TEST2DEC15answerforstudent SOLUTION

UNIVERSITI TEKNOLOGI MARA PULAU PINANG (UiTMPP) FACULTY OF MECHANICAL ENGINEERING MEC 420 – DYNAMICS SEPT 2015  – DEC 2015 TEST 2 (MARKING SCHEME)

Q1

25 /25

Q2

25/25

TOTAL

50/50

TOTAL MARKS: 50 marks

QUESTION 1 Link AB in the mechanism shown in Figure Q1 has an angular velocity of 8 rad/s (CCW) constant. At the instant shown,  AB is vertical and BC is in horizontal position. Determine the, (i) velocity and acceleration of point C , (ii) angular velocity and angular acceleration of link CD. VC

VB ωCB, αCB

VC/B ωCD, αCD αAB=0

VC=VB + VC/B Where; VB= ω AB

r  AB

=8k

VC= ωCD

r CD CD

= ωCDk

[(0.8/tan30)i + 0.8j] = 1.38ω 1.38 ωCD j –  j – 0.8  0.8ω ωCDi m/s

= ωCBk

(-0.3i) = -0.3ω -0.3ωCB j m/s

VC/B= ωCB

r C/B C/B

(0.8j ) = -6.4 i m/s

therefore; 1.38ω 1.38ωCD j –  j – 0.8  0.8ω ωCDi = -6.4 i - 0.3ω 0.3 ωCB j

UNIVERSITI TEKNOLOGI MARA PULAU PINANG (UiTMPP) FACULTY OF MECHANICAL ENGINEERING Equating i components; -0.8ωCD = -6.4 ωCD =

8 rad/s

Equating j components; 1.38(8) = - 0.3 ωCB ωCB =

or

- 36.8 rad/s

Therefore, VC= 1.38(8)j – 0.8(8)i = 11j  – 6.4i m/s

 Acceleration aC=aB + aC/B

or

Where; aC(t) = αCD

r C

aC(n) = ωCD (ωCD

= αCD k

(1.38i + 0.8j)= 1.38α CD j – 0.8αCDi m/s2

r C) = -(ωCD)2 (r C) = -(8)2

(1.38i + 0.8j) = -88.3i - 51.2j m/s 2

r B) = -(ω AB)2 (r B) = -(8)2

(0.8j) = -51.2j m/s 2

aB(t) = 0 m/s 2 aB(n) = ω AB (ω AB aC/B(t) = αCB

r C/B

= -0.30αCB j m/s2

= αCBk (-0.30i)

aC/B(n)= ωCB (ωCB r C/B )= - (-36.8)2 (-0.3i)

= 406.3i m/s2

therefore; 1.38αCD j – 0.8αCDi - 88.3i - 51.2j = -51.2j -0.30αCB j +406.3i Equating i components;  – 0.8αCD - 88.3 = 406.3i ,

αCD =

-618.25 rad/s 2

Equating j components; 1.38(-618.25) - 51.2 = -51.2 - 0. 30αCB,

αCB=

2844 rad/s 2

aC = 1.38(-618.25)j  – 0.8(-618.25)i - 88.3i - 51.2j = -904j + 406i cm/s 2

UNIVERSITI TEKNOLOGI MARA PULAU PINANG (UiTMPP) FACULTY OF MECHANICAL ENGINEERING QUESTION 2 Disk A rotates at the constant rate of ω2 = 5 rad/s with respect to arm OA as shown in Figure Q2.  At the same time, arm OA also rotates at the constant rate of ω1= 1 rad/s about point O. Determine the velocity and acceleration of point B on the disc. (All dimensions are in mm)  Y

ω1

Z X FIGURE Q2 B ω2

(25 Marks)

UNIVERSITI TEKNOLOGI MARA PULAU PINANG (UiTMPP) FACULTY OF MECHANICAL ENGINEERING Q2. RB = R A+ RB/A ; where; R A= R1 u1 +ω1^ R1 u1 = 0+ 1j^(0.4i-0.25j) = -0.4k m/s RB/A= R2 u2 +ω2^ R2 u2 = 0+( 1j+5k)^(0.15i) = -0.5k+0.75j m/s RB= -0.4k -0.5k + 0.75j = -0.9k+0.75j m/s

RB = R A+ RB/A ; where; R A= R1 u1 +2ω1^ R1 u1 +α1^ R1 u1 + ω1 ^(ω1^ R1 u1 ) Where α1= ω1 j + ω1 j = 0 + 0 =0, so R A = 0 + 0 + 0 + (1j^-0.4k) = -0.4i m/s2

RB/A= R2 u2 +2ω2^ R2 u2 + α2^R2 u2 + ω2(ω2^ R2 u2), Where α2= ω1 j + ω1 j + ω2k + ω2k = 0 + 0 + 0 + 5( 1j+5k)^k = 5i

, so

RB/A= 0 + 0 + 5i^ 0.15i + ( 1j+5k)^ (-0.5k+0.75j) = -0.5i -3.75i

= - 4.25i m/s 2

Therefore, RB= -0.4i -4.25i = -4.65i m/s 2