SO
d
d
t
t
m
t
t
(
(
LUTION
PROB
L
A compo
called th
e
pendulu
m
from A t
o
and coin
c
L
EM 19.52
und pendulu
m
e
center of s
u
m
is equal to t
h
o
the mass ce
n
c
ides with the
m
is defined a
s
u
spension. S
h
h
e period of
a
n
ter G is
GA
=
center of per
c
s
a rigid slab
w
h
ow that the
a
simple pend
u
2
/kr
=
. Point
A
c
ussion define
w
hich oscilla
t
period of os
c
u
lum of lengt
h
A
is defined a
d in Problem
t
es about a fi
x
c
illation of a
h
OA, where
t
s the center o
f
17.66.
x
ed Point O,
compound
t
he distance
f
oscillation
PROBLEM 19.53
A rigid slab oscillates about a fixed Point O. Show that the smallest period of oscillation
occurs when the distance r from Point O to the mass center G is equal to .k
SOLUTION
See Solution to Problem 19.52 for derivation of
gr
n
n
g
rr
g
ω
For smallest ,
n
τ
we must have 2
k
r
r+ as a minimum:
()
2
2
k
r
dr k
+
P
R
Sh
o
osc
i
SO
Sa
m
Th
u
Le
n
h
h
h
e
a
a
Q
R
OBLEM 1
9
o
w that if the
i
llation is the
s
LUTION
m
e derivation
a
u
s,
n
gth of the eq
u
9
.54
compound p
s
ame as befor
e
a
s in Problem
θ
u
ivalent simpl
e
endulum of
P
e
and the ne
w
19.52 with
r
2
gR
R
k
θ
θ
+
=
+
e
pendulum i
s
P
roblem 19.5
2
w
center of osc
i
replaced by
0
=
s
2
is suspende
i
llation is loc
a
.R
d from A ins
t
a
ted at O.
t
ead of O, t
h
h
e period of
u
u
o
o
e
e
n
n
r
r
a
S
O
Eq
u
O
LUTION
u
ation of mot
i
si
n
i
on.
PRO
B
The 8-k
g
constan
t
release
d
value o
f
2
1
12
0.04167
0.04
0
n
t
I
ml
I
a
αθ
α
θθ
=
=
=
=
=
=
≈
B
LEM 19.5
5
g
uniform ba
r
t
500 N/
m
k=
d
, determine (
a
f
the spring co
2
1(8)(0.2
5
12
kg m
0
.04
θ
⎛⎞
⎜⎟
⎝⎠
⋅
5
r
AB is hinge
d
m
.
If end A
a
) the frequen
c
nstant k for
w
2
5
0)
d
at C and is a
t
is given a
c
y of small o
s
w
hich oscillati
o
t
tached at A t
o
small displ
a
s
cillations, (b)
o
ns will occur
o
a spring of
a
cement and
the smallest
.
S
O
O
LUTION
PROBL
E
Two unifo
r
together t
o
constraine
d
equilibriu
m
Determine
t
22
l
dl
=
+
E
M 19.56
r
m rods, eac
h
o
form an
L
d
by two sp
r
m
in a vert
i
t
he frequency
22
5
44
l
l
=
h
of mass m
a
L
-shaped asse
r
ings, each o
i
cal plane i
n
of small osci
l
a
nd length l,
mbly. The
a
f constant k
,
n
the positi
l
lations of the
are welded
a
ssembly is
,
and is in
on shown.
system.
o
m
m
M
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.57
A uniform disk of radius r and mass m can roll without slipping on a cylindrical
surface and is attached to bar ABC of length L and negligible mass. The bar is
attached at point A to a spring of constant k and can rotate freely about point B
in the vertical plane. Knowing that end A is given a small displacement and
released, determine the frequency of the resulting vibration in terms of m, L, k,
and g.
SOLUTION
()
eff:44 2
DD
kL r mrL mLr
MM mgr
θ
θ
θθ
Σ=Σ − − = +
30
42
mLr kLr
mgr
θθ
⎛⎞
+
+=
⎜⎟
⎝⎠
242
33
n
g
k
Lm
ω
=+
12 4
Hz
23 3
n
kg
fmL
π
=+
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.58
A 1300-kg sports car has a center of gravity G located a distance h above a line
connecting the front and rear axles. The car is suspended from cables that are
attached to the front and rear axles as shown. Knowing that the periods of
oscillation are 4.04 s when L = 4 m and 3.54 s when L = 3 m, determine h and the
centroidal radius of gyration.
SOLUTION
Let the mass center of the car be displaced a small distance x to the right. The mass center is moves on a
circular arc of radius L − h, so that x = (L − h) sin
θ
, where
θ
is the counterclockwise rotation of the car.
From kinematics
()
t
aLh
α
θθ
==−
The moment of the weight force about O is
0()sinMmgLh
θ
=− −
0()
t
M
ILhma
α
=+−
2
()sin ()mg L h I m L h
θ
θθ
−− =+−
Dividing by m and transposing terms yields
22
[()]()sin0kLh gLh
θθ
+
−+− =
For small angle
θ
, sin
θ
θ
≈
22
22
22
() 0
()
()
0()
nn
gL h
kLh
gL h
kLh
θθ
θωθ ω
−
+=
+−
−
+= =
+−
22
2
() ()
n
g
kLh Lh
ω
+
−= −
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.58 (Continued)
Using the two different values (L1 and L2) for L,
22
11
2
1
() ()
n
g
kLh Lh
ω
+−= −
(1)
22
22
2
2
() ()
n
g
kLh Lh
ω
+−= −
(2)
Subtracting Eq. (2) from Eq. (1) to eliminate 2,k
()
12
22
1222 22
12 12
22
12 12
()( )
2( )
nn nn
gL gL gg
Lh L h h
LL LLhABh
ωω ωω
⎛⎞
−−−= − − −
⎜⎟
⎜⎟
⎝⎠
−− − =−
where 12
22
12
nn
g
LgL
A
ω
ω
=−
and 22
12
22
12
12
2( )
nn
gg
B
LLA
hLL B
ωω
=−
−−
=
−
−
Data: 2
12
4 m, 3 m, 9.81 m/sLLg===
1
1
2
2
2
22
22
22
22
1.55524 rad/s
4.04 s
22
1.77491 rad/s
3.54 s
(9.81)(4) (9.81)(3) 6.8812 m
(1.55524) (1.77491)
9.81 9.81 0.94190 m
(1.55524) (1.77491)
(4) (3) 6.8812 0.11228 m
2(4 3) 0.94190
n
n
A
B
h
π
π
ωτ
ππ
ωτ
== =
== =
=−=
=−=
−−
==
−−
0.1123 mh=
12
3.88772 m 2.88772 mLh L h−= −=
From Eq. (1), 22
2
(9.81)(3.88772)
(3.88772) (1.55524)
k+=
22
0.65336 mk= 0.808 mk=
Checking, using Eq. (2), 22
2
22
(9.81)(2.88772)
(2.88772) (1.77491)
0.65339 m
k
k
+=
=
SO
Eq
u
o
9
9
=
LUTION
u
ation of moti
o
o
n.
G
M
Σ=
Σ
PRO
B
A 6-lb
have a
throug
h
oscilla
t
eff
():
G
M
θω
Σ
−
+
B
LEM 19.
5
slender rod
i
torsional spr
h
180° about
t
t
ion, (b) the
m
2
2
0
n
n
KI
θ
θ
θ
ω
θω
=
=
5
9
i
s suspended
ing constant
t
he vertical a
n
m
aximum velo
c
2
0
n
K
I
K
I
θ
θ
+=
=
from a steel
1.5 ft lb/
r
K
=
⋅
n
d released, d
e
c
ity of end A
o
wire which i
r
ad.
If the ro
d
e
termine (a) t
h
o
f the rod.
s known to
d
is rotated
h
e period of
S
O
Ki
n
O
LUTION
n
ematics:
P
A
n
d
t
b
P
ROBLEM
A
uniform di
s
n
egligible ma
s
d
isplaced 2°
fr
t
he maximum
b
earing at A,
(
I
=
t
=
=
α
=
19.60
s
k of radius
r
s
s which can
r
fr
om the posit
i
velocity of P
o
(
b) is riveted t
o
2
1
2mr
θ
=
250 mm=
i
s
r
otate freely i
n
i
on shown an
d
o
int A, assum
i
o
the rod at A
.
s
attached at
A
n
a vertical p
l
d
released, d
e
i
ng that the di
.
A
to a 650-m
m
l
ane about B.
e
termine the
m
sk (a) is free
t
m
rod AB of
If the rod is
m
agnitude of
t
o rotate in a
o
y
e
r
r
l
l