Fenomenos

91962_07_s17_p0641-0724

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•17–1. Determine the moment of inertia Iy for the slender 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 terms of 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

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91962_07_s17_p0641-0724

6/8/09

3:32 PM

Page 642

© 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 x axis. 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