Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 11 - Exercises and Problems - Page 202: 51


a) .25 rads/s b) 6.5 kJ

Work Step by Step

a) Using conservation of angular momentum, we find: $I_0\omega_0=I_f\omega_f \\ ((4,800)(8.5^2)+\frac{1}{2}(15,000)(8.5)^2)(.15)=(\frac{1}{2}(15,000)(8.5)^2)(\omega)\\ \omega=.25 \ rads/s$ b) We know that work is equal to the change in kinetic energy, so it follows: $W=\frac{1}{2}I\omega_f^2-\frac{1}{2}I\omega_i^2$ $W=\frac{1}{2}(\frac{1}{2}(15,000)(8.5)^2)(.25)^2-\frac{1}{2}((4,800)(8.5^2)+\frac{1}{2}(15,000)(8.5)^2)(.15)^2 \approx 6.5 \ kJ$
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