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 551: 46

Answer

$h=\frac{7}{5}r$

Work Step by Step

We can determine the required height as follows: According to the impulse and angular momentum principle $H_{A_1}+r\int Fdt=H_{A_2}$ $\implies I_A\omega_1+h\int Pdt=I_A\omega_2$ $\implies (\frac{2}{5}mr^2+mr^2)(0)+P\Delta th=()\omega_2$ This simplifies to: $P\Delta th=(\frac{2}{5}mr^2+mr^2)\omega_2$ [eq(1)] We know that $mv_1+\int Fdt=mv_2$ $\implies m(0)+P\Delta t=mv_2$ $\implies P\Delta t=mr\omega_2$ [eq(2)] We plug in this value from eq(2) into eq(1) to obtain: $mr\omega_2h=(\frac{2}{5}mr^2+mr^2)\omega_2$ This simplifies to: $h=\frac{7}{5}r$
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