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91962_07_s17_p0641-0724

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Page 641

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

•17–1. Determine the moment of inertia Iy for theslender rod. The rod’s density r and cross-sectional area A are constant. Express the result in terms of the rod’s total mass m.

z

l

y

Iy = =

L M
0 L l

x2 dm
A

x2 (r A dx)

x

1 = r A l3 3 m = rAl Thus, Iy = 1 m l2 3 Ans.

17–2. The right circular cone is formed by revolving the shaded area around the x axis. Determine the moment of inertia Ix and express the result in termsof the total mass m of the cone. The cone has a constant density r.

y y
r –x h

r

dm = r dV = r(p y dx) m = L 0
h

dIx = = =

1 2 y dm 2

1 2 y (rp y2 dx) 2

r(p) ¢

r2 2 r2 1 1 ≤ x dx = rp ¢ 2 ≤ a b h3 = rp r2h 3 3 h2 h
2

x

h

1 r4 r(p)a 4 b x4 dx 2 h
h

Ix = Thus, Ix =

1 1 r4 r(p)a 4 bx4 dx = rp r4 h 10 h 0 L 2 Ans.

3 m r2 10

641 91962_07_s17_p0641-0724

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© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

17–3. The paraboloid is formed by revolving the shaded area around the xaxis. Determine the radius of gyration kx. The density of the material is r = 5 Mg>m3.

y y2 50x 100 mm

dm = r p y2 dx = r p (50x) dx Ix = 1 2 1 y dm = 2L 0 L2 50 1 b c x3 d 2 3 0
2 200

x

= r pa

50 x {p r (50x)} dx
200 mm

200

= rp a L

502 b (200)3 6
0 L 200

m =

= rp a kx =

200 1 = r p (50)c x2 d 2 0

dm =

p r (50x) dx

Ix 50 = (200) = 57.7 mm Am A3

50 b(200)2 2

Ans.

*17–4. The frustum is formed by rotating the shaded area around the x axis. Determine the moment of inertia Ix and express the result in terms of the total mass m of the frustum. The frustum has a constant density r.
b

y y
b –x a

b 2b x

dm = r dV = rpy2 dx = rp A 1 1 dIx = dmy2 = rpy4 dx 2 2 dIx =

1 b4 4 b4 6 b4 4b4 rp A 4 x4 + 3 x3 + 2 x2 + x + b4 B dx a 2 a a a La

b2 2 2b2 x + x + b2 B dx a a2

z a

Ix =

dIx = =

1 b4 4b4 6 b4 4 b4 rp x + b4 B dx A x4 + 3 x3 + 2 x2 + a 2 L a4 a a 0 31 rpab4 10
0 L a

m =

L m

dm = rp

93 2 Ix = mb 70

A 2 x2 +
b2 a

2b2 7 x + b2 B dx = rpab2 a 3

Ans.

642

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© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rightsreserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

•17–5. The paraboloid is formed by revolving the shaded area around the x axis. Determine the moment of inertia about the x axis and express the result in terms of the total mass m of theparaboloid. The material has a constant density r. dm = r dV = r (p y2 dx) 1 1 dm y2 = r p y4 dx 2 2 1 a4 r p a 2 bx2 dx 2 h 0 L
h

y – y2 = a x h a x
2

d Ix =

h

Ix = =

1 p ra4 h 6
h

m = = Ix =

1 r p a2 h 2 1 ma2 3

a2 1 r pa bx dx 2 h 0 L

Ans.

17–6. The hemisphere is formed by rotating the shaded area around the y axis. Determine the moment of inertia Iy and expressthe result in terms of the total mass m of the hemisphere. The material has a constant density r.

y

x2

y2

r2

m =

L V

r dV = r

= rpc r2 y -

1 3 r 2 y d = rp r3 3 3 0

0 L

r

p x2 dy = rp

0 L

r

(r2 - y2)dy

x

Iy = = Thus, Iy =

rp 4 y5 r 4rp 5 2 cr y - r2 y3 + d = r 2 3 5 0 15 2 m r2 5

L 2 m 1

(dm) x2 =

r r r rp px4 dy = (r2 - y2)2 dy 2 L 2...
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