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


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.
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