Physics (10th Edition)

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

Chapter 9 - Rotational Dynamics - Problems - Page 247: 50


The kinetic energy of rod A is $3.26J$ and the kinetic energy of rod B is $1.06J$

Work Step by Step

Rod A has a particle of mass $0.66kg$ attached to its free end. The particle is at a distance of $0.75m$ from the axis, so $$I_1=0.66\times0.75^2=0.37kg.m^2$$ Rod B has its mass equally distributed along its length, with the axis at one end, so $$I_2=\frac{1}{3}ML^2=\frac{1}{3}\times0.66\times0.75^2=0.12kg.m^2$$ We have $\omega=4.2rad/s$. Therefore, the kinetic energies of 2 rods are - Rod A: $$KE=\frac{1}{2}I_1\omega^2=3.26J$$ - Rod B: $$KE=\frac{1}{2}I_2\omega^2=1.06J$$
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