Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 11 - Exercises and Problems - Page 201: 45


5.5 meters per second

Work Step by Step

We consider the maximum force: $\omega = \sqrt{\frac{F}{ml^2}}\sqrt{\frac{300}{60\times12}}=.6455$ This corresponds to a maximum speed of 7.75, which is less than eight, so the speed requirement is met. Thus, we must find the initial velocity that corresponds to this value of omega: Using conservation of angular momentum, we find: $v = 5.5 \ m/s$
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