Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 5 - Section 5.3 - Circular Motion - Example - Page 78: 5.7

Answer

$ 7.9 \ m/s$

Work Step by Step

The minimum speed to stay at the top of the track is when the normal force equals 0, meaning that the gravitational force equals the centripetal force: $\frac{mv^2}{r} = mg \\ v = \sqrt{gr}= \sqrt{ 9.81 \times 6.3 } = 7.9 \ m/s$
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