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PROBLEM 19.109 (Continued)
Then
2
224
112 1
fff
nnn
ωωω
ωωω
⎡⎤
⎛⎞ ⎛⎞⎛⎞
⎢⎥
−=−+=
⎜⎟ ⎜⎟⎜⎟
⎢⎥
⎝⎠ ⎝⎠⎝⎠
⎣⎦
f
ω
fn
l
f
l
S
O
(a
)
Fo
r
O
LUTION
)
r
small angle
s
s
cos 1.
θ
≈
A
c
PROBLE
The 2.75-lb
3-lb collar
x
C
sin
m
δω
=
Determine
(
b
e applied t
o
sin
x
y
F
T
F
θ
Σ=
−=
Σ=
c
celeration in
T
mx
=
=
M 19.110
bob of a sim
p
C. The col
l
,
f
t
ω
with an
(
a) the amplit
u
o
collar C to
m
cos
x
ma
mx
Tmg
θ
−
=
the y directio
n
sin
mg
mg
θ
−
p
le pendulum
l
ar is force
d
amplitude
δ
m
u
de of the m
o
m
aintain the
m
0
=
n
is second or
d
of length l =
2
d
to move a
= 0.4 in. an
d
o
tion of the b
o
m
otion.
d
er and is ne
g
2
4 in. is susp
e
ccording to
d
a frequency
f
o
b, (b) the fo
r
g
lected.
e
nded from a
the relation
f
f
= 0.5 Hz.
r
ce that must
(b)
PROB
L
c
a
x
F
Σ
sin
F
T
θ
−
L
EM 19.110
2
cm
f
x
δ
ω
==−
cc
ma=
cc
ma=
(Continue
sin
f
ft
ω
xx
−
d)
S
O
(a)
O
LUTION
Support
m
m
otion.
a
δ
PROBL
E
An 18-lb
b
moving su
p
that the ac
c
6rad
/
f
ω
=
amplitude
o
sin
m
a
δω
==
2
sin
m
f
a
ω
ω
⎛⎞
⎜⎟
=−⎜⎟
⎝⎠
E
M 19.111
b
lock A slides
p
port B
b
y m
e
c
eleration of t
h
/
s,
determine
o
f the fluctuat
i
f
t
ω
f
t
ω
in a vertical
e
ans of a spri
n
h
e support is
(a) the maxi
m
i
ng force exer
t
frictionless
s
n
g AB of con
s
sin
mf
aa
t
ω
=
m
um displac
e
t
ed by the spr
i
s
lot and is co
n
s
tant
8lb/
f
k=
,
t
where
m
a
=
e
ment of blo
c
i
ng on the bl
o
n
nected to a
f
t.
Knowing
2
5 ft/s
=
and
c
k A, (b) the
o
ck.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.112
A variable-speed motor is rigidly attached to a beam BC. When the
speed of the motor is less than 600 rpm or more than 1200 rpm, a
small object placed at A is observed to remain in contact with the
beam. For speeds between 600 and 1200 rpm the object is observed
to “dance” and actually to lose contact with the beam. Determine the
speed at which resonance will occur.
SOLUTION
Let m be the unbalanced mass and r the eccentricity of the unbalanced mass. The vertical force exerted on
the beam due to the rotating unbalanced mass is
2sin sin
f
fm f
P
mr t P t
ω
ωω
==
Then from Eq. 19.33,
()()
2
22
11
f
m
ff
nn
mr
P
kk
m
x
ω
ωω
ωω
==
−−
For simple harmonic motion, the acceleration is
()
4
2
2
1
f
f
n
mr
k
mfm
ax
ω
ω
ω
ω
=− =
−
When the object loses contact with the beam, the acceleration ||
m
a is greater than g.
Let 1600 rpm 62.832 rad/s.
ω
==
()
44
4
1
1
122
|| 1
1
n
n
mr U
mr
kk
m
aU
ω
ω
ω
ω
==
−
−
(1)
where 1.
n
U
ω
ω
=
Let 21
1200 rpm 125.664 rad/s 2
ω
ω
== =
()
44
4
2
2
(2 )
222
|| 41
1
n
n
mr U
mr
kk
m
aU
ω
ω
ω
ω
==
−
−
(2)
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.112 (Continued)
Dividing Eq. (1) by Eq. (2),
2
22
2
2
1
11
41
1or161641
16(1 )
17
20 17 20
17 20 1.08465
20 17
n
n
UUU
U
UU
ωωω ω
ω
−
=−=−
−
==
===
(1.08465)(600 rpm)
n
ω
= 651 rpm
n
ω
=
P
R
As
th
e
re
a
of
t
S
O
Us
R
OBLEM 1
the rotationa
l
e
unbalance o
f
a
ched, the am
p
t
he rotor and
i
O
LUTION
e the equatio
n
9.114
l
speed of a s
p
f
its 15-kg rot
o
p
litude of the
i
ts axis of rot
a
n
derived in P
r
p
ring-support
e
o
r first increa
vibration ap
p
a
tion. (Hint:
U
r
oblem 19.11
3
2
f
m
ω
e
d 100-kg mo
t
ses and then
d
p
roaches 3.3
m
U
se the formul
a
3
.
2
t
or is increase
d
ecreases. It i
s
m
m. Determin
a
derived in P
d, the amplit
u
s
observed th
a
e the distanc
e
roblem 19.11
3
u
de of the vib
r
a
t as very hig
h
e
between the
3
.)
r
ation due to
h
speeds are
mass center
Copyrig
h
ht
© McGra
w
P
A
2
2
m
w
r
o
w
-Hill Educ
a
P
ROBLEM
A
motor of
w
2
5 lb/in. The
m
otion is obs
e
w
eight of the r
o
o
tor and the a
x
a
tion. Permis
s
19.115
w
eight 40 lb
motor is con
s
e
rved to be 0
.
o
tor is 9 lb, d
e
x
is of the sha
ft
s
ion require
d
is supporte
d
s
trained to m
o
.
05 in. at a s
p
e
termine the
d
ft
.
d
for reprodu
d
by four s
p
o
ve vertically,
p
eed of 120
0
d
istance betw
e
ction or disp
p
rings, each
o
and the amp
l
0
rpm. Know
i
e
en the mass c
lay.
o
f constant
l
itude of its
i
ng that the
enter of the
PROBLEM 19.116
A motor weighing 400 lb is supported by springs having a total constant of 1200 lb/in. The unbalance of the
rotor is equivalent to a 1-oz weight located 8 in. from the axis of rotation. Determine the range of allowable
values of the motor speed if the amplitude of the vibration is not to exceed 0.06 in.
SOLUTION
Let mass of motor, unbalance mass, eccentricityMm r== =
2
2
400 12.4224 lb s /ft
32.2
11 0.001941 lb s /ft
16 32.2
8 in. 0.66667 ft 1200 lb/in. 14,400 lb/ft
M
m
rk
== ⋅
⎛⎞⎛ ⎞
==⋅
⎜⎟⎜ ⎟
⎝⎠⎝ ⎠
== = =
k
