ME 274 – Spring 2010 SOLUTION Final Examination …ME 274 – Spring 2010 SOLUTION Final...
Transcript of ME 274 – Spring 2010 SOLUTION Final Examination …ME 274 – Spring 2010 SOLUTION Final...
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