Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 8 - Rotational Motion - Problems: 32

Answer

$\tau = 1.5\times 10^4~m\cdot N$

Work Step by Step

We can find the angular acceleration: $\alpha = \frac{\Delta \omega}{t} = \frac{0.68~rad/s}{34~s} = 0.020~rad/s^2$ We then find the moment of inertia: $I = \frac{1}{2}MR^2 = \frac{1}{2}(31,000~kg)(7.0~m)^2$ $I = 759,500 ~kg\cdot m^2$ We can then use this angular acceleration and the moment of inertia to find the torque: $\tau = I \alpha$ $\tau = (759,500 ~kg\cdot m^2)(0.020~rad/s^2)$ $\tau = 1.5\times 10^4~m\cdot N$
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