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 294: 111


$10.8ft/s$, $11.3ft\cdot lb$

Work Step by Step

According to the principle of conservation of angular momentum $r_Am_A(v_A)_t=r_Bm_B(v_B)_t$ We plug in the known values to obtain: $2(\frac{8}{32.2})\times 5=1\times (\frac{8}{32.2})\times (v_B)_t$ This simplifies to: $(v_B)_t=10ft/s$ Now, $v_B=\sqrt{v_t^2+v_{B}^2}$ $\implies v_B=\sqrt{(4)^2+(10)^2}$ $\implies v_B=10.8ft/s$ Now we apply the principle of work and energy to calculate the required work done $\frac{1}{2}m_Av_A^2+\Sigma U_{A\rightarrow B}=\frac{1}{2}m_Bv_B^2$ We plug in the known values to obtain: $\frac{1}{2}(\frac{8}{32.2})(5)^2+\Sigma U_{A\rightarrow B}=\frac{1}{2}(\frac{8}{32.2})(10.8)^2$ This simplifies to: $\Sigma U_{A\rightarrow B}=11.3ft\cdot lb$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.