University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 10 - Dynamics of Rotational Motion - Problems - Exercises - Page 332: 10.53

Answer

τ = 2.4 × 10−12 N ⋅ m.

Work Step by Step

Ω = Δφ/ Δt = 1.745 × 10^−8 rad/ 1.8 × 104 s = 9.694 × 10−13 rad/s. Ω = τ/ Iω so τ = ΩIω = ΩMR2ω. Putting in the numbers gives τ = (9.694 × 10−13 rad/s)(2.0 kg)(2.5 × 10−2 m)2 (2.01 × 103 rad/s) = 2.4 × 10−12 N ⋅ m.
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