Physics (10th Edition)

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

Chapter 8 - Rotational Kinematics - Problems - Page 216: 68


The magnitude of the total acceleration is $380m/s^2$

Work Step by Step

The radius of the rotation circle is the distance between the racket top and the shoulder $r=1.5m$ We have $\alpha=160rad/s^2$. At the top of the serve, we also know $\omega=14rad/s$, so - The racket's tangential acceleration: $a_T=r\alpha=240m/s^2$ - The racket's centripetal acceleration: $a_c=r\omega^2=294m/s^2$ Since these two acceleration vectors are perpendicular with each other, the sum of them, which is the total acceleration, is $$a=\sqrt{a_T^2+a_c^2}=380m/s^2$$
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