Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 12 - Rotation of a Rigid Body - Exercises and Problems: 34

Answer

$KE = 0.375~J$

Work Step by Step

The kinetic energy of the can is the sum of the translational kinetic energy and the rotational kinetic energy. $KE = KE_{trans}+KE_{rot}$ $KE = \frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2$ $KE = \frac{1}{2}Mv^2+\frac{1}{2}(\frac{1}{2}MR^2)(\frac{v}{R})^2$ $KE = \frac{1}{2}Mv^2+\frac{1}{4}Mv^2$ $KE = \frac{3}{4}Mv^2$ $KE = \frac{3}{4}(0.50~kg)(1.0~m/s)^2$ $KE = 0.375~J$
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