PROBLEM 14.1
A 30-g bullet is fired with a horizontal velocity of 450 m/s and
becomes embedded in block B which has a mass of 3 kg. After the
impact, block B slides on 30-kg carrier C until it impacts the end of
the carrier. Knowing the impact between B and C is perfectly plastic
and the coefficient of kinetic friction between B and C is 0.2,
determine (a) the velocity of the bullet and B after the first impact,
(b) the final velocity of the carrier.
SOLUTION
For convenience, label the bullet as particle A of the system of three particles A, B, and C.
(a)
Impact between A and B: Use conservation of linear momentum of A and B. Assume that the time
period is so short that any impulse due to the friction force between B and C may be neglected.
Σ mv1 + Σ Imp1 2 = Σ mv 2
Components
:
m A v0 + 0 = (m A + mB )v′
v′ =
m Av0
(30 × 10−3 kg)(450 m/s)
=
= 4.4554 m/s
m A + mB
(30 × 10−3 kg + 3 kg
v′ = 4.46 m/s
(b)

Final velocity of the carrier: Particles A, B, and C have the same velocity v′′ to the left. Use
conservation of linear momentum of all three particles. The friction forces between B and C are internal
forces. Neglect friction at the wheels of the carrier.
Σ mv 2 + Σ Imp 23 = Σ mv3
Components
:
(m A + mB )v′ + 0 = ( mA + mB + mC )v
v′′ =
=
mA v0
(m A + mB )v′
=
mA + mB + mC m A + mB + mC
(30 × 10−3 kg)(450 m/s)
= 0.4087 m/s
30 × 10−3 kg + 3 kg + 30 kg
v′′ = 0.409 m/s

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843
PROBLEM 14.12
For the system of particles of Problem 14.11, determine (a) the
components vx and vz of the velocity of particle B for which the angular
momentum HO of the system about O is parallel to the z axis, (b) the value
of HO.
PROBLEM 14.11 A system consists of three particles A, B, and C. We
know that WA = 5lb, WB = 4 lb, and WC = 3 lb and that the velocities of
the particles expressed in ft/s are, respectively, vA = 2i + 3j − 2k ,
vB = vx i + v y j + vz k , and vC = −3i − 2 j + k. Determine (a) the
components vx and vz of the velocity of particle B for which the angular
momentum HO of the system about O is parallel to the x axis, (b) the value
of HO.
SOLUTION
i
H O = Σri × mvi = Σmi xi
(vi ) x
i j k
i
5
4
= 0 5 4 + 4
g
g
2 3 −2
vx
=
HO =
(a)
(b)
j
yi
(vi ) y
k
zi
(vi ) z
j k
i
j k
3
4 3 +
8 6 0
g
2 vz
−3 −2 1
1
[5(−10 − 12) + 4(4vz − 6) + 3(6 − 0)]i
g
+
1
[5(8 − 0) + 4(3vx − 4vz ) + 3(0 − 8)]j
g
+
1
[5(0 − 10) + 4(8 − 4vx ) + 3(−16 + 18)]k
g
1
[(16vz − 116)i + (12vz − 16vz + 16) j + (−16vx − 12)k ]
g
(1)
For H O to be parallel to the z axis, we must have H x = H y = 0:
H x = 0: 16vz − 116 = 0
vz = 7.25 ft/s 
H y = 0: 12vx − 16(7.25) + 16 = 0
vx = 8.33 ft/s 
Substituting into Eq. (1):
HO =
1
[−16(8.33) − 12]k
32.2
H O = −(4.51 ft ⋅ lb ⋅ s)k 
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859
PROBLEM 14.32
In Problem 14.4, determine the energy lost as the bullet (a) passes
through block A, (b) becomes embedded in block B.
SOLUTION
The masses are m for the bullet and m A and mB for the blocks.
The bullet passes through block A and embeds in block B. Momentum is conserved.
Initial momentum:
mv0 + mA (0) + mB (0) = mv0
Final momentum:
mvB + mA vA + mB vB
Equating,
mv0 = mvB + mA vA + mB vB
m=
mA vA + mB vB (6)(5) + (4.95)(9)
=
= 0.0500 lb
1500 − 9
v0 − vB
The bullet passes through block A. Momentum is conserved.
Initial momentum:
mv0 + m A (0) = mv0
Final momentum:
mv1 + mA vA
Equating,
mv0 = mv1 + mA vA
mv0 − mA vA (0.0500)(1500) − (6)(5)
=
= 900 ft/s
0.0500
m
0.05
m=
= 1.5528 × 10−3 lb ⋅ s 2 /ft
The masses are:
32.2
6
mA =
= 0.18633 lb ⋅ s 2/ft
32.2
4.95
mB =
= 0.153727 lb ⋅ s2 /ft
32.2
(a) Bullet passes through block A. Kinetic energies:
1
1
Before:
T0 = mv02 = (1.5528 × 10−3 )(1500) 2 = 1746.9 ft ⋅ lb
2
2
1 2 1
1
1
After: T1 = mv1 + mA v 2A = (1.5528 × 10−3 )(900) 2 + (0.18633)(5)2 = 631.2 ft ⋅ lb
2
2
2
2
T0 − T1 = 1746.9 − 631.2 = 1115.7 ft ⋅ lb
energy lost = 1116 ft ⋅ lb 
Lost:
v1 =
(b)
Bullet becomes embedded in block B. Kinetic energies:
1
1
Before:
T2 = mv12 = (1.5528 × 10−3 )(900) 2 = 628.9 ft ⋅ lb
2
2
1
1
T3 = ( m + mB )vB2 = (0.15528)(9)2 = 6.29 ft ⋅ lb
After:
2
2
T2 − T3 = 628.9 − 6.29 = 622.6 ft ⋅ lb
energy lost = 623 ft ⋅ lb 
Lost:
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886
PROBLEM 14.38
Two hemispheres are held together by a cord which maintains a spring under
compression (the spring is not attached to the hemispheres). The potential
energy of the compressed spring is 120 J and the assembly has an initial
velocity v 0 of magnitude v0 = 8 m/s. Knowing that the cord is severed when
θ = 30°, causing the hemispheres to fly apart, determine the resulting
velocity of each hemisphere.
SOLUTION
Use a frame of reference moving with the mass center.
Conservation of momentum:
0 = −m Av′A + mB vB′
v′A =
mB
vB′
mA
V=
1
1
m A (v′A ) 2 + mB (vB′ ) 2
2
2
Conservation of energy:
2
m

1
1
= m A  B vB′  + mB (vB′ ) 2
2
2
 mA 
m (m + mB )
(vB′ ) 2
= B A
2m A
vB′ =
Data:
2m AV
mB ( mA + mB )
m A = 2.5 kg mB = 1.5 kg
V = 120 J
vB′ =
v′A =
(2)(2.5)(120)
= 10
(1.5)(4.0)
1.5
(10) = 6
2.5
vB′ = 10 m/s
v′A = 6 m/s
30°
30°
Velocities of A and B.
vA = [8 m/s
] + [6 m/s
vB = [8 m/s
] + [10 m/s
30°]
30°]
vA = 4.11 m/s
vB = 17.39 m/s
46.9° 
16.7° 
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893
PROBLEM 15.1
The brake drum is attached to a larger flywheel that is not shown. The
motion of the brake drum is defined by the relation θ = 36t − 1.6t 2 ,
where θ is expressed in radians and t in seconds. Determine (a) the
angular velocity at t = 2 s, (b) the number of revolutions executed by the
brake drum before coming to rest.
SOLUTION
Given:
θ = 36t − 1.6t 2
radians
Differentiate to obtain the angular velocity.
ω=
dθ
= 36 − 3.2t
dt
(a)
At t = 2 s,
ω = 36 − (3.2)(2)
(b)
When the rotor stops,
ω = 0.
0 = 36 − 3.2t
rad/s
ω = 29.6 rad/s 
t = 11.25 s
θ = (36)(11.25) − (1.6)(11.25) 2 = 202.5 radians
In revolutions,
θ=
202.5
2π
θ = 32.2 rev 
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1009
PROBLEM 15.11
In Problem 15.10, determine the velocity and acceleration
of corner B, assuming that the angular velocity is 9 rad/s
and increases at the rate of 45 rad/s 2 .
PROBLEM 15.10 The bent rod ABCDE rotates about a
line joining Points A and E with a constant angular velocity
of 9 rad/s. Knowing that the rotation is clockwise as viewed
from E, determine the velocity and acceleration of corner C.
SOLUTION
EA2 = 0.42 + 0.42 + 0.22
EA = 0.6 m
rB/A = −(0.25 m) j

EA = −(0.4 m)i + (0.4 m) j + (0.2 m)k

EA  −0.4i + 0.4 j + 0.2k  1
λ EA =
=
 = ( −2i + 2 j + k )
EA 
0.6
 3
1
ω = ω AE λ EA = (9 rad/s) (−2 i + 2 j + k )
3
ω = −(6 rad/s)i + (6 rad/s) j + (3 rad/s)k
v B = ω × rB/A = (−6i + 6 j + 3k ) × (−0.25) j = 1.5k + 0.75i
v B = (0.75 m/s)i + (1.5 m/s)k 
1
α = α AE λ EA = (45 rad/s 2 ) (−2i + 2 j + k )
3
2
α = −(30 rad/s )i + (30 rad/s 2 ) j + (15 rad/s 2 )k
a B = α × rB/A + ω × (ω × rB/A ) = α × rB/A + ω × v B
i
j
k
i
j k
30 15 + −6 6 3
a B = −30
0 −0.25 0
0.75 0 1.5
= 3.75i + 7.5k + 9i + (2.25 + 9) j − 4.5k
a B = (12.75 m/s 2 )i + (11.25 m/s 2 ) j + (3 m/s 2 )k 
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1019
PROBLEM 15.18
A series of small machine components being moved by a
conveyor belt pass over a 120 mm radius idler pulley. At the
instant shown, the velocity of Point A is 300 mm/s to the left
and its acceleration is 180 mm/s2 to the right. Determine (a) the
angular velocity and angular acceleration of the idler pulley,
(b) the total acceleration of the machine component at B.
SOLUTION
vB = v A = 300 mm/s
rB = 120 mm
(aB )t = a A = 180 mm/s
(a)
(b)
vB = ω rB ,
ω=
vB 300
=
= 2.5 rad/s
rB 120
(aB )t = α rB ,
α=
(aB )t 180
=
= 1.5 rad/s
rB
120
ω = 2.50 rad/s

α = 1.500 rad/s 2

(aB ) n = rBω 2 = (120)(2.5) 2 = 750 mm/s 2
aB = (aB )t2 + (aB ) n2 = (180) 2 + (750) 2 = 771 mm/s 2
tan β =
750
,
180
β = 76.5°
a B = 771 mm/s 2
76.5° 
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1026