Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 19 - Planar Kinetics of a Rigid Body: Impulse and Momentum - Section 19.4 - Eccentric Impact - Problems - Page 550: 40

Answer

$\omega_1=1 ~rad/s$

Work Step by Step

The required angular velocity can be determined as follows: We apply the conservation of angular momentum $I_{P_{\circ}}\omega_{\circ}=I_{P_{1}}\omega_1~~~$[eq(1)] We know that $I_{P_1}=\frac{ml^2}{12}+mr^2$ $\implies I_{P_{1}}=\frac{ml^2}{12}+m(\frac{l}{2})^2$ $\implies I_{P_{1}}=\frac{ml^2}{12}+\frac{ml^2}{4}=\frac{ml^2}{3}$ We plug in the known values in eq(1) to obtain: $\frac{ml^2}{12}\omega_{\circ}=\frac{ml^2}{3}\omega_1$ This simplifies to: $\omega_1=\frac{\omega_{\circ}}{4}$ $\implies \omega_1=\frac{4}{4}= 1 ~rad/s$
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