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 4 - Kinematics in Two Dimensions - Exercises and Problems - Page 107: 35

Answer

The pebble's speed is 5.7 m/s and the pebble's acceleration is $107~m/s^2$

Work Step by Step

The wheel rotates 3 revolutions each second. We can find the angular speed of the wheel. $\omega = (2\pi~rad/rev)(3~rev/s)$ $\omega = 6\pi~rad/s$ We can find the speed of the pebble. $v = \omega ~r$ $v = (6\pi~rad/s)(0.30~m)$ $v = 5.7~m/s$ We can find the centripetal acceleration of the pebble. $a_c = \omega^2~r$ $a_c = (6\pi)^2(0.30~m)$ $a_c = 107~m/s^2$ The pebble's speed is 5.7 m/s and the pebble's acceleration is $107~m/s^2$
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