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 534: 5


$\int\Sigma Mdt=0.833Kg\cdot m^2/s$

Work Step by Step

We can determine the required angular impulse as follows: $\Sigma \int Mdt=I\omega~~~$[eq(1)] We know that $I=\frac{1}{12}mr_r L_r^2+2(\frac{1}{2}mr^2+mr^2_{A/G})$ We plug in the known values to obtain: $I=\frac{1}{12}(1)(0.58)^2+2[\frac{1}{2}(1)(0.01)^2+1(0.3)^2]$ This simplifies to: $I=0.2081kg\cdot m^2$ We plug in the known values in eq(1) to obtain: $\int\Sigma Mdt=I_{axle}\omega=(0.2081)(4)=0.833Kg\cdot m^2/s$
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.