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PROBLEM 19.80 (Continued)
Conservation of energy.
222 22 2
11 2 2 0
11 1 1 1
:00
22 3 2 2 2
AB m m AB m
l
TV T V mr ml kr mg
θθ
⎛⎞
+=+ + +=+ +
⎜⎟
⎝⎠
For simple harmonic motion,
22222 2
0
2
2
11
23 2
mnm
AB n m AB m
AB
l
mt m l kr m g
kr m gl
θωθ
ωθ θ
=
⎛⎞⎛⎞
+=+
⎜⎟⎜⎟
⎝⎠⎝⎠
+
S
LUTION
P
A
coll
con
B c
equ
dis
vib
OBLEM 1
lender 10-kg
ars of negli
tant
1.5
k=
an slide freel
librium whe
lacement an
ations.
.81
bar AB of
ible weight.
N/m
and ca
y on a vertic
bar AB is v
released,
(
1
2
Vkl
11
212
T
ength
0.
l
Collar A is
slide on a h
al rod. Kno
rtical and tha
etermine the
)
2
4
l
mg
⎛
+⎜
⎜
⎝
2
2
2
l
lm
⎛⎞
+⎜⎟
⎝⎠
m
is conn
attached to
rizontal rod,
ing that the
collar A is
period of t
⎞
⎟
⎟
⎠
2
θ
⎤
⎥
⎥
⎦
cted to two
a spring of
while collar
ystem is in
iven a small
e resulting
PROBLEM 19.82
A slender 5-kg bar AB of length l = 0.6 m is connected to two collars,
each of mass 2.5 kg. Collar A is attached to a spring of constant
k = 1.5 kN/m and can slide on a horizontal rod, while collar B can
slide freely on a vertical rod. Knowing that the system is in
equilibrium when bar AB is vertical and that collar A is given a small
displacement and released, determine the period of the resulting
vibrations.
SOLUTION
Let m be the mass of rod and C
m be mass of each collar
The
22 2 2
24 2
C
kl mgl l
Vmg
θθ
⎛⎞
=+ +
⎜⎟
⎜⎟
⎝⎠
()
22
2
11
23 2
C
l
Tm ml
θ
⎛⎞
=+
⎜⎟
⎜⎟
⎝⎠
2
mgl m gl
n
S
o
LUTION
ition
1
P
A
k
at
p
1
2
1(
2
1
)12
GAB
TI
m
=
=
OBLEM
800-g rod
12 N/m
is
C. Knowing
riod of small
2
2
1
)2
1(0.8)(
12
AB m
m
+
=
9.83
B is bolted
ttached to th
that the dis
scillations o
2
2
2
.6) 0.024
Bm
lr
θ
⎛⎞
−
⎜⎟
⎝⎠
=
o a 1.2-kg d
center of th
rolls witho
the system.
disk
2
1()
2
gm
G
I
θ
+
⋅
sk. A spring
disk at A an
t sliding, d
22
disk
1
2
mr
+
of constant
to the wall
termine the
2
m
I
m
m
(
0
1
(
0
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.83 (Continued)
2
2
2
11 2 2
222
22 2
2
2
2
1[0.750 2.354]
2
1(3.104) N m
2
11
(0.1385) 0 0 (3.104)
22
(3.104 N m)
(0.1385 kg m )
22.41 s
m
m
mnm
mn m
n
V
TV T V
θ
θωθ
θω θ
ω
−
=+
⋅
+=+
=
+=+
⋅
=⋅
=
22
22.41
n
n
π
τω
== 1.327 s
n
τ
=
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.84
Three identical 3.6-kg uniform slender bars are connected by pins as
shown and can move in a vertical plane. Knowing that bar BC is given
a small displacement and released, determine the period of vibration of
the system.
SOLUTION
22
12
0, 2
222
l
VVmgl mg
θ
⎛⎞
== +⎜⎟
⎜⎟
⎝⎠
2
mgl
=
2
22 2
21
11
0, 2
223
ml
TTml
θ
⎤
== +
⎥
⎦
22
5
6
ml
=
2
222 22
56
:,
65
nnn
ml g
mgl l
θωθ ωω
== =
()
()
22
1.586 s
9.81 m/s
6
0.75 m
5
n
π
τω
== =
SO
o
P
A
rot
ro
LUTION
ition
1
OBLEM 1
4-oz sphere
te in a verti
.
.85
and a 10-o
al plane abo
sphere C ar
t an axis at
attached to
. Determine
he ends of a
the period o
20-oz rod
small oscilla
which can
ions of the
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.85 (Continued)
osition 2
2
2
2
2
2
2
2
2
2
2
0
58 1
(1cos) (1cos) (1cos)
12 12 8
1cos 2sin 22
14 5 10 8 20 1 (lb ft)
16 12 16 12 16 8 2
[ 0.3646 0.4167 0.1563] 2
0.2084
2
mC mAC m
mm
m
m
m
m
T
VW W W
V
V
V
θθ
θθ
θ
θ
θ
θ
=
=− − + − + −
−= ≈
⎡⎤
⎛⎞⎛⎞⎛⎞⎛⎞⎛⎞⎛⎞
=− + + ⋅
⎢⎥
⎜⎟⎜⎟⎜⎟⎜⎟⎜⎟⎜⎟
⎝⎠⎝⎠⎝⎠⎝⎠⎝⎠⎝⎠
⎣⎦
=− + +
=
Conservation of energy. 22
11 2 2
1 0.2084
: (0.01775) 0 0
22
mm
TV T V
θ
+=+ +=+
Simple harmonic motion.
20.2084 11.738
0.01775
22
11.738
mnm
n
n
n
θωθ
ω
ππ
τω
=
==
==
1.834 s
n
τ
=
SO
o
LUTION
ition
1
PR
A 1
whi
Kn
sma
1
1
2
T
BLEM 1
-lb uniform
h is welded
wing that th
ll oscillations
B
WW
=
2
disk
()
Am
I
+
.86
od CD is we
o the center
disks roll w
of the syste
disk
W
disk
2
1(
Wr
g
⎛⎞
⎜⎟
⎝⎠
ded at C to
of two 20-l
thout sliding
.
2
)
m
shaft of neg
uniform dis
determine t
igible mass
s A and B.
e period of
5
5
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.86 (Continued)
Small angles:
2
2
2
2
2
2
1cos 2sin 22
1
22
1(10) (1.5)
2
1(15)
2
mm
m
m
CD
m
m
VWl
θ
θ
θ
θ
−= ≈
=
=
=
Conservation of energy and simple harmonic motion.
11 2 2
22 2
2
11
(70) 0 0 (15)
22
15
70
mnm
nm m
n
TV T V
g
g
θωθ
θθ
ω
+=+
=
+=+
=
Period of oscillations. 270
2(15)(32.2)
n
n
π
τπ
ω
== 2.39 s
n
τ
=