Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 5 - Circular Motion; Gravitation - General Problems: 77

Answer

Yes, it is possible. $v = \sqrt{gr}$ is the minimum speed.

Work Step by Step

If the speed is fast enough such that, at the top of the swing, the gravitational force is equal to the centripetal force, then the water will not fall out of the bucket. Therefore: $m\frac{v^2}{r} = mg$ $\frac{v^2}{r} = g$ $v = \sqrt{gr}$, where $r$ is the radius of the circle. Therefore, $v = \sqrt{gr}$ is the minimum speed.
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.