Physics (10th Edition)

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

Chapter 8 - Rotational Kinematics - Problems - Page 215: 52


The angle is 1 radian.

Work Step by Step

We call the constant angular acceleration $\alpha$ and the angle the drill has moved through $\theta$. It starts from rest, so $\omega_0=0$ The drill's final angular speed is $\omega^2=0^2+2\alpha\theta=2\alpha\theta$ The radius of the rotation circle at the mentioned point is $r$. At that point, we have $$a_c=r\omega^2=2r\alpha\theta$$ $$a_T=r\alpha$$ It is said that $a_c=2a_T$. Therefore, $$2r\alpha\theta=2r\alpha$$ $$\theta=1rad$$
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