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 539: 26


$\omega=9 rad/s$

Work Step by Step

We can determine the required angular velocity as follows: We apply the principle of impulse and momentum $0+\Sigma\int_{t_1}^{t_2}Mdt=I_{AB}\omega+mv_{AB}r_{AB}+I_{BC}\omega+mv_{BC} r_{BC}$ [eq(1)] We know that $I_{AB}=I_{BC}=\frac{1}{12}ml^2=(\frac{1}{12})(9)(1)^2=0.75Kg\cdot m^2$ and $r_{AB}=\frac{1}{2}=0.5m$ Similarly $r_{BC}=\sqrt{(1)^2+(0.5)^2}=1.118m$ $v_{AB}=r_{AB}\omega=0.5\omega$ $v_{BC}=r_{BC}\omega=1.118\omega$ Now, we plug in the known values in eq(1) to obtain: $\int_{0}^{t}15t^2dt=0.75\omega+9(0.5\omega)(0.5)+0.75\omega+9(1.118\omega)(1.118)$ This simplifies to: $\omega=9 rad/s$
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