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 538: 21


$\omega=12.72 ~rad/s$

Work Step by Step

We can determine the required angular velocity as follows: We know that $I_A\omega_1+\Sigma \int_{t_1}^{t_2} M_A dt=I_A\omega_2~~~$[eq(1)] Now $I_A=m(K_{\circ}^2+r^2)$ $\implies I_A=(\frac{30}{32.2})[(0.45)^2+(0.9)^2]$ $\implies I_A=0.9433 slug\cdot ft^2$ and $\omega_1=0$ Similarly $\Sigma \int_{t_1=0}^{t_2=4} M_A dt=0.6 Pt$ $\Sigma \int_{t_1=0}^{t_2=4} M_A dt=(0.6)(5)(4)=12lb\cdot s$ We plug in the known values in eq(1) to obtain: $0+12=0.9433\omega$ This simplifies to: $\omega=12.72 rad/s$
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