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: 48



Work Step by Step

The required velocity can be determined as follows: We know that $e=\frac{(v_b)_2-(v_B)_2}{(v_B)_1-(v_b)1}$ $\implies 0.8=\frac{(v_b)_2-(v_B)_2}{0-12}$ $\implies (v_b)_2=(v_B)_2-9.6$ [eq(1)] We also know that the angular momentum is conserved about A $\implies H_{A_1}=H_{A_2}$ $\implies m_b(v_b)_1 r_b=I_A(\omega_B)_2+m_b(v_b)_2r_b$ [eq(2)] As $I_A=\frac{m_rl^2}{12}+m_r r^2$ $\implies I_A=\frac{4}{32.2}\times \frac{(3)^2}{12}+\frac{4}{32.2}(1.5)^2=0.37267 slug/ft^2$ We plug in the known values in eq (2) to obtain: $\frac{2}{32.2}\times 12\times 3=0.37267(\omega_B)_2+(\frac{2}{32.2}\times (v_b)_2\times 3)$ This simplifies to: $2.236=0.12422(v_B)_2+0.18633(v_b)_2$ [eq(3)] We plug in the values from eq(1) into eq(3) to obtain: $2.36=0.12422(v_B)_2+0.18633(v_B)_2-9.6$ This simplifies to: $(v_B)_2=12.959ft/s$ Now, from eq(1), we obtain: $(v_b)_2=12.959-9.6=3.36~ft/s$
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