PROBLEM 19.74
A connecting rod is supported by a knife edge at Point A; the period of its small
oscillations is observed to be 1.03 s. Knowing that the distance ra is 6 in. determine the
centroidal radius of gyration of the connecting rod.
SOLUTION
P
osition 1 Displacement is maximum.
2
11
1
0, (1 cos ) 2
am am
TVmgr mgr
θ
θ
==− ≈
P
osition 2 Velocity is maximum.
2222 22
2
()
111 1
222 2
Gm am
Gmam m
vr
T mv I mr mk
θ
θ
θθ
=
=+= +

For simple harmonic motion, mnm
θ
ωθ
=
Conservation of energy. 11 2 2
TV T V
+
=+
()
22222
11
00
22
am a n m
mgr m r k
θωθ
+=++
Data:
2
22 2
2
22
1.03 s 6.1002 rad/s
1.03
6 in. 0.5 ft 32.2 ft/s
(32.2)(0.5) (0.5) 0.43265 0.25 0.18265 ft
(6.1002)
nn
n
a
rg
k
π
π
τω
τ
====
== =
=−=−=
0.42738 ftk= 5.13 in.k=
S
Fi
D
P
o
s
P
o
s
P
A
d
f
O
LUTION
n
d
n
ω
as a fun
c
a
tum at
2
:
s
ition
1
s
ition
2
P
ROBLE
M
A
uniform ro
d
istance c ab
o
f
or which the
c
tion of c.
1c
o
−
M
19.75
d
AB
can ro
t
o
ve the mass
c
frequency of
t
11
1
2
2
1
0
(1
o
s2sin
2
m
m
TV
Vmgc
Vmgc
θ
θ
θ
=
=
−
=
=
2
2
1
2
C
m
TI
θ
=
t
ate in a vert
i
c
enter G of th
e
t
he motion wi
2
2
cos )
22
m
mm
m
mgh
θ
θ
θ
=
−
≈
2
m
i
cal plane ab
o
e
rod. For sm
a
ll be maximu
m
o
ut a horizon
t
a
ll oscillation
s
m
.
t
al axis at C
s
, determine t
h
located at a
h
e value of c
S
O
We
P
o
s
P
s
s
i
u
O
LUTION
denote by m
t
s
ition
1
Max
i
t
he mass of h
a
i
mum deflecti
PROB
L
A homog
e
about a
oscillatio
n
small osc
i
a
lf the wire.
ons:
1
1
0,
T
V
l
=
L
EM 19.76
e
neous wire
o
frictionless
p
n
s when
β
=
i
llations is
2
τ
1
cos
(
2
l
V
mg
l
=−
o
f length 2l is
b
p
in at B. D
e
0,
determine
0
.
(
)
m
m
g
θ
β
−−
b
ent as show
n
e
noting by
τ
the angle
β
cos(
2
m
l
g
θβ
+
n
and allowed
0
τ
the perio
d
for which th
e
)
to oscillate
d
of small
e
period of
2
c
=
2
c
β
NPROBLEM 19.76 (Continued)
Conservation of energy. 11 2 2
TV T V+=+
222
11
cos cos cos
23
mm
mgl mgl ml mgl
β
βθ θ β
−+ =−
2222
11
cos 4
β
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.77
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 to a spring of constant k and can rotate freely in the vertical plane about
Point B. Knowing that end A is given a small displacement and released,
determine the frequency of the resulting oscillations in terms of m, L, k, and g.
SOLUTION
22
1
22 4
L
L
Vk mg
θ
θ
⎛⎞
=+
⎜⎟
⎝⎠
22 2 22
2
11
24 22
4
LmrL
Tm r
θ
θ
⎛⎞ ⎛⎞
=+
⎜⎟ ⎜⎟
⎜⎟ ⎜⎟
⎝⎠ ⎝⎠

22
3
16
mL
θ
=
2
2
284
3
16
mgL
kL
nmL
ω
+
=
22
3
kg
mL
⎛⎞
=+
⎜⎟
⎝⎠
12 4
23 3
n
kg
fmL
π
=+
S
M
A
O
LUTION
a
ss and mome
n
p
proximation.
n
t of inertia o
f
PROB
L
Two uni
welded t
o
each spr
i
and rele
a
f
one rod.
m
1
12
I
m
l
=
L
EM 19.78
form rods, e
a
o
gether to for
m
i
ng is k = 0.6
a
sed, determi
n
1.2
32.
2
W
m
g
==
2
1(0.037
12
l
=
a
ch of weigh
t
m
the assemb
lb/in. and th
a
n
e the frequen
c
0.037267 l
b
2
=
2
8
267) 12
⎛⎞
=
⎜⎟
⎝⎠
t
W = 1.2 lb
ly shown. Kn
o
a
t end A is g
i
c
y of the resu
l
2
b
s/ft⋅
3
1.38026 10
−
×
and length l
o
wing that th
e
i
ven a small
d
l
ting motion.
3
2
lb s ft⋅⋅
= 8 in., are
e
constant of
d
isplacement
r
2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.78 (Continued)
P
osition 2 2
2
2
2
2
2
0
1
(1 cos ) 2
222
(1.2)(0.66667) 1 1 0.66667
(2) (7.2)
22 2 2
0.6
mm
mm
m
T
Wl l
Vk
θθ
θθ
θ
=
⎛⎞⎛ ⎞
=− − + ⎜⎟⎜ ⎟
⎝⎠⎝ ⎠
⎛⎞⎛⎞⎛ ⎞
≈− +
⎜⎟⎜⎟⎜ ⎟
⎝⎠⎝⎠⎝ ⎠
≈
Conservation of energy. 11 2 2
TV T V
+
=+
32 2
3.4506 10 0 0 0.6
13.186
mm
mm
θ
θ
θ
θ
−
×+=+
=
Simple harmonic motion. mnm
θ
ωθ
=
13.186 rad/s
n
ω
=
Frequency. 2
n
n
f
ω
π
= 2.10 Hz
n
f=
S
(a
)
O
LUTION
)
P
osition
1
P
osition
2
1
2
P
A
is
m
p
e
c
y
1
0T
V
=
ROBLEM
1
15-lb unifor
m
attached to
a
m
oved 0.4 in.
e
riod of vibra
y
linder.
1st
1(
2
V
k
r
δ
=+
1
9.79
m
cylinder ca
n
a
spring AB a
s
down the i
n
tion, (b) the
m
2
)
m
r
θ
n
roll withou
t
s
shown. If th
e
n
cline and re
l
m
aximum ve
l
t
sliding on a
n
e
center of th
l
eased, deter
m
l
ocity of the
c
n
incline and
e cylinder is
m
ine (a) the
c
enter of the
PROBLEM 19.79 (Continued)
Substituting Eq. (2) into Eq. (1)
22 2 2
mmm
mnmm m nm
kr I mv
vr r
θθ
θ
ωθ θ ωθ
=+
===

12 0.715 s
m
⎜⎟
⎝⎠
⎝⎠
m
S
O
P
o
s
(
(
O
LUTION
s
ition
1
PR
O
A 3-
k
const
posit
i
relea
s
O
BLEM 19.
k
g slender ro
d
ant 280 N/m
i
on shown. If
s
ed, determin
e
0.08 m
0.3 m
r
l
=
=
1disk
1
2
TI
θ
=
80
d
AB is bolte
is attached
end B of the
e
the period o
f
2
rod
1()
2
mA
I
θθ
+
d to a 5-kg
u
to the disk
a
rod is given
f
vibration of
t
2
m
θ
u
niform disk.
a
nd is unstre
t
a small displ
a
t
he system.
A spring of
t
ched in the
a
cement and