Physics (10th Edition)

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

Chapter 9 - Rotational Dynamics - Problems - Page 247: 49


a) $v_1=12m/s$, $v_2=9m/s$ and $v_3=18m/s$ b) $KE=1080J$ c) $I=60kg.m^2$ d) $KE_R=1080J$

Work Step by Step

We have $\omega=6rad/s$ a) The tangential speed of each object is: $v_1=r_1\omega=12m/s$ $v_2=r_2\omega=9m/s$ $v_3=r_3\omega=18m/s$ b) The total kinetic energy of this system is $$KE=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2+\frac{1}{2}m_3v_3^2=1080J$$ c) The moment of inertia of the system is $$I=I_1+I_2+I_3=m_1r_1^2+m_2r_2^2+m_3r_3^2$$ $$I=60kg.m^2$$ d) The rotational kinetic energy of this system is $$KE_R=\frac{1}{2}I\omega^2=1080J$$
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