Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 10 - Exercises and Problems - Page 192: 28


a) .001166 b) 1.23 Nm

Work Step by Step

We add the rotational inertia of the mass spread out in the rim and the mass spread out over the whole disk to obtain: a) $I = mr^2+\frac{1}{2}mr^2 = \frac{3}{2}mr^2=\frac{3}{2}(.054)(.12)^2=\fbox{.001166 }$ b) We first must find the value of the angular acceleration: $\alpha= \frac{\omega^2}{2\theta}=\frac{3456}{\pi}=1055.9$ Thus, we find torque: $\tau=\alpha I = 1055.9 \times .001166 = \fbox{1.23 Nm}$
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