PROBLEM 9.120
The area shown is revolved about the x axis to form a homogeneous
solid of revolution of mass m. Using direct integration, express the
mass moment of inertia of the solid with respect to the x axis in
terms of m and h.
SOLUTION
We have
y=
2h − h
x+h
a
so that
r=
h
( x + a)
a
For the element shown:
dm = ρπ r 2 dx dI x =
1 2
r dm
2
2
ªh
º
= ρπ « ( x + a) » dx
¬a
¼
h2
1
h2
2
(
x
a
)
dx
[( x + a)3 ]0a
+
=
ρπ
0
3
a2
a2
1
h2
7
= ρπ 2 (8a 3 − a3 ) = ρπ ah 2
3
3
a
³
Then
m = dm =
Now
I x = dI x =
³
³
a
³
1 2
1
r ( ρπ r 2 dx) = ρπ
2
2
ρπ
³
aªh
0
4
º
« a ( x + a ) » dx
¬
¼
4
1
1h
1
h4
5 a
[(
x
a
)
]
(32a5 − a 5 )
+
=
ρπ ×
ρπ
0
2
5 a4
10
a4
31
= ρπ ah 4
10
=
From above:
Then
3
7
ρπ ah 2 = m
Ix =
31 § 3 · 2 93 2
m h =
mh
10 ¨© 7 ¸¹
70
or
I x = 1.329 mh 2 W
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1561
PROBLEM 9.141
The machine element shown is fabricated from steel. Determine the
mass moment of inertia of the assembly with respect to (a) the x axis,
(b) the y axis, (c) the z axis. (The density of steel is 7850 kg/m3 .)
SOLUTION
First compute the mass of each component. We have
m = ρSTV
Then
m1 = (7850 kg/m3 )(π (0.08 m) 2 (0.04 m)]
= 6.31334 kg
m2 = (7850 kg/m3 )[π (0.02 m) 2 (0.06 m)] = 0.59188 kg
m3 = (7850 kg/m3 )[π (0.02 m)2 (0.04 m)] = 0.39458 kg
Using Figure 9.28 and the parallel-axis theorem, we have
(a)
I x = ( I x )1 + ( I x )2 − ( I x )3
­1
½
= ® (6.31334 kg)[3(0.08)2 + (0.04) 2 ] m 2 + (6.31334 kg)(0.02 m) 2 ¾
12
¯
¿
­1
½
+ ® (0.59188 kg)[3(0.02) 2 + (0.06)2 ] m 2 + (0.59188 kg)(0.03 m) 2 ¾
12
¯
¿
­1
½
− ® (0.39458 kg)[3(0.02)2 + (0.04) 2 ] m 2 + (0.39458 kg)(0.02 m) 2 ¾
12
¯
¿
= [(10.94312 + 2.52534) + (0.23675 + 0.53269)
− (0.09207 + 0.15783)] × 10−3 kg ⋅ m 2
= (13.46846 + 0.76944 − 0.24990) × 10−3 kg ⋅ m 2
= 13.98800 × 10−3 kg ⋅ m 2
or I x = 13.99 × 10−3 kg ⋅ m 2 W
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1601
PROBLEM 9.141 (Continued)
(b)
I y = ( I y )1 + ( I y )2 − ( I y )3
ª1
º
= « (6.31334 kg)(0.08 m) 2 »
¬2
¼
ª1
º
+ « (0.59188 kg)(0.02 m) 2 + (0.59188 kg)(0.04 m) 2 »
2
¬
¼
ª1
º
− « (0.39458 kg)(0.02 m 2 ) + (0.39458 kg)(0.04 m) 2 »
¬2
¼
= [(20.20269) + (0.11838 + 0.94701)
− (0.07892 + 0.63133)] × 10−3 kg ⋅ m 2
= (20.20269 + 1.06539 − 0.71025) × 10−3 kg ⋅ m 2
= 20.55783 × 10−3 kg ⋅ m 2
or
(c)
I y = 20.6 × 10−3 kg ⋅ m 2 W
I z = ( I z )1 + ( I z ) 2 − ( I z )3
­1
½
= ® (6.31334 kg)[3(0.08)2 + (0.04) 2 ] m 2 + (6.31334 kg)(0.02 m) 2 ¾
¯12
¿
­1
½
+ ® (0.59188 kg)[3(0.02) 2 + (0.06)2 ] m 2 + (0.59188 kg)[(0.04) 2 + (0.03) 2 ] m 2 ¾
¯12
¿
­1
½
− ® (0.39458 kg)[3(0.02) 2 + (0.04) 2 ] m 2 + (0.03958 kg)[(0.04)2 + (0.02) 2 ] m 2 ¾
12
¯
¿
= [(10.94312 + 2.52534) + (0.23675 + 1.47970)
− (0.09207 + 0.78916)] × 10−3 kg ⋅ m 2
= (13.46846 + 1.71645 − 0.88123) × 10−3 kg ⋅ m 2
= 14.30368 × 10−3 kg ⋅ m 2
or
I z = 14.30 × 10−3 kg ⋅ m 2 W
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1602
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. (,!## &$'%!) % ## (! *) ()(+ PROBLEM 9.167
The thin bent plate shown is of uniform density and weight W. Determine
its mass moment of inertia with respect to the line joining the origin and
Point A.
SOLUTION
First note that
and that
m1 = m2 =
λ
A
=
1
3
1W
2 g
( + + )
Using Figure 9.28 and the parallel-axis theorem, we have
I x = ( I x )1 + ( I x )2
ª 1 §1W · 2 1W
=« ¨
¸a +
2 g
¬« 12 © 2 g ¹
2
§a· º
¨ ¸ »
© 2 ¹ ¼»
ª§ a ·2 § a ·2 º ½°
«¨ ¸ + ¨ ¸ » ¾
¬«© 2 ¹ © 2 ¹ ¼» °¿
1 W ª§ 1 1 · 2 § 1 1 · 2 º 1 W 2
a
=
«¨ + ¸ a + ¨ + ¸ a » =
2 g ¬© 12 4 ¹
©6 2¹ ¼ 2 g
­° 1 § 1 W
+® ¨
°¯12 © 2 g
· 2
1W
2
¸ (a + a ) +
2 g
¹
I y = ( I y )1 + ( I y )2
­° 1 § 1 W ·
1W
2
2
=® ¨
¸ (a + a ) +
12 2 g ¹
2 g
¯° ©
ª§ a ·2 § a ·2 º ½°
«¨ ¸ + ¨ ¸ » ¾
«¬© 2 ¹ © 2 ¹ »¼ ¿°
­° 1 § 1 W · 2 1 W ª 2 § a ·2 º °½
+® ¨
«(a) + ¨ ¸ » ¾
¸a +
2 g «¬
© 2 ¹ »¼ °¿
°¯12 © 2 g ¹
1 W ª§ 1 1 · 2 § 1 5 · 2 º W 2
=
a + ¨ + ¸a » = a
« +
2 g ¬¨© 6 2 ¸¹
© 12 4 ¹ ¼ g
I z = ( I z )1 + ( I z ) 2
ª 1 §1W · 2 1W
=« ¨
¸a +
2 g
¬«12 © 2 g ¹
2
§a· º
¨ ¸ »
© 2 ¹ ¼»
2
· 2 1 W ª 2 § a · º °½
+
+
a
(
a
)
«
¸
¨ 2 ¸ »¾
2 g ¬«
© ¹ ¼» °¿
¹
1 W ª§ 1 1 · 2 § 1 5 · 2 º 5 W 2
=
+
a + ¨ + ¸a » =
a
«
2 g ¬¨© 12 4 ¸¹
© 12 4 ¹ ¼ 6 g
­° 1 § 1 W
+® ¨
¯°12 © 2 g
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1649
PROBLEM 9.167 (Continued)
Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of both components are zero because
of symmetry. Also, y1 = 0
0
1W
§a· 1W 2
I xy = Σ( I x′y ′ + mx y ) = m2 x2 y2 =
(a) ¨ ¸ =
a
Then
2 g
©2¹ 4 g
0
1 W § a ·§ a · 1 W 2
I yz = Σ( I y′z′ + my z ) = m2 y2 z2 =
a
¸=
2 g ¨© 2 ¸¨
¹© 2 ¹ 8 g
0
I zx = Σ( I z ′x′ + mz x ) = m1z1 x1 + m2 z2 x2
=
1 W § a ·§ a · 1 W § a ·
3W 2
(a ) =
a
+
2 g ¨© 2 ¸¹ ¨© 2 ¸¹ 2 g ¨© 2 ¸¹
8 g
Substituting into Equation (9.46)
I
A
= I x λx2 + I y λ y2 + I z λz2 − 2 I xy λx λ y − 2 I yz λ y λz − 2 I zx λz λx
Noting that
λx2 = λ y2 = λz2 = λx λ y = λ y λz = λz λx =
1
3
We have
I
A
§ 1 W 2 1 W 2 3 W 2 ·º
1 ª1 W 2 W 2 5 W 2
= «
a + a +
a − 2¨
a +
a +
a ¸»
3 ¬2 g
g
6 g
8 g
8 g
©4 g
¹¼
1 ª14
§ 3 ·º W
= « − 2 ¨ ¸» a 2
3¬ 6
© 4 ¹¼ g
or
I
A
=
5 W 2
a W
18 g
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
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1650