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 552: 50


$\omega_1=7.173 rad/s$

Work Step by Step

The required angular velocity can be determined as follows: First, we apply the equation of motion in normal direction $\Sigma F_n=ma_n$ [eq(1)] As $a_n=\omega_2^2 r=0.2\omega_2^2$ and $Wcos\theta=20(0.2)\omega_2^2$ $\implies cos\theta=\frac{0.2-0.03}{0.2}=0.85$ Now from eq(1), we obtain: $20(9.81)(0.85)=4\omega_2^2$ This simplifies to: $\omega_2=6.456 rad/s$ Now according to conservation of angular momentum $mv_1r_1+I\omega_1=mv_2r_2+I\omega_2$ [eq(2)] We know that $v=0.2\omega$ and $I=\frac{1}{2}mr^2=\frac{1}{2}(20)(0.2)^2=0.4Kg/m^2$ Now we plug in the known values in eq(2) to obtain: $20(0.2\omega_1)(0.17)+0.4\omega_1=20(0.2\omega_2)(0.2)+0.4\omega_2$ This simplifies to: $\omega_1=1.2\omega_2$ $\implies \omega_1=1.111\times 6.456=7.173 rad/s$
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