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 548: 33



Work Step by Step

We can determine the required angular velocity as follows: According to the conservation of angular momentum $I_{z1}\omega_1=I_{z_2}\omega_2$ [eq(1)] We know that $I_{z1}=I_{body}+I_{arms}+I_{dumbbells}$ $\implies I_{z1}=\frac{1}{2}m_br_b^2+2(\frac{m_al_a^2}{12}+m_ar_a^2)+2(m_dr_d^2)$ We plug in the known values to obtain: $I_{z1}=\frac{1}{2}(68)(0.2)^2+2[\frac{6(0.65)^2}{12}+6(\frac{0.65}{2}+0.2)^2]+2[10(0.65+0.2)^2]$ This simplifies to: $I_{z1}=19.54Kg/m^2$ Similarly $I_{z2}=I_{body}+I_{dumbbells}$ $\implies I_{z2}=\frac{1}{2}m_br_b^2+2(m_dr_d^2)$ $\implies I_{z2}=\frac{1}{2}(80)(\frac{0.45}{2})^2+2(10)(0.3)^2=3.825Kg/m^2$ Now, we plug in the known values in eq(1) to obtain: $19.54(0.5)=3.825\omega_2$ This simplifies to: $\omega_2=2.55~rev/s$
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