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: 73

Answer

The minimum speed is 9.2 m/s.

Work Step by Step

If the centripetal acceleration is at least equal to $g$, then the people on the ride will not fall out. Therefore: $\frac{v^2}{r} = g$ $v = \sqrt{gr} = \sqrt{(9.80 ~m/s^2)(8.6~m)}$ $v = 9.2 ~m/s$ The minimum speed is 9.2 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.