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Problem 21.80 The moments of inertia of gears Aand
Bare IA=0.014 slug-ft2and IB=0.100 slug-ft2. Gear
Ais attached to a torsional spring with constant k=
2 ft-lb/rad. What is the frequency of angular vibrations
of the gears relative to their equilibrium position?
AB
5 lb
10 in
3 in
0.8333 ft, and RM=3in.=0.25 ft, and W=5 lb. The kinetic
energy of the system is
W
sum of the energy stored in the spring plus the energy gain due to the
increase in the height of the 5 lb weight:
RB˙
y=RMθB=−RMRA
M=6.114 rad/s,
is the equation of motion about the unstretched spring position. Note
that
˜
θ=θA−F
ω2,
from which the canonical form (see Eq. (21.4)) of the equation of
Problem 21.81 The 5-lb weight in Problem 21.80 is
raised 0.5 in. from its equilibrium position and released
from rest at t=0. Determine the counterclockwise
angular position of gear Brelative to its equilibrium
position as a function of time.
6 in
A
B
10 in
3 in
Solution: From the solution to Problem 21.80, the equation of
motion for gear Ais
As in the solution to Problem 21.80, the equilibrium angular position
θAassociated with the equilibrium position of the weight is
=−0.2778 rad,
from which the initial conditions are
θA
782
Problem 21.82 The mass of the slender bar is m. The
spring is unstretched when the bar is vertical. The light
collar Cslides on the smooth vertical bar so that the
spring remains horizontal. Determine the frequency of
small vibrations of the bar.
k
l
C
2(1−cos θ) +1
2k(L sin θ)2.
The system is conservative,
Take the time derivative:
dt2+ω2θ=0,
Problem 21.83 A homogeneous hemisphere of radius
Rand mass mrests on a level surface. If you rotate the
hemisphere slightly from its equilibrium position and
release it, what is the frequency of its vibrations?
R
Ip=ICM +m(R −h)2=83
320 mR2+5
82
mR2=13
20 mR2,
Problem 21.84 The frequency of the spring-mass
oscillator is measured and determined to be 4.00 Hz.
The oscillator is then placed in a barrel of oil, and
its frequency is determined to be 3.80 Hz. What is the
logarithmic decrement of vibrations of the mass when
the oscillator is immersed in oil?
k
10 kg
ωd=2πfd=23.88 rad/s.
Problem 21.85 Consider the oscillator immersed in oil
described in Problem 21.84. If the mass is displaced
0.1 m to the right of its equilibrium position and released
from rest, what is its position relative to the equilibrium
position as a function of time?
k
10 kg
784
Problem 21.86 The stepped disk weighs 20 lb, and its
moment of inertia is I=0.6 slug-ft2. It rolls on the
horizontal surface. If c=8 lb-s/ft, what is the frequency
of vibration of the disk?
16 lb/ft
c
16 in.
Problem 21.87 The stepped disk described in Prob-
lem 21.86 is initially in equilibrium, and at t=0itis
given a clockwise angular velocity of 1 rad/s. Deter-
mine the position of the center of the disk relative to its
equilibrium position as a function of time.
16 lb/ft
8 in
c
16 i
n
Since d2<ω
2, the system is sub-critically damped. The solution is of
the form
x=e−dt(A sin ωdt+Bcos ωdt).
Problem 21.88 The stepped disk described in Prob-
lem 21.86 is initially in equilibrium, and at t=0itis
given a clockwise angular velocity of 1 rad/s. Determine
the position of the center of the disk relative to its equi-
librium position as a function of time if c=16 lb-s/ft.
16 lb/ft
8 in
c
16 i
n
786
Problem 21.89 The 22-kg platen Prests on four roller
bearings that can be modeled as 1-kg homogeneous
cylinders with 30-mm radii. The spring constant is
k=900 N/m. The platen is subjected to a force F(t) =
100 sin 3tN. What is the magnitude of the platen’s
steady-state horizontal vibration?
kP
F(t)
From kinematics, θB=−xB
R,
The platen: The sum of the forces on the platen are
FP=−kx +4FB+F(t).
From Newton’s second law,
mP
d2xP
dt2=−kxp+4FB+F(t).
Substitute for FBand rearrange:
mp
d2xp
dt2+kxP+2mB+IB
R2d2xB
dt2=F(t).
and a(t) =F(t)
M=4.255 sin 3t(m/s2).
(ω2−32)=0.1452 m
Problem 21.90 At t=0, the platen described in
Problem 21.89 is 0.1 m to the right of its equilibrium
position and is moving to the right at 2 m/s. Determine
the platen’s position relative to its equilibrium position
kP
F(t)
M=4.255 sin 3tm/s2.
The solution is in the form x=xh+xp, where the homogenous
B=0.1, and
788
Problem 21.91 The moments of inertia of gears Aand
Bare IA=0.014 slug-ft2and IB=0.100 slug-ft2. Gear
Ais connected to a torsional spring with constant k=
2 ft-lb/rad. The bearing supporting gear Bincorporates
a damping element that exerts a resisting moment on
gear Bof magnitude 1.5(dθB/dt) ft-lb, where dθB/dt
is the angular velocity of gear Bin rad/s. What is the
frequency of angular vibration of the gears?
AB
10 in
3 in
Newton’s second law applied to the weight,
W
RA2
k
M=8.412 (rad/s)2.
The system is sub critically damped, since d2<ω
2, from which
Problem 21.92 The 5-lb weight in Problem 21.91 is
raised 0.5 in. from its equilibrium position and released
from rest at t=0. Determine the counterclockwise
transformation ˜
θB=θB−θeq , from which, by substitution,
ωd=0.2437.
Problem 21.93 The base and mass mare initially sta-
tionary. The base is then subjected to a vertical displace-
ment hsin ωitrelative to its original position. What is
the magnitude of the resulting steady-state vibration of
the mass mrelative to the base?
m
k
790
Problem 21.94* The mass of the trailer, not including
its wheels and axle, is m, and the spring constant of its
suspension is k. To analyze the suspension’s behavior,
an engineer assumes that the height of the road surface
relative to its mean height is hsin(2π/λ). Assume that
the trailer’s wheels remain on the road and its horizontal
component of velocity is v. Neglect the damping due to
the suspension’s shock absorbers.
(a) Determine the magnitude of the trailer’s vertical
steady-state vibration relative to the road surface.
(b) At what velocity vdoes resonance occur?
x
λ
m=8.787 rad/s, and
=a02πv
sin 2πv
Problem 21.95* The trailer in Problem 21.94, not
including its wheels and axle, weighs 1000 lb. The
spring constant of its suspension is k=2400 lb/ft, and
the damping coefficient due to its shock absorbers is
(b) From Example 21.7, and the solution to Problem 21.94, the
canonical form of the equation of motion is
792
Problem 21.96* A disk with moment of inertia I
rotates about a fixed shaft and is attached to a torsional
spring with constant k. The angle θmeasures the angular
position of the disk relative to its position when the
spring is unstretched. The disk is initially stationary
θp=Ap+Bpe−t,
where Apand Bpare constants that you must determine.
θ
Solution: The sum of the moments on the disk are M=−kθ +
M(t). From the equation of angular motion, Id2θ
dt2=−kθ +M(t). For
dt2+ω2θp=Bpe−t+ω2(Ap+Bpe−t)=M0
I(1−e−t).
Rearrange:
Apply the initial conditions: at t=0, θ0=0 and
A=−M0
I1
ω(1+ω2),
B=−M0
