978-0073398242 Chapter 19 Solution Manual Part 22

subject Type Homework Help
subject Pages 9
subject Words 1367
subject Authors Brian Self, David Mazurek, E. Johnston, Ferdinand Beer, Phillip Cornwell

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page-pf1
(b)
Rearrangi
n
Substitute
n
g Equation (
1
m
PROB
L
1
), we have
2
2
dx dx
m
cdt
dt
+
dx
dt
L
EM 19.154
P=
;
t
P
A
ec
λ
=
+
(Continue
2
dx
dt =−
d)
t
Ae
λ
λ
page-pf2
Copyri
g
g
h
t
© McGra
w
PRO
B
Draw t
h
corresp
o
w
-Hill Educ
a
B
LEM 19.1
5
h
e electrical
a
o
nding to the
f
a
tion. Permi
s
5
5
a
nalogue of t
h
f
ree bodies m
s
sion require
d
h
e mechanical
and A.)
d
for reprod
u
system sho
w
u
ction or dis
p
w
n. (Hint: Dr
a
p
lay.
a
w the loops
page-pf3
SO
LUTION
PROB
L
Draw th
e
correspo
L
EM 19.15
e
electrical a
n
nding to the
fr
6
n
alogue of th
e
fr
ee bodies m
a
e
mechanical
a
nd A.)
system show
n
n
. (Hint: Dra
w
w
the loops
page-pf4
S
O
(a)
O
LUTION
Mechanic
a
Point A:
P
R
W
r
of
t
a
l syste
m
.
R
OBLEM 1
r
ite the differ
e
t
he Point A, (
b
FΣ=
9.157
e
ntial equatio
n
b
) the charges
0:
n
s defining (
a
on the capac
i
a
) the displac
e
i
tors of the el
e
c
e
ments of the
e
ctrical analo
g
()
A
m
d
c
xx
dt
+
mass m and
g
ue.
0
A
kx+=
page-pf5
PROBLEM 19.157 (Continued)
Substituting into the results from Part (a), the analogous electrical characteristics,
d
⎛⎞
page-pf6
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.158
Write the differential equations defining (a) the displacements of the masses m1
and m2, (b) the charges on the capacitors of the electrical analogue.
SOLUTION
(a) Displacements at masses m1 and m2
2
11
1111212
2()0
dx dx
mckxkxx
dt
dt +++=
2
22
22221
2()0
dx dx
mckxx
dt
dt ++=
(b) Electrical analogues.
We let: 11
qidt= 22
qidt=
Thus, 1
1
dq
idt
= 2
2
dq
idt
=
2
11112
11
2
12
()
0
dq dq q q q
LR
dt C C
dt
+
++ =
2
2221
22
2
2
0
dq dq q q
LR
dt C
dt
+
+=
page-pf7
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.159
An automobile wheel-and-tire assembly of total weight 47 lb is attached to a mounting
plate of negligible weight which is suspended from a steel wire. The torsional spring
constant of the wire is known to be 0.40 lb in./rad.K
=
The wheel is rotated through
90° about the vertical and then released. Knowing that the period of oscillation is
observed to be 30 s, determine the centroidal mass moment of inertia and the
centroidal radius of gyration of the wheel-and-tire assembly.
SOLUTION
Torsional spring constant: 3
0.40 lb in/rad 33.333 10 lb ft/radK
=⋅=×
Let the wheel-and-tire assembly be rotated through the small angle .
θ
The moment that the wire exerts on the
assembly is
M
K
θ
=
eff :
0
M
MI KII
K
I
α
θαθ
θθ
Σ=Σ = = =
+=


2
n
K
I
ω
=
(1)
Frequency: 110.033333 H
30 s z
f
τ
== =
2 (2 )(0.033333) 0.20944 rad/s
nf
ωπ π
== =
From Eq. (1),
3
22
33.333 10 lb ft/rad
(0.20944 rad/s)
n
K
I
ω
×⋅
==
Centroidal mass moment of inertia: 2
0.75990 lb s ftI
=
⋅⋅
2
0.760 lb s ftI=⋅
Mass: 2
2
47 lb 1.4596 lb s /ft
32.2 ft/s
W
mg
== =
Centroidal radius of gyration:
2
22
2
0.75990 lb s ft 0.52061 ft
1.4596 lb s /ft
I
km
⋅⋅
== =
0.7215 ftk= 8.66 in.k=
page-pf8
S
O
P
R
T
h
re
m
c
o
O
LUTION
R
OBLEM
1
h
e period of v
m
oved, the p
e
o
nstant of the
s
1
9.160
ibration of th
e
e
riod is obse
r
s
pring.
1
1
2
W
m
τ
ω
=
=
e
system sho
w
r
ved to be 0.
5
1
1
11
3
0
2
2
A
W
g
τ
π
π
ω
ω
τ
+=
=
w
n is observe
d
5
s. Determin
e
0
.6 s
23.333
0.6
π
π
==
d
to be 0.6 s.
A
e
(a) the wei
g
rad/s
π
A
fter cylinde
r
g
ht of cylind
e
r
B has been
e
r A, (b) the
page-pf9
S
O
Sm
a
P
o
s
O
LUTION
a
ll oscillation
s
s
ition
1
s
:
PROB
Disks A
C is att
a
b
etwee
n
system.
(1
B
hr
=
BB AA
rr
θ
θ
=

1
1
(
2
C
Tm
=
LEM 19.1
6
and B weigh
a
ched to the
n
the disks,
d
cos ) 2
B
B
m
r
θ
θ
2
1
(
)2
Bm B
rI
θθ
+
6
1
30 lb and 12
l
rim of disk
B
d
etermine th
e
2
B
θ
2
1
2
B
mA
A
r
Ir
θ
+
l
b, respective
l
B
. Assuming
e
period of
s
2
m
θ
l
y, and a smal
that no slip
p
s
mall oscillat
i
l 5-lb block
p
ing occurs
i
ons of the
page-pfa
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.161 (Continued)
P
osition 2 2
2
2
0
2
C
Cm
T
Vmgh
mg
θ
=
=
=
Conservation of energy and simple harmonic motion.
11 2 2
2
222
2
2
2
22
100
22
22
()
2
5 (32.2 ft/s )
(12 30) 6
5ft
212
12.39 s
mnm
CBm
BA
Bnm
C
C
n
BA
B
C
n
n
TV T V
mgr
mmr
m
mg
mm
r
m
θωθ
θ
ωθ
ω
ω
ω
+=+
=
⎡⎤
⎛⎞
+=+
++
⎜⎟
⎢⎥
⎝⎠
⎣⎦
=+
+
=+⎛⎞
+⎜⎟
⎝⎠
=
Period of small oscillations. 22
12.39
n
n
π
π
τω
== 1.785 s
n
τ
=

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