ME 274 – Spring 2010 SOLUTION Final Examination PROBLEM NO. 1 – 20 pts Given: The ball shown strikes a smooth wall with a speed of v1.

The coefficient of restitution between the ball and the wall is e.

Find: Determine:

a) The magnitude (v2) and direction (θ2) of the ball’s velocity after impact.

b) The percentage of the ball’s kinetic energy lost through the impact event.

Use v1 = 10 m/s, θ1 = 30°, and e = 0.75 in your analysis.

!1

!2

v2

v1

ME 274 – Spring 2010 SOLUTION Final Examination PROBLEM NO. 2 – 20 pts Given: A T-shaped bar consisting of bar OA (m, L) welded to the end of bar BC (m,

L) is pinned to ground at O. Point A is attached to a spring with a stiffness k and unstretched length u

! . A constant HORIZONTAL force FB is applied to point B. The T-bar is initially at rest with OA being horizontal.

Find: Determine the angular velocity of the T-bar at position 2 when OA is vertical. Use the following parameters for your analysis: m = 5 kg, L = 2 m, k = 30 N/m, u

! = 0.8 m, FB = 60 N .

ME 274 – Spring 2010 SOLUTION Final Examination PROBLEM NO. 3 – 20 pts Given: A homogeneous square plate (of mass m) is pinned to ground at O. A bullet

(mass mB) strikes the stationary plate at corner C with a speed vB. The bullet becomes embedded in the plate immediately after the impact and sticking to the plate at C. All motion of the system is in a HORIZONTAL PLANE.

Find: Determine the angular velocity of the plate immediately after impact.

Use the following parameters in your analysis: L = 2, m = 10 kg, mB = 0.05 kg, and vB = 200 m/s.

1. FBD 2. Angular impulse-momentum M

O! = 0 " HO( )2= H

O( )1

HO( )

1= IO!1k + mBrB /O " vB1

= 0 + mB L / 2( )i # L j$%

&' " vB1 j( ) = mBvB1L / 2( )k

O

B

m

vB C

mB

2

L

L

2

L

HORIZONTAL PLANE

O

B

Ox

Oy

HO( )2= IO!2 k + mBrB /O " vB2

= IO!2 k + mB L / 2( )i # L j$%

&' " vB2x i + vB2y j( )

= IO!2 + mBL vB2x + vB2y / 2( )$%

&' k

where

IO = IG + mL

2

!"#

$%&2

=m

12L2+ L

2( ) + mL2

4=5mL

2

12P.A.T .( )

Therefore,

5mL2

12!2 + mBL vB2x +

vB2y

2

"#$

%&'=mBvB1L

2

3. Kinematics After impact, O and B are on the same rigid body:

vB2 = vO +!

2" rB /O = !2 k( ) " L / 2( )i # L j$

%&'

= L!2 i + L / 2( )!2 j$%

&'

Therefore,

vBx2 = L!2

vBy2 = L / 2( )!2

4. Solve Combining:

5mL2

12!2 + mB

L L!2 +L!24

"#$

%&'=mBvB1L

2

5

4L2 m

3+ m

B

"#$

%&'!2 =

mBvB1L

2(

!2 =mB

mB+ m / 3

"

#$%

&'2v

B1

5L

ME 274 – Spring 2010 SOLUTION Final Examination PROBLEM NO. 4 – 20 pts Given: Block E, having a mass of m, is attached by two springs (having stiffnesses of

k and 2k) to moving supports C and D, as shown in the figure below. Supports C and D are given prescribed motions of xC t( ) = C sin!t and xD t( ) = Dcos!t , respectively. Note that xC is measured positively to the right, whereas x

D is measured positively to the left, as indicated in the figure.

The springs are unstretched when x = xC = xD = 0 . Find: For this problem:

a) Draw the free body diagram (FBD) of block E.

b) Based on your FBD in a), derive the equation of motion (EOM) of block E in terms of the displacement x.

c) If the particular solution of the EOM in b) is written as: xP t( ) = Asin!t + Bcos!t , derive expressions for A and B.

d) If the solution found in c) is written as xP t( ) = X sin !t "#( ) , determine the numerical value for X. Use the following parameters in your analysis: m = 12kg , k = 300N / meter , ! = 15rad / sec , C = 20mm and D = 40mm .

FBD Newton

Fx! = "k x " xC( ) " 2k x + xD( ) = m!!x # m!!x + 3kx = kxC " 2kxD

k

x+

2k

C D E

m

xC t( ) = C sin!t xD t( ) = Dcos!t

k x ! xC( ) 2k x + xD( )

Particular solution

xP t( ) = Asin!t + Bcos!t

!!xP t( ) = "!2Asin!t "!

2Bcos!t

Substituting into EOM:

!m"2+ 3k( )Asin"t + !m"

2+ 3k( )Bcos"t = kCsin"t ! 2kDcos"t #

cos"t : !m"2+ 3k( )B = !2kD # B =

!2kD

!m"2+ 3k

sin"t : !m"2+ 3k( )A = kC # A =

kC

!m"2+ 3k

xP t( ) = X sin !t "#( ) $

X = A2+ B

2=

kC

"m! 2+ 3k

%&'

()*2

+"2kD

"m! 2+ 3k

%&'

()*2

=k

"m! 2+ 3k

C2+ 4D

2