Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 8 - Rotational Motion - Problems - Page 222: 8


4300 rev.

Work Step by Step

4300 revolutions. Each time the tire makes one revolution, it travels a distance $\pi D$ along the floor, where D is the diameter. $$N rev\frac{\pi 68 \times 10^{-2}m}{1 \; rev}=9200 m$$ Solve for $N \approx 4300 \; rev $.
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