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 535: 11


$\omega_2=53.7~ rad/s$

Work Step by Step

We can determine the required angular velocity as follows: We know that $I_{\circ}\omega_1+\Sigma \int_{t_1}^{t_2} M_{\circ} dt=I_{\circ}\omega_2~~~$[eq(1)] As $I_{\circ}=\frac{1}{2}mr^2$ $\implies I_{\circ}=\frac{1}{2}(\frac{8}{32.2})(0.6)^2=0.0447slug\cdot ft^2$ and $\omega_1=0$ Now $\int_{t_1}^{t_2} M_{\circ}=T_{B}rt-T_A rt=(5)(4)(0.6)-(4)(4)(0.6)=2.4lb\cdot ft/s$ We plug in the known values in eq(1) to obtain: $0+2.4=0.0447\omega_2$ This simplifies to: $\omega_2=53.7 rad/s$
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.