College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 5 - Problems - Page 189: 73


The drill rotates through $(1.80\times 10^6)^{\circ}$

Work Step by Step

We can find the angular displacement in $1.00~s$: $\Delta \theta = \omega ~t$ $\Delta \theta = (3.14\times 10^4~rad/s)(1.00~s)$ $\Delta \theta = 3.14\times 10^4~rad$ We can express the angular displacement in degrees: $\Delta \theta = (3.14\times 10^4~rad) \times \frac{180^{\circ}}{\pi~rad}$ $\Delta \theta = (1.80\times 10^6)^{\circ}$ The drill rotates through $(1.80\times 10^6)^{\circ}$.
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