Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 8 - Rotational Motion - Problems: 51

Answer

The translational speed at the bottom of the incline is 9.70 m/s.

Work Step by Step

As the cylinder rolls down the incline, the initial potential energy will be converted into translational and rotational kinetic energy. We can use conservation of energy to solve this question: $KE_t + KE_r = PE$ $\frac{1}{2}mv^2+ \frac{1}{2}I\omega^2 = mgh$ $\frac{1}{2}mv^2+ \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2 = mgh$ $\frac{1}{2}mv^2+ \frac{1}{4}mv^2 = mgh$ $\frac{3}{4}v^2 = gh$ $v = \sqrt{\frac{4gh}{3}} = \sqrt{\frac{(4)(9.80~m/s^)(7.20~m)}{3}}$ $v = 9.70~m/s$ Therefore, the translational speed at the bottom of the incline is 9.70 m/s.
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