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 554: 55

Answer

$h_B=0.9797ft$

Work Step by Step

The required height can be determined as follows: According to the conservation of energy $T_1+V_1=T_2+V_2$ [eq(1)] As $T_1=\frac{1}{2}I\omega_1^2=\frac{1}{2}I(0)^2=0J$ $V_1=mgh_1=15(1)=15J$ and $T_2=\frac{1}{2}I\omega_2^2$ $\implies T_2=\frac{1}{2}(\frac{mL^2}{3})\omega_2^2$ $\implies T_2=\frac{mL^2}{6}\omega_2^2=\frac{15(2)^2}{32.2\times 6}\omega_2^2=0.3105\omega_2^2$ $V_2=mgh_2=mg(0)=0J$ We plug in the known values in eq(1) to obtain: $0+15=0.3105\omega_2^2+0$ This simplifies to: $\omega_2=6.949 rad/s$ We know that $e=\frac{(v_g)_3-(v_B)_3}{(v_B)_2-(v_g)_2}$ $\implies 0.7=\frac{0-\omega_3 r}{\omega_2 r-0}$ $\implies \omega_3=4.8643 rad/s$ Now, we apply the conservation of energy for the rebound motion $T_3+V_3=T_4+V_4$ [eq(2)] As $T_3=\frac{1}{2}I\omega_3^2$ $\implies T_3=\frac{1}{2}(\frac{mL^2}{3})\omega_3^2$ $\implies T_3=\frac{1}{2}(\frac{15}{32.2})(\frac{(2)^2}{3})(4.8643)^2$ $\implies T_3=7.348J$ $V_3=mgh_3=mg(0)=0J$ $T_4=\frac{1}{2}I\omega_4^2=\frac{1}{2}I(0)^2=0J$ and $V_4=mgh_4=15h_4$ Now we plug in the known values in eq(2) to obtain: $7.348+0=0+15h_4$ $\implies h_4=0.489ft$ We know that $\frac{h_4}{L/2}=\frac{hB}{L}$ $\implies h_B=2h_4$ $\implies h_B=2(0.489ft)=0.9797ft$
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