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: 9

Answer

a. $1.05\times10^{-1}\frac{rad}{s}$ b. $1.75\times10^{-3}\frac{rad}{s}$ c. $1.45\times10^{-4}\frac{rad}{s}$ d. $0$

Work Step by Step

a. $\omega = \frac{2 \pi \; rad}{60 \; s}= \frac{\pi}{30}\frac {rad}{s}\approx1.05\times10^{-1}\frac{rad}{s}$ b. $\omega = \frac{2 \pi \; rad}{3600 \; s}= \frac{\pi}{1800}\frac {rad}{s}\approx1.75\times10^{-3}\frac{rad}{s}$ c. $\omega = \frac{2 \pi \; rad}{43200 \; s}= \frac{\pi}{21600}\frac {rad}{s}\approx1.45\times10^{-4}\frac{rad}{s}$ d. The angular acceleration is zero because the angular velocity is constant.
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