22–21.
If the wire AB is subjected to a tension of 20 lb, determine
the equation which describes the motion when the 5-lb
weight is displaced 2 in. horizontally and released from rest.
SOLUTION
d
+
ΣFx=m ax
; –2T
x
L
=mx
$
x
$
+
2T
Lm
x=0
P=
A
2T
Lm
=
A2(20)
6
(
5
32.2
)
=6.55 rad
>
s
x=A sin pt +B cos pt
x=
1
6
ft at
t=0
, Thus B=
1
6
=0.167
v=A p cos pt –B p sin pt
v=0
at
t=0
, Thus
A=0
So that
x=0.167 cos 6.55t
Ans.
6 ft
A
1211
22–22.
The bar has a length land mass m.It is supported at its ends
by rollers of negligible mass.If it is given a small displacement
and released, determine the natural frequency of vibration.
SOLUTION
c
2
u
$
+3g(4R2–l2)1
2
6R2–l2u=0
+©MO=IOa;mgaBR2–l2
4bu=-maR2–1
6l2bu
$
AB
R
l
C
6R
–l
22–23.
The 20-kg disk, is pinned at its mass center O and supports
the 4-kg block A. If the belt which passes over the disk is
not allowed to slip at its contacting surface, determine the
natural period of vibration of the system.
SOLUTION
k = 50 N/m
300 mm
k 200 N/
m
O
22–25.
If the disk in Prob. 22–24 has a mass of 10 kg, determine the
natural frequency of vibration. Hint: Assume that the initial
stretch in each spring is
d
O.
SOLUTION
2p
2p
k 80 N/m
k 80 N/m
O
150 mm
22–26.
A flywheel of mass m, which has a radius of gyration about
its center of mass of , is suspended from a circular shaft
that has a torsional resistance of . If the flywheel
is given a small angular displacement of and released,
determine the natural period of oscillation.
u
M=Cu
kO
a
Comparing this equation to the standard equation, the natural circular frequency of
the wheel is
Thus, the natural period of the oscillation is
Ans.t=2p
vn
=2pkOAm
C
vn=AC
mkO2=1
kOAC
m
u
$
+C
mkO2u=0
+©MO=IOa;–Cu=mkO2u
$
L
O
22–27.
SOLUTION
Thus, for small ,
c
Thus,
u
$
+11.926u=0
+©MO=IOa; 6(3) –
C
9+5(2)u
D
(2) =a6
32.2 ba3u
$b(3)
u
The 6-lb weight is attached to the rods of negligible mass.
Determine the natural frequency of vibration of the weight
when it is displaced slightly from the equilibrium position
and released.
2ft
k=5lb/ft
O
*22–28.
AB
2.50 m
1.83 m
O
G2
G1
The platform AB when empty has a mass of 400 kg, center
of mass at , and natural period of oscillation .
If a car,having a mass of 1.2 Mg and center of mass at ,is
placed on the platform, the natural period of oscillation
becomes . Determine the moment of inertia of
the car about an axis passing through .G2
t2=3.16 s
G2
t1=2.38 sG1
1218
22–29.
The plate of mass m is supported by three symmetrically
placed cords of length l as shown. If the plate is given a
slight rotation about a vertical axis through its center and
released, determine the natural period of oscillation.
Ans:
t=2p
A
l
2g
SOLUTION
ΣMz=Iz
sin
f
K
f
u
$
+
6T
Rm
f=0
ΣFz=0
3T cos
f
–mg =0
f
=0
, T=
mg
3
, f=
R
l
u
u
$
+6
Rm
amg
3ba
R
l
u
b
=0
u
$
+
2g
l
u=0
t=2p
v
n
=2p
A
l
2g Ans.
l
l
22–30.
SOLUTION
Ans.x
$
+
333x
=
0
3x
$+1000x=0
1.5(2x
#)x
$+1000xx
#=0
T+V=1.5x
#2+500x2
V=1
2(500)x2+1
2(500)x2
T=1
2(3)x
#2
T+V=const.
k500 N/mk500 N/m
3kg
Determine the differential equation of motion of the 3-kg block
when it is displaced slightly and released. The surface is smooth
and the springs are originally unstretched.
1220
22–31.
For small ,,then
Ans.t=2p
vn
=2p
17.047
=1.52 s
u
$
+17.047u=0
sin u=uu
1.4079 (2u
#
)u
$
+48(sin u)u
#
=0
T+V=1.4079u
#2+48(1 –cos u)
V=8(4)(1.5)(1 –cos u)=48(1 –cos u)
T=1
2(2.8157)(u
#
)2=1.4079 u
#2
T+V=const
h=y(1 –cos u)
+1
32.2 c1
12(2)(8)(2)2+2(8)(2)2d=2.8157 slug #ft2
IO=1
32.2 c1
12(2)(8)(2)2+2(8)(1)2d
Determine the natural period of vibration of the pendulum.
Consider the two rods to be slender, each having a weight
of 8 .lb>ft
O
2ft
Ans:
t=1.52 s
*22–32.
Determine
the natural period of vibration of the 10-lb
semicircular disk.
SOLUTION
Datum at initial level of center of gravity of disk.
F
or small
Ans.t=0.970 s
t=2p
vn
=2pAIIC
Wr =2pA0.05056
10(0.212)
=0.05056 slug #ft2
=0.02483 +10
32.2(0.5 –0.212)2
IIC =IG+m(r–r)2
IG=0.02483 slug #ft2
IA=IG+mr2
r=4(0.5)
3p
=0.212 ft
u
$
+Wr
IIC
u=0
u, sin u=u
E
#
=u
#
(IICu
$
+Wr sin u)=0
=1
2IIC(u
#
)2+Wr(1 –cos u)
E=T+V
¢=r
(1 –cos u)
0.5ft
22–33.
If the 20-kg wheel is displaced a small amount and released,
determine the natural period of vibration. The radius of
gyration of the wheel is
kG=0.36 m
. The wheel rolls
without slipping.
k
500 N
/
m
G
0.5 m
SOLUTION
1223
22–34.
Determine the differential equation of motion of the 3-kg
spool.
Assume that it does not slip at the surface
of
contact as it oscillates.The radius of gyration of the spool
about its center of mass is
kG=125 mm.
SOLUTION
Kinematics:
Since no slipping occurs, hence . Also,
Ans.u
$
+468u=0
0.076875u
#
(u
$
+468.29u)=0 Since 0.076875uZ0
0.076875u
#
u
$
+36uu
#
=0
=0.03844u
#2+18u2
E=1
2[(3)(0.125)2]u
#2+1
2(3)(0.1u)2+1
2(400)(0.3u)2=const.
E=T+V
vG=0.1u
#
.
sF=0.3
0.1SG=0.3usG=0.1u
k400N/m
G
200mm
100 mm
Ans:
u
$
+468u=0
1224
22–35.
Determine the natural period of vibration of the 3-kg
sphere. Neglect the mass of the rod and the size of
the sphere.
SOLUTION
By statics,
Thus,
Ans.t=2p
vn
=2p
12.91 =0.487 s
vn=2166.67 =12.91 rad>s
u
$
+166.67u=0
3(0.3)2u
$
+500(0.3)2u=0
dst =3(9.81)
500
T=3(9.81) N
T(0.3) =3(9.81)(0.3)
E=u
#
[(3(0.3)2u
$
+500(dst +0.3u)(0.3) –3(9.81)(0.3)] =0
=1
2(3)(0.3u
#
)2+1
2(500)(dst +0.3u)2–3(9.81)(0.3u)
E=T+V
300 mm300 mm
k500 N/m
O
Ans:
t=0.487 s
1225
Ans:
f=1.17 Hz
*22–36.
If the lower end of the 6-kg slender rod is displaced a small
amount and released from rest, determine the natural
frequency of vibration. Each spring has a stiffness of
k=200 N>m
and is unstretched when the rod is hanging
vertically.
SOLUTION
2p
2p
O
k
k
2 m
2 m
22–37.
The disk has a weight of 30 lb and rolls without slipping on
the horizontal surface as it oscillates about its equilibrium
position. If the disk is displaced, by rolling it counterclockwise
0.2 rad, determine the equation which describes its
oscillatory motion and the natural period when it is released.
SOLUTION
0.5 ft
k 80 lb/ft
22–38.
The machine has a mass mand is uniformly supported by
four springs,each having a stiffness k.Determine the natural
period of vertical vibration.
SOLUTION
G
1228
22–39.
SOLUTION
Ans.f=vn
2p
=8.025
2p
=1.28 Hz
vn=8.025 rad>s
vn2=64.40
0.0699vn2u2
max =4.5 u2
max
T
max =V
max
V
max =W¢G=4(1.5)(0.75u2
max)
=0.0699 vn2u2
max
=1
2[1
3(4(1.5)
32.2 )(1.5)2]vn2u2
max
Tmax =1
2IAv2
max
¢G=1
2¢=0.75u2
max
=0.75(4u2
max
2)
0.75(1 –1+f2
2)
¢=0.75(1 –cos f)
f=1.5umax
0.75
The slender rod has a weight of . If it is supported in
the horizontal plane by a ball-and-socket joint at Aand a
cable at B, determine the natural frequency of vibration
when the end Bis given a small horizontal displacement
and then released.
4lb>ft
0.75 ft
1.5 ft
A
B
Ans:
f=1.28 Hz
*22–40.
If the slender rod has a weight of 5 lb, determine the natural
frequency of vibration. The springs are originally
unstretched.
SOLUTION
the compression of the spring at its ends can be approximated as and
.Thus, the elastic potential energy when the rod is at this position is
.The datum is set at the
rod’s mass center when the rod is at its original position. When the rod undergoes a
small angular displacement ,its mass center is abovethe datum
hence its gravitational potential energy is .Since is small,
can be approximated by the first two terms of the power series,that is,
.Thus,
The mass moment inertia of the rod about point Ois
.The kinetic energy is
The total energy of the system is
[1]
Time Derivative:Taking the time derivative of Eq.[1], we have
Since , then
[2]
From Eq.[2], , thus,.Applying Eq.22–14, we have
Ans.f=p
2p
=13.06
2p
=2.08 Hz
p=13.06 rad/sp2=170.66
u
$
+170.66u=0
0.1553u
$
+26.5u=0
u
#
Z0
u
#
A
0.1553 u
$
+26.5u
B
=0
0.1553 u
#
u
$
+26.5uu
#
=0
U=T+V=0.07764u
#2+13.25u2
T=1
2IOv2=1
2(0.1553) u
#2=0.07764u
#2
+5
32.2
A
0.52
B
=0.1553 slug #ft2
IO=1
12 a5
32.2 b
A
32
B
V=V
e+V
g=12u2+1.25u2=13.25u2
V
g=2.5c1–a1–u2
2bd=1.25u2
cos u=1–u2
2
cos u
uV
g=5[0.5(1 –cos u)]
0.5(1 –cos u)ftu
V
e=1
1+1
x2=1u
x1=2u
k=4lb/ft
O
1ft