Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 10 - Rotation - Problems - Page 289: 35a

Answer

The rotational inertia of a rigid body depends on how its mass is distributed. EXPRESS since the rotational inertia of a cylinder is } $I=\frac{1}{2} M R^{2} $ (Table 10-2($\mathrm{c}$) its rotational kinetic energy is $$ K=\frac{1}{2} I \omega^{2}=\frac{1}{4} M R^{2} \omega^{2} $$ For the smaller cylinder, we have $$K_{1}=\frac{1}{4}(1.25 \mathrm{kg})(0.25 \mathrm{m})^{2}(235 \mathrm{rad} / \mathrm{s})^{2}=1.08 \times 10^{3} \mathrm{J}$$

Work Step by Step

The rotational inertia of a rigid body depends on how its mass is distributed. EXPRESS since the rotational inertia of a cylinder is } $I=\frac{1}{2} M R^{2} $ (Table 10-2($\mathrm{c}$) its rotational kinetic energy is $$ K=\frac{1}{2} I \omega^{2}=\frac{1}{4} M R^{2} \omega^{2} $$ For the smaller cylinder, we have $$K_{1}=\frac{1}{4}(1.25 \mathrm{kg})(0.25 \mathrm{m})^{2}(235 \mathrm{rad} / \mathrm{s})^{2}=1.08 \times 10^{3} \mathrm{J}$$
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