University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 10 - Dynamics of Rotational Motion - Problems - Exercises - Page 329: 10.8

Answer

(a) $v = 0.868~m/s$ (b) $a = 4.06~m/s^2$

Work Step by Step

(a) We can find the angular acceleration. $\tau = I\alpha$ $\alpha = \frac{\tau}{I}$ $\alpha = \frac{R~F}{\frac{1}{2}MR^2}$ $\alpha = \frac{2~F}{MR}$ $\alpha = \frac{(2)(30.0~N)}{(40.0~kg)(0.200~m)}$ $\alpha = 7.50~rad/s^2~$ We can find the angular velocity after the disk turns through 0.200 revolution. $\omega^2 = \omega_0^2+2\alpha \theta$ $\omega = \sqrt{2\alpha \theta}$ $\omega = \sqrt{(2)(7.50~rad/s^2)(2\pi~rad/rev)(0.200~rev)}$ $\omega = 4.34~rad/s$ We can find the tangential velocity. $v = \omega ~R$ $v = (4.34~rad/s)(0.200~m)$ $v = 0.868~m/s$ (b) We can find the tangential acceleration. $a_t = \alpha ~R$ $a_t = (7.50~rad/s^2)(0.200~m)$ $a_t = 1.50~m/s^2$ We can find the centripetal acceleration. $a_c = \omega^2~R$ $a_c = (4.34~rad/s)^2(0.200~m)$ $a_c = 3.77~m/s^2$ We can find the magnitude of the resultant acceleration. $a = \sqrt{(a_t)^2+(a_c)^2}$ $a = \sqrt{(1.50~m/s^2)^2+(3.77~m/s^2)^2}$ $a = 4.06~m/s^2$
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