College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 8 - Problems - Page 315: 55


A torque of $~1.48~N\cdot m~$ must be applied.

Work Step by Step

We can find the rotational inertia of the system about this axis: $I = \sum M_i~R_i^2$ $I = (\sum M_i)~(0.375~m)^2$ $I = (4.0~kg+3.0~kg+5.0~kg+2.0~kg)~(0.375~m)^2$ $I = 1.97~kg~m^2$ We can find the required torque: $\tau = I~\alpha$ $\tau = (1.97~kg~m^2)(0.75~rad/s^2)$ $\tau = 1.48~N\cdot m$ A torque of $~1.48~N\cdot m~$ must be applied.
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