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PROBLEM 17.112 (Continued)
moments about B:
22 211
L
122211
4244412
2
5
24 mL
0
12
5
v
L
0
2.4 v
L
2212 2
vv v v
PROBLEM 17.113
The slender rod AB of length L 1 m forms an angle β 30° with the vertical as
it strikes the frictionless surface shown with a vertical velocity
1
2v
m/s and no
angular velocity. Knowing that the coefficient of restitution between the rod and
the ground is e 0.8, determine the angular velocity of the rod immediately after
the impact.
PROBLEM 17.113 (Continued)
PROBLEM 17.114
The trapeze/lanyard air drop (t/LAD) launch is a proposed innovative method for airborne launch of a
payload-carrying rocket. The release sequence involves several steps as shown in (1) where the payload rocket is
shown at various instances during the launch. To investigate the first step of this process where the rocket
body drops freely from the carrier aircraft until the 2-m lanyard stops the vertical motion of B, a trial rocket is
tested as shown in (2). The rocket can be considered a uniform 1-m by 7-m rectangle with a mass of 4000 kg.
Knowing that the rocket is released from rest and falls vertically 2 m before the lanyard becomes taut,
determine the angular velocity of the rocket immediately after the lanyard is taut.
PROBLEM 17.114 (Continued)
For impact use the principle of impulse and momentum.
1
Syst. Momenta
12
Syst. Ext. Imp.
2
Syst. Momenta
1
y
Kinematics.
BB
vv
y
Substitute from Eqs. (1) and (3) into Eq. (2).
2
1
(3.5) [ (3.5) ]mv I m
22
PROBLEM 17.115
The uniform rectangular block shown is moving along a
frictionless surface with a velocity
1
v
when it strikes a
small obstruction at B. Assuming that the impact between
corner A and obstruction B is perfectly plastic, determine
the magnitude of the velocity
1
v
for which the maximum
angle
through which the block will rotate is 30.
PROBLEM 17.115 (Continued)
Angular velocity after impact 1
222
3
2( )
vb
ab
ω
(1)
The motion after impact is a rotation about corner B.
33
sin( 30 ) 0.11180sin56.565 0.093301m
00
a
hd
v
Potential energy:
23
2
mgb
VVmgh
222
23
222
1() 0
6
ma b T
Principle of conservation of energy:
2333
TVTV
222
1() 0
mgb
222 2 2
( ) (0.200) (0.100)
ab
2
50.974 (rad/s)
2
7.1396 rad/s
Magnitude of initial velocity.
2
12( )
3
ab
1
(3)(0.100)
v
1
2.38 m/sv
PROBLEM 17.116
The 40-kg gymnast drops from her maximum height of h = 0.5 m
straight down to the bar as shown. Her hands hit the bar and clasp onto
it, and her body remains straight in the position shown. Her center of
mass is 0.75 meters away from her hands, and her mass moment of
inertia about her center of mass is 7.5 kg·m2. Assuming that friction
between the bar and her hands is negligible and that she remains in the
same position throughout the swing, determine her angular velocity
when she swings around to θ = 135°.
42
7.5 40 0.75
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 17.117
A slender rod of mass m and length L is released from rest in the position
shown and hits edge D. Assuming perfectly plastic impact at D, determine
for
0.6 ,bL
(a) the angular velocity of the rod immediately after the
impact, (b) the maximum angle through which the rod will rotate after the
impact.
PROBLEM 17.117 (Continued)
(a) Angular velocity. 32
553
14 14 2
g
L
3
0.437
g
L
sin
10
L
h
4
444
sin
10
mgL
Vmgh
mgL mgL
