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SO
Ki
n
LUTION
n
ematics:
PR
O
Tw
o
sho
w
dete
r
O
BLEM 1
9
o
uniform rod
s
w
n. Knowing
r
mine the per
i
2
2
2
A
AC
AC
rr
θθ
θ
θ
θ
θ
=
=
=
9
.87
s
AB and CD,
that the mas
s
i
od of small o
s
C
C
C
θ
each of lengt
h
s
of gear C is
s
cillations of
t
h
l and mass
m
m and that t
h
t
he system.
m
, are attache
d
h
e mass of g
e
d
to gears as
e
ar A is 4m,
PROBLEM 19.87 (Continued)
P
osition 2 1
1
0
(1 cos ) (1 cos 2 )
22
mm
T
lmgl
Vmg
θ
θ
=
=−+−
2
2
mm
θθ
22
12 2
rl
n
n
gl
ω
Copyrig
h
ht
© McGra
w
P
R
T
w
as
s
de
t
1
2
A
1
1
1(
4
2
1(
2
1
12
1
2
1
2
A
C
A
B
I
Im
Im
T
m
T
m
=
=
=
=
=
w
-Hill Educ
a
R
OBLEM 1
w
o uniform ro
d
s
hown. Kno
w
t
ermine the p
e
2
A
mC
2
2
2
2
2
22
4
)(2 ) 8
1
)( ) 2
8
4
2
5
10 3
CD
mr
m
m
rm
r
m
lI
r
m
r
m
rl
=
=
=
⎡⎛⎞
+
⎢⎜⎟
⎜⎟
⎢⎝⎠
⎣
⎡
⎤
+
⎢⎥
⎣
⎦
a
tion. Permis
s
9.88
d
s AB and C
D
w
ing that the
m
e
riod of small
mAB
m
2
2
2
22
2
1
1
12
4
12 3
0
m
m
r
r
ml
ll
l
V
θ
=
+++
⎤
=
⎦
s
ion require
d
D
, each of len
g
m
ass of gear C
oscillations o
f
2
m
CD m
2
22
4
0
m
l
l
θ
⎤
+⎥
⎥
⎦
d
for reprodu
g
th l and mas
s
is m and that
f
the system.
2
AB
⎝
ction or disp
s
m, are attac
h
the mass of g
e
2
2
mC
D
⎟
⎠
lay.
h
ed to gears
e
ar A is 4m,
2
D
m
⎜⎟
⎝⎠
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.88 (Continued)
P
osition 2 2
2
0
(1 cos ) (1 cos 2 )
22
mm
T
lmgl
Vmg
θ
θ
=
=− − + −
For small angles,
2
2
22
2
2
2
2
11 2 2
1cos 2sin 22
1cos2 2sin 2
2
22 2
13
22
mm
m
mmm
m
m
m
mnm
lmgl
Vmg
mgl
TV T V
θθ
θ
θθθ
θ
θ
θ
θ
ωθ
−= ≈
−= ≈
=− +
=
+=+ =
2222 2
3
22
22
5
3
22
15 13
10 0 0
23 22
10
9
60 10
mn m
n
mr l mgl
gl
rl
gl
rl
θ
ωθ
ω
⎡⎤
++=+
⎢⎥
⎣⎦
=+
=+
22
26010
29
n
n
rl
gl
π
τπ
ω
+
==
PROBLEM 19.89
An inverted pendulum consisting of a rigid bar ABC of length l and mass m is
supported by a pin and bracket at C. A spring of constant k is attached to the bar
at B and is undeformed when the bar is in the vertical position shown. Determine
(a) the frequency of small oscillations, (b) the smallest value of a for which these
oscillations will occur.
SOLUTION
Moment of inertia: 2
1
12
I
ml=
P
osition c Maximum deflection. Let rod AC rotate through angle .
m
θ
The spring stretches an amount
sin
mm
xa
θ
=
and the center of gravity moves down an amount
2
1
2
22 2
22
1
(1 cos )
2
1
2
1(sin ) (1 cos )
11
222
11
22
0
mm
mm
mm
mm
m
l
y
Vkxmgy
l
ka mg
l
ka mg
ka mgl
T
θ
θ
θ
θθ
θ
−= −
=+
=−−
⎛⎞⎛ ⎞
≈−
⎜⎟⎜ ⎟
⎝⎠⎝ ⎠
⎛⎞
=−
⎜⎟
⎝⎠
=
P
osition d Maximum velocity:
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.89 (Continued)
Kinetic energy: 22
2
11
22
Tmv I
θ
=+
2
22
22
222
2
11
2212
11
23
11
23
0
nm
l
mml
ml
ml
V
θ
θ
θ
ωθ
⎡
⎤
⎛⎞
⎢
⎥
=+
⎜⎟
⎢
⎥
⎝⎠
⎣
⎦
⎡⎤
=⎢⎥
⎣⎦
⎛⎞
=⎜⎟
⎝⎠
=
Conservation of energy: 11 2 2
22222
2
2
2
11 11
022 23
63
2
mnm
n
TV T V
ka mgl ml
ka mgl
ml
θ
ωθ
ω
+=+
⎛⎞⎛⎞
+− =
⎜⎟⎜⎟
⎝⎠⎝⎠
−
=
(a) Frequency: 2
n
f
ω
π
=
22
1(6 3 )/2
2
f
ka mgl ml
π
=−
(b) Smallest value of a for oscillations. f is real for 2
63ka mgl>
2
mgl
ak
> min 2
mgl
ak
=
SO
P
o
s
Co
n
But
LUTION
s
ition 1:
n
servation of
e
for simple h
a
1
0,T=
2
2
e
nergy
11
TV+
a
rmonic motio
1
V
=
dis
k
AB
22
:TV=+
n,
m
v
PROBL
E
Two 12-lb
shown. K
n
that the di
s
vibration o
2
1
2
m
kx
=
k
m
2
1
02
m
kx
+
=
:
nm
x
ω
=
E
M 19.90
uniform dis
k
n
owing that th
e
s
ks roll with
o
f the system.
d
1(3
2
AB
mm+
k
s are attache
d
e
constant of
t
o
ut sliding, de
t
2
d
isk
)
m
v
d
to the 20-l
b
t
he spring is
3
t
ermine the f
r
b
rod AB as
3
0 lb/in. and
r
equency of
S
O
M
a
O
LUTION
a
sses and mo
m
P
osi
t
6
r=
m
ents of inerti
a
PROB
L
The 20-l
b
disks roll
the syste
m
t
ion
2
6
in.
a
.
m
L
EM 19.91
b
rod
AB
is at
t
without slidi
m
.
8
32
1
2
AB
AB
m
m
I
I
m
==
==
t
ached to two
ng, determin
e
2
8
0.24845
l
.2
1(0.2
4
2
AA
m
r
=
=
8-lb disks as
e
the frequen
c
P
ositi
o
2
2
l
bs/ft
6
4
845) 12
⋅
⎛⎞
⎜⎟
⎝⎠
shown. Kno
w
c
y of small o
s
o
n
1
w
ing that the
s
cillations of
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.91 (Continued)
Conservation of energy. 11 2 2
2
2
2
2
80 80
0 cos 0.101795
12 12
65.491(1 cos )
1
65.491 2
32.745
5.7224
mm
mm
m
m
mm
TV T V
θθ
θ
θ
θ
θ
θθ
+=+
−= −
=−
⎛⎞
≈⎜⎟
⎝⎠
=
=
Simple harmonic motion. mnm
θ
ωθ
=
5.7224 rad/s
n
ω
=
Frequency. 5.7224
22
n
n
f
ω
π
π
== 0.911 Hz
n
f=
Copyri
g
g
h
t
© McGra
w
2
n
⎝
w
-Hill Educ
a
PROBL
E
A half se
c
casters A
a
m/8. Kno
w
released a
n
oscillation
s
2
15
r
ππ
⎟
⎠
a
tion. Permi
s
E
M 19.92
c
tion of a uni
f
a
nd B, each o
f
w
ing that the
n
d that no
s
s
.
s
sion require
d
f
orm cylinde
r
f
which is a
u
half cylinde
r
s
lipping occu
r
d
for reprod
u
r
of radius r
a
u
niform cylin
d
r
is rotated t
h
r
s, determine
u
ction or dis
p
a
nd mass m
r
d
er of radius
r
h
rough a sma
l
the frequen
c
n
p
lay.
r
ests on two
r
/4 and mass
l
l angle and
c
y of small
r
