PROBLEM 19.162
The block shown is depressed 1.2 in. from its equilibrium position and released.
Knowing that after 10 cycles the maximum displacement of the block is 0.5 in.,
determine (a) the damping factor c/c, (b) the value of the coefficient of viscous damping.
(Hint: See Problems 19.129 and 19.130.)
SOLUTION
From Problems 19.130 and 19.129:
()
2
2
1ln
1
c
c
c
c
n
nk c
c
x
kx
π
+
⎛⎞
⎛⎞ =
⎜⎟
⎜⎟
⎝⎠⎝⎠−
where number of cycles 10k==
c
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.162 (Continued)
(b) Critical damping coefficient. 2(Eq. 19.41)
c
k
cm
m
=
or
()
2
9 lb
32.2 ft/s
2
2 (8 lb/ft)
2.991 lb s/ft
c
c
c
ckm
c
c
=
=
=⋅
From Part (a), 0.01393
(0.01393)(2.991)
c
c
c
c
=
=
Coefficient of viscous damping. 0.0417 lb s/ftc=⋅
SO
Fro
m
Dat
Th
e
Copyrig
h
LUTION
m
Equation (
1
a:
e
cord become
ht
© McGra
w
1
9.31 and
19.
3
s slack if
m
x
−
w
-Hill Educ
a
PRO
B
An 0.8
–
consta
n
accord
i
the ma
b
ecom
e
F
Σ
3
3):
′
m
x
m
k
n
ω
m
δ
−
exceeds
st
δ
=
a
tion. Permis
s
B
LEM 19.1
6
–
lb ball is con
n
n
t
5lb/ft.k=
i
ng to the rel
a
ximum allo
w
e
slack.
(
F
ma k
δ
=
2
2
1
f
n
m
m
ω
ω
δ
=⎛⎞
−
⎜⎟
⎝⎠
0.8
32.2
0.024845
l
W
m
g
=
=
=
5lb/ft
k
=
5
0.02484
5
14.186 ra
d
n
k
m
=
=
=
st
,
δ
where
0.8 lb
5lb/ft
W
k=
=
s
ion require
d
6
3
n
ected to a pa
Knowing t
h
a
tion
m
δ
δ
=
w
able circular
)xmx
δ
−=
2
l
bs/ft⋅
8i
n
m
δ
=
5
d
/s
0.16 ft
=
d
for reprodu
ddle by mean
s
h
at the pad
d
sin ,
f
t
ω
whe
r
frequency
ω
k
x
m
⎛
+⎜
⎝
n
. 0.66667 ft=
ction or disp
s
of an elastic
d
le is move
d
r
e
8 in.,
m
δ
=
f
ω
if the cor
d
k
x
m
δ
⎞=
⎟
⎠
lay.
cord AB of
d
vertically
determine
d
is not to
PROBLEM 19.163 (Continued)
Then
()
2
0.66667 0.66667 0.16
1f
n
ω
ω
−<
−
22
ff
ωω
⎛⎞ ⎛⎞
f
f
PROBLEM 19.164
A 3-kg slender rod AB is bolted to a 5-kg uniform disk. A dashpot of
damping coefficient 9N s/mc
=
⋅ is attached to the disk as shown.
Determine (a) the differential equation of motion for small oscillations,
(b) the damping factor /.
c
cc
SOLUTION
Data:
222
disk disk
100 mm 0.100 m, 400 mm 0.400 m
11
(5 kg)(0.100 m) 0.025 kg m
22
rl
Imr
== ==
== =⋅
Equation of motion: Let the disk and rod assembly be rotated through a small counterclockwise angle
θ
disk
22
AB d AB AB
Wx Fr I I m
θθ θ
−−= ++
⎜⎟
⎝⎠
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.164 (Continued)
where sin
2
(29.43 N)(0.2 m) sin
5.886 sin N m
5.886
AB AB
l
Wx W
θ
θ
θ
θ
=−
=
=⋅
≈−
Damping force: d
Fcr
θ
=
22
(9 N s m)(0.100 m) 0.09
d
F
rcr C
θ
θθθ
==⋅⋅ = =
Inertia:
2
22
disk 0.025 0.040 (3)(0.2) 0.185 kg m
2
5.886 0.09 0.185
0.185 0.09 5.886 0
0
AB AB
l
IIm
MCK
θθ θ
θθ θ
θθθ
⎛⎞
++ = + + = ⋅
⎜⎟
⎝⎠
−−=
++ =
++=
5.886 5.6406 rad/s
0.185
n
K
M
ω
== =
2
2 (2)(0.185)(5.6406) 2.087
c
CM
ω
== =
0.09
2.087
cc
cC
cC
== 0.0431
c
c
c=
SO
Sm
a
(a)
LUTION
a
ll angles:
Newton’s
L
L
aw:
sin ,
c
6
f
12
6
f
12
18
f
12
A
C
B
y
y
y
θ
θ
δ
δ
δ
≈
⎛
=
⎜
⎝
⎛
=
⎜
⎝
⎛
=
⎜
⎝
00
(
M
MΣ=Σ
f
−
PROBLE
M
A 4-lb unif
o
and is conn
e
equation of
rod will for
m
0.9 in. dow
n
c
os 1
f
t2
f
t2
3
f
t2
θ
θ
θ
θ
θ
θ
θ
≈
⎞=
⎟
⎠
⎞=
⎟
⎠
⎞=
⎟
⎠
eff
)
M
19.165
o
rm rod is sup
e
cted to a das
h
motion for s
m
m
with the ho
r
n
and released
.
ported by a p
i
h
pot at B. Det
e
m
all oscillatio
r
izontal 5 s aft
.
i
n at O and a
e
rmine (a) the
ns, (b) the a
n
er end B has
b
spring at A,
differential
n
gle that the
b
een pushed
α
=
PROBLEM 19.165 (Continued)
Equation (2) becomes
79 0
12 4 4
774
0.07246
12 12 32.2
99
(0.15) 0.3375
44
51.25
44
k
m
m
c
k
θθθ
⎛⎞ ⎛⎞⎛ ⎞
++=
⎜⎟ ⎜⎟⎜ ⎟
⎝⎠ ⎝⎠⎝ ⎠
⎛⎞⎛ ⎞
==
⎜⎟⎜ ⎟
⎝⎠⎝ ⎠
⎛⎞
==
⎜⎟
⎝⎠
==
0.07246 0.3375 1.25 0
θθθ
+
+=
(b) Substituting t
e
λ
into the above differential equation,
00
0
(0) 0 2.329 sin 3.439 cos
3.439
tan 2.329
0.9755 rad
0.05 0.06039 rad
sin (0.9755)
θ
θφ θφ
φ
φ
θ
==− +
=
=
==
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.165 (Continued)
Substituting into Eq. (3), 2.329
0.06039 sin (3.439 0.9752)
t
et
θ
−
=+
At
5 s,t= (2.329)(5)
11.645
6
0.06039 sin[(3.439)(5) 0.9752]
0.06039 sin (18.1702)
(0.06039)(8.7627 10 )( 0.6283)
e
e
θ
−
−
−
=+
=
=×−
6
0.332 10 rad
−
=− × 6
19.05 10 degrees
θ
−
=− ×
S
O
To
U
n
Fo
r
Sp
r
N
a
Fr
e
Vi
s
Cr
i
D
a
U
n
St
a
Copyri
g
O
LUTION
tal mass:
n
balance:
r
cing frequen
c
r
ing constant:
a
tural frequen
c
e
quency ratio:
s
cous dampin
g
i
tical dampin
g
a
mping factor:
n
balance force
a
tic deflection
g
h
t
© McGra
w
c
y:
c
y:
g
coefficient:
g
coefficient:
:
:
w
-Hill Educ
a
40
0
M=
23
g
100
m
r
=
=
800
83.
7
f
ω
=
=
(4)(150 10×
38.
7
n
k
m
ω
=
=
2.1
6
f
n
ω
ω
=
65
0
c=
2
30,
c
c=
=
0.2
0
c
c
c=
16.
1
m
P
mr
ω
=
=
st
26.
9
m
P
k
δ
=
=
a
tion. Permi
s
0
kg
g
0.023 kg
mm 0.100
m
=
=
rpm
7
76 rad/s
3
N/m) 600=
×
3
600 10
400
7
30 rad/s
k
m
×
=
6
31
0
0N s/m⋅
2(60
0
984 N s/m
kM =
⋅
0
978
2
(0.023)(
0
1
424 N
f
ω
=
3
6
16.1424
600 10
9
04 10 m
−
=×
×
s
sion require
d
m
3
10 N/m
×
3
3
0
10 )(400)×
0
.100)(83.77
6
d
for reprod
u
2
6
)
u
ction or dis
p
p
lay.
a
.