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 537: 19



Work Step by Step

We can determine the required speed as follows: According to the principle of impulse and momentum $I_{\circ}\omega_1+\Sigma\int_{t_1}^{t_2} M_{\circ}dt=I_{\circ}\omega_2+m_Av_A r_A$ [eq(1)] We know that $I_{\circ}=mk_{\circ}^2=15(0.11)^2=0.1815Kg\cdot m^2$ Similarly $\omega_1=0$ and $v_{A_2}=0.2 \omega_2$ $\implies \omega_2=5v_A$ $\Sigma\int_{t_1=0}^{t_2=3}M_{\circ} dt=0.075Ft-0.2m_A gt$ $\Sigma\int_{t_1=0}^{t_2=3}M_{\circ} dt=0.075(2000)(3)-0.2\times 40\times 9.81\times 3=214.56N\cdot s$ We plug in the known values in eq(1) to obtain: $0+214.56=0.1815(5v_A)+40(v_A)(0.2)$ This simplifies to: $v_A=24.1~m/s$
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.