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 549: 35


$\omega_2=0.0906 rad/s$

Work Step by Step

We can determine the required angular velocity as follows: According to the conservation of angular momentum $H_{C_1}=H_{C_2}$ $\implies 2I_{d}\omega_1=I_r\omega_2+2I_d\omega_2+2m_dv_d r_d~~~$[eq(1)] We know that $I_d=\frac{1}{2}m_dr_d^2$ $\implies I_d=\frac{1}{2}(4)(0.15)^2=0.045Kg/m^2$ and $I_r=\frac{m_r l_r^2}{12}$ $\implies I_r=\frac{2(1.5)^2}{12}=0.375Kg/m^2$ $v_d=\omega_2 r$ $\implies v_d=(0.75)\omega_2$ We plug in the known values in eq(1) to obtain: $2(0.045)(5)=(0.375)\omega_2+2(0.045)\omega_2+2(4)(0.75\omega)(0.75)$ This simplifies to: $\omega_2=0.0906 rad/s$
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