo ntal ) acceleration o f the car, wh ich is o nl y poss ibl e if the tangential and the norm al acccle rati ons have the sa me mag nitude and the vec tor parallel og ram formed by them is a squarc whosc diagonal has thc sa nK magni tude as the accclerati on of the car:

that is, ((

U

J2

I?'

- = /'. -

from whi ch As the normal accclerati on IS

," nt (

rL

)"

= /"

a H'

77

300 Cr eative Physics Problem s with Solu tions

------------~-----------------------------------------------------

from here

a nd with thi s the angul ar di spl aceme nt req uired after starlin g is

1 a

cp = "2 . Ii'

I'¥

2

RI

(

= "2

)

rad

= 28.65

o .

First solution of Problem 7. With superposition of velocities (coo rdinate tran sformation). The "absolute" velocity Va o f the point with respect to the g round is the sum of' its re lati ve velocity Vr el with respect to the coordinate frame travellin g alo ng with the centre of the disc and the translati o nal velocity Vtr of the coordin ate fr ame:

(1) At point B, thi s is a lso true for the magnitu des of the velocities (s ince in iti a lly the velocity vectors arc parallel):

R

Vn =va + 2 w , where Rw = Va since the di sc rolls without slip ping. Thus the speed of po int B is

3

Vn = "2 va . Let us app ly (I) to point A. As seen from the figure, the mag nitudes of the absolu te ve locity and the translational velocity are eq ua l, VA,e] = Rw = Va = Vt r , and they encl ose an ang le cp , equal to the angu lar di splacement of the radiu s. From the isosce les tri an g le, VA

Va = 2vacos '2cp = 2vacos ( 2R' t)

(2)

.

At the time in stant in question , Vo ) 3 2vacos ( 2R' L = "2 va.

He nce cos ( ; ; . L)

= ~,

a nd the time in quest io n is 2R

t=_

Va

78

3

0.2

4

4m s

III

.a rccos - = --/_ .a rccos 0 .75 = 0. 36 s.

6.1 J\in ematics

6 M ech anics Solu tion s

;;:.:.--

Second solution of Problem 7. With an illstalltaneous ax is of rotatioll. The fi gure shows th at the insta ntaneo us ax is of rotation is the point P where the di sc touches the ground. The speed of point A is thu s

where rA

= 2 R cos ( ~ )

Va

and

W =- .

R

p

The express ion (2) is obtai ned again by substitut ing ((! = wi = 7'f t : VA

Va. t ) . -Vo = 2vacos ( VO. {) = 2R cos ( -

2R

R

2R

'

and hence the requirement of the probl em leads to the sa me res ult as above .

Solution of Prohlem 8. a) Accordin g to the co nditi on V A = V D . Sin ce th e mag nitude of the veloc ity for a point on the circum ference 0 ::; VA ::; 2v , there wi II he momen tary instants when it is equ al to VjJ, whi ch is always greater than zero and al ways le ss than 2v (its direc ti on will obvi ously be di ffere nt). Thi s requirement is fulfill ed whe n r A w = rjJw , where TA and I"jJ are the momentary radii of ro tati on draw n to A and B respectively, which are the di stances of A res p. B meas ured from the point th at is stationary at the mome nt (the point th at is exact ly touchin g the ground ). Since the angular speed is the same for eac h point of a ri gid di sc, in order to fulfil the co nd iti on TA

= TjJ = T

should be true. From the fig ure it can be see n where the po ints sho ul d be to fulfil thi s conditi on. (From the fi gure it is clear that thi s state is reached in the seco nd qu arter of the angul ar displacement firs t. ) Trian gles OAP and PAB are simil ar because they an: isosce les and their angles on the base are co mm on. Let ((! stand for the angul ar di spl acement of the rad iu s A belongin g to point A and C\' for its suppl eme ntary angle . The moment ary rad iu s of rotatio n fro m the big triangle is 0:

,. = 2Rsin - . 2

and from th e slll all tri angle

R

T= - - - .

4s in ~

p

79

300 Creative Physics Problems with Solutions

Therefore the right sides are eq ual:

R

. a

2R sm-= - - .

4 s in %

2

From he re - independe ntl y of the rad ius - the following value is acq uired for the si ne of the ha lf of the angular position: .

2

1

a

8'

sm"2 from whi ch

. a 1 J2 = -=-. 2 vis 4

SIl1-

From this

a =2 . arcsin

J2 4 =41.41

0

.

So the first ang ul ar displacement th at fulfils the requirements o f the probl em is

7r -

a = 180 0

-

41.41 0 = 138.59 0 = 2.42 rad.

The time elapsed in the meantime is


tI = -

w

= - - =4.84s. v

b) The req uested speed is V A = V D =rw

a v a =2Rsin - . - = 2sin 2 R 2

·V

= -

2

vis

m

m

·0.5 - = 0.354 - . s s

c) T he di stance travell ed by the centre is 81

= R
Solution of Problem 9. a) Let us examine the phenomenon from a refere nce fra me fixed to the uniforml y mov ing cart. At thi s frame the bit of mud leaves the rim in the direc ti on of the tangent, that is, in this system the initial velocity of the projection e ncl oses an angle of a = 60 0 with the hori zo ntal. As the bit of mud fall s back at the same he ight as it starts from , the distance between these po ints is the distance of the proj ection : X m ax

= 2Rsin a .

So in the reference frame of the cart : 2 .

2Va sll1 a cosa 2R ' ----'''--------- = sin a , 9 where Va is the peripheral veloc ity of the rilll and the travelling velocity of the cart as we ll. From here

_f!g -

Va -

80

2

- - - )0.6 111.9. 8.1 m /s -_

cos a

0.5

~ 3.43 -m . v/ 11772 . m 2/ s 2 ~ s

6.1 K in ematics

Mechanics S olutions ~ IN THE FRAME OF TH E GROUND

_.A ------- - --

,, ,,

,

I I

Ai

I

,, I

,,

.. ..

I

:...

~ R ·s in a

s

R ·sin a

:

b) The time of the oblique proj ection of the bit of mud is

t=

2vo sin a 9

=

2 · 3.43 m/ s · 0.866 = 0.61 s. 9.81 111/S 2

!Ii this time the point of the rim where the bit of mud detaches from tra vels through an arc of i=vol=3.43 m/s ·0.61 s=2 .09 m;::::;;2.1 m due to the skid-free rollin g moti o n, from thi s arc i' th at the bit of mud fl ew along, th at is, , 27r i = R-:3 = 1. 257 met res, should be subtracted, so the length of the requested arc is

6.i = i - i' = 2.09 m - 1. 257

111

= 0.833 m;::::;; 0. 8 m.

c) The moti on of the cart is skid-free ro llin g, the axles of its wheels move exactl y at velocity Vo = 3.43 m is, so the in the time between the detachment and the fallin g back of the bit of mud the cart trave ls a di stance of

s = voi=3.43 m /s ·0.61 s = 2.09 m;::::;; 2.1 m , which is obvi ously equal to i . (The speci al angle of projecti on enabl es us to reach the conclu sio n faster. Namely, in the cOordin ate sys tem fixed to the gro und the bit of mud travels exactl y at veloc ity v," = vocos 60 0 + .+ Va = 3va/2 in the horizo ntal direc ti on, whi ch IS 3/2 times the veloc ity of the carl. Therefore the hori zo ntal displ ace ment of the bit of mud is 3/2 times that of the cart , th at is,

2R'sina + s

.1

. 3 2 R Slna+s=-s. 2 From thi s, the distance trave ll ed by the cart is: 5

= 4Rsina = 4·0.6 m· 0.866 = 2.1 111 .) 81

300 C rea ti ve P hysics Problem s \\'i t h Soillt ions

------------~--.------------------------------------~-----------------

Solution of Problem HI, a) W ith a good appro xima ti on the ball oo n in th e prob l ell per forms a harill oni c osc ill atory moti on, si nce it s accelerat ion is pro porti onal to it s he igh; Illeasured from the altitu de H , and has opposite di recti on. T hrough this anal ogy the kin eill ati c equ ati ons o f th e ball oo n arc sim il ar to th ose o f a harm oni c motio n. The k inemati cs o f th e harm on ic moti on is uniqu ely determin ed by its amp l itude A and i ts max im al acce lera ti on (L ''' il X' T hi s exact data is gi ven in th e prob lem:

A= lJ

and

[n th e li rst q ues ti on th e speed at the altitude H i s ju st th e max im al speed of the osc ill ati on. T he max ima l ve loc it y and m ax im al acce lerati on o f th e harm onic moti on can be ex pressed usin g the angul ar freq uency :

From here, th e an gul ar frequ ency ca n be expressed usin g th e know n data:

w-

f¥

"t< IX ---A '

So , usin g th e previous notati ons, the unkn ow n speed at altitude /I is:

b) T he speed at altitude

rr /2 can

be determin ed by the k now n formula:

u=wJ A 2 _y2, so / '=

ffj

/I 2

_

(~)

2 =

J O.7S(LII H .

c) T he tim e in qu esti on is one quarter o f th e peri od o f th e harm oni c oscill at ion , so :

T

'if t--4---<271 -Iw - 2

{£---l

a ll .

Anoth er solution ( for the ques ti ons co nce rn ing th e speed) is based on th e workenergy theore m. Sin ce th e net fo rce ac tin g on th e ba ll oon is linearly dec reasin g to zero w ith th e alt itud e, th e force averaged over th e altitu de is ju st ha l f of th e m ax im al force .

82

6.1 Kin e ma.tics

Mechan ics SO ilitioll S

6

~

The change of the kinetic energy is eq ual to the work done hy th is force: I II ILO

-

2

- [J

=

F

1 'J -IILV2 '

thLlS , the speed at height Hi s:

moo __ __ __ ____ _ 2

u ( ff ) = Ja nH . From ground level to H / 2 the average force is:

H

2

O+T = 0. 75mao, ----=--'I7I CL

H

y

2 so the speed at height II /2 is

V= J O.7 5a o H . The simp les t, most elemen tary way to determine the time needed for the balloon to rise is to apply the analogy to the harmo ni c osci ll atio n.

Solution of Prohlem 11. First we have to determine the time it takes the bullet , moving at a co nstan t horizontal velocity compone nt, to arri ve at the trajectory of the fa lling ball. During this time, in a ve rti cal direction , the bullet performs a free fall and loses its height of Ii, . However, because the bullet hits the ball, the ball covers the distance (It - hi ) + hi. The time in question is therefore simply the diA'ere nce between the time of th e free fall of the ball and the time of the flight of the bullet.

It takes the bullet 11

50 111 d = - = - - = 0.5 v 100 !.!..'s

5

to reach the tra jectory of the fa llin g ba ll. During this time it loses a height of 1

.)

It 1 = -;;.g·[i

1 = -;;.

· 9.8 1

III

52

· 0. 25

.)

S - ;:::j

.

1.23111 .

The time of the free fall of the ball , until be in g hit by the bu ll et , is: 1=

2( h-h'+ llj ) g

2 ·1 1. 23 m 9.81 ;S

= 1. 51 s . 81

300 Creative Phy sics Problems with Solutions

So the time difference in question is: f::lt = t - h = l. 51 s - 0. 5 s = l.01 s ~ 1 s .

The parametric solution of the problem can be significantly simplified by considerino to the fact that h - hi = h / 2. With this in mind , the time of the fall of the ball is:

9

and the time difference between the start of the fall and the gun fire is:

h

d2 v2

d

-+-- ~

9

V ·

Solution of Problem 12. Let us write down all the requirements for the two mot ions . The projectile and the object sliding on the slope arrive at point P at the same instant. Thus , the vertical velocity component of the projectile is equal to the vertical compone nt of the average velocity of the other object. Since the object on the slope performs uniformly accelerated motion , its average velocity is half of its final velocity, so Va

cos 'p =

v

"2 cos O! ,

where 'P is the angle of the throw (above the horizontal), and O! is the angle of the slope. The two objects arrive at P at the same speed, i.e. , Va = V, so we can simpli fy using the initial speed of the throw and the final speed of the sliding: cos O! cOS 'P =

~2-'

(1)

The two angles should certainly obey this relation, but it does not mean that an y O! could solve the problem. A further requirement should al so be satisfied: the total time of the th row should be equal to the time of the sliding . (Indeed , if only equation ( I) was satisfied then the x coordinates o f the two objects would be equal at the instant when the object on the sl ope gets to P , but the y coordinate of the I a \ \, projectile would not be 0.) From this \ requirement we get that t t lirow = t s li de , so: 2va s in 'P Va 9 g sin O! '

~~"\

84

6. 1 I\in em at ics

6 Mech a.nics Solu tions

::.:--

which mean s th at the ang le s o bey another rel ation as well . Only o ne pair o f ang les ca n satisfy all the requireme nts o f the pro ble m. After s implifi cation:

1

2sin (j! = - .- . sm 0'

(2)

Now we solve the system o f equ ati o ns ( 1-2) . Takin g the square of equati o n (2), and using a trigon ome tri c equation to c ha nge the s ine functi o ns to cos in es, w hi c h appear in (I), we get th at: .) 1 4 . (1 - cos - (j!) = 2 1 - cos 0' Using (I ) to substitute cosO'=2 , cos(j! :

1

?

4 . (1 - cos - (j!) =

1 - 4cos

2

(j!

After multipl y in g w ith the de nomin ator: 4 . ( 1 - cos 2 (j! )( 1 - 4 cos 2 (j!) = l. Then performing the multiplic ati o ns: 4 - 4 cos 2 (j! -1 6cos 2 (j! + 16cos 4 (j! = 1 After arrang ing the term s: 16 cos 4 (j!

-

20 cos 2 (j! + 3 = O.

The solution of thi s equ ati o n for cos 2 (j! is: 20 ± V400 - 4 · 16 · 3 {1.0 75.69 2. 16 = 0.1 743 1

2

cos

(j!=

Only the seco nd root is meanin g ful , and s ince we obtain that: cos (j!

= + J O.17 431 = 0 .4 1750

-+

(j!

(j!

is an ac ute angle in thi s problem ,

= arccos O. 41750 = 65 .32 0.

From equati o n ( I ) the an g le o f the slope can al so be de termined : cosO'

= 2cos(j! =

2 ·0.4 175 = 0.835 ,

so the an g le o f the s lope is 0'

= a rccos O. 835 = 33 .38° .

The pheno me non desc ribed in th e probl e m occurs o nl y at thi s va lue.

First solution of Problem 13. We c an sta rt by us in g the tw o co nditi o ns w hi c h state that the the two pi eces reac h the g round at the same mome nt, and that o ne o f them fall s back to the pl ace at w hi c h it was pro jec ted. From the first co nditi o n it can be co nc luded that only the ho ri zo ntal co mpo ne nts o f the ir ve lociti s chan ge , thu s th e he ights o f th e trajectories o f bo th pieces are the same as the he ight o f the path o f th e proj ec tile if it 85

300 Creative P hysics Prob lems with Solu tion s

does not ex pl ode, from the second we can draw the concl usion that the two pieces alwayS move in the same ' plane, whi ch is the pl ane of the trajectory of the origin al projecti le motion. The ce ntre of mass of the sys tem of splinters - according to the theorem of the mot ion of the centre of mass - moves along the trajectory of the ori gin al imaginary projectil e whi ch does not ex pl ode, and reac hes the ground at the range of the original projec til e, at the moment when the splin ters reach the ground. Let us descri be the motion of the projectile which does not explode. The verti cal component of the initi al vel oci ty is Vy = vasi n a = 200 m is, and its horizontal component is Vx = Vacosa = 346.4 m i s, The time whil e the projectil e ascends is tase = Vy I 9 = 20 s, thus the time of the projectile moti on is tpro = 2tase = 40 s, The total range of the projectil e moti on is Xmax = Vx tpro ::::: = 13856 m , and the height to whi ch the projectil e ascends is v~/2g = 2000 m. The vertical component 1.1y of the new veloc ity 1.1 of the splinter which goes back to the place of the projection after the ex pl os ion does not change, and its hori zontal co mponent Ul x can be calcul ated fro m the fi nal speeds given in the probl em. Because the vertical moti on is not influenced by the ex pl os ion the magnitude of the vertical component of the fin al veloc ity of the splinter which hits the ground at the original pos iti on is the same as the mag nitude of the vertical co mponent of the initi al velocity of the projec tile, which is: Va si n a

m

= 200 -

s

.

y

150 I

5km

x 10 km

15 km

The speed of the impact is given: Vi mp = 250 m is, from these data using the Pythagorea n theorem the constant hori zontal component of the veloc ity can be calculated:

Let us denote the time elapsed between the moment of the projec ti on and the exp los ion by t l . The time elapsed between the expl osion and the impac t is t = tpro - t 1 · During thi s time the splinter which goes back to the pl ace of projec tion covers a distance of Xl = Ul x (tpro - t 1) along the hori zontal , while the common centre of mass covers a hori zo ntal di stance of X2 = Va:" (tpro - t1) ' The sum of these distances is exactl y the total range of the projec tile moti on, thu s: XIlJaX = Xl +X2 = Ux(t pro - t1) +Vax(t pro -tJ) = m m m = 150 - (40 s-tJ) +34 6.4 -(40 s- t 1 ) = 346.4 - ·40 5 (= 13856m). s 5 s

86

~Ja ni cs

6.1 Kin ema.tics

Solu t ions

The soluti o n o f the equati o n g ives the time be tween the proj ecti o n and the expl os io n: t :::: 12. 087 s . J Thus the he ight of the ex pl os io n is: 1

2

m

1

22

m

h I = vovtl - -gt 1 = 200 - · 12.087 s - - . 10 2' · 12.087 s . 2 s 2 s

= 1686 .9 m.

Second solution of Problem 13. The pro ble m can be eas il y solved if we use the fac t that phenome na in mec hanics are revers ibl e. If the ve loc ity of the impact was re fl ected , that is , if the splinte r was proj ected with the veloc ity oppos it e to the the veloc ity at w hich the splinter hits the gro un d the n the splinte r, w hi c h fa ll s back to the initi al pos iti o n, would mo ve fro m the pl ace of the proj ec ti o n to the place o f the ex po los io n alo ng the same trajectory up on w hi c h it moves whe n it fall s back to the pl ace o f the proj ecti o n from the positi o n o f the ex pl os io n. Thu s the paths o f the proj ectile, which d oes no t explode, and the splinter w hi ch fall s back , intersect eac h othe r at the place whe re the . projectil e e xpl odes. There fo re, we o nl y have to write the equa tio ns o f both path s a nd to solve the equati o n system to y. The equati o n o f the path (o f the centre o f mass) o f the orig in al projectile is Y = tan a l

.x -

2

9

2 2 .X ,

2v o cosa l

The equation o f the path o f the splinte r whi ch fa ll s bac k to the pl ace of the proj ec ti o n, and which is " projected bac k", is

Y = tan a2

.x -

9 2 2 2 .X , 2v'i '/np cos a 2

where al = 30° , tan a2 = 200/ 150, cosa2 = 150/250. The trivial soluti o n o f the equ ati o n sys te m is x = 0, whic h is the initial po int o f both paths, and the soluti o n whi ch we wo uld like to find whe n x =1= 0 ca n be calc ul ated fro m the linear equati o n, whic h we ga in if x is cancelled . tan a l -

9

2V6 cos 2 a l

x= tan a 2

9

2 ? x. 2v i n> p COS- a2

-

Substitutin g the num eri ca l d ata: ° tan 30 -

1

200

2.400 2 cos 2 30°

x= -

150

-

10 2 . 2502

x.

(1 50) 2 250

The soluti o n fo r x :

x= 4187111, and substitutin g thi s res ult into the first equati o n the he ight at which the ex pl os io n occurred is:

I

11 =Y1

° 10 2 =tan30 · 4187111 - 2 . 400 2 cos 2 30 0 ·4 187 111= 1686.9111 .

(The ve loc ity of the splinte r w hi c h moves fo rward and the rati o o f the masses o f the SPlinters c ann ot be calc ul ated fro m the g ive n d ata.)

87

300 C reMi\'(' Ph,\'s ics P ro b/el/l s \\'ith So /utioll s

------------~----------------------------------------------------------

Solution of Problem 14. Let lin e (J be the path of the bullet , point A is the lai r oj' the ga mekeeper, (The descent of the bullet ca n be neglec ted in thi s phase o f th e moti on,) AU=d = 1111 , During it s fli ght. the bullet produees sphe r_ B v a ll y propagating sound waves at each po int ica • of th e path , The enveloping surface to th ese at a gi ve n moment is a conica l sur face w hose genera tri ces mark th e max imu ill di stan ce reached at th e moment in co ncern , A fter time t , when the sp heri ca l wave of th e bUllct A is at di stance cI from poi nt C, then the bUllet itself is at a distance of vi from th is place (point D show n in the li gure), Point f) is the ape x of the enveloping co ni cal surface (beca use the spherical wave start s from thi s poi nt exactl y now),

•

The sound ca n he sensed when th e en vel oping co ni ca l surface reaches A, Then for the se miape x angle of the co ne on one hand:

a

,

o

SI11 0 =

CA

ct

C f)

uI

(' v

- - = - =- ,

on the other hand

, S lll O

AB AD

d

= - - =- , ,I'

where x is the unknown di stance , From these ,(' =

v

C

,d =

G80 Ill / s :~4 0

Ill /s

'I

III

= 2 Ill.

6.2 Dynamics Solution of Problem 15. Let us aSSLlme that the object is pointlike, T he object should reac h the top of the semicircle, after w hich it s motion will be a hori zo ntal proj ecti on, If th e objec t is to reach the top of th e sem icircle. the norm al foree should not beco me zero anywhere oth er th an at th e top, According to Newton ' s second law :

III l] + i-;

P:

exe rt ed hy th e track

= lil a,

w hi ch at th e top of th e semi circle takes the form o f : ')

III {)

88

+ [\-=

(' -

111 - . /'

( Il

6 .2 Dy nam ics

6. Mechallics Soilitio lls

-----

where v 2 / I' is the ce ntripetal accd eration of the object. (As the vectors are all ve rti cal, their magnitudes ca n be used in the equ at io n. ) The object will reac h the top of the elllicircle if the normal force remains pos iti ve or becomes zero there. Since the ex treme ~asc shou ld be in vesti gated (because that is when the vel oc ity Vo and therefore the distance of horizontal projection is minimum ) let us substitute I\' = 0 into equ ati on ( I): v-'J

mg = 111 .J2. ,. . from whi ch

VIl =vrg.

(2)

Thi s hori zo ntal ve loci ty is maintained throughout the horizontal project ion until the object hits the track in it s origina l pos iti on. The time of fall equals the time of a freefall from a he ight of h = 2,.: l

=

(2h = f4i . vr; v-;;

The hori zo ntal co mpone nt of the obj ec t's disp lacement d urin g the fall is: .1'

= vo t =

.;rg ff =

21'

It is interesti ng to see that the hori zontal disp lace ment equal s the height from wh ich the object fall s. This di stan ce is the shortest poss ible distance of the horizontal track that we wanted to fi nd. The ve locity given to the object in its initial pos iti o n in the ex treme case can be ca lc ulat ed usin g the conservati on of energy: 1

?

"2 1lLV -

=

1

?

mg2,. + "2mvo,

hence

V= J ,.g+ v'6 = ) 4I'g + ,.g= j5;:g

(= Vsvo).

Solution of Problem 16. a) Co nsiderin g the requirement s of the problem, let us start the reaso nin g at the end. The object leavi ng the track at !3 becomes a projec til e with a 8 va g··· .. •. .... . hori zo nt al initial ve loc it y of (1/3. Since it ..... IS to trave l throu gh C, it necd s to cover .. ··· Fa !R ......................... . a hori zo nt al dista nce of d wh ile fallin g A ~~--"'--.....__ __ ___________ _________ ~c.o through a vert ica l di stance of n be low the FA 0 d C Point B. T he eq uati ons for the displacement of the prujec til e arc

·!f'm

('/J I =cI

and K

89

300 Creative P hysics Problem s wit h Solution s

------------~-----------------------------------------------------

Hence, with the elimin ati on of t, Va

=d ·

;;t;..

Given the speed at B, the law of energy conservation ca n be used to determin e the speed at J{ : 1 2 1 ? 2mv I< = mg( h + R ) + 2 mvn· With the substituti on of the above ex pression for Va, the speed at X is V I<

=

2g (h + R )+

2;

d2

m 9 m 2 ·10 !.4 I1l 2 ' 10 -(2 m + 1m)+ s- =10 .25-. 52 2 ·1 Jl1 5

=

b) Since the obj ect mu st slide all the way along the track to get to point force cannot decrease to zero anywhere along the track . Th at im poses a the initi al speed of the projectil e motion. Thus the horizontal di stance d enough to be reac hed durin g the time of the fall. The norma l force Fa trac k at point B is obtained from the eq uati on fo r the centri petal fo rce:

B , the normal lower lim it on has to be lo ng exerted by the

2

2

d - 1) > O. Fa = m -V - 1n g = 1n g ( --R 2R 2 -

(1)

The value of zero corresponds to the minimum value of d , thus d ll1i ll = Rh = l. 41 111. c) At the po int A , the normal force of the track changes suddenl y fro m the value of zero along the line segment J{ A to the value FA required by circul ar moti on. At po int A of the circ ul ar arc, V

2

FA = m ~ . The value of V A is obtained fro m the law of energy conservat ion aga in, appli ed between the po ints A and B separated by a verti cal di stance of R: mv 2

'm v 2

mgR +--a = _ _A . 2 2 Hence the square of the speed and the normal fo rce at A are

2 VA

2

d g = 2R g + 2R )

2

FA= m g ( dR22+2 ) = 0.5 kg .10m2 ( 9m 2

5

2· 1 m 2

+ 2) = 32 .5 N.

From ( I) and (2), m

Fa = FA -3mg= 32.5 1- 0. 5 kg · 10 ? = 27.5 1 s-

90

(2)

--

6. Mech anic' So /ll tions

6.2 D Y II
Solution of Problem 17. Let us describe the in stantaneous positi on of the objL:ct with the angle a between the vertical and the radiu s drawn to the sill all objL:ct. WhL:rL: thL: wall e nd s, the object undergoes projectile Illotion with an initi al angle of Ci as WL:II. The om iss ion of' so me part of the wa ll does not lead to the failure of the trick if thL: downward part of the parabola smoothl y fit s to the circle agai n. From thi s, it is derived that the mi ss in g part of the circul ar track mu st be sy mmetri cal along the verti ca l diameter of the circle. Thi s condition can be considered as the horizon tal compo nent of the di splaccment of the object durin g the time of the asce nt of the object (half of the range of' the projectile moti on) and is the same as the half of the chord whi ch belo ngs to thL: arc cut off the circle. Let us determ in e the ce ntral angle subte nded by the arc cut ofl Let the magnitude of thi s angle be 20. . The time of the ascent of the projecti Ie moti on is: vsin a

ta= - - - . 9

The di stance covered during thi s time is: vcosa · La =

v cos 0 . v s i11 0.

=R sin o

9

Rg vThe initial speed of the project ile moti on ca n be ca lculated using the work-energy theorem: From thi s cosa =

-? .

1

1

'J

-mgR ( 1 +COSo. ) = -mv 2 - - m vij, 2 2 Rg from which: v2=v6- 2gR(1 +cosa) , and cosa= 2 ( Vo - 2Rg 1 +co c.~) This is a qu adrati c eq uat ion for the cos ine of half of th e asked angle : 2

COS o. +

In our case that:

V2 0

2R g

(1 - ;~g

)

coso. +

.

~ = O.

400

2.8 .1 6.9. 8 = 2.5, whi ch can be substituted into our L:l)u ati on, such 1

'J

cos - 0. - l. 5 cos 0' + 2' = 0 , from whi ch 2 so luti ons are gai ned for coso' : casa l = 1,

and

caso.2 = 0. 5.

91

300 Crea.tive P hysics Prob lem s wit h Solu t ion s

------------~------------------------------------------------------

The soluti o ns fo r the central a ng le of the arc which is to cut off are: epl = 20:1 = 0° and o

..

..

ep2 = 20:2 = 120 , so the le ngth o f the c Ircul ar path IS:

~

2n

= R ep = - R = 17.1S m . 3

Solution of Problem 18. T he object mo ves o n a verti cal c irc ular path , d uring whic h the te ns io n in the strin g decreases. The max imum te ns io n occ urs at the bottom of the c irc le, i.e. at the start o f the mo tio n. If the string does not break at the start, the re are two possibiliti es: I . If the initi al veloc ity is e no ugh to make the o bject to move arou nd the circle w itho ut the string beco ming s lac k, the strin g w ill never break. 2. If, ho wever the o bj ec t leaves the c irc ul ar path at a g ive n po int (w here the strin g becomes slack), i; w ill unde rgo projectil e motio n until it reaches the c irc ul ar path again , where the string w ill become te nse. In thi s case, ass uming th at the strin g is not stretchabl e, the tens ion in the strin g w ill be extre me ly hi g h (well above the max imum 40N) and because it is pulled sudde nl y, the string will break . In thi s case, the soluti o n becomes the point of intersecti o n o f the parabolic path a nd the circle. I . The te ns io n ac tin g at the start can be dete rmined us in g Newto n' s second law: J(

y

v2 -mg= 1n L'

whi ch y ie lds J(

= m

V2 ) (196m2 /s2 m) ( ~L + g = O.S kg S.6m + 10 2"s =

x L

=22.SN < 40N , so the string does not break at the start of the mo ti o n.

2. T he point in whic h the proj ectile moti o n of the • Vo o bjec t starts is the point where the strin g becomes s lac k, i.e. te ns io n J( becomes zero . Let ep be the angle fo rmed by the string and the vertical in the above point. Ap pl ying New to n' s second law to the o bj ect in that po int, we o btain :

v2 mg cos ep + J( = mT '

(1)

where /\- = O. T he veloc ity ( v ) of the obj ec t in that point can be de te rmined us in g the wo rk-kinetic energy theorem :

1 2

2

1 2

2

- mgL(l +cosep) = - m v - - m vo (the wo rk of the te ns io n is zero) , he nce

v = v5 - 2gL(1 +cosep) . 2

S ubstitutin g equ ati o n (2) and J( = 0 into equ ati o n ( 1) a nd di vidin g by m g ive:

v2

gcosep= l - 2g( 1 +cosep), 92

(2)

Nkc::.: h~ a.J_l_ i c_s_S_o_h_lt_io_'_1S_________________________________________ 6 _.2__ D~Y_ I1_ a_ m_i_cs

~

which yi e lds V

2

3 coscp = ~ - 2 ,

gL

hence th e cos ine o f th e ang le th at g ives the posi ti o n in w hi ch th e o bj ec t leaves th e c irc le is: coscp SO

V6

=-

3gL

2

- -= 3

142 111 2 /S2

2

2

- -

3·lO m /s ·5 .6m

3

= 0. 5 .

(3 )

'

the ang le fo rm ed by th e strin g a nd the ve rtical turn s out to be:

cp = 60° . The launch s peed of th e proj ectil e mo ti o n can be calcul ated by s ubstitutin g the value o f COS ip into equ ati o n (2):

v == VV6

- 2gL (1 + 0. 5) = V1 4 2 m 2 /s 2 -

2 2 · 10 m /s . 5. 6 m .1. 5 =

58

m

s

= 5 .29 m. s

(4)

Accordin g to th e fig ure , th e coord in ates o f th e po int w here th e obj ec t le aves th e circle are: Xo = - L sin cp,

Yo = L coscp , because the an g le above the hori zo nta l at w hic h the obj ect is laun c hed equal s th e ang le formed by the strin g and th e verti cal. The equ ati o n o f th e c ircle in th e coordin ate-syste m de fin ed in the fi g ure is:

X2+y2 =L2. The coordinates o f th e o bject du rin g its proj ec til e mo ti o n as a fun c ti o n o f time are:

x

= Xo +vcoscp · t = -

(5)

L sin cp+ vcoscp ' t,

. 1 2 L . 1 2 Y=Yo +vslll cp· t - gt = coscp+vs lll cp . t - gt.

2

(6 )

2

Inserting these coordi nates into th e equati o n o f th e c irc le , we get:

L 2sin 2 cp - 2Lvsin cp cos cpt + v 2 cos 2 cpe + L 2 cos 2 cp + v 2s in 2 cpt 2 + + 2Lvcoscpsin cp · t - 2L coscp · ~ gl2 -

2

~g2t4+ 4

2vsi n cp ~gt3 = L 2. 2

Usin g th at L 2 sin 2 cp+ L 2 coS2 cp = L 2, v 2es in 2 cp+v 2e cos 2 cp = v 2 t 2 a nd th at th e sUm o f th e second a nd seventh te rm eq uals zero , o ur equati o n s impli fie s to:

v 2t 2 +

~g2 t4 4

L coscpge - vsin cpgt 3 = O.

DiViding by t 2 =1= 0, we obta in :

~g2t2 _ vsin cpgt + 4

(v 2 - L coscpg) = O.

93

300 C rciH il 'c Ph,I's ics Prob/clIls wit/l So /llti o n s

------------~----------------------------------------------------------

This is a quadrati c eq uati o n for I. w hose so luti on is: 1,:2 si 11 2 yy '2 + .(;2 ( Leos y[) - t,2) l ' sill yY ± , = --------~------71--,)----------------

2[r after silllrlifying the rraction by 9 . rart o r the exrression under the squ are root can he written as: thu s

1= 2. I' sill y ± J( L9 -

t,:2

cos Y) l'OSY .

9 Substituting known va l ues anu using equations (3) anu (4), WL: get:

1=2 .

5.29

~ · 0.866 ± '

,~~2 ,.

/ (5.6 1l1'1O ;;':f - 28

V

·0.5) ·0. 5

10 .!4 ~-

hence

a ile! Substituting till: va lue or II into the .1' and y coordinates o f the object as givL:n in equation s (5) and (6) givL:s the position or the object at thL: Illoment when thL: string breaks: :1:

= (- 5.6 III ·0. 866 + 5.292 . 1. 8:.l3· 0.5) III = - ( 1. 8 I96 + 4 .85 0) III ~ 0, Y = (5 .6 · 0. 5 + 5.292·0.866·1 .833 - ;) , 1.833 2 ) III = - 5.6 \11 = - L, L

which means that thL: object comes back illlo its initial ro sition (a t distance L bL:low thL: point of suspL:n sion) whL:n thL: string bn:ak.

Solution of Problem 19. The positi o n o f thL: object can hL: dL:scribed hy angle (\. w hi ch is the angle L:nclosed by th e verti ca l and the radiu s co nnecting the centre of thL: sL: lllicircle and the objL:ct as shown on the figure . The acceleration of thL: objL:ct can bL: reso l vL:d into tangential and normal co mronL:nt s. which arc a, = gS ill Cl and (J /I u2 / I? , whL:re u is the in stantaneou s vdocit y o f the object in thL: g ivL:n position . As th e two co mpo nents are pL:rpL:ndicular, th L: nl,;t acceleration or the object can be ca lculat L:d as:

=

Usi ng the conservation of energy , we call L:x pn:ss thL: velocit y o f the object In tL:nn s of angle ct:

R

1119 Rei mg

94

= -1 lI lU-. 'J

(,OSCl')

2

6. ]\lecha llics So /uti olls

6.2 DY llamics

----

which yie lds v2

= 2gR ( 1 - cos O') .

suhstitutin g thi s into the formul a for acce lerati on anu ass umlll g that the acceleration should be two th ird s of g . we get: a = gVSin 2n + 4(1 -

('OSO) 2

after some alge bra, we get a quadrati c equation for 27 cos

2

(\ -

72 coso

= ~. g.

(,OSCI :

+ ·11 = 0,

the only so luti on to the equatio n that has a phy sica l meaning is: coso

=

J21

12 -

9

= 0. 8242.

The net acceleration and the tan gent line form an angle 'P , for wh ich we get: 9 sm C\

cos

-/ .- = 1. 5s ill C\ =0 .8LI!.l6 ,

2g 3

--7

'P= 31. 8° .

The angle enc losed hy the acceleration and the hori zo ntal is

E:

= 'P + Q = 66.3° .

Solution of Prohlcm 20. The object leaves the slope when the normal reacti on becomes zero. In a our case thi s may happen along the part of the path wh ich is co ncave downward , with the chan ce of it v, happenin g heing greatest at the lowest points. Let R us assume th at the object is a point -lik e one. The crit ica l positi o n is the inflexion point of the path , so if the normal reac ti on hecomes zero there for a moment , then it will increase abruptly to a high R value becau se of the opposite curvature. Let us cons ider thi s case as the critical case, because if the , mg a object reac hes thi s po int at a greater speed than the Speeu whi ch bel o ngs to the crit ical case the n it will Surel y leave a finite segment of the path . .The qu esti on ca n be stated in the fol low in g way as well: What sho uld the initi al speed 01 an object whi ch sliues down on a smooth circ ul ar path of radiu s R be, so th at it it leaves th e path just as it desce nds from a hei ght o f "/ 2 '? Let th e speed of the object at the top of the path be V I, and U2 after descendin g a h~lght of 11 / 2 . During th e moti o n of the objec t the resultant force is the vec tor sum 01 the normal reacti o n and the grav itati o nal force. At the mome nt when the normal reacti on ceases the resultant force is equ al too the gravitational force . At thi s moment

-

95

300 Creative Physics Problems with Solutions

the object moves along the circular path of radius R making the centripetal forc e equ al to the radial component of the gravitational force. Thus for this position:

v~ m g cos a = m R ' After calculating the speed the law of con serva_ tion of energy is applied : 1

2

1

2

m g6h = 2mV2 - 2mvI ' where 6h

= h/ 2.

From this

v2 =Vgh+ vi , and

vI=Vv~ - gh.

Expressing V2 from the first equation and substituting it into the last one: VI

= VRg cosa- gh.

Reading from the figure

cos a =

R - h/2 h = 1--. R 2R

using the value of cosa and substituting the data of the problem , the asked speed becomes :

( 3h) = 4.43 -ms .

Rg 1 - 2R

Remark. The fact that the circular paths join 'without break ' does not mean that the object passes ' smoothly ', because at the connection the normal reaction abru ptly increases from zero to a very high value, allowing an elast ic ball to be 'ejected ' at the inflection point. Here the path cannot be considered as straight, unlike, for example, a sinus curve at its inflection. Let the mass of the sliding object be 1 kg: the normal reaction exerted on it at the inflection point of the curve increases from zero to a certain value of N for which N - mg cosa = mv 2 / R , from which

2

N=m(gcos a + v / R)=lkg·

2 ( 9.81m/s 2' S+-54 4.43 m/s 2) =ll.77 N. Solution of Problem 21. At the moment when the object detaches from the slope, the reaction force of the slope becomes zero , so Newton ' s second law in radial direction is: v2

.. -.... - . --.~~~~...--

96

fR "'\\\

m g cosa=m R'

6.2 Dy nam ics

6. Mech a.nics Solu tion s

-----

According to the work-energy theorem:

mg[R( l - cosO')

1 + h] = 2' 17W2 ,

frorn w hi ch th e square o f th e speed is

v

2

= 2g[R(1 -

cosO')

and, according to th e give n data , th e cosine of cos 0'

+ h],

is

0'

',iR

3

R

4

= -" - = - .

Plugging in all these formu las into New ton 's second la w (ye t in parame tri c form ), we get:

2gR( 1 -cosO') +2hg R .

gCOSO' =

. From this the altitud e h measured from the co mmo n ho ri zo ntal tangent of the arcs is:

h=

( ~cos

0' -

1)

R,

and substituting the value of coso', we get that

h=

(~ .~ -

1) . R =

( ~ - ~ ) . R = ~ R.

The who le distance covered by th e small objec t o n th e s lo pe is th e s LI m o f th e distan ce s covered on the concave and convex parts:

The first part of the d istance can be determined from ha nd R.:

81 = R

7 = R· arccos - = 0.5 8

~ h = R . arccos ( 1 - ~) In'

0 .505;:::0 0.253

= R· arccos ( 1 -

~)

=

Ill.

The second di stance, acco rding to th e data given in th e problem , is: 3

82 = R O' = R · a r ccos - = 0. 5 4

Ill'

0.72 27 = 0.361

111.

ThLis the smal l object covered a tot al distance of

8=81+ 82= 0.2 53 m +0.361m=0.6141l1 =6 1A

Clll ,

on the slope, which has the parametric form

(

3

7) .

8 = R

97

300 C reative Phy sics Problem s with Solutions

Solution of Problem 22. Let us solve this problem in general for an arbitrary initial angle 0 < a o < 90 ° .

r

a

b)

0)

Figure a) shows the forces acting on the bob. The resultant force of gravitational force and tension j{ points towards the inside of the circle, its tangential compone nt being equal to that of the gravitational force. Figure b) shows the accelerations of the bob, which can be derived from figure a) by dividing each force by the mass m of the bob. The magnitude of the net acceleration can be written into the form of:

mg

a=va;1 +a~ = (Vr2)2 + (gsina)2 .

(1)

The velocity of the bob can be calculated using the conservation of energy formul a. While the angle which the cord makes with the vertical decreases from a o to a , the bob ' s height decreases by : I ~h l

= r(coSQo -

cosa).

Applying the conservation of energy formu la and inputting the initial velocity as zero, we obtain:

1

2

mg l ~h l = mgr(coSQo -cos a ) = 2mv ,

hence v 2 = 2gr (cosao - cos a ). Insertin g this into equation (I), we get:

a = 9V[2(cosa -

2

cosao)]2 +s in a

= gVSin 2 a

+ 4[cos 2 a - 2 cosacosao +cos 2 a o]

0:;:

(2 )

98

6.2 Dy namics

6. Mech an ics Solutions

:..---,[he bob will reac h its minimum accelerati o n whe n the ex press io n under the squ are roo t takes its mini m um value . T hi s express ion is the second degree fun c ti on of cosa: 3cos 2 00- 8cos a ocosa + (1 + 4cos 2 (00) .

(3)

,[he genera l fo rm of a qu adratic functio n of x is:

ax 2 + bx+c.

By calculating the zeros of thi s fun ctio n, yo u will fin d the x -inte rcepts of a parabo la, with an ax is para ll e l to the y -a xis. T hese intercepts are sy mme trical to the ax is (a nd the vertex) of the parabo la, therefore the x -coord in ate of the minimum poi nt of the parabola is the ari thmeti c mean o f the two zeros:

[0

Xll!i ll = (X l

+X2)/2,

where

and Calculatin g the ir arithmetic mean g ives : xn 1in

b 2a

== - - .

Let us apply thi s to the sec ond de gree e xpress io n o f cosa . T he angl e at whi c h the bob reaches its minimum acce lerati on is g i ve n by: 8coso~ o 4 cosall!i ll = --6-- = "3cosao .

Using this calc ul ati o n in o ur case the initi a l a ng le acceleration is reac hed wh e n: 4 12 cosalHi ll = "3 . 2

IS

(5)

000

= 45 ° a nd the mIJ1imum

2V2

=-

3- = 0.9428

from whi ch the ang le formed by the cord and the ve rti cal is:

The questio n set in the probl e m is therefore answered . . Let us calc ulate so me add iti o nal d ata regard in g the minimum accelerati o n. Su bsti tutIng equation (5) fo r eq uati o n (2), we arri ve at a minimum accelerati o n of:

/ 16cos 2 ao 4cos 2 ao 2 a m ill =g · y l +3 . 9 -8 · 3 +4cos 000= =

~.

J

(48 - 96 + 36) cos6 +9 =

~ . )9 -

12cos 2 a .

(6)

99

300 C r ea til"(' Pi/.I'sit·s Probielll s witli So illti o lls

----------~-----------------------------------------------------

In our caSL:

011 =

I j O,

so

(,0';011 =

Ji ')

In sL:rling

this into eq uation (6) . ror 11lL: Illinillluill -valuL: or thL: accL:ie rat ion WL: gL:l:

(;\ a

(lllI i ll

==

:39

9 - 12·

'1)2:3 III = - ·CJ=0 . j77CJ~5 . 8--:- . :3 . . S2 ( '1

Angle <,? rOf'llll':U by the IlllnllllUIll accL:leration and the tangent o r tilL: path is: ,l'}sillCl

cos' = - - = y Ji .9j

Sill 19. 17°

1}/ :3 vo)

= 0. 577:3.

1--I

\

l~,/

thu s y = 5.1.7:3° , whi lL: thL: angle forlllL:u hy thL: minimum acceleration and Ihe horizontal is:

Solution of Problem 23. a) WhL:n thL: hlocks mOVL: with the samL: acceleration , their acceleration equa ls thL: accL:leration of thL:ir centre or mass. which can hL: dL:tL:rJn illed casi ly. ThL: sum or the cxterna l rorce s acting on the sys tL:m eq ual s th e mass or the systL:1ll tilllL:s the accL:leratio n or thL: cent rL: or mass:

Aply in g this to thL: systL:m or blocks , WL: obta in :

Us in g

1111

= /112 = III

si mplilies our equation to: 211lg s illo -

( /11

+ /1 2)lIIgCOS(\ = '11110 ""1111'<"

rrom wh ich WL: Ii nel: .

/11

+ /' 2

a = ,I'} (s IIl Ct ------- cosn)= '1

III

.

- 0

10 ~( S I1l 1 o

s-

0 .,1 .-0 • • III - - cos Jo ) =0.()56 ~ . '1 s-

Note that thL: upper hlock ( ir il was not cO llilectL:d to thL: lowL:r block) wou ld not star! to Illove on the inclin L:d plane by itselr or wou ld stop alkr bL:ing pus heu down wards because the rriction acting on it is greater than l1I,C} sill o. But as the rriction actin g oil the lower block is sma ll. the 10wL:r block moves down pullin g thL: upper b lock with it. b) ThL: L:xtL:nsio n or the sprin g could bL: dL:terlllinL:d by app lying Newton 's SL:CO l1d law to either block and substitulin g th e va lue or acceleration calcu lated above . L ei LIS now , however. calculatL: thL: extL:nsion uirect ly , using on ly the givcn uata and not tllC accL:lL:rat ion.

100

6.2 DYlla mics

6. j\/cc/Iil Jlics Sol utio Jl s

.:.:----

Newton' s seco nd l aw app lied to th e lower block gives: IIIgsil1 Cl - 1'-2I11gC05Cl - Dt:!./

= liLa ,

,vhile apply ing Newton's seco nd la w to the upper bl ock. we obtain :

+ f)t:!./ -

IlIgsi l1 0

III I IIgC050

= ma o

subtractin g th e lirst equati on rrom th e seco nd , th e accelerati on and mgs in O' cance l out:

2Dt:!.I-11 I Il1g cos Cl + Jlzlllg cos Cl

= O.

fro J11 whi ch we get :

t:!.l

= -IIf)I " /l[ (I -

- 1' 2 -- .

2

co. Cl

'

inserting the gi ve n data. we fi nd :

t:!./

=

:3 ko' 111 O? / · 10 --:;- . .-:.:.. cos 15° = 0.0 1448 ~ 1.45 cm. 200 :\ III s~ 2

Solution of Prohl em 24. Co nsider th e moti on or th e sys tem . According to Newton 's second law . wi th th e notati ons or th e fi gure,

(1) where FI i s th e rorce or stati c rriction , F2 is the rorce or kinetic friction, and a., is th e acceleration or sys tem. Note th at there are tw o rriction rorces o r oppos in g direc tion s, ac ting on the sys tem at the same time . The first is oppos in g th e mot ion, whi le the second , acting in the direc tion of th e moti on, advances translati on and retard s ro tation . The rri cti on rorce Fl is obtained by co nsidering the torque s ac tin g on th e cy linder: ~

1

.,(( "

I IR =C:W=-1\ /R- - . 2 R

v

Hence

and

IS = /' (lIIg -

1\' sin o).

•

With th e fri ction forces substituted in ( I ):

(2) The ten sion force 1\' ac tin g on the string ca n be determined lI sin g Newton's second law applied to the cy linder: ~ ~

/\' cos n - FI

= 1\ 1(/ ., .

101

300 Creative Physics Problems with Solutions

With the expression for

F1

substituted ,

J(

J(=

can be expressed:

3Ma s

.

2 cosa

If this result is written in (2) , the only unknown that remains is the acceleration:

3M a, 1 ( p,mg - {l - - " sin a = - M a s + m+NI)a s . 2cosa

2

Hence the acceleration of the centre of mass of the cylinder (and the acceleration of the plate) is

a, = "

2{lm

g.

3M( 1 + {l t a n a)+ 2m

According to the given information , m = 2NI and t an a = 13/3, thus:

as =

4p, 3(1 + {lta n a) +4

=

l.6

3+0 .4 .j3 + 4

m 111 ·10 - = 2.08 - . s2 s2

Hence the distance covered and the time taken are V

2

8= _ 0.g= 2a s

4m 2 /s2 2 · 2.08 m /s

2

=0.98 m

28

t= - =

and

Vo

2·0.98 111 / =0.98s.

2m s

Solution of Problem 25. a) The hoop will get higher if there is frict ion , since in that case a part of its rotational energy (or poss ibly all of it) is also converted to gravitational potential energy. This statement can also be proved true by considering the forces acting on the hoop: as the hoop ascends, the speed of its centre of mass decreases. To decrease its angular velocity, there is an uphill friction force acting on the hoop, which represen ts a lifting force on the hoop. It is instructive to compare the heights quantitatively : if the hoop rolls up the slope

1mv6 + 18w 2 2 2 mg mg 1 2 2 E t r a n s ] 2 m v O vo without friction its rise will only be 6h 2 = - - - = - - = - . mg mg 2g .

.....

.

6Ekin

wIthout shpplllg, It wIll rIse to a heIght of 6h 1 = - - - =

v2 0 9

whi le

The ratios of the two heights for a hoop and for a sphere are , respectively ,

6h

7

1 - 1 .4 . - 5 6h- 2

If there is only a weak friction on the slope and the hoop rolls with or without slipp ing, it will rise to a height 6h , such that

6h 2 < 6h < 6h 1 , and part of the energy is dissipated . b) If r < R , the hoop reaching the slope will roll on up the slope without a collision . On arrival , the speed of its centre of mass is vo and its angular speed is wo = voir. 102

6. Nlechani cs Solutions

6.2 DY Ila.mics

In the absence of friction (since there is no external torque act in g on the hoop) , the angular momentum of the hoop, and thus the angula r speed too, wi ll rl:main co nstant all the way . The hoop wil l slide up to a height of h = v6/ 29 ,

where its speed is reduced to zero , and where it wi ll then slide back to the base of the slope. It aITives at the base of the slope at a ve locity of Vn again , but in the opposite direction. Its sense of rotation does not change. Therefore, the hoop arri ving back to the horizontal plane keeps sliding, and wil l con tinue to slide until the torque of the friction forces adjust the angular velocity to a va lue of w = V/ 1' ' in tune ' with the instantaneous velocity and in the appropriate direction. In princip le, this may occur in threc dinerent ways. Fricti on may reverse the sense of ro tati on , it may reverse the direction or velocity , or, in a special case it may reducc both to zero at the same time. The result will depend on the magni tude of rotational inertia. To solve the problem , this question need s to be investigated. The diagrams below represent the three cases graph ica ll y. (t = 0 at the in sta nt when the hoop arri ves back down to the base of the slope.)

Calculations: The acceleration of the centre of mass of the hoop after it has sl id back down on the horizontal plane: a = {lg = const. The angular acceleration of the hoop: (3 =

-

{lmg/'

= _ {lg = con st.

ln T2

T

The speed and angu lar speed of the centre of the hoop expressed in terms of time: v

= -Vo + at = -Vo + {l gt ,

and

W

= W o + (3t = Wo -

~lg

-

r

t.

Sliding stops in a time l 1' such that -V O + ~l gt1

=wo/' - {lgl1 ,

hence, and

Un 3. 5 m/s t [ =- = 2 = 1. 755 . fig 0.2 ·10 lll /s

1m

300 Creative Physics Problems with Solutions

The speed of the centre of the hoop at the same time instant is Vo V = -vo + /-ig =0 , {tog that is, we are dealing with the special case in which the hoop stops exactly whe n its angular speed also decreases to zero. Therefore it will not reverse either its veloci ty Or angular velocity after its return. Thus the hoop will stop in 1.75 s , which means that at 2.4 s, the time instant in question, it will be in the same position at the point of the horizontal plane where its coordinate is 1 m 1 m 22 2 8 = X = -vOtl + -{tgt l = - 3.5- ·1.75 s+ - ·0.2 · 10 -2 ·1.75 s =~ - 3.06 m, 2 s 2 s with the origin at the base of the slope and the positive values of x under the inclined plane. Remark. The case of a solid sphere is more complicated because the sphere will , after a while , continue to roll without slipping while its angular velocity is reversed . In th is case, the given time interval is then made up of a decelerating part and a part of uni for m motion . w = 0 at 2 Vo t] = - - = 0.7s , 5 {tg when the velocity is v = Vo - /-igt l = 2.1 m/s. The sphere rolls while slipping for this much time and then continues on unifomly. The distance rolled while slipping is 8 1 = vot l - /-igt l = 1.96 m , and then it rolls without slipping until the interval of 2.4 s is over, covering a distance of 8 2 = v(t - t l ) = 3.57 m. The total distance covered fro m the base of the slope during the 2.4 s is therefore 8 = 5.53 m. (The x axis now poin ts the other way.) A case of an object with cylindrical symmetry rolling back towards the slope may only occur if its rotational inertia is even greater than that of the hoop. (For example, a reel rolling with its narrow part on a track) .

Solution of Problem 26. The motion of the block is retarded by the kinetic fri ction force and the force that is exerted by the dropped object parallel to their surface of contact. The force of kinetic friction is

S=

{t[(,

where [( is the normal force exerted by the ground on the block . This force has a small value before the collision and will therefore not change the speed of the bl oc k significantly during the ' infinitesimally short' duration of the collision. When the small mass falls onto the object, however, the normal force increases considerably, and thus friction can reduce the speed of the block by a finite value even 'in an instant' . Quantitatively: The vertical component of the change in momentum of the small object equals the total impulse of the vertical forces acting on it:

LFy .!:"t = !:,. mv l v · 104

6. Mechanics Solu tions

6 .2 Dy nam ics

----

In detail , since the small object loses its vertica l veloci ty,

(mg - N)6t = - mVl, where N is the mean magnitude of the normal component of the force exerted by the block on the small obj ect (equal to the magnitude of the normal force acting on the block). Hence 7nvl N=mg + - - ' . 6t The resultant vertical force acting on the block is therefore M 9 + N - ]( = 0, since the block does not accelerate vert icall y. Thus the magnitude of the fr iction force acting on the block is K S = jJJ(

r~; ;gW ,k~~_~l

= ~i(N + M g).

v,

-

u

I IConsider the system comprisin g the block ' and the object dropped onto it. The Mg change in momentum in the x direction is made up of the change in momentum IvI 6 u of the block and m 6 v h = mu of the small object, where u is the veloc ity of the block after the co lli sion . From the law of impulse and momentum applied to the x componen ts, with the coordinate axis pointing in the directi on of the velocity V2 of the block: I

I

-M(N + M g)6 t = M 6 u+mli. With the value of N substituted : - ~i

7TW1) 6 t=M 6 u + mu. (mg+Mg+-6L

Multiplied by 6t: -~i(mg 6 t+Mg6t + mV l )

= M 6u+mu.

Since the collision is momentary , 6 t ---> 0, the only term of the expression in brackets not vanishing is mVl . (Considerin g that u = V2 + 6 u) , the change in velocity of the block is

6U= _ m(~iVl + V2) = _lkg . (0.4.1 0m /s+2m/s) = - lm/s. m+M 1 kg+5kg Thus the velocity of the block after the collis ion is

u = V2 +

A

LJ.1i

=

IvI V2 - ~imVl m = 1 -. m+M s

Solution of Problem 27. According to Newton's second law app li ed to the radial COll1ponents of the forces actin g on th e ball,

lOS

300 C /'cat i,'c P h,l's ics Prob lcllls \\'itJ l Solut io ll s w hen.: [ \' is th e te nsion rorce act in g o n th e strin g. theorem. IIIg/COS(I

=1

Accord ing to th e work-energy

'J

-1))1;-,

2

Fro m th e two equ ati ons, th e te nsion rorcc is

Ks ina Kcosa

(1)

CJ.

K

A t th e ang le o r th e strin g to th e verti ca l ror w hi ch th e stati c fri cti on roree is a max imum ,

K mg

New to n's seco nd l aw appli ed to ' he block at the time in stant w hen the bl ock start s to sl ide (th at is, w hen it has not yet acce lerated hut w hen th e stati c fri cti on force has reac hed it s max imum ) is /1 ,( I\ [g+ [\' coso) - [\'si n o = O.

(2)

W ith the ex press ion ( I ) o f /\' w ritt en in (2), and rea rran ged : 11 ,1\ / g

+ 3/1, IlIgC05'2 0 = 3171gcoso sin o.

Div ided hy 9 and w ith cos in es expressed in term s o f sines :

/1,, 111

.1'

+ 31L , 1II -

3lL,llLsin 2 0

= 3m sin (1 J ] - sin 2 o.

Rearrange ment lead s to an equ ati on th at is quadrati c in sin 2 (1. With the nota tio n = sin 2 0 int rodu ced :

1\/ .) +9 ] .1:+/1,2( -') 1\/ 2 + 61\/ + 9 ) =0 . 9 ( ,1 + 1' 2) , .C.2 - [.6/1,2 ( -+3 III

111-

III

W ith th e substituti o n o f th e g i ve n num eri ca l data , the equati o n beco mes simpler :

9,36.(2 - 10.2.1 + 1 = 0 , to.2 ± J I02 2 - ,I· 9.36 2·9.36 th at is,

a nd

.

.1:2

'J

= S Ill- 02 = 0. 1089.

T he va lu es ob tain ed fo r th e sine o f th e angle arc sin o 1 = JO.9808 = 0990:)6,

a n ci

S ill el2

=

J O.I089

T he bl ock wi ll start to sli de w hen th e str in g enc loses an angle o f (,) J

w ith th e vert ica l.

106

= arcsin 0.990:36 = 82 .038° ~ 82°

= 0 .3:3004.

--

6.2 Dy na.m ics

6. Mechanics So llI t io lI s

( An y an gle (\ in the interva l (12

= a rcs in 0.:3:300
J 9.27° <

0

< 82.0:38° =

OJ ,

wo ul d sati s fy the co nditi o n of slidin g. prov ided th at the block is held fi xed until the strin g reaches th at angle . However, under th e IlI'ese nt co nditi ons the bl oc k will start to slide at 82.038° . so any oth er angle is irre levant for answerin g the quest io n. ) Remark . The lin e of acti on of the ten sion force in the strin g docs not pass throu gh the centre of mass of the bl ock , thu s it also re prese nts a torque about the centre of masS . As a res ult . the ' front' o f th e bl oc k is pressed harder ag ain st the groun d than its rear end . an d the valu e of the stati c fri cti on force may beco me uncert ain. Howe ver, since the so luti o n is ind er e nde nt of the len gth of the pendulum and the onl y variabl e that the torqu e dere nds on in ad diti o n to the angle and the magnitude of the force is the hei ght Ii of th e rod, it ca n be made neg li gibly small. If the bl ock is lo ng eno ugh relative to the rod. th e e n'ec t of the torque becomes ve ry small. (Note th at the size of the ba ll imposes a limitati o n on short enin g the strin g: the di ameter of the ball mu st remain neg lig ible relative to the len gth of the string, otherwise rotati onal kineti c e nergy should be con sidered in the work-e nergy theorem , too.) Solu tion of Prohlem 28. The batten docs not acce lerat e in the verti ca l directi on, so Newton' s seco nd law for thi s direc ti on is:

where G is the grav it ati onal force exe rted on the ballen , J\' I and J\'2 are the abso lute va lues of the normal reacti ons exerted by the cy linders. The ballen does not rotate, thu s th e equ ati on for the torques cal cul ated for an axis thro ugh the ce ntre of mass is:

Based o n the above equ ati ons. the mag nitudes of the normal reacti ons are:

/\' = G (L -.c) . 1

21..,

'

a ll el

Because. accordin g to the pro blem , th e cy linde rs ro tate quickl y, the re lati ve speed of the ballen and the surface of the cy linders is not zero, causin g dy nami c fri cti on to be exerted betwee n the m and givin g the m each mag nitudes of: a nd and altho ugh the y arc oppos ite, both arc exerted to ward s the centre of the batten. Newton's seco nd law fo r the hor izo ntal forces is:

107

300 Creative Phy sics Problem s with Solutions

Substituting the values of the normal reactions:

Kd =

?

/-L(1\2 -

- {LG

X

L=

ma .

The force is proportional to the displacement and oppositely directed so the batte n will undergo simple harmonic motion. Therefore, the net force exerted on the batten of mass m is:

'""' F ~

=-

{L m g . x

L'

L

and considering the force law of SHM which is : F = -mw 2 x we can fin d the angular frequency and the period of the motion , which are:

w = )/-Lg/ L ,

and

Our result holds true unless the maximum speed of the batten does not reach the speed of any point on the surface of the rotating cylinder.

Solution of Problem 29. It is known that if the direction of the force exerted on an object placed onto a surface is such that its line of action is inside the so called ' fric tion al cone ' , then the object is not able to move regardless of the magnitude of the forc e. Th is is because the static frictional force , which is exerted in the plane of the surfaces in contact, is always greater than that component of the external force which is parallel to the surfaces in contact. (Of course the gravitational force is also meant to be part o f th is external force .)

I

tge·K I

tge·K

The figure on the left shows the forces exerted on a small object of negligible weight placed onto a horizontal plane. Let us apply Newton ' s second law to the boundary case when the object is in equilibrium. The equations for the components of the forces whic h are parallel and perpendicular to the surfaces are the following:

F sin c: - /-LJ(

= 0,

F cosc: - J( = O. 108

(1) (2)

--

6. Mechanics So lu tions

G.2 D Yll Cllll ics

Fsin e: = ~!Fcos e:, SO ~!

= tan e:,

where e: is the so called frictional angle, and the abo ve menti oned fri ct ional cone is the cone which is drawn around the normal to the plane, with the fricti onal angle being hall' of the vertex angle of the co ne. From the first eq uati on it can be see n that if thc line of act ion of the force is olltside th e cone (F1 ) , th e obj ect must move because the parallel co mponent of FI is greater th an the frictional force (a i= 0). If the lin e of actio n is in side the cone (is ), the fri ctional force is always equal in magnitude and oppos itel y directed as the parallel comro ne nt of the force , thu s the obj ec t remain s at rest. Ir th e lin e of ac ti on is along th e generator of the cone than the boundary case of the eq uilibrium is ga ined. The figure in the left shows the forces exerted on the objec t and the fri cti onal co ne. The object can be pulled up along the plane if the line of ac tion of the r esulwlI{ of the gravitation al force mg and the pul lin g force, which makes an angle of (J with the plane, is outside the frictional co ne . From the figure it can be see n that th at if the object is pulled ur uniformly the lin e of ac ti on of the resultant force must co incide with the generator of the co ne . Because the angle between the normal of the rlane and the generator of the co ne is e:, the angle between the pul lin g force and the plane in the positive direction can on ly be 0 OS!3« 90 - 0' ) (if !3 = 90 0 - 0' , the n th e needed pulli ng force is mg) and in the negative directi o n it can only be OS !3 « 90 0 -e: )

t hus

OS!3« 90 0 - arctan/l )

These angles are independe nt of the mass of the obj ec t. Naturally if {J ~ 90 0 - e:. then F ~ 00. If the absolute value of the angle is greater than thi s there is no force with which the object can be pulled (or pu shed ) up along the plane. The e ndpoi nts of the applicable pulling forces , F, are along th e line which makes an angle of 90 0 + E with the inclined plane.

Solution of Problem 30. Newton ' s seco nd law applied to the coin states th at S = :::: ma , where S is the force of stati c friction. The co in start s to slip on the turntable when its acceleration reac hes a value at whi c h the force of static fri ct io n reaches it s rnaximum: S,tlax = ~img and can not increase any more. After time t the acceleration of the coin has two compo ne nts: the co nstant tangentia l acceleration and thc increas ing centripetal acceleration . Knowing th at these compo nents are perpe ndicul ar, we can use Pythagoras ' theorem to find the magnitude of the net accelerati on, which is:

109

300 Creat ive Phy sics Problems with So lu tions

where w is the ang ul ar velocity of the turntable a fter time t. Writing the angular velocity as w = {3t , we find th at the acce lerati o n of the coin is:

In sertin g thi s into the law of mo ti o n of the co in , we get:

~img = m1'{3 j {3 2t 4 + I , from which the time e lapsed ca n be calculated as:

Solution of Problem 31. Le t us consider the thin-walled cylinders first. The frict ion force Fh due to the rod obvi o usly points in the directi o n of the motion of the rod. Let us take the fricti o n force Fh acting at the ground in the same direction. (After the so luti o n of the equati o n system the s ig n of Fh wi ll te ll us the real direction .) The law of the moti o n of the centre of mass for the cylinder with the components of the forces (as they happen to be collin ear vectors) is:

(1) where a C Ill is the acceleration of the centre o f mass of the cylinder. The law for the mome nts of forces is

(Fh - Fh )1' = 8 c rn {3 .

(2)

The kinematic constraint for no skiddin g is

(3) substituting the ex press ion for a C lIl from (3) into (I) the followin g re lationshi p is acquired:

(4),

Fh + Fh =m1'{3 then by multipl y ing (4) by 8

C1l

,/ m the

follow in g rel ationship is acquired:

8 cm ( Fh + Fh ) - - = 8 C Ill 1'{3 1n finally by multipl yin g (2) by

l'

the following equatio n is acquired:

(Fi' - Fh )1'2

= 8 crn 1'{3

The ri ght sides of (5) and (6) are equal , so the le ft sides are also equal:

(Fh

+ Fh )8~" =(Fj, -

Fh)1'2.

From here the magnitude of the friction force exerted by the gro und is

Fh 110

(5) ,

m1' 2 - 8 cmF + 1n 1'2 j,. C III

=8

(6) .

6 .2 Dyna.mics

6. Mecha.nics Solutions

-----

Let us make use of the fact that the rotat ional inertia of a thin-walled cyl inder for its 2 axis is B ell! = 7nT :

meanin g there is no fr icti on force between the tab le and the cy linders, and the minimum coefficient of friction required is ~i = O. So in our case the cylinders rol l with out sk idd in g even on a comp lete ly smooth surface. Let us turn to the relationship between the cy lin ders and the rod now. The rod also moves on the cylinders without skidding , meaning the velocity and the acce leration of the topm ost points of the cy linders are the same as those of the rod. But if there is no friction on the ground, the friction forces acting on the points that are touching the rod should be the same for both cy linders, considering this is the on ly way their accelerations can be the same. This is true despite the fact that as a result of the motion of the rod the normal forces Nl and N2 acting between the rod and the cy linders change continuous ly. The magn itude of Nl starts from /1I1g/4 and its maximum va lue is 31\1[g/4 at the end of the process while for N2 the oppos ite holds : it decreases from 3Mg/4 to Mg/4, while Nl +N2 = Mg is alw ays true. Obvious ly , this can on ly be true if the coeffic ient of static fricti on between the cy linders and the rod is big enough to provide the relevant friction force Fh even in the case of the minimum normal force . The eq uati on of the motion of the rod is F -2Fh =Ma.

(7)

The law of the motion of the centre of mass for either cyl in der is Frl a C lIl ==-.

7n

The rolling on both surfaces without skidding requires that a C rll = a/2 holds . So a

Ftl

2

7n

M

m

Mg

F m

that is, 2Ftl = ma. Subst itutin g th is into (7) gives

F-ma= Nla, that is, F

a=--m+M

(=3~) ,

and

s= 2(mm+ M) F .

With the given numerical values Fr= . I

12 N·1 ko b =15N 2(1 kg +3 kg) . .

III

:JOO C'/'C'ati l'C' P h.l·sics Proll le' lIl :; II-ith Soluti o ll s

------~--------------------------------

So for till': minimum codlic ient of fri cti o n hetween th e roo ano the cy linoers !- ~r, r r, I r r, 2111 r 6 . I' ",j" = - . - = --- = - - = - - - . -- = - = 0.2 ,

l\1 g / I

_\,,,j,,

if .r;~

1\1 .r;

III

+ 1\1

30

J\I .r;

III

10 -::; is useo. s-

Basco on rel at ionship

l'

= ~. the final speed of the roo is

F /J 2- - - · ' 2 - =

L'==

III

+M

3

m ;3 111 III 2·:3 - ·2 · = 3. 16 - . s" 3 s

Solution of Prohl~m 32. Since th e bri ck slips on the cart. the net force actin g on it is the kineti c friction: S = 1(/1I.r;. Thercf'ore the brick has an acceleration of (1 1 = 11g whost: oirecti on is the saille as that o f the acce leration of the carl. The nex t step is to oet ermint: the acceleration of the carl. Let us apply New ton's seco no la w to the object that pulls the cart and tll the cart it se lf. Let the masses of the object and cart o be 111 2 and 111:3 respecti vely.

-

s

o

111 2.r; - /\.

K

= 1112(/,

/\" - 1'III)g =

)0

111 :; 11.

If we add th e two equations. /\" (w hich is th e tension in the cord ) cancels out. thu s th e acceleration of the cart (and of the object) turn s out to be : 111 2 (1=

1112

IWII

+ 171:3

'g.

Quite interestingly , the bri c k is asslIllled to be pointlike in the problem . thcrefore tht: din'erence in the distances covered by the ca rt and bri ck is the length o f th e cart (I:::: = 0.'1 Ill ): I

- (I I

'2

2

-

L :1 - (III

2

=1

'

thu s 1112 - 1" 11 I

'2

1112

+ Ill :;

.) 1 'J . gl- - -2 /'.ql -

= I.

H ence the coe nicient o f th e kinetic friction between the brick and cart is:

1' =

111 2.r;( 2 - 21(11/ 2 + Il l:;) ( 1111

11 2

+

1112

+ l1L:l)g I 2

=

5 · 10·0.6·1- 2· 0.1· (5 + 3) = 0.11. (2 + 5 + 3)· 10·0.6,1

6. j\{cc/I;l lJics SO /lJtio lJ S

6.2 DY ll a mics

Solution of Prohlem 33. Newton 's seco nd law app li ed to the small sphere gi ves :

( I) where 1\- is th e magnitude or the normal rorces exe rt ed

by th e spheres on each oth er. U sin g th e noti on th at the accelerati ons o r th e spheres arc th e sa me, and apply ing New ton's second law to the hollo w sphere , we obt ain :

(2)

j\l .cJ + /\- - h,2= j\/ u.

Let us add eq uati o ns ( I ) and (2) to determ in e the accelerat io n or the objec ts:

(1/1 + j\f) rJ - /..- ('"2

= (Ill + j\ / )11 . F(N)

hence

80

/..- (12

(3)

u=9 - - - - · 111 +1\/

Inserting thi s int o equ ati o n ( I ) and so l vin g ror the required normal force. we get:

_= /ll er; -

1\

50

(/,"('2) = - /II- - /..-c .-) =

u) = III rJ - rJ + - - 177

+ .\ /

lIZ

') = -III- /..-1 '-') = -]k-t'= -1 ·0. 1 III + 117 2 2

+ j\l

~

--

v(m/s)

')

O ~~~~--~---+-

40

'V-

m "2::;2

T he run cti o n o f force versus veloc it y is therefore a parabola. At th e moment w hen the spheres are dropped, th e ve loc ity and air-d rag are zero. meani ng that the spheres arc in freefa ll (w hi ch is a state o r we ightl ess ness), and th at the normal force /\- is also zero. Duri ng th eir fall th e sph ere s accelerate cau sin g th eir ve loc it y and air-drag (w hi ch is opposi te rrom thei r veloci ty) to acce lerate as we ll . Th is, in turn . ca uses their acceleration to decrease. A rter a w hile th ey reach the termin al ve loc ity (asslimin g constant air den sity and co nstant gravitati o nal rorce) at w hi ch the result ant force is zero . Sin ce th e grav it ati ona l force ac tin g o n th e sys tem is eq ual to the air-drag:

( III

+ _\1 )9 = /..- V~'''X'

thu s III

('111ftX

==

+ j\ /

---rJ= /,-

11/

+ III

- h-'-9=

8 kg + 8 kg

111

0. 1 1,,2~ - 2

52

-=---,-,--=- . 10 -

111

= 40 - . 5

At th at veloc it y th e magnitude of th e normal force is:

11 3

300 Creative Physics Problems with Solu tions

Solution of Problem 34. a) The equa_ ti o n o f the mOli o n o f th e small body in the bo tt o m and top pos iti o ns, co nside rin g the re lati o nship between the te ns ions and the resultant of the normal force N exerted by the surface a nd the gravitational force mg (and re placin g 1 by R), is : 2

2!\'

-111,g ' s in o.

= m. ~ )

and

.

'

mg,s Il1 O' + !\ =

In

v2 R)

from which , by eliminatin g the te nsions in the bottom and top pos iti o ns, the fo ll owing equation is acquired: 111,v2

mg's in O'+ -

( 1Tw2 ) · sin o. R-o =2 · ---mg 11. .

Reorganizin g: 111,v2

111,v 2

3mg.sin O' - 2 - - = _ __0

11.

R )

from which vO=V2v2- 3gR- sin O'=

111 2 III III 2.9- 3 ·9. 81 2'· 0.2111·0 .5=3.88- . 2 s s s

J

b) The work- kinetic energy theorem for the original and final states is 1

2

1

2

-mg 2R'sin O'- ~tmgR7r ' cosO' = -m.v - -m vo' 2 2

that is, From thi s the coefficient o f friction is ~t

=

V6 _v 2 2g11.· 7r cos 0'

2 - - t a n 0' 7r )

with numerical va lues ~t=

3.88 2 - 3 2 2 ° - - tan30 =0 .1 995::::::0.2. 2·0.2·7r · 9 .81· cos30° 7r

c) After releas ing the string, the body moves upwards on the slope along a straigh t line and covers di stance x . Applying the work- kin eti c e nergy theore m aga in for the initial and final states:

-mg(R +x)sin O'- ~t mg· ( ~R7r + X) cosO'= O- ~mv6' 11 4

---

6. Mechanics Solutions

6.2 DY/l am ics

From thi s r-

V6 - (571' JL cosn + 2 sin 0 )Rg

., -

=

2(s iIl O + II C050)g

-

-

3.88 2 - (5 71' ·0 .2· COS 30° + 2· sin 30° ) ·0 .2·9.81 m ~ 0. 59 Ill . 2 (s in30° + 0.2 cos300) . 9.8 1

(Those who calcu lated with 9 = 10 m/s2, acquired Vn = 3.87 111 /s for the initi al vd oc ity, J.L:;::; 0.1 8 for the coe ffi cient of fri ction and x = 0.62 metres ror the distan ce covt.:red.)

Solution of Problem 35. Ass ume the ball is a pointm ass as suggested by th e fi gure. Accordin g to the work-e nergy theore m, the kineti c e nergy or the ball in po int A t.:(ju als the sum of the work done by the grav itati onal force and the sprin g: 1

2

1

2

/l/,g8 + -/]( /::" / ) =-mu ,

2

2

where /::,.1 =..c- L , x and L are the t.: xtend ed and relaxt.:d Icn gth s or th t.: sprin g. Wt.: ust.:d the notion th at in its vert ical pos ition the sprin g will reach its I'd axed len gth . Lt.: ngth .1' is prov ided by the Pyt hagorea n theorem:

=

JOC

2

2

= jO.8 m2 - 0.48 m 2 = VO.4096 m 2 = 0.6 1Ill. The verti cal d isplace ment of the ball is 8 = 0 A - 0 B = 0.8 Ill - 0 .48 III = 0.32 Ill . Us i ng x

-

OB

1

this, the speed turn s out to be: 2g8 + D (:r- L )2,

V=

In

substi tuting known values, we find : V=

. m 75N/ Ill . 2 2·10? · 0.32 III + (0.64 111 - 0. 32 Ill ) s3.2 kg

III = 2.97 -.

s

Let us now determine the force t.:xerted on the shell by the ball . At po in t A thc onl y forces ac ting on the ball are the grav itati onal and normal forces, sin ct.: th t.: sprin g is in its relax ed state at th at moment. Appl yin g New ton' s second law to the ball at th at point , We obta in: J:: + 6 = rnii, With the pos iti ve direc tion pointin g up wa rds , we have: ,)

u!\' - IIIg = 111 - ,

r

from wh ich the normal rorce ( !\') is: !\'

= mg +

v 'J/1Ll'

III

= 3.2 kg · l 0 -2 + 3.2 kg · S

') 2.9 _(-oJ nr' ) /s0.8 m

= 67 .3 i\ . li S

300 C rea.t ive P hy sics Problem s wit h S olution s

T hi s is the same as the mag nitude of the fo rce exerted by the ball o n the shell , s ince it a nd the normal force are bo th ac ti o n-reacti o n fo rces .

Solution of Problem 36. T he equ ati o n of mo ti o n of the toy car in the directio n of the relevant radius is

mgcostp+ N

v2

=m R '

At the mome nt of detaching N = 0 , and accord ing to the data give n in the problem, for the cosine of angle tp

ho = ?.

h

R/2

,

cos 't' = -R- = 0 ., 5 1/')

that is, tp = 60° . At the mo me nt whe n the norm al fo rce N ceases , the equ at io n of moti o n of the toy car in the radi al direc ti o n is

v2 m g costp = m R '

(1 )

Fro m thi s, the remainin g speed of th e toy car at the mo me nt of de tac hin g is:

v 2 = Rg cos tp .

(2)

From thi s posi ti o n a n o blique proj ecti o n w ith thi s initi al vel oc ity and an initi al angle of

a = tp = 60° sta rts. a) T he he ig ht of the pl ace the car starts fro m ca n be acquired fro m the work-kinetic e nergy theore m : 1

mg(h o - h)

= 2mv2.

(3)

1 S ubstituting (2) into (3) gives m g(h o - h) = 2mgR cos tp . From this

1 3 1 R ho = h + 2Rcostp = 2 R + 2R costp = '2 (3 + costp) = 1.75R = 56 cm . b) T he max imum heig ht is acquired if the max imum height Yrn ax of the obli que projectio n is added to he ight h whe re the car detac hes from the trac k: h 1llax = h +

Y l1Iax

= h+

v 2 sin 2 tp 3 Rgcostp· sin 2 tp = - R + --"-----'----'2g 2 2g

=!!: ( 3 + Sill tp , sin2 tp ) = 32 2 2 2

11 6

(3+ ~ . v'3. v'3 ) cm = 54 cm. 2

2

2

--

6. Mechani cs Solutions

6.2 Dynamics

In short: the toy car starts from height ho = 56 cm , detaches from the track at height h:=; 48 cm and reaches a maximum height h lllax = 54 cm. (The total height of the track is 64 cm.)

Solution of Problem 37. The centre of mass of the wheel undergoes pro.~ ~ jectile motion after leaving the ramp. Although it isn ' t stated clearly in the probl em, x is the maximum height reached by the wheel measured from the right end of the ramp . The direction of the wheel ' s velocity at the right end of the ramp forms an ang le (3 with the horizontal. The maximum height reached by a projectile is given by the formula: sin 2 ,8 x2g , .. ,?"

.......

._ v5

where Vo is the speed of the wheel' s centre of mass at the right end of the ram p. Let us determine this speed. Accord in g to the work-kinetic energy theorem:

mg[(R - 1')(1 - cos a) - (R - 1')( 1 - cos ,8 )]

1

1

2

2

= - m v6 + _mT 2 w 2 = mV6,

where l' is the radius of the wheel, w is the angular velocity of the wheel when leav in g the ramp , which is w = VOlT. Thus V

2

I:::,h = (R - 1')[( l - cos a) - (1- cos ,8) ] = ~, 9

and

2

Va 9

= (R - 1') [1 - cos a -I + cos,8] = (R - 1') (cos,8 - coSet ).

Substitutin g this into the formula for x , we have: 1

'J

x = 2 sin- ,8( cos,8 - cosa)( R - 1'). It can be see n that the answer depends on the radius of the wheel, which is not given. Let us therefore assume that the radius of the small wheel is small eno ugh relative to R to be neglected , i.e. 1'« R . In that case our formu la for x simplifies to: 1

x = -sin 2 ,8(cos,8 - cosa )R , 2

Insertin g given data, we obtain

x=

~ . sin 2 300(cos 30° -

cos 600)· 1 m = 2 = 0.5·0.25 · (0. 866 - 0. 5) ·lm = 0.04575 m;::::; 4.58 cm . 117

JOU C rcatil'c Ph,I's ics PI'O/) /C' IIlS lI·it/, S o/utiollS

Solution of Prohlem 38. a) ThL: L:quation o f thL: motion o f thL: car is 'J

(' -

IIJ.CJ -

.\'

=

/1/ -

.

(}

whL:re j\' is the normal forcL: ( it s magnitude is L:qual to the unknown forcL: eXL:rted Oil the hridge ) and (} is thL: radiu s of curva turL: helonging to the g i ve n point of the pa th. The essence of the problem is determining thi s value. The relati o nship hetween the radiu s of cu rva ture and the co rrespondin g normal accL: lerat ion is: (/,,=-. {}

In ou r case neither normal acceleration nor the radiu s o f curvaWre is known. Luckily. in a simple caSL: we can determine thL: sL: va lu es through elemen tary co nsiderati ons. T he parabola of ob lique projectile motion comes in handy. it ca n be used to model the arc of th e hridge and therel'ore its c urva ture as wel l. All we need to do. is lind thL: angle and thL: initial VL: locit y w ith w hi c h a thrown stone foll ows the lon g itudin al sec tional line the bridge . In thL: case o f a thro w n slOne normal acceleration can be detL:rmined easi ly. and thL:refore if th e stone is launched sui tab ly. the radiu s of the curva ture of our bridge w ill be in our grip. Let us determine the radiu s this way: the range of the projectile motion is the same as the span of the hridge (d):

or

2L'0 sill Cl coso and the height of the projectile motion is the sa ille as the hei ght of the hi ghest poin t of the bridge relative to the level of the hanks:

,=

2

.

2

('os in n

29 ThL: ratio o f the two is

frolll w hi ch

[a ll n

I cusn

rI

Sill n

"

I"

= - = 0.2

(\ = a rc t 1\ 11 0.2 = 1 I .3 I 0

d

~ ~Voy

The ve rtical cOlllponent o f the initial velocit y o f the stone is ('II II

vox and the hori zo ntal cOlllponen t of the ve locit y (10. ,

11 8

.

IS

I'll " I () III _ III = - - = - - =00 talln 0.2 s s

= /2rJ/, 20" = 10 :J

III

--

6. MeclJ
6.2 Dyna. mics

Becau st.: at th e hi g hes t po int o f th t.: pat h the vt.:l oc ity o f the sto ne is ho ri zontal , its norm al acce krati o n is th e sam e as acct.: lerati o n it self ( w hi ch is 09 thro ug ho ut th e m oti o n), therefore

.)

.)

u=09= ~

(1 "

'11 -

0=~=?5 ~ 9 - 01ll .

(}

With thi s th e unkn ow n norm al force is N = 1I1

~ ' = 1000 k ( g - ~'U?ar)

( .10 -c lOO -) 250

' 0

.

(}

0

b) Th e r adi us of curv ature at the pa raho la po int that hel o ngs to :\/4 o f th e hori zo nta l d istance can he determin t.:d fro m tht.: fac t th at th e hori zo ntal veloc it y comp o nent o f th e th row n sto ne is co nstant, th erefore 3/ 4 of th e tim e of th e proj t.:c til e fli g ht helo ngs to 3/ LI of th e ho ri zo nta l di stance:

III

~ =8 400N . S2

, ,,

3 Vn sill Q

2

9

The normal acce lerati o n o f th e sto nc at thi s po int of th e path is

VT

a" = gcosip = - , (}1

and acc ordin g to th e fi g ure, the radiu s o f c urv ature belon g in g to thi s po int is 01 =

~

ui

- -- = Y COS ' '1 Y

uf =VI-

-

g . !CL "/11

gv .t. '

where U I is th c in stant aneo us ve loc ity o f the sto ne at th e po int in co ncern at tim e instant /1, Th c tim e in co nce rn is .1: 1

= -

LI

3 rI = - - = L1 1'.('

l'.,.

300 ,1· 50

111

111 /S

;3 = 2

S.

With thi s, th t.: slJuart.: o f th e ve loc it y of th e stone at th t.: po int in co ncern is

.)

.)

.)

.) 111 -

Ilr

= [50- +( l O- J5) - l?"" =2525 s-

T.

and it s in sta nt aneo us vel oc ity is

1' 1 = V 2525

lll /S;:::O

50.25

mg

111 /5 .

Therefore , th t.: un k no w n r ad iu s o f c ur vaturt.: is

01 = -

~

d

o9L',

50. 25° = --10· 50

III

= 25 3.77

III ;:::0

254

Ill.

11 9

300 Creati ve Physics Problem s with Solutions

------------~-- ----------------------------------------------------

With these, the equation of the motion 01' the car can already be determined. Let li S take into consideration that the car moves at uniform velocity throughout its motion therefore it has only normal acceleration. The relat ionship between gravitati onal forc~ mg, the so-called supporting force T, the norm al force N and the sta tic fricti onal force Ff is shown in the figure . From this , force T (which is the reaction to the forc e exerted on the bridge by the car) can be determined immediately using the cosine rul e:

T=

(mg)2

2

V2 ) + ( m---.l

v cos'p = m - 2mg· m---.l

{!

= 1000 kg

{!

160000 100 + - 254 2

-

400 2·· 10·0.995 = 8434 .47 N. 254

If only the norm al force exerted by the car is to be found , then the absolute value of N should be determined. This is acquired from the equat ion of the motion of the car in the normal direction :

mgcos'P- N

2

ar = m~, v

(! l

where cos'p =

V,c/VI = 50/50.25 = 0.9950 .

N=7n(gCOS'P-

From thi s

V~ar) =1000( 10 .0.995- 400)

N=8375.2N. 254 The resultant force of the static frictional forces th at act on the lyres is acquired fro m the equation of the motion in the tangential direct ion. Tak ing into consideration that the car moves uniformly : S - mgsin 'P = 0, so (!I

S = mgsin 'P = mgy'l- cos 2 'P . Therefore the magnitude of the friction al force is F f = 1000 kg·lO ~ . \.11 s

- 0. 995 2 = 998 .75 N,

and obviously T = y'8375.2 2 + 998.75 2 = 8434.5 N.

First solution of Problem 39. a) Let J( be the force exerted by the rod and let F be the tension in the thread. At the vertical position of the thread the ve locity of the ball A is maximal , thus its acceleration is zero, maA = 0, so the equat ion of motion at that instant is: (1) J( - mg - F = O. At the same time the equation of moti on of the ba ll B

F-7ng

IS:

= man,

(2)

and the energy conservation law gives the equation:

7ngL 120

1 2 = 2· -mvn = mVn2 · 2

(3)

--

6.2 Dy nam ics

6. Mechan ics Solu tion s

(Here we have used the fact th.at VA = -V [J s ince the ce ntre o r mass moves along a vertical lIn e, and the velocIty of ball B is hOri zontal at the mome nt In vesli gatL:d.) From thi s equ at ion the ve loci ti es of the ball s ca n be determi ncd. From equ atio n (3) the vel oc ities of the ba ll s are: III

s Since m aA = 0 the x co mponen t of the aCL:e leration of the ball [3 is zero, whil e its total acce leratio n is the centripetal acce lerati on

K

h-~where (! IS the rad iu s of curvature of the ball ' s trajectory at the lowest poi nt. But thi s acce leration can be determined in a diA'ere nt way as well.

I I

ac =

171 ·0 + ma/3 171+171

a [J

= -2

--> a[J

I

\\\\:''',:a,:t.

::+-

// / //////

F

(5)

= 2ac ·

I I I

F

\ \

zero, ands instant, the masses the ball s are equball al. TAhu s, At thi the of accelerat ion of the is the accelerati on of thc ce ntre of mass is the average of the acce lerati on of the ball s:

". ~i

VB

'{

..

mg

On the other hand , at thi s in stant , the ve loc ity of the ce ntre of mass is Us = 0 since th L: two balls of equ al masses move at oppos ite ve loc it ies. Thi s means that the cent re of mass C is the in stantan eous ce ntre o r rotat io n and that the angul ar speed of the thread IS: V[J

2v/3

w - -~- - -L- .

(6)

2

Since the ce ntre of mass 5 accelerates up wards, the acce lerati on of the ball B is the sum of its acce leration relati ve to 5 and the acce lerati on of S. Us in g (5) , we get: a[J = ac + a rel = ac +

L

'2w

2

=

a/3

2

+

L

2

'2w .

Thus the acce leration of the ball B , ca lculated in the seco nd way , is: 2

4v~

CL [J = L w = - .

L

(7)

The accelerat ions (4) and (7) , de termin ed in the two diAe rent ways, are eq ual: v2

-.fL (!

4v 2 L '

------.ll

Which gives the va lue of the rad iu s of curvat ure :

L 0=-= 4 0.4 111 .

~

12 1

300 Creative P hysics Problem s with Solu tions

I

'~

x

~

:

A : VA - 1- (:'.",,-, - - --'\. ~~~,,~-~, I

\

i

····i .... m

,

L

2

\,

I I

Oc

\

\

C

./

\

a nd

v1 =

I I I

y

" .. 2

\

Summarizing, the accelerations of the two balls are:

an = -

v2 gL an=....lI =-y-=4g. )

/ \

/ \

\,

°B

,

/

/

12

(o r in parametric form :

/

\

16 m s- 2 m =400.4 m s2 '

12

/

4

/

""

/

"'41--'-'~lblr ~...­

b) From equation (2) the te ns io n in the thread

/

is:

VB

F= m(g + an)=2 kg (10

:l~ +40 ~ )

=100 N,

and from equation ( I) the forc e exerted by the rod is:

J(=mg +F=20 N+ 100 N= 120 N. Second solution of Problem 39. One may know that the trajectory of the ball B is an

b2 , where a

e lliptic are, and the radiu s o f curvature at the endpoint of its major axis is 12 = -

a is the semimajor axis, and b is the semi minor axis, which are in our present case a = Land b = L/2. Using these equations the radius of curvature can be determined immed iate ly: £2

-!:.

12- 4 - L - 4' With thi s knowledge, the accele rati o n can be calculated directly from equation (4), while the o ther results can be obtained using the same way as before. Third solution of Problem 39. Pa rt a) o f the probl e m can be solved in a shorter way as well . Whe n the thread is vertical , the ball A has zero acceleration, thus the reference fram e fixed to the ball is an inerti a l o ne. The acceleration of ball B is the same in all inerti a l re fere nce frames, so we immedi ate ly kn ow that:

an=

2

Vre l T

(2)2 4 2 4 42 11, 2 =~= Vn = ' 52 =40 m L L 1. 6 m s2 '

where Vre l is the speed of ba ll B re lative to ba ll A . In the first so luti o n we saw that at the vertical position of the thread of the ve lociti es of the ba ll s have equal mag ni tude and oppos ite direct io ns, due to the equal masses of the ball s. It means that Vrel is just two times as much as the speed o f a ny o f the balls w ith respect to the gro und. 122

6. Mechanics Solu tions

6.2 D y na mics

----

Althou gh it was not required in the probl em, we deduced the equati on of the orbit of the ball B . Because of the simil arity of the two ri ght triangles in the fi gure :

x

JL2 _y2

L

L

4x 2 +y2

= L 2,

2"

from which:

folloWS , whi ch ca n also be written in the form 4x2

y2

x2

y2

L2 + L2 = 1, or (L /2)2 + L2 = 1. This means th at the orb it is an ellipse with semi maj or ax is a = L / 2 and semi min or ax is b:= L: 4x2

y2

-

x2

y2

+ - = 1 or + =1 L2 L2 ' (0.8 111 )2 (1.6 m)2 .

Solution of Problem 40. First, let us exa min e whether or not the marble ca n roll along the path at all (ass umin g th at it rolls without slidin g) . In order for it to do thi s, it is necessary that the speed of its ce ntre of mass at the top of the circul ar path exceeds the critical speed Vnit for which: V

2

mg + J(=ma " =m-- , R -1' where the norm al reac ti on

J(

becomes zero, makin g the va lue of the criti cal speed: Vc ri t

= J (R - 1' )g

In order to answer thi s, we have to appl y the work-ene rgy theorem : 1 2 1 2 2 2 W , 2 2 5 from this we ca n ex press the speed , usin g the work-energy theore m, because the marbl e rolls without slidin g 1'W = v, thu s:

mg [3R -(2R -r)] =- mv + - · - mr

V

2

=

710 g(R + 1' ) > ( R

.

2

-7' )g= V cr it"

which is definitely greater th an the criti cal speed, meanin g th at even at the topm os t point the norm al reacti on is not zero.

mg

123

JOO C r ea t i " e P III'sics Proble1l1.s lI' i tl! Soillt ioll s

------------~-------------------------------------------------------------

Fro m now on, kt us neg lec t ,. w ith respec t to R sin ce th ei r rati o is 1,/ R = 0.005 There are three forces exe rt ed on th e marhle: th e grav it ati onal force , the fri cti onal ro rc~ and th e normal reac ti o n. The (irs t o ne is co nsta nt. T he seco nd one is opposite to the d irecti o n o f th e moti o n un til the marb le reaches th e bott om o f the pa th . The thi rd one is co nstan t o n the slope and alo ng the circ ul ar pa rt it decreases , until it reac hes zero at the lowest po in t. T hen, whe n the Ill arbk movcs up. it increases for a whi k and i ts direct ion becomes the sa me as the d i rec ti on of the mot io n, however, it then decreases again un ti l il beco mes zero at th e top of the path . The n, once aga in , it s di rec ti o n changes . The norm al reac ti o n is al ways perpe ndi cul ar to th e path , alo ng th e slope it is co nstant , its ma xim unl is at th e bott o m o f th e c irc ul ar pat h and th en it mo noton ic decreases until the top of th e path . T he fri ctio nal coeO icie nt i n ques ti on ensures th at the ro llin g m arble does not sl ide , and de pends on the latt er two forces . In depe nds on th ese two forces becau se the fri ctio nal force. w hi ch causes th e appro pri ate angul ar acce lerat ion p and w hi ch depend s on th e pos iti o n o f th e m arble, ca n be calc ul ated as the produc t of th e norma l reaetion and th e minimum o f the fri cti onal coe Oicie nt. As a res ult, we have to determin e the norm al react ion A' as a fun cti o n o f the pos iti on o f th e marble, th e fr icti onal force S and th e max imum o f th eir rati o.

beca use the coe Oic ient of fr icti on ca n be greater th an thi s, but not slll al ler. Thi s mea ll s that the m ax imu m o f th e rati o S I /\' is th e lowe r l imit o f th e fr icti onal coe nicient. It is wort h exa minin g th e moti on w hi ch occ urs until the top o f th e path since it is there th at the translational moti on is th e reverse o f th e upward tra nslati o nal moti on 01' th c marble (altho ugh th e rotati onal moti on is not). As a re sult, th e asked fri ctio na l coeOic ients are th e sa me at th e same heig ht. Le t us desc ribe th e moti on o f th e marb le in term s o f th e angle y enc losed by the vertica l and th e radiu s co nnec tin g the ce ntre o f th e c ircle and th e marble . By app lyi ng New to n' s second law , th e fri cti o nal force ca n be calc ul ated from the rad i al compon en ts , w hi le th e norm al reac ti on ca n be ca lcul ated from th e tange nti al com ponents. Thi s can also he do ne hy usi ng th e eq uati on wr itten for th e ro tati onal moti o n and th e work -energy th eo rem . For th e ta nge nti al co m po nc nts: IIIgs ill -;: - S and

= IIl et "

(1)

(I,

(:2 )

S,.=8 ,d= 8--. I' w here 8 is th e mo ment o f in erti a of th e m arble abo ut i ts ce ntre, and com ponent o f th e accele ratio n o f th e ce ntre o f Ill ass o f th e marhl e.

al

is th e tange nt ia l

For th e radial compo nent s:

. /\ -

I1 Ir;COS y

and III f)!':,."

.

124

= lila" = 1

'J

u2

III

(3)

R'

1

'J

= -2 Ill u' + -2 8,-,)' .

( I)

---

6, MecllAllics Soi llt io n s

6,2 Dyna mics

Here 6. h is the height dine rence whi ch equa ls h - H + n. cos'P, and w U2 / r2 , Us in g these

2

211~g(h-n.+n.cos 'P )=

7nr2 + 8

,)

r-

' 1},

i s equal to

(4/ )

FroIll equati on (2)

5r2 u'l

=

8 '

whi ch can be w ritten int o equ ati on ( I ) and we ga in mgsill 'P - 5 the fri ct ional force is:

5 From equ ati on

51'2

= m-8

fro m w hich

= mgsi n

8 + 1111'2 '

(.J') ')

u-

= 2g(h -

R + R cos
mr2

8

+

)'

117 '} '-

whi ch ean Ix; w ritt en into equ ati on (:l) and th e normal reac ti on is:

!\'

[ (flh)

= IlIIj 2 - - 1 ' '

rn}, 2

8

+ m},2

+ 8 + 317l7,2 ' cos 'P ] 8 + mr2

the necessary fri cti onal coenicient is: ,

5

8

mgslll 'P~

!( = !\' =

'I1~y [2 (i!.I? - 1) ~ + 88 +31111~2 COS ((')] 8+ 1111' + 111 1' y

S ill

- 2(i!. If


- 1) .'.'H.!..'.2+ (1 + 31111,2) COS ,1l , <3 y

The max imum o f the above fun c ti on has to be determin ed , The appro xim ate va lue o f th e ma ximulll ca n be gain ed by pl ottin g the graph , or from the deri va ti ve o f th e fun cti on a more pun ctu al valu e ca n be ca lcul ated quite eas ily, De notin g the co nstant s by A and B th e fu ncti on ca n be w ritt en in thi s form: !(=

Let the num erat or of th e fun cti on be rUl e, Thu s th e deri v ati ve is:

(!'. )'= C'

II' V -:- [(,V i

v2

'It

si n 'P , A + B cos'P and th e den om in ator be v and appl y th e q uoti ent

cos 'P( A

+ lJ cos
sill 'P( - B sin 'P) (A +BCOS'P)2

T he fun cti on may have a max imum (or minimum ) w here the der iva ti ve is ze ro, thu s th e fo llow in g equ ati on mu st be so l ved :

A cos

=0 125

300 C reati ve Physics Problems wit h Solu tions

Because

i 11

Its soluti o n is:

2

B cos

A

3 +~ e ('

).

2 * - 1

(The seco nd deriv ati ve shows that thi s is a ma ximum . The second derivative here is negative, which means that the first derivati ve changes from a pos itive to a negat ive, meaning the function first increases, and then decreases.) The value of the asked fri ctio na l coefficient, with parameters and substitutin g costp, can be expressed as:

e Iih =

Using that

{i

3, and

2

e=

2

Smr , 2

'J. 1n7, 2

=

5

V4(mr2)2(2)2 -

= - - = 0.1 898 ~ 0.19.

(t mr2 +3mr2 )2 Jlll

2/11,,. 2/ 5

Or, determining the value of cos tp : cos

3 +~ ( ) = - 0. 85 , from whic h

2 3-1

the frictional coefficient from the expression which contains the angle is: sin

10 +8 .5 cos
=

0. 527 10 +8.5·(- 0 .85)

=0 .1 8 99 ~ 0.19.

It was ex pected to gain quite a large angle
Solution of Problem 41. a) The speed of the ball is c ha nged by the force compone ntS of the forces ac ting on it that are parallel with the track, so the speed ca n be ma ximu n1 126

---

13.2 DYllalll ics

6. Meclwnics Sol ution s

only where components by the track the absolute are equal :

thei r sum is zero . On ly gravitational force and spring force may have para llel with the track (as the ball slides witho ut fri cti on, the force exe rted is perpendicular to the track) , so at the place where speed has a maxi mum , va lues of the parallel components of grav itational force and sprin g force l~ prill g cos5 =

rng cos(3 .

From thi s eq uat ion the mass of the ball can be determined if spri ng force and angle 5 are known. The former can be determined from the elongation of the sp ring . the laller from the geometrical data of the arrangement. At the place where ve locit y is maximum , the length of the spring can be determined usi ng the Pythagorean theore m: l2 = j (R - R sin (3)2 + (2R - R cos (3 )2 = R j6 - 2 sin (3 - 4 cos /3

= Rj(1 -

= 0.2 III V6 -

sill /3 )2 + (2 - cos 8 )2

=

2 sin 3 L] 0 - 4 cos 34 ° = 2.5.02 C ill .

The elongati on of the spring is b.l 2

= l2 -

lo

= 25.02 cm -

20 cm

= 5.02 cm ,

with it, the mag nitude of spri ng force is N

f ,prillg

= Rb.l = 100~

·

III

5.02 cm = 5.02 N.

R

Angle 5 can be determined from angle 'Y shown in the figure: cos 'y =

R (l - sin /3 ) l2

=

0.2m·(1 -sin 34 ° ) 0.25 m

.' F.PriOg

1_---=2c.:

R' - - - ---.I

=

= 0.3523 .

From thi s, 'Y = 69.4° , so 5 = 180 0

-

n = 76.6° .

(/3 + 'Y ) = 180° - (34° + 69. L

With these the mass of the ball is

m=

F , prill u cos5

'" gcos /3

=

5 N . cos76.6° 9.81 ~~ ·cos34 °

=143 g .

b) The maximum speed of the ba ll ca n be determined from the work-k ineti c energy theorem. The work done by the force exe rted by the track is zero (beca use it is perpendi cu lar to the track and therefore to ve loc it y at any time ), the kinet ic elll.:rgy of the ball is given by the work of the gravitati ona l force and of the spring force:

127

300 C reative Physics Problems w ith Solution s

The still unknown data can be determined by using the geometrical data of the arrangement:

hI =2R=40cm. h 2 = R (l - sin !3)

= 8.82 cm. I] = J R2 + (2RF = RV5 = 0.2 m · 2.236 = 44 .7 cm.

From here 61 1 = h - 10 = 44.7 cm - 20 cm = 24.7 cm. 61 2 = 12 - 10 = 25.02 cm - 20 cm = 5.02 cm . With these

100N/m m N 2·9.81-· (0.4 m - 0 .0882 m) + k [(0.247 m)2 - (0.0502 mF ] = 6.86- . kg 0.143 g s

First solution of Problem 42. (Using basic mathematics.) Let us assume that the object is very small to make sure that its rotational inertia is zero. Let us al so assume that the action of hitting the disk is determined only by the length of the string and not by the shape of the object. However, the size of the object cannot be zero because that would make the tension in the string increase to infinity , and would subsequently cause the string to break well before the moment of impact. While the disk is fixed and the string is not stretchable, the velocity of the object remains perpendicular to the stri ng throughout the motion. This means that the te nsion in the string does not do work and that the speed of the object is constant: Ivl = 0 .4 m/s . Al l we have to do is simply calculate the distance covered by the object until it hi ts the disk and then use the formula t = s/v to find the time elapsed until the impact. a) Let us consider a small time interval in which the string moves so that the angle enclosed by its in itial and final positions is 6<.p . The radii drawn to the points where the string touches the disk in its initial and final positions also from angle 6 <.p as show n. In the considered time interval part of the string, whose length is l' 6 <.p, gets wrapped around the disk, therefore the length of the unwrapped string decreases by i1s L 1611= 1" 6 <.p. In this time interval the length of the path of the object is : 6 s = l · 6 <.p. Isolating 6 <.p from the first equation and inserting it into the second , we obtain: l 6 s = - · 6l . l'

128

--

6.2 Dynam ics

6. Mechan ics S olutions

The LOtal di stance cove red by the obj ec t is the sum of the length of these path elements:

"

S

." 1

1 "

1

= """"' 6. s = """"' - 6.1 = -r """"' 1 . 6.1 = - L ~ ~ r ~ 2r j

i= ]

i=]

" . L L z . ;: . ;:

L2

= n2

8= '11,

.

i= 1

This res ult can be found usin g that 6.l i inside the sum equ als:

8

2

.

Z

= L in

and 1 = i · Lin . Thu s, the ex press ion

L2 l +n n 2 . - 2- . n

~

( 1 1)

= L - 2n 2 + "2 .

If n tend s to infinity, the express ion in the brac ket tend s to 1/2 , therefore the total length of the path tends to: L2 0.8 2 m 2 s= 2r

=

2 · 0.2m

= 1.6111.

Therefore, the time elapsed until the objec t hits the di sk is:

L2 T= =4 s. 2 T'U

b) The tension in the strin g prov ides the centripetal accelerati on of the obj ec t, thu s ~

F= m~- , where 1 is the in stantaneo us va lue of the radius o f curv ature of the object' s path. time. path , point

Therefore, we need to find the length o f the unwrapped string as a fun cti on of To do thi s, let us calcul ate the sum written in the prev ious part not for the whole but for an arbitrary 1 < L. In th at case the di stance covered by the objec t until th at is:

1

S

=~.

1

L L

L I· 6. l = ~ L i . ;: . ;: = 'II,

II.

L2

i= k

i= ' ·

"

Tn 2 .

Li= i=k

=~. [ (n + k )(n - k + 1) ] = rn2

2

2

2 k n _ C [ k k 1] -_ -L 22 [n +nk -nk - f. ; 2+ + ] - 1--+-+2 2

rn

2

2r

n

n

n·

If n -> 00 and k -> 00, then the last two term s in the brac ket tend to zero. N ote th at the seco nd term multipli ed by L 2 gi ves

kL

where -

is the un wrapped length o f the strin g, so for the length of the path , we have:

1 n 8 :::: _

2l'

(L2_ l2) .

129

300 C l'eati\ 'c Pln's ics Pl'Oble llls \vitl , Solutio ns

-----------~---------------------------------------------------------

As the end of th e string ll10 ves at co nstant s[Jeed . we obtain: I

.)

.)

s = - ( L- - I- )=t'I '21'

'

from w hich the length of th e unwrappcd strin g (or th e in stantan eo us radiu s of c ur va ture) is:

1= J U - 2,.1'I. Therefore the ten sion in th e strin g as a fun cti on o f time ca n be exp re ssed as:

F

,)

,)

1111'-

III 1' -

=-

1-

O.G kg · ()..I '2 '~'1"

= ---;==.=== L'2 - '2,.('/

CUl2

2 ,0 .2

Ill :? -

Ill ' (l. LI ~ ./

,

Second solution of Prohlem 42. (Us in g hi gher mathematics. ) Let the inlinites im al angle rotated be el y . While rota tin g thro ugh this angle. th e leng th o f th e strin g decreases hy ,. · dy- . so the c hange in the len gth o f the unwrapped string is negati ve: ell = - ,. . el y . The len gth o r th e inlinitesirnal part o r the path is d ., = I ·el y . Ex terrninatin g d tp g i ves: I el 8= - - · eli . I'

The total len gth o r the path is:

u

j

8= -

'l L'2 - dl = - = l. G Ill. ,.

2,.

L

Therefore for the tirne elapsed until the impact , we obtai n: 1~2

T=-

2rv

= <15.

The in stantaneou s rad iu s o f curvature is I, thererore the tension that provides Ihe ce nt ripeta l accelerat ion has magnitude li l t ) Le t us ex press 1 as a fun ction of tilllc. To do thi s, let the upper limit o r integrali o n be I in stead of 0 and let the di stan ce co ve rcd be .'; = v i:

/1.

1

8= -

.I

l

.)

.J

- dl =-:- ( L- - I- )=vt. ./ ,. 2,. I.

From which:

1= V L'2 - 2,. ul. The rorce as a fun ct io n o f tirne is th erefore: .)

F'

= ro;=;;:lI='v=-== L2 -

2 1'1'1

Solution of Prohlem 43. The siring remain s perpendicular to th e in stant aneoUS ve loc it y o f th e obj eci attached 10 it s end , thai is, 10 Ih e tangent o f its traj ec tory : it is 110

6.2 Dy na mics

6. Nl echallics So illtions

----

always in the line of the normal to the trajectory. The normal force is provided by the combined act ion of the string and the force of gra vi ty. At the position of the object where the normal component of the grav it ationa l force alone is able to supply the centripetal force required, the string wi ll become slack. While the free part of the string winding 011 the se mi -cy linder is taut, it remain s tangential to the cy lindri cal surface. Notice that when the string bec oilles slack , the part that has already wound up on the cylinder is longer th an a quarter circle, that is, th e object attached to the e nd has risen higher than the lowe rmost point of the semi-cyl inder. It is only during the ri sing part of the motion that the force of gravity has a compone nt that can pull the object along the direction of the string (that is , toward s the point of suspens ion). The position of the small x object at th e en.d of the string ca;1 be described in terms a of the angle (1 enc losed by rcos a the free (s traight ) part of the s sin a string and the horizontal. The same angle is enclosed between the ve rtical and the ramg dius drawn to the point where Y the line of the string touches the cylindrical surface. If the ;c axis of the coord inate frame is attached to the flat horizontal face of the semi-cy linder (in a direction perpendicular to the axis of the cylinder) and the y axis is set vertical , then the U coord inate of the point at the end of the string is

u= r , cOS(1 - 8 · S111 (1 .

(1)

The work-energy theorem ca n be used to express the instantaneous speed of the object in term s of u: mgy

1

=-

?

Inv- .

2 Division by III and the substitution of y from ( I ) gives the following expression for the square of the speed: ? . v - = 2g(,. . COSet - s· S lI1 0') . (2) At the time in stant when the string becomes slack , the tension force in the string is zero, and the centr ipetal force required for moving in an orbit of instantaneous radiu s of Curvature .'j is provided by the component of the gra vitati ona l force in the direction of the strin c· o' .)

.

v-

IIlg·511l0'=171 -

.s

.

Hence

.

?j g ,

S 'SIIIG= V -

and with the use of (2):

.s . s ill (1

= 2( r cosO' -

.ss ill (1).

131

300 Creat ive Physics Problems w ith Solu tion s

------------~-----------------------------------------------------

Division by cosn g ives

s tan n = 21' - 2stann. The tangent of the angle correspond in g to the position s = 0.961': 21' 21'

tann =

- = 3s

3·0.961'

III

question

IS

obta in ed with

= 0.694,

a nd the ang le is

n

= arc tan 0.694 = 34 .8° = 0.607 radians.

The total length of the string is therefore

L

= s + ( n + ~) l' = 0.48 m + (0.607 + ~) ·0.5 m = 3.141' = 1.57 m ,

that is, the total le ngth of the string in this case is equal to the le ngth o f the semicircle.

Solution of Problem 44. The fact th at the string connecting the two blocks is long means that it is longer tha n the he ight of the table (h) , therefore the block of mass m2 moves vertically all the way down and will not be pulled in the horizonta l d irection by block 7nl before hitting the gro und. In the first phase o f the motion the objects m1 move with consta nt accelerati o n, whi c h can be --J· . .----------~f·,· determined by using the laws of moti on of the :m 2 \ two objec ts:

.

h

7n2g - J(

I(

~

=

7n2a,

= 7n I a ,

where J( is the tension in the strin g, a is the mag nitude of the acce lerati on of both objects. (They move with the same accelerati o n because the strin g is not stretch ab le.) Addi ng the above equations , we get that the magnitude of the acceleration of the two objec ts is:

x

a=

7n2

'g.

7nl+7n2

The time taken by block

7n2

to hit the gro und is:

The velocity gai ned by the first block in that time is: v

= V2a h =

2gh .

7n2

+m2 In the second phase of the moti o n the second block is at rest on the gro un d , while the 7nl

first block moves towards the edge of the table at constant velocity . After leavin g the table, the first block undergoes hori zo ntal projection. Knowing th at the di sta nce covered in the vertical direction is h , we can calcu late the time of fall as: tl = )2h/ g.

132

6. Mecha.nics Solutions

6.2 D Y llCl.m ics

----The distance covered in the horizontal direction in that time is: x = vt1

=

2gh ·

'I7'/.2

m1 + 'rn2

.

fill -

9

=211 1771

rn2 +m2

Note that our result is independent of g . Substituting given data, we find .r =

: :;: 2V5/5 m;::::; 0.896111. Investigating the parametric formu la gained for distance .r, we can see that as the ratio of masses (mdm2) increases from 0 to CXJ, distance .r decreases from 2h to O.

Solution of Problem 45. Let us assume that the masses of the pendulums are negligible (or are included in the masses of the carts) . The pendulums should be held in their previously calculated positions and shou ld be released together with the carts, otherwise they would swing, which would have an unwanted eA'ect on the movement s of the carts. The positions in question cou ld also be defined as th e positions about which the pendulum s wou ld oscillate. (After these oscillations were damped , the pendulum s would remain in these positions, but that would require a cord and track far too long .) The first task is to determine the acceleration of the carts. Applying Newton' s seco nd law to each of the two carts (let 1\' be the tension in the cord), we obtain:

'/TI2gsin o: - !\' = m2 Cl .

Addin g the two equations, we get th at the acceleration is: a=

'17'/..)

-

ml +m2

gs in o: .

A pendulum remains stationary relative to a cart if the acceleration of its bob equals the acce leration of the cart. The acceleration of the bob is caused by the net force of the gravitation and tension . According to the figure , the angle enc losed by the vertical and the pendulum in cart m 1 that moves horizontally can be written as: tan cpl

a

77'/.2

9

'/TIl +m2

=- =

9

.

5111 0:.

In case of cart m2 that moves down th e inclined plane, let Us apply Newton 's second law to the bob of the pendulum in direction s that are parallel and perpendicular to the inclined plane. If the mass of the bob is m, we obtain: mgs in o:- X sin .::= 'rng

1Thsin o: , 771 1

+'rn 2

rngcoso: - !\, cos.:: = 0,

(1)

(2)

a

133

300 Creative Physics Problems with Solutions

------------~--- ---------------------------------------------------

where E is the angle formed by the pendulum and the normal to the inclined pl ane. Isol atin g the tension fro m the second equation, we have: I(

cosa COSE

=mg-- ,

substituting this into the first equation, we find: .

m2~na

mgsm a - mgtanEcosa = mg------ml+m2 Dividing the equation by mgcosa, we get: tanE = tana -

m2 ml +m2

tana =

ml ml +m2

tana

The tangent of the angle enclosed by the vertical and the pendulum in cart m2 ( t.p 2 == = a - E) can be determined using the compound-angle formula: tan t.p2 = tan (a - E) =

tana - tanE . . 1 + tan a . tan E

Substituting the expression for tanE, we find: tant.p2=

1TI2SmaCOsa . m l +m2cos 2 a

Substituting known values, we get that the acceleration of the carts is:

51 a = 125 9

m

= 4 s2 '

the angle enclosed by the vertical and the pendulum in the cart moving horizontall y is:

51 tant.pl = = 0.408 , 125 the angle enclosed by the vertical and the pendulum in the cart moving down the plane IS :

51 = 0.4322 , ll 8

t.p2 = 23.37°

tan t.p2 = -

the angle formed by this pendulum and the normal to the inclined plane is: 6

tanE = -- = 0. 24 25 '

-+

E

= 13.5.°

First solution of Problem 46. If the string is pulled with constant acceleratio n, it exerts a torque on the cube that te nds to tilt the cube. Thus the base edge of the cube on the front is lifted off the plane, and the cube slides on its rear base edge. The torq ues of the string , the normal force of the ground and the force of gravity are in equilibriu m.

134

6.2 Dyna.mics

6. Mecha.nics Solution s

:--------Let F denote th e force of th e string, its vertical component being P y , and let J( be he normal force of the tabl e . Based o n the figure , Newton's second law can be used to t et up equations for forces in the horizontal and vertical direct io ns and for torques about ~he centre of th e c ube. With the notations of the figure , J( -

Fy - G = 0, Fy cot a. =3G ,

]( xcosa. + P y(l- x) cosO' = 3Gxcosa..

(1) (2 ) (3 )

Furthermore,

I I - - x =-tan a. . 2 2

(4)

a From (2) and (4),

(5) From (I) and (3),

(F7J +G) x +f.y(l- x ) - 3Gx =0.

(6)

Then the mag nitude of the vertical component of the string force can be obtai ned from (5) and (6):

3 F,!. =-G. 4 The force press in g on the gro und is equal in magnitude to K:

K=F,,+G= ~mo=140N . ., 4" Thus the force exerted by the strin g is

F=

/ F2

V

y

+ (3G)2 =

+ 9G2 = ~v'I7G = ~. v'I7. 80 N = 247 .4 N. VI~G2 16 4 4

The base of the cube is li fted through an angle of

p '! 1 a. = arct a n ' = a rctan- = 14° 3G 4

135

JOO C rcMil"(' P h.l·sics Problclll s lI' itl l Soil it io lis

Second solution of Prohlcm 46. Fro lll th e fi gure:

= :.IC . I'~I = :~(.' t ;111 II.

I'~,

F, :IG F = -·-= - - .

T he torques ahout th e ce ntre o f Illass :

~ 1ll' ( 1-0 I.'l TS

0

- ? _ C\

)-

I \ 'l~' TS1ll (.1-.J

0

- n

)

= 0.

With th e substituti on of the above ex press ions for rand 1\' . and tran sforillati ons to silllpi i fy,

JSill (.I.j O-

:20)

= (COS(l + :~Si ll O)s ill ( 15° -

0 ).

With the app lic atio n o f th e add iti on forillula :

2cos 2 (1. Hence

= 8s in ocosn .

]

Lall O = 4'

a lld

Thu s

and

F

:3G

=- = 2 · 1 1.:~:\. (,OS(\

Solution of Prohlem 47. a) In equilibriulll (uniform moti on in a strai ght line), the net torque o f all forces Illu st be O. Thi s occ urs w hen there is no fri ction , w hi ch is only poss ihle if the string is strai gh t and if its ex tension passes through th e centre of masS o f the di sc. (i .e. th e strin g mu st be hori zo nt al until the disc reaches it s uniform speed). From that point onwa rd s. th e cart may eve n reduce it s speed. and th e slacke ned stri ng may take an y shape. w hile the di sc w ill co ntinu e to tra vel uniforml y. (S uch an ideal case. howeve r. w ill neve r occ ur in rea lit y. th e onl y rea li sti c case is b).) b) I f there is fr iction , its torque need s to be co unteracted by th e turque o f the strin g. (The fo rce of grav it y and the normal force have no torques about the centre o f mass.) Co nsider th e torques 01 th e forces arou nd the po int where th e di sc tou ches th e ground. a point Ill ov in g along w ith unifor lll velocit y in the inertial frame. Si nce neithe r the fri cti on force S nor the norilla l force o f the ground ha ve torques arou nd th at point. the 136

6.2 D y /J a mics

6. j\ {cciJ a llics So /l/tio l/ s

to rqut: of the strin g mu st also bt: zero . This is only poss ible if the ex tension of the string passt:s through th t: point wht:rt: the di sc touches tht: gro und. Tht: li gurt: shows that tht: angle of in clinati o n in ques ti on sati sfi es the eq uati o n

.

Sill 0.

=

R 21? + 2 17 sill o

.

A quadrati c equa ti on is obtained for sin n: 2s iIl L O+2"; iIl O - 1 =0,

and ht:n ct: ,

- 2 ±v'4 +8 sill o.' = -----'---4

)3

1

2

2'

and tht:
FI +F2 m 1 + 171 2

The force exerted by the sp rin g sca le ca n be ca lcul ated by appl ying Newton 's second law for any of tht: two object s. For exa mple , if !\' is the forc e exerted on the object on the left handside. which has a mass of 1711 , then:

From whi ch

- l O N · l 0k g-20~·2 k g

140 N = - 11. 67 N . LO kg+ 2 kg 12 Thi s is the forct: exerted by the spring on tht: obj ec t of mass In ] . The obj ec t of mass Tnl t:xt:rt s a forct: of 11 .67 N on the sprin g. b) T ht: fact that F I and 1-2 arc swappt:d mean s that now PI is exerted on 7712 and F2 is ext:rtt:d o n /1/ I (in their orig inal directi on). So now the strin g bt:twee n the objects and thc sprin g mu st ht: substitutt:d by a rod , otherwi se we ca n onl y pull the sprin g balance. The system mows into the sa me directi on with the same accelerati o n. In thi s case th e forct: t: xt:rted by th t: sprin g scak o n th t: ohjt:ct of mass 1//1 is:

-----=------=- = - -

20;\· I Ok bo· -( - IO N · ? k o-) 17/1

+ 111 2

_ _ _----"_-'--_ _ _-----'b::..:..

= 18.33 N .

10kg+2 kg 137

300 Creative P hysics Problems with Solutions

-----

Thu s,. the fo.rce ~x erted on the sprin g by th~ obj ect of mass mI. is : - l S.33 J . (The negatI ve spnn g force means th at the sprIng IS compressed .) c) If the mass of the tw obj ects are equal, namely 6 kg, then the results in case of a) and b) are: 0 1(=

F2 m - F 1m m +m

=

F2 - F 1 2

=

- 10 N- 20

= - 15

2

and

F - Po 1 2 = + 15N 2 ' respectively . So the force exerted on the sprin g by the obj ect of mass m 1 in case a) is + 15N, in case b) it is - 15N. If the mass of any of the obj ects is zero th an the sprin g balance reads the force exerted on that side whi ch is 10 ,or 20 1(=

Solution of Problem 49. a) Appl yin g the work-kinetic energy theorem to the bloc k' s initial and fin al states, we get:

he nce

2mgsin a .6.1max =

D

=

2 · 30 N · 0 .5 3 SON / m =8

111

=0 .375111 .

The vertical di spl ace ment is there fore: 2mgsin 2 a 0 .3 75 .6.y = .6.1· sin a = D = - 2- = 0. l S75 m ~ 0 .19 m = 19 cm. b) If fri ctio n can be neglec ted , the equilibrium pos ition is de fin ed by the equati on:

D.6.1 1 = mg sin a , thus

.6.1 1 =

X

=

mgsin a

D

.6. I III ax _ = - - = 0 .l S70 m ~ 19 cm . 2

However, if fri ction is very small but can not be neg lec ted , the eq uili brium position of the bl ock is not a po int, but a s mall interval. In thi s casc Newton' s second law applied to the bl oc k being in equilibrium takes the form o f: mg sin a - D.6.1 ± ~£ mg co sa = O. Let .6.1 = x be the distance betwee n the bloc k' s orig in al and equili brium positi ons. The interval for x is defin ed by the inequ aliti es : mgsin a - Mmgcosa:::; D x:::; mgsin a + ~£mg cosa . fro m whi ch we have: mg ) D( si n a- ~£ cosa

138

:::;x:::;

mg D (si n a+ ~£ cosa) ,

6.2 Dynamics

Mechani cs Solutions

~

stitutin g known values , we find:

SU b

O.19 - 0 .323{t:S x:S O.19 + 0 .323 p. f the block is not simply placed onto the inclined plane in an equilihrium position , t reaches its final position from a state of motion as it does in our case, the exact ~~ace where it stops inside the above interval can also be determined , but that is a more difficult task to do . I

Solution of Problem 50. The vert ical displacement of the body that slides down the slope is the same in both cases, but in the case of a moving slope it requires twice as much time. According to the quadratic distance relationship, the vert ical acceleration component of the body is four times as Illuch in the case of the stat ionary slope as in the case of the moving slope because for I the distances travelled vert ical ly in the two I F • IN cases: Sy

=

1

2

2CLllitl

=

1

..

I

mg

2

2 CL211( 2td )

from which a ty

= 4a2.
Let us state the equation of the motion of the bodies, the constraining condition and the equation that expresses the special condition of the problem. Let us handle the horizontal and vertical components of the motion of the slidin g body separately. As from here on the acceleration of the small body is used in the case of the moving slope , we will omit the indices 2. Using the notations of the figure, for the x component of the motion of the small body (1) , N sin 0: - (tN cos 0: = ma." for the y component of the motion m g - N coso: - {tN sin o: = may

(2),

for the wedge:

F - N sin 0: + {LN cos 0' =

j\/ A

(3)

the constraint condition:

ay = (a.,. - A ) t an 0:

(4)

the condition of the problem: 4CL II

= g(sin 0: -

( LCOS 0:) S illO'

(5).

Constraint condition (4) and condition (5) can be understood from the following figure s:

139

300 C reative Physics Pro blem s with Solutions

------------~------------------------------------------------------

~--------s--------~

The vertical and horizontal distances trave lled by the small body are Sy and s"' , respectively , the horizontal distance travelled by the wedge in the meantime is S. The height of the wedge is Sy and its base is S x - S. With these the tangent of the an gle of inclination of the wedge is Sy

= ----S- '

t an a

SX -

and the magnitude of vertical displacement is Sy

= ( s ," - S) t a n a.

From these , in the case of the moving slope the use of the quadratic distance law gives the following for the vertical acceleration of the body:

after simplifying by t 2 /2 (4) is acquired. The right side of equation (5) is the vertical component of the acceleration g (sin a - f.Lcosa) of the body sliding down on a stationary slope: a · sin a . We have 5 independent equations for the 5 unknowns . From (5) a y is expressed and substituted into (2) and (4) , in (2) N is factored out: mg - N(cosa+ f.L sin a)

~ (sin a

= m 2.( sin a

- f.Lcosa ) sin a

4

= (a," -

- f.L cosa ) sin a ,

(2')

A) tan a ,

(4')

from (2 ') the value of N is expressed , (4') is divided by tan a : 1 - ~ (sin a - f.L cosa ) sin a

N= 1ng ---2~--------~----

cosa+ f.L sin a

2.(sin a 4

~. cosa) cosa = a x - A,

from which the acce lerati o n of the wedge is A = a x -

140

'

~ (s in a

- f.L cosa ) cosa .

6.2 Dynamics

6. Mech anics Solu t ions

:----

substitutin g the va lue of N into ( I) and simplifyin g by m gives the foll owing for ax : 9

sin ex-f,lcosex ( s i n ex - ~Lcosex. ) _ . 1Sll1 ex -ax, cos ex + ~L SIl1 ex 4

From (I ) ax is substitu ted into (3):

F = !VI A +ma x , (which could h?ve been acquired fro m the th eo ~e m for .the m o ti o~ of the centre of masS directly ) fin ally by substltutlll g the va lue of A as fun ctI on 01 ax first and then by substitutin g the value of ax from the parameters of the probl em the requested force is acquired :

F:::: M [ax -

~(sin ex - ~LCOSex ) cos ex ] +ma,,; = (M +m) a," - M~(sin ex-~LCOSex) COS ex =

:::: (M+m)g

s in ex-~Lcosex (

.

cos ex + ~L S Il1 ex

1-

sinex-~Lcosex

4

·sin ex

)

9

-M-(s in ex-~Lcosex)cosex,

4

reorgani zin g:

F =g(sin ex-f,lcosex) [ (AI +m )

4-(sin ex-f,lcosex)sin ex M ] ( .) - -cosex . 4 4 cos ex + ~LS ll1 ex

With numerical values: m

[

F = 9. 81 ~ (0.5 - 0.2· 0.866) 2 kg

4 - (0.5 - 0.2· 0.866) 0.5 1 kg ] 4( 0.866 + 0. 2.0 .5) - 4 0.866 = 5.67

First solution of Problem 51. First we will solve thi s proble m usin g Newton's second law . Let us apply it to the three obj ec ts in two perpendi cular directi ons. Since the inclined pl ane al so moves, these directi ons should be the horizontal (x) and vertical (y) direction s. Let N be the normal force exerted by the inclin ed pl ane on bl ock m l , [( be the te nsion in the cord , A be the acce lerati on of the inclined pl ane, aL" and al y be the hori zontal and vertic al components of the accelerati on of bl ock ml respectively , a2 x and a2y be the horizontal and verti ca l components of the accelerati on of bl oc k m2 respectively. Obviously a2x and A are equ al, therefore we have six unkn ow ns. The seventh unknown, th at we are to determine, is time t but thi s can be ca lcul ated usin g kinemati cal equ ati ons. To solve the pro blem, we need to un de rsta nd th at the two blocks will be nearest to each other when the length of the cord on eac h side of the pull ey is h/2. The di stance between the two blocks - th at takes its max imum ( h ) at the beg innin g of the moti on and at the moment when the seco nd bl oc k reac hes the pull ey - decreases in the first part of the moti on and the n increases. The di stance will take its minimum value when the POsitions of the bl ocks are sy mmetrica l about the bi sector of the top angle of the inclined plane. In th at pos iti on the two parts of the cord form an isosceles tri angle, therefore to

14 1

300 C l'cilti,'c Physics Pl'Oblel1l s " 'i/ il So illtion s

------------~---------------------------------------------------------

so lve the prob km , wt: sim ply nt:t:d to dt:tt: rmin e th e till1t: take n by the han gin g hlock to movt: a di stanct: of 82y = II / 1 with accderati o n (fly' whi ch is: /=

(1)

Tht: rt:i"ore, wt: havt: to lin d lI'2y alk r sL:lti ng up th t: fo ll owing sy stt:m of eq uation s, Equ ati on (2) is Nnv to n's st:Cll nd law applied y to tht: in clin t:d pl ant: and bl oc k /lL2 in th t: horizo nt al direc ti on (ass umin g th at th t:i r hor izo nt al accekrati ons art: equ al), eq uatio ns (3) and (4) arc New to n's st:co nd law appli ed to bl oc k 11 1 1 in the hor izo ntal and vt:rti ca l dirt:ct io n rt:s pec ti vt:l y, equ ati o n (5) is Newto n' s st:co nd law app li t:d to bl ock /11 2 in th t: verti ca l di rt:c ti o n, t:q uati on (6) m,g is th t: kin t:mati ca l rt:strainin g co nditi on betwt:e n bl oc k III I and th e in c lin t:d plane, whik equ ati o n (7) co mt:s from the fac t th at th t: kn gth of tht: cord is co nstant :

= (J\! + 1l72)A = /Ill (/1.1' j\' cos n - /\' siIl 0 = 111 1 (/ I!I / \' - 11/ 29 = 1I12 Ul!l (liy = (((1.1' + A) la ll o j\' sill o

- /\' coso

(2) (3)

A sill o - /\' cas u

11119 -

( I)

(:j)

(G)

111//

(/2.1/

(7)

== -,-

SIII O

Equa tion s (2)- (5) ca n bt: dt:ri vt:d direc tl y fro m th t: li gurt: abow, but t:quati o ns (6) anu (7) nt:t:d to bt: ex plained, Tht: re is a kinematical restrainin g co nditi on bt: tween bloc k 1111 and the in clin ed pla ne because the bl ock has to rt: main o n th t: in clin t:d plan e, The li gure bd ow shows the initi al and fin al (al'tt:r timt: /) pos it ions o f th t: systt:m , Whil e bl ock 1111 mows 8 11/ dow n vt:rti call y, it cowrs a distanct: o f 81 ,1' in th t: horizo nt al direc ti o n, In thi s ti me the in clint:d plan e mows a di stan ce of S, It ca n bt: st:t: n fro m the ligurt: that the tangt:nt of th t: angk of i ncl inati on is tht: rati o of k ngth s S ly and 5 + s 1,1' , so: l a ll (!

S i ll = -'- , 5 + 81,1'

,"U\

Ass umin g th at th t: initi al ve loc it it:s of eac h obj t:c t are zt:ro and th at th e accek r· at io ns art: constants, we find th at th t: ra tio of di stances are equ al to the rati o of the correspondi ng accd erati o ns: La l1 o =

142

(fl !l 1 .1 ')

'2 I't 1-

I

+

')

'2 (1 1.1' /-

A + o l ,1'

6.2 Dynamics

6. M eciJ a llics So ill t io ll s

-----

{[ I y = (([ I.,. + A ) t a n (1 , as was statl:d in equation (6). As thl: kngth of the cord remains constanl, the displacement s of the two hlocks rclativl.: 10 Ihl: in c lined plane mu SI hI.: equal : Slr,,1 = S2 rel. Using that Slrpl = s l y/ sin o, we ha vl.: .';Iy

.'; 2 y = sin o ' AssLiming again Ihat the inilial ve locilil.:s afe Zl.:ro and Ihl.: acce leration s arl: constant , we find th at: (/I !f

(/~ y

= Sill O

as was slatl.:d in l:quati o n (7 ). Let us now solve the system of equalion s. Multipl ying I.:quation (3) hy coso and equati on (4) by sin o , we obtain :

a\

Ns ill o coso - /\. cos 2 (\ 1I11.9 sin Ct - N cosCtsin 0' - /\' sin 2 0

= 77~lCLLrCOSO' , = m lCLl y sin 0 ' .

After adding Ih esl.: Iwo l:quati ons , WI.: lind thaI N sill Ctcos o cancels out , and by factoring out }\', its l:odlil:il:nt wi ll he sin 2 Cl. + C05 2 0 = I : IiIl.9 sill O - /\. = 1711(/1.,.(;050 + i7l1CLl ys in o . Let us solvl: equation (5) for }\' and suhstitute it into the ahove equat ion: 1111.9 si n (\ -

1n2C/2y -

1112.9 = IiI lal.,. coso +

In ]

CLly

sin a

(8)

.

The ri gh t hand s ides of equatio ns (2 ) and (3) are eq ual:

(J\I + m)A

= IIIICLl.r ,

thus 1771

A=

1\/ + 1112 insertin g this int o I.:quation (6) , Wl: have: (/I y =

(

(/1,. + .

1711

1\/ + 1712

a l ,.

.

)

0.1,.,

ta ll o=

.

1\}

+

m]

+ in2 tall O · Cl I,.·

1\/ + m2

.

Which yil:lds al'= .1 1\}

/\I + lJh a l ii 1\} + '1I12 coso . _- = - - 'CL II ' +1712 tallCl. /\} + IIII +m2 siIl O. I

+ 11/1

LeI us now insl:rt ((1.1' as expre ssed ahovl: and CL I!! = CL2ys in o from equation (7) into equation (H) . Thl: only unkno wn in thi s l:quation wi ll be CL2y : . . 1\/ + 1112 cos 2 0 . 2 ml .9S III Cl. - 1I12(}211 - m2.9= 11I 1(l21/ 5In o · \ · - .- - + m' ICi2!! slll O. . . 1 / + ml + 11l2 SII1 0 143

300 Creati ve Physics Proble ms with Solu t ions

----------~~---------------------------------------------------

After so me algebra, we find : . a- 1n2 ) g= [ ml ( (1nlSm

. 2 a ) + 1n2] ' a2v' M + m2 cos 2 a+sm NI +m l +m2 .

After writing the coeffic ient of a2y as one fraction , we have: (mlsin a- m 2)g= 2 m l (M +m2)cos a+ml(M +m l +m2 )sin 2 a+m2( M +m l +m2) = M +m l +m2 ·a2y . Let us write the second term of the numerator as : 2 2 ml(M +m l +m2)sin 2 a = ml( NI +m2)sin a+mls in a. Factoring out ml(M + m2) from the first term of the numerator and the first term of the ex pression above, we obtain ml(M +m2)(s in 2 a+cos 2 a) =ml (M + m 2), which simplifies our equation to: 2 . ) m l (M +m2) +misin a+m2( M +m j +m2) (1nl sln a - m2 9 = "1 . a2y ' l Vl + ml + 1n2 Solving for a2!1 gives:

Substituting this into equation ( I ), the time in question takes the form of:

h [ml(M +m2) + m isin 2 a+ m2(M +m j + m 2)] (mlsin a -m2)(M + ml +m2) ' 9 Substituting the given data, we fi nd :

t=

2 1 m [7 kg(2 kg + 1 kg) +49 kg · 0.36 + 1 kg (2 kg+ 7 kg+ 1 kg)] (7kg ·0 .6 - 1 kg)(2 kg + 7kg+ 1 kg) · 10 ~

= 0.389 S;::,j 0. 39 s.

Second solution of Problem 51. Let us solve the problem usin g conservation laws. The mechanical energy of the system is conserved : h) hI 22 1 2 1 2 mlgh=mlg(h - - sin a +m2 g-2 +-2m1(VX1 +v1,.)+-m2V2,,+ -(M +m2) V , 2 • 2 • 2

144

--

6.2 DYJlam ics

6. M echan ics S olu tions which y ields:

m ]Q -h25 .1110. -

h . 2

1 2

2

'm 2g - = -(V I '

..

1

? 'J +v-'JI !J ) + -12 7T1 ?V,)" + -1 ( j\! + 111 ·,) 11-. - - .' 2 -

(1)

where V is th e hori zo ntal ve loc ity o f th e inc lined plan e (and of block /11 2 )' T he lin ear momentum o f th e system i s conserved :

= ('1772 + J\f ) \I.

1171 V I.I

(2)

Since th e length of the cord is co nstant (sec first so luti on): _

Vly

V?, - - - . - .1 sin o. As bl oc k

1n ]

(3)

remains on the inclin ed plane (sec first so luti on):

(4)

vl y = (v l.l + V) Lan o.. Let us now in sert gi ve n data:

70 J . 0.6 - 10 J

= 7 kg( V ?, + v L) + 1 kg · V~'I + 3 kg· V 2 , 7 v I.I = 3V, V I !J

(1' )

Vl y = 0.6V2.'1 '

(2') (3' )

= (V J. I + V)0 .75 .

(il ' )

From equ at io n (2' ), we have V 1.1

= ~ V,

(2")

which is then substituted tion (4') to get :

Vly

into equ a-

= ( "73 + "77) v· 43 = 710 . 4"3 V,

inserting thi s into eq uation

(3' )

( "

il )

a

gives:

10 3 - · - V=0.6 V2 7 4 .'I' thus

V

4 7

= -3 . -10 ·0 .6V2 !J'

(5)

substitutin g thi s into equati on (2" ), we find: 3 4

Vl .

."

Substitutin g eq uati o ns (6),

7

4

= -7 . -3 . -10 ·0 .6v?- !J = -10 ·a .6v·)-!J .

(6)

(3' ) and (5) int o equat io n (1' ), we ob tain an eq uat io n for

1J2y :

. 32

16 2 . 2 2 . 16 49 = 7· -100 . 0.36v 211 + 7· 0.36v 21J + V211+ 3· - . 100 ·0.36 1'2.11' · .. 9 .

.'J

145

300 C rea.ti ve Physics Problem s with S olution s

After some algebra, we find: 2 V 2y

therefore the verti cal ve locity of block V2v

.

=

3200 m 2 = 48 6.4~ ' m2

is:

J

3200 = 2.5649 m 486 .4 s

Using this , the time in question turns out to be:

2s hIm t = _ 21_.v = - = . =0.3 8978s::::::; 0.39 s. V2y V2y 2.5649 'if

Ksina

---

-~--

-

----

...

--

,,Kcosa 2 K -h

a

tIMgiii';;'~" ~:

Solution of Problem 52. a) First let us write down the equations of motion for the sphere as well as for the wedge, and the relation between their accelerations The data referring to the sphere is denoted by lowercase letters , while the data ones corresponding to the wedge is denoted by uppercase letters. According to the fig ures a) and b), we get:

= ma Ksincx = MA

mg - K coscx

:~

1 h

a=Ata ncx

3ig-a Fig. a)

(1)

(2) (3)

The last equation comes from the restriction th at the sphere touches the top surface of the wedge durin g the ir motion. The distances covered by the sphere and by the wedge in time tare:

1 2

6. s = -at 2

6.S=~Ae , 2

since all the forces , and consequently the accelerations are constant. According to figure b),

6. s = 6.S· tancx ,

a

Fig. b)

thus 1

2

1

2

-at = - At ·tan cx 22

from which , by dividing with

~t2 , equation 2

'

(3) is obtained.

The mass ratio M 1m is obtained by eliminating the quantities A and K from the system of equations. From equation (3) : a

A= - - . tan cx

146

6.2 Dy nam ics

6. Mecha n ics S olutions

----

Plugg in g thi s into equati on (2), A- ca n be expressed :

a A-= M - - - t a n a · sin a

(4)

Insertin g thi s into the equ ati on ( I), we get: m g - Iv!

a

. ' cos a = m a o t a n a · sm a

Rearran ging thi s, and usin g the identity cosa/sin a

= l/

t a n a, we ob tain :

'mg= ( -M 2 - +171- ) ·a , ta n a

from which the accelerati o n of the sphere is: a=

gtan 2 a

mg ~+m t a n 2 0:

M '/I I,

(5)

+ tan 2 a '

Accordin g to the fi gure a), the co nd iti on ass urin g that the wedge does not tilt , ex pressed in terms o f torques relative to the ce ntre o f mass , reads as foll ows:

2 / . 1 / h h - !\ h sm ct+ - !\ cosa-- < (NJg+ I\ cosa) - - - . tan a 3 tan a 3 3 r

Here, we have used the fact th at in the margin al case the line o f ac ti on o f the force exerted by the ground on the wedge is shifted to the ri ght edge o f the wedge . After simplify in g this equ ati o n by h / 3 and multiplyin g by t a n a, we get: 2[( sin a · t a n a+ [( cosa:S M 9 + [( cosa . Omitting the same terms on both sides, and insertin g the ex press ion of [( from equation (4), we get:

2M

a

.

. sm a· tan a:S1I.1g, t a n a · sm a so the maximal accelerati on of the sphere is:

a < fl .

- 2

Putting thi s into (5) we obtain the desired mass rati o M/ m: a=

gta n 2 a 9
from whi ch

T1/

l\f 2tan 2 a:S - + t a n 2 a, 1n

and fin ally the relati on

M

- > tan 2 a 'm -

IS

obtain ed . T hu s the rati o 11.1/m and the angle a should sati s fy the inequality above.

147

JUG C reatin' P in'sics Pro i)icllI.5 II'itll Solut io ll S

h) II' (\

= GO ° . the n

,\I

-

::::

.,

(<1 11 -

GO

° = J is the co nditi o n. T hu s in the case o r -1\1

=

1/1

1/1

the wedge does not till. Accordin g to the co nservat io n o r mec hani ca l energy: .

1

II l!j / S 11 J()

.

= -

2

.) 1/ I I ' -

+ -:2 1 \1 II - . 1

.,

T he re latio n o r th e sreeds is silllil ar to that of the accele ratio ns: ('

(, =

11(<1110

11= - - . (a 11 0

al1d usin g th is in the rrev ious equ ati on we get : .)

2/111)/ si 11 (1 =

1' -

III ('"

+ 1\1 --.-)La 11 -

=

(1

(

1/1

.\-1) + --. )- . V .-) . l a wn

Fro m thi s the sreed o f the srhere after cove rin g a di stance / along the wedge is:

u=

2 ·9 .8 1 lll s - 2 ·D.2 lll · s in60 ° ------~-----

2g/sil1(1 1

+ -'-"-.,-

I+

IIllill l -()

-V--t.illl- (j O O

;::::: D. S2

111

s

Solution of Prohlem 53. O ne r oss ibl e meth od wo uld be to suspend a small ball o a long cord th at hangs rro lll the top o f th e box. Let us wait until the osc ill ation of r e ndulum stor s and meas ure angle .r , whi ch is the an gle e ncl osed by the cord and line th at is r erpe ndi cul ar to the top o f th e box. Th e bob of the renduluIll is at n:st w respect to th e hox. if it s accelerati on equal s the acce lerati on o r the box re lati ve to ground. T he acce lerati o n o r the hox is give n by the ex rress io n: u

= g (s il1 C1 -

11(' OS (1 )

The acce leratio n o r the bo b is ca used by the net force o f the gra vit ati o nal force ac ve rti ca ll y and the ten s io n exert ed hy th e cord .

Let us use th e fi gure ahove to arrl y Newto n's seco nd law to the hob in the d irecti th at are r arall e l and perpe nd ic ul ar to the in c lined pl ane. 148

6. ]\Jedl a lli es So luti o ll s

6. 2 Dynamics

---

In th e [lcrpcndicu l ar direction we ha ve: m g sin o - P ('OS.I;= O, while in th c parallcl direction we obtai n:

mgs in o - F sin .l' = /Ilg(s in o hence

III g(,05C1 -

-

I L(,0 5 0. ) ,

F eos .!' = O.

From which wc lind: P sill .l'

= I"ng ('os Cl

F ('oS .1' =

IIlg ('os (1

.

Di vidin g th c lirst cq uati on by th e second , we get: Lan.r = II. Th is way the coeflic ient of kineti c fri ction is determined. There are sc veral methods th at could be used to determil1l: the angle or th e in cl ined pl ane. Since we have three un know ns (It , 0., F) in the above equat ions. one more measurement is needs to be taken. The first possible meth od wo ul d be to meas ure the period o r the pendulum th at sw in gs th rough a small angle. The [le ndu lulll in side the box acce leratin g down the inc li ned plane behaves as if it was placed int o a uniform grav itati onal field in which the gravitati onal acce lera tion

=.r; -

a

was gt (I, where is th e acceleration or the box. The connection between this apparent grav it ati onal acce lerati on and the tension is gi ve n by th e eq uati on: 12 ['-'J = (my cos o )-'J + (/(/ 1I9 ('OS Cl )-.)= ? lIl-g ,

where gl is th e Illagnit ude o f th e a[lparent gra vi tat ional acceleration , ror wh ich we get:

gl=gcoso~ , and th ererorc the [leri od of th e pendu lu lll is:

T = 27i

g cos(\' ~ '

from whi ch the angle or the inclin ed plane ca n be calc ul ated as : 47i 2

0.=arc('os gJl + IL2 · T 2 '

For exalll[lle, ir the mass or the bob is III = 1 kg , angle .1' is measured to be 11.3 0 • the len gth or th c cord is 1= 1 III and th e [leriod is round to be T = 2.5 s, th e coefl-ic ient of kineti c fri ct ion wo uld be: Ii = La ll 11. :)° = 0.2 , whil e th e angle or the in clined plane wou ld be: (1

= a rccos

1 9.8 1v'IJf,I

LI7i

2

. -G.25

= 50.85

°

.

149

300 Creative Phy sics Problem s with Solu tions

------------~----------------------------------------------------

the acceleration of the box would be: m ( sm50. . m a=9 .81 2' 85 ° -0.2cos50. 85 0) =6 .372" s s

and finally the tensi o n in the cord (in th e pendulum 's stati onary state) wo uld be:

F = (mgcosa)J1 + f.i2 = 9.81 N cos50. 85°V1 + 0.04 = 6.32 N.

A seco nd possible method is to measure the ten sion. In thi s case the mass of the bo of the pendulum should be meas ured be fore gettin g into the box. Usi ng the figure ahove we obtain:

AB = A C - B C = mgsin a - mg(sina - j.icosa) = I-icosa. Thu s

2

2

2')

2

~

F = (mgcosa) + (j.imgcosa) = (mg) - cos - a · (1 + j.i ), which yields 2

cos a hence cos a = therefore

F2

= (mg )2 ( 1 + f.i 2) ,

F mgJ1+l-i 2

F

Fcos x

mgV1 + t a n x

mg

= -----;===7i= 2

Fcosx a = a r ccos - -- . mg

Substituting the values give n in the example, we find : a = ar ccos

6.32· cosll. 3° ° = 50.52 , 9.81

which equals approxim ately the previou s result. Note that these methods can only be used if the value o f 9 is kn ow n and we are su about the fact that the bottom and top o f the box are parall el to the inclined pl ane. I any other case the quantiti es in questi o n can not be determined .

Solution of Problem 54. New ton 's seco nd law ap plied to the force components parall e l to the string is 2 /' V J\ - rngcosa = 1TLT'

(1

where J{ is the tensio n ac ting in the string. Accordi n to the work-energy theorem (applied to a ge neral cas of an initial angular di splacement o f aD), 1

mgl(cosa - cosao) = -mv 2 . 2 150

(2

--

6. M ech a llics Sol ut ioll S

G.2 D .),118111 ics

Fr olll (I ) and (2) : /\' - mgcoso.

a

= 2/ng( coso. - cosO:o) ,

K

and hence the tens ion !\' is !\'

= 711g(3 C05 0. -

2coso.o).

h

The torque of the strin g is the onl y torque th at the grip at the lower end needs to counteract: it mu st be able to exe rt an equal torqu e in the oppos ite direc ti on. The fi gure shows that the torque of the te nsion force with respec t to the lower end is T

= J( h sill o. = mg( (3coso. -

mg

2coso:o) . h · sin ct.

Since the pendulum started from an initi al pos iti on of 0.0 = 90° , COSClO = 0 and the torque in questi on is sin 20 T = 3mgh · sill o.cosCI. = 3m g l l -. 2 This torque is a max imum if sin 20: is a max imum , that is, at 2n = 90° , whi ch means a=4So . Thu s o 3 T = 3mghsin 4S o cos ,IS = 2 mgh = 9 N Ill . lllax

It is worth notin g that the ten sion force and the torque are both independe nt o f th e len gth of the string.

Solution of Problem 55. a) Sin ce the external forces ac tin g on the syste m are all ve rtical , the ce ntre of mass of the sys tem will not move hori zontall y. When the ball has Ic rt the hemisphere, it must move verti call y upwards: since the ball slides all the way along the hemi sphere, it ca nn ot ha ve a hori zo nt al velocity component at separati on. Otherwise, its horizontal veloc ity would be equ al to that of the cart , whi ch would mean a di spl acement of the centre of mass of the syste m to the left or to the ri ght. A fter separati on, th e motion of the ball is verti ca l projec ti on, and the cart is bro ught to rest again . Thu s, all the energy is conce ntrated on the ball aga in , and the ball will rise bac k to its initial height. Let X denote the di spl ace me nt of th e cart and let x de note the hori zo ntal co mponent of the displ acement of the ball. As show n in the figure , the di stance o f the ce ntre o f mass o f

R

m

x

, :R-XiX I- ':

:

212 '

:------1 x : X

:

15 1

300 C reati ve Physics Problem s with S olutions

the cart from the common centre of mass is X / 2 , and the di stance of the ball from there is R - X / 2 metres. By definition of the centre of mass ,

N!X 2

=m

(R- X) 2

'

and hence the displaxement of the cart is

X =~R M+m

to the right, whil e the horizontal displacement of the ball is

x=

2M R M+m

to the left. With nume rical data:

x=

2·2 kg ·0.5 m = 0.8 m. 2kg + 0.5 kg

b) The normal force acting at the lowermost point of the path of the ball is obtained by applying Newton ' s second law:

v2

N -mg= m~

(1)

R '

where N is the normal force and V i e I is the speed of the ball relative to the c art. At that time instant the cart represents an inertial reference frame, therefore Newton 's law is valid in it as well. From the work-energy theorem ,

(2) where v and V are the speeds of the ball and the cart, respectively, relative to the ground. S in ce horizontal momentum is conserved ,

(3)

m v = M V. Finally , the re lation ship of the speeds measured in the two reference fram es is v i eI

= v+ V.

The solution of the simultaneo us equations for the magnitudes of the qua nti ties involved is

V=

2m 2g( h + R ) and v =M V = 2Mg(h + R ) / ,,{2 + mN! m M +m Mm+m2 h +R ) N lIl ax = ( 2 M '~+ ln g=

m NT = ( 2 · 2 kg· 0.5 kg + 0 .25 kg2 . 0. 5 m + 0. 5 m + 0. 5 k) g ·10 - 2 = 30 . 2 kg

152

0.5 m

8

--

6.2 Dy na.mics

6. Mecha.nics Solu tion s

Solution of Problem 56. a) When point

A hits the ground, the rods form a strai ght

o~

line. As rods have constant length , the two ends of the system mu st have zero ve locity at th at moment, so the block of mass mz stops. Since onl y co nservat ive forces act in thi s situ ati on, the initial grav itat ional . potenti al energy of the rods will be transformed into the rotat ional kinet ic energy of co A co the rods, which is independent of the mass ~A~=:=:=:=::$;~=:=:=:=:=:::!~L..J._ of the bl ock . At the moment when point A hits the ground , the rotational kinetic energies of the two rods are equa l. Assumin g that the rotati onal inertia of a rod about its end is e = m 1L 2 / 3 , the rotational kinetic e nergy of one rod is:

0

-

..

I

Be

The law of conservation of e nergy will take the form of: L . 2mlg- sm a 2

= 2 · -1 . -1 1n 1 L 2 w 2 2 3

'

where L /2'sin a is the distance of the centre o f mass of each rod from the ground in their initi al pos iti on. As a = 60 0 , we know that sin a = V3/ 2 . Substitutin g thi s into the equation and isolating the angu lar ve locity, we find:

W =j3~V;. We then use this to express the veloc ity of point A when hittin g the ground : VA

= Lw =

j39~'/3

2

=

.81 I11 S - ·0.5111·)3 -3·9- -- - -- = 3.57 -111

2

s

Note that the result does not depend on the masses of either the rods or the block. y

o

x

a 153

JOO C I'Cilt i l"(' PI I.n;ics Pl'UiJlc lll s lI ' i tiJ SOllltioll s

------------~----------------------------------------------------------

b) T he accelerati o n o r mass 11/2 at th e mo ment w hen po int A hits th e ground is m ade up o r two dinc rent co mpo nent s: th e first o ne is th e hori zo ntal co m ponent or th e accelerati on o r po int .4 ((I .r )' the seco nd one is th e hor izo nt al co mpo nent of the accelerati o n of poi nt /3 re lati ve to poi nt A . Let th e refere nce frame be att ac hed to po int B , w hose vel oc it y beco mes zero , whe n po int A hi ts th e gro und . In this rrame rod A!3 (a nd th ererore poi nt rI) ro t a t l~S in an acc elerat ed moti on around point f] . T he hori zo ntal co mpo nent o f the accele ration or point A mu st be th e centripetal accelerati o n, so (I .r

= (/" = PA2 l L

If th e reference frame is now all ac hed to po int A. the horizo nt al acce lerati o n of po int [J (a nd th erefore that o f mass 11 / '2) wi ll have th e sa llle mag nitu de as th e one calculated

above but w ill be in oppos it e di rec ti o ns. As th e hori zo ntal accelerati on of point rI in thi s situati o n has magnitude u ,. = (I " = I '~\ I L an d po int s towa rd s po int 0, th erefore the total acce lerati on o f po int /J and mass 1112 at th e moment w hen l 'U 0 i s:

=

.)

0 /3

= LO .r = 21'- I L

Subs titutin g k now n va lu es gi ves :

{f/3=2o r = 2·

;ly.J;iL '2L

r.)

_'J

_

co ('7 ~ s'2

-0V " 'y-U.<1

Note th at thi s res ult is independent o f the masses and o f the len gth s of th e rods . (T he force ac tin g in hin ge U is

ill

r = 1112(1 n = '2 kg· 50 .97 s? = LO l. U4 ~.)

Solution of Prohlem 57 . a) Du r in g the 1lI1llbiing o f the tri angle. th e initi al positional energy tra nsfers int o ro tati onal ki neti c en ergy. The sid e h li es on th e gro und. so only th e ed ges na nd c have non zero pos iti onal energy. Thi s energy (relati ve to th e ground. as zero le ve l) can be ex pressed w ith th e help o f th e height o f th e ce ntre o f Ill ass of the rod s. N oti ce that thi s hei ght is Il c = n l '2 for both rod s. So , appl y in g th e co nservatio n o f energy for th e initial state and th e fin al slat e o f th e tri angle (j ust before th e tr iangle hit s gro und ). we ge t: l .)

(III" + lII cl .q!t c

= 2 (8" + 8 ,. ),..<;-.

w here 8 " and 8 e are th e moment s of inerti a o f the rods (( and c w ith respec t to the edge h, as ro tati on ax is. It is we ll know n th at th e mo ment o f inert ia o f a thin rod abo ut an ax is perpe ndicular I 2 to th e rod at o ne o f its endpo int s is (-) = J"II . In our case thi s app li es for th e side (I :

8"

I

'J

= :3 "1,,((-.

T he mOl11 ent o f inert ia of th e rod c is still to be det ermin ed. It ca n be obt ain ed in a tri cky way. Firs t le t us place th e rod o f le ngth c and of Ill ass III,. perpe ndi c ul arl y to th e ax is. T hen let us 'co mpress' the rod. keepin g it s mass un changed. to a size w hi ch is equ al to th e d istance of th e endpoi nt /3 o f the ohl iqu e rod fro m th e axis:

154

--

6. j\[cc!I
6.2 D y na mics

B

Bl ________ _____ Jr= o

c

B

b

_______ :~s;;g

0

t

The mo ment o f in erti a o f th e co mpressed rod is G,. =

~1I1,./,2,

w here /' = C' SillQ = a

is the di stance o f po int 13 from th e ax is. Fin all y, let us shear th e rod in such a way that its len gth beco mes aga in the orig in al len gth , and every po int o f th c rod moves parali el to the ax is. T hi s tra nsformati on does not cha nge th e mo mcnt o f in erti a, thu s: G ,.

1 = }1Il ,.C- SII1- O = "3111,.([-. ]

')

.

'J

'J

Thi s re sult could have bee n ob tained also by the definiti on o f th e moment o f inerti a: '\'

G,. = L

.

III

'J

.

i (Ii S ll1 (1)- =

.)

S 111 -

'\' (I

L

.)

III

iii

.

.)

= S111- 0

1 .) 1 "311/,.e = "3 771 (.0-. 'J

.

PUllin g th ese res ult s int o th e equati o n o f energy co nserva ti on, and ex press in g th e mass of each rod by th e mass o f rod c, we get th at:

so

8 -::1I1 (·fJ(I ;) We ha ve used th e fac t th at velocit y /' =

II..J

(I

=

= -8

:3

'5("

J·5

') 'J 111,.o-",r

thu s

III "

=

.

.)

---> ~1g=a,-,.r

3

'5 111,. . Co nc ludin g. th e va luc o f th e final

o f th e vcrt ex /J is:

['=uw= )3yaJ3· 10 111 S- 1 . O.3 111 =3

111 S

and it is di rected ver ti ca ll y dow nwards. b) In thi s case thc ce nt re o f mass o f the tri ang le m ay move onl y in verti ca l d irec ti o n, since ali the extern al fo rces are vcrti ca l. T he verti ca l forces ca nn ot g i ve ri se to rotati on in hori zo ntal pl ane ( i. e. about verti ca l ax is) ei th er. thu s th e rod b ly in g on th c gro und just transl ates para ll c ll y. A t th e mome nt w hcn th c tri ang le hit s the ground none o f its point s have horizonta l veloci ty d ue to th e co nse rva ti o n o f horizo nt al momentum. Thu s, accordin g to the law o f energy co nserva ti on. th e angul ar ve loc ity .iu st before hitt ing th e ground . as we ll as th e veloc ity o f po int F3 at thi s instant co in c ide w ith th e res ult obt ain ed 1n case a). 0.4 The pl ace w here /J hit s th e gro und is now by 1L/3 = ."" 0.] 111 c lose r to th e initi al pos it io n of sidc b th en y 111 case a). To ob tain thi s th e pos iti on o f th e ce ntre o f 0.3 0.2 111ass has to he determ ined . Accord in g to Py th agoras' th core m. side b o r th e trian gle has a len gt h O. I 111. Le t us usc a coordin ate

155

300 Creati ve Physics Prob lems wit h Solu tions

syste m whose x ax is passes thro ugh the ve rtex B and parall e l to the edge b, as it i show n in the fi g ure. The n the y coordi nate of the ce ntre of mass of the tri ang le is si1np l s the weig hted average of the y coord in ates o f the centres of the three rods, so : y

y=

4·0.3 + 3·0.1 5 + 5·0.15 m=0 .2m. 4+3+5

Here the numbe rs 3, 4 and 5 are the ' we ights' propo rti o na l to the masses of the rods ' a nd 0.3 m, 0.1 5 m are the y coord inates of the ce ntres o f the rods. W hile in case a) the vertex B hits the gro un d d 1 = a = 0.3 m far from the initial ve rtical plane of the tri angle, now the centre of mass re mains in thi s pl a ne, so the vertex B hits the gro und o nl y d 2 = 0.2 m fa r from the plane. (The d istance betwee n the places where B hits the ground in the two cases is d' = 0. 1 m.)

Solution of Problem 58. Since no hori zo nta l extern al force ac ts and the parts of the syste m are originall y at rest, the centre o f m ass o f the struc ture ca n move on ly verti ca ll y. Let 51, 52 and 5 stand for the ce ntres o f mass of the two rods a nd the whole system, respecti vely . The n based o n the Pythagorea n theore m 5 15 2 =

(t ) (t ) 2_

2 = ) 0. 25 m 2

-

0.09 1112 = VO. 16 m 2 = 0.4 m.

From the rati o of the masses of the rods (as 1nl : 1n2 = II : l2):

-- -l2 1m 5 1 5=515 ? · - - = 0.4 m·-- =0.25 m . - it + l2 l. 6 111

In the ho ri zo ntal pos itio n o f the rods, based on the fi gure:

Q

5 1' 5'2 = -l2 - -it 2 2

11

= 0 .5 111 -

O.3 m

= 0 .2 m ,

accordin g to the coord in ate of the centre of mass (w ith 51 in the o ri gin ):

- 0. 125 5 1' 5' -- 0 + l2 . 5~ 5~ -_ 0.2 m II + l2 l. 6

111 ,

the di stance betwee n the j o int a nd the centre of mass o f the syste m is

Q' 5' =

Q' S'1

S' S~

Q5~ + 5~ 5'

= 0.3111 + 0.125

= 0.425 III

Based o n the fi gure, the req uested d isplacement of the j o int is

6. x= Q'5' - 5 1 5= 0 .425 111 - 0.25111 = 0.175

156

III

111,

6. M ech a.n ics Solu tio ll S

6.2 D.' · Il <1m ics

;...---

that is, the hori zo ntal di spl ace me nt of the commo n point of the rods is 17 .5 c m in the directi on of the tumblin g, so point Q reaches the ground 17. 5 c m to the left o f th e foot of rod 11 ' Determinin g the speed: If in the hori zo ntal pos iti on o f th e rods the speed then the rod s ca n o nl y ro tate around their free end s rods and in the hori zo ntal directi o n the speed of the The angular speeds of the two rods are therefore u W j = /; ' and W2 =

of the co mm o n point (joint ) is u, beca use u is perpendi cul ar to the free endpoint s of th e rod s is zero .

v

G'

The law of co nserva ti on of mec hani cal e nergy : IJ 1n192

IJ

+ m292 =

1

?

'281 Wi

1

2

+ '2 8 2W2 '

where the rotati onal inerti as of the rods are calcul ated for the ir endpoint s. If Q stand s for the linear dcnsity o f the rods (' the Ill ass of a unit len gth ') , then

With these, our equ ati o n can bc written in thc foll ow in g form: 2 2 11 Ij 1 1 2v 1 1 2 u I j oo · -+ I?no · -=-·-lj o·1 --:-+_ . _[.){) . l.--:- . d 2 _ t:.1 2 2 3 ~ 1 If 2 3 -~ 2 I~ ,

by dividing by Q , mUltipl yin g by 2, aftcr co mbinin g the like term s the follo win g equation is acquired:

after fact orin g out

31dl l + 12 ) ' 9 = (I I + 12 ) , v 2 , and from thi s the reques ted speed of impact is v = j31 1 9 =

J

3· 0.6 m ·9.81

Ill /S

2

= 4.2

111 S

Solution of Problem 59 . The fi gure shows that the vert ica l pl ane co ntainin g the rod is perpendi cul ar to the ax is of the cy linder, so the rod will move in that pl ane. The change in grav itati onal potenti al energy ca n be de tcrmined by co mparin g the initial and fin al states. Since the sys tem is co nserv ati ve , the oppos ite of th at change equal s the total kin eti c e nergy gained . When point B reaches the cy linder, the rod and th e radiu s dra wn to it are perpendicular to each other. As seen in the fig ure, they e nclose a trian gle with sides

o B = R,

13 A = 2 R,

OA = VR2 + 4R2 = RVs. 157

300 Creative Physics Prob lem s with Solu tions

------------~------------------------------------------------------

.P:

//l

/') l ,./

./

j

:// / l

,

/ , ,,

,

I:

,, ,

, ,,

"

,'/

,

! "

I

/

rB='4R .

l i'

! ,

,

'', ,,,

,, ,,

, rs=J17R

i rA=2J.5 R

:A VA o The distances of the centre of mass S of the rod from the horizontal pl a ne in the initial and final pos iti o ns are

hI

= R.j2 2

and h

2

= R V5 5 '

s ince h2 : R = R: R V5 . E ne rgy is co nserved:

mg (h 1 - h 2 ) =

1 2 21 mv s2 + 28 sw .

The in stantaneous speed Vs of the centre of mass (S) is determined by the in stanta neous angu lar speed a nd the distance TJ] = CB of the in stantan eous axi s of rotation C of the rod from the po int S. Since the ve locity of point A is horizo nta l and that of point B -- at tha t time instant -- is tange nti a l to the cylinder, the instantaneous ax is of rotation is the intersectio n of the perpendicul ars drawn to these veloc iti es . From geometry, the in stantaneous radius TA

thu s

15 8

TA

= CA

:RV5 = 2R:R,

drawn to point A satisfies

6.2 Dy na.mics

6. Mecha.nics S olu tions

:.----

fh e instantaneous rad iu s of rotation of po int B is obta ined from the Pyth agorean theorem: Fin all y, the in stantaneous radiu s 1'S = CS at the centre of mass is

Since th e speed to be fo un d is VD = 1'DW, the angul ar speed remain s to be determined. fhe energy eq uati on in term s of the unkn ow n angular speed and the rad iu s TS ob tained above IS m gR

15)

v'2- (2 5

1 . 2 W2 +1 · -1 m4R 2 W2 = -mR 26 2 2 = -m17R W . 2 2 12 3

The soluti on o f the eq uati on for w is

Thus the speed of the point B of the rod at the time instant in qu esti on is Va

m

= 4R w = 2.19 - . s

Solution of Problem 60. Let us ass ume th at the rod is thin and homogeneous and moves in a verti cal pl ane perpendic ul ar to the edge of the table. The peg makes the rod rotate about its bottom end without translati o n until it leaves the tabl e. Durin g this time, the centre of mass of the rod ga ins a hori zo ntal ve loc ity because of the force exerted by the peg, and maintains thi s ve loci ty aft er leavin g the tabl e, sin ce during that time there is no force actin g hori zontally. The rod loses co ntac t with the peg (a nd due to its being slim al so with the edge of the tab le) when the direction of the acce lerati on of its centre of mass becomes verti cal. From that moment the hori zo ntal componen t of the veloc ity of the centre of mass remain s constant. The rod itse lf undergoes both tran slati on and rotat ion. Its ce ntre of mass moves on the path of a hori zontal projection , x While the rod itse lf ro tates abo ut its ce ntre of mass x-~ +~ cos