Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 8 - Dynamics II: Motion in a Plane - Exercises and Problems - Page 201: 52

Answer

(a) $\omega_{min} = \sqrt{\frac{g}{L}}$ (b) $\omega_{min} = 29.9~rpm$

Work Step by Step

(a) We can find the angular velocity $\omega_{min}$ such that the force of gravity is equal to the centripetal force at the top. $m\omega_{min}^2~L = mg$ $\omega_{min} = \sqrt{\frac{g}{L}}$ (b) We can find $\omega_{min}$ in units of rad/s $\omega_{min} = \sqrt{\frac{g}{L}}$ $\omega_{min} = \sqrt{\frac{9.80~m/s^2}{1.0~m}}$ $\omega_{min} = 3.13~rad/s$ We can convert $\omega_{min}$ to units of rpm. $\omega_{min} = (3.13~rad/s)(\frac{1~rev}{2\pi~rad})(\frac{60~s}{1~min})$ $\omega_{min} = 29.9~rpm$
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.