Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 15 - Kinetics of a Particle: Impulse and Momentum - Section 15.7 - Principle of Angular Impulse and Momentum - Problems - Page 292: 103


$v_2=9.22ft/s$, $\Sigma U_{1\rightarrow}2=3.04ft\cdot lb$

Work Step by Step

We can determine the required speed and work done as follows: According to the principle of conservation of angular momentum $m_B(v_B)r_1=m_B(v_{\theta})_2r_2$ $\implies 6\times 3=2(v_{\theta})_2$ $\implies (v_{\theta})_2=9ft/s$ Now we can calculate the speed as $v_2=\sqrt{(v_{\theta})_2^2+(v_r)_2^2}$ We plug in the known values to obtain: $v_2=9.22ft/s$ Now according to the principle of work and energy $\frac{1}{2}m_Bv_1^2+\Sigma U_{1\rightarrow 2}=\frac{1}{2}m_Bv_2^2$ We plug in the known values to obtain: $\frac{1}{2}(\frac{4}{32.2})(6)^2+\Sigma U_{1\rightarrow 2}=(\frac{4}{32.2})(9.22)^2$ This simplifies to: $\Sigma U_{1\rightarrow}2=3.04ft\cdot lb$
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