Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 5 - Section 5.3 - Circular Motion - Example - Page 81: 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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