978-0073398242 Chapter 19 Solution Manual Part 7

subject Type Homework Help
subject Pages 9
subject Words 1340
subject Authors Brian Self, David Mazurek, E. Johnston, Ferdinand Beer, Phillip Cornwell

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page-pf1
PROBLEM 19.46
A three-bladed wind turbine used for research is supported on a shaft so that it is
free to rotate about O. One technique to determine the centroidal mass moment of
inertia of an object is to place a known weight at a known distance from the axis
of rotation and to measure the frequency of oscillations after releasing it from rest
with a small initial angle. In this case, a weight of Wadd = 50 lb is attached to one
of the blades at a distance R = 20 ft from the axis of rotation. Knowing that when
the blade with the added weight is displaced slightly from the vertical axis, the
system is found to have a period of 7.6 s, determine the centroidal mass moment
of inertia of the 3-bladed rotor.
SOLUTION
Let the turbine rotor be turned counterclockwise through a small angle
θ
. The moment of the added weight
about Point O is
22
add
2
add
0
nn
WR
I
mR
θωθ ω
+= =
+

page-pf2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.46 (Continued)
Solving for ,I
2
add
add
2
n
WR
mR
ω
=− (1)
Data: add
2
add
add 2
20 ft, 50 lb
50 lb 1.5528 lb s /ft
32.2 ft/s
RW
W
mg
==
== =
Period and frequency: 7.6 s
11
Hz
7.6
2
2 0.82673 rad/s
7.6
n
f
f
τ
τ
π
ωπ
=
==
===
From Eq. (1), 22
2
(50lb)(20ft) (1.5528 lb s /ft)(20 ft)
(0.82673 rad/s)
1463.10 621.12
I=−
=−
2
842 lb s ftI=⋅
page-pf3
Copyrig
h
ht
© McGra
w
PROBL
E
A connect
i
oscillation
s
edge at Po
i
1
0
ab
rr+=
gyration
.
k
w
-Hill Educ
a
E
M 19.47
i
ng rod is s
u
s
is observed
i
nt B and the
p
0
in.
determin
e
.
B
a
tion. Permis
s
u
pported by
to be 0.87 s.
p
eriod of sma
l
e
(a) the locat
i
B
bb
s
ion require
d
a knife-edge
The rod is th
l
l oscillations
i
on of the ma
s
d
for reprodu
at Point A;
en inverted a
n
is observed t
o
s
s center G, (
b
ction or disp
the period
o
n
d supported
o
be 0.78 s. K
n
b
) the centroi
d
lay.
o
f its small
by a knife-
n
owing that
d
al radius of
page-pf4
PROBLEM 19.47 (Continued)
page-pf5
S
O
O
LUTION
P
A
fr
i
o
s
p
e
ROBLEM
1
semicircular
i
ctionless pin
s
cillations of t
h
e
riod.
1
9.48
hole is cut
at its geom
e
h
e plate, (b) t
h
b
=
in a uniform
e
tric center
O
h
e length of
a
2
2
2
2
16
rr
r
π
π
⎛⎞
⎜⎟
⎜⎟
⎝⎠
=
square plate
O
. Determine
a
simple pend
u
2
4
30.0
2
2
r
r
π
π
=
which is at
t
(a) the peri
o
u
lum which h
2
1440 m
t
ached to a
o
d of small
as the same
page-pf6
P
A
n
t
h
(
b
P
ROBLEM
A
uniform dis
k
egligible mas
h
e period of s
b
) if the rod is
19.49
k
of radius
r
s, which can
mall oscillati
o
riveted to th
e
250=
mm is
rotate freely
o
ns (a) if the
e
disk at A.
attached at
A
in a vertical
p
disk is free t
o
A
to a 650-m
m
p
lane about
B
o
rotate in a
b
m
rod AB of
B
. Determine
b
earing at A,
page-pf7
Copyrig
h
ht
© McGra
w
w
-Hill Educ
a
PROB
a
tion. Permis
s
LEM 19.49
3.
7
s
ion require
d
(Continue
d
7
49
d
for reprodu
d
)
ction or disp
n
lay.
page-pf8
S
O
Eq
u
O
LUTION
u
ation of mot
i
Σ
si
n
PROB
L
A small
750
m
L=
the rod is
i
on.
(
A
A
M
M
Σ
n
θθ
L
EM 19.50
collar of m
a
m
m.
Determi
n
given a small
eff
):
2
R
L
W
αθ
=
a
ss 1 kg is
r
n
e (a) the dist
a
initial displa
c
sin
s
2
C
L
Wd
θ
,() 2
tR
L
a
θα
=
r
igidly attac
h
a
nce d to max
i
c
ement, (b) th
e
s
in
R
I
m
θα
=+
,(
)
2
t
La
α
θ
=
h
ed to a 3-k
g
i
mize the fre
q
e
correspondi
n
()
2
RtR
L
m
a
m
+
)
C
dd
αθ
==

g
uniform ro
q
uency of osci
n
g period of o
()
CtC
m
da
d of length
llation when
scillation.
page-pf9
PROBLEM 19.50 (Continued)
Natural frequency.
(
)
()
2
3
22
2
22
2
3()
C
R
C
R
m
LL
m
nm
L
m
dg dg
Ld
d
ω
++
==
+
+
(a) To maximize the frequency, we need to take the derivative with respect to d and set it equal to zero.
(
)
(
)
222 3
2
222
22 2
()(1) (2)
10
() ()
L
n
dLd d d
gdd Ld
ω
+−+
=
=
+
4.6476
n
n
n
page-pfa
PROBLEM 19.51
A thin homogeneous wire is bent into the shape of an isosceles
triangle of sides b, b, and 1.6b. Determine the period of small
oscillations if the wire (a) is suspended from point A as shown,
(b) is suspended from point B.
SOLUTION
()()
2
2
21.6
11.6
20.6
33.6 3.6 12
A
b
mm
I
bb
⎡⎤
⎛⎞
=+ +
⎜⎟ ⎢⎥
⎝⎠
⎣⎦
2
11
25 mb=
(
)
20.3
3.6 6
bb b
yb
=
=
13
0.6 630
b
A
Gb b=−=
(b)
0, 6.67
25 6 n
n
mg b
g
θθτ
ω
+===
⎜⎟
⎜⎟
⎝⎠

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