Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 9 - Rotational Dynamics - Problems - Page 247: 53



Work Step by Step

The sphere is rotating with the axis at its center, so $I=\frac{2}{5}MR^2$ The rotational kinetic energy of the sphere is $$KE_R=\frac{1}{2}\times\frac{2}{5}MR^2\omega^2=\frac{1}{5}MR^2\omega^2$$ The sphere's linear speed is also its tangential speed at the outer edge, so we have $$v=R\omega$$ The total KE is $$KE=\frac{1}{2}Mv^2+KE_R=\frac{1}{2}MR^2\omega^2+\frac{1}{5}MR^2\omega^2$$ $$KE=\frac{7}{10}MR^2\omega^2$$ Thus: $$\frac{KE_R}{KE}=\frac{2}{7}$$
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