Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 11 - Exercises and Problems - Page 208: 46

Answer

a) $ \omega_f=2.5 \ rads/s$ b) $W=3.4 \times 10^{-3}J$

Work Step by Step

a) We know that angular momentum is conserved. Since the old angular speed was 2.3 radians per second, we use the new moment of inertia to find the new angular speed: $L_0=L_f \\ I\omega_f = I\omega_0 + m\omega_0r^2 \\ .0154\omega_f=.0154(2.3)+(.0195)(2.3)(.25)^2 \\ \omega_f=2.5 \ rads/s$ b) Thus, we can find the work done by finding the change in energy: $W = \frac{1}{2}I\omega_f^2 -\frac{1}{2}I\omega_0^2+\frac{1}{2}mr^2\omega_0$ Plugging in the known values, we find: $W=3.4 \times 10^{-3}J$
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