Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 8 - Dynamics II: Motion in a Plane - Exercises and Problems: 21

Answer

The speed at the top is 19.8 m/s.

Work Step by Step

Note that at the top of the loop, the normal force from the track is directed straight down. We can use an equation for the centripetal force to find the speed. $\sum F = \frac{mv^2}{r}$ $F_N+mg = \frac{mv^2}{r}$ $mg+mg = \frac{mv^2}{r}$ $v^2 = 2~g~r$ $v = \sqrt{2~g~r}$ $v = \sqrt{(2)(9.80~m/s^2)(20~m)}$ $v = 19.8~m/s$ The speed at the top is 19.8 m/s.
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