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 19.2 - Principle of Impulse and Momentum - Problems - Page 537: 17



Work Step by Step

The required velocity can be determined as follows: According to the principle of impulse and momentum $I_A\omega_1+\Sigma \int_{t_1}^{t_2}M_A dt=I_A\omega_2$ [eq(1)] We know that $I_A=mK_G^2+mr^2=(\frac{100}{32.2})(0.75)^2+(\frac{100}{32.2})(0.5)^2=2.5233 slug\cdot ft$ and $\omega_2=\frac{v}{r}=\frac{v}{0.5}=2v$ $\omega_1=0$ $\implies v_{G_1}=0$ Similarly $\Sigma \int_{t_1=0}^{t_2=3}M_A dt=Tt(2r)-\mu_k Ntr$ $\implies \Sigma \int_{t_1=0}^{t_2=3}M_A dt= 50(3)(2\times 0.5)-0.1(100)(3)(0.5)=135lb\cdot s$ We plug in the known values in eq(1) to obtain: $0+135=2.5233(2v_{G_2})$ This simplifies to: $v_{G_2}=26.8~ft/s$
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