Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 8 - Rotational Kinematics - Problems - Page 212: 3

Answer

a) $\overline\omega=+7.3\times10^{-5}rad/s$ b) $\overline\omega=+1.99\times10^{-7}rad/s$

Work Step by Step

a) The Earth finishes 1 revolution, which is $\Delta \theta=2\pi rad$, on its axis in $\Delta t=24h=8.64\times10^4s$. Therefore, $$\overline\omega=\frac{\Delta\theta}{\Delta t}=+7.3\times10^{-5}rad/s$$ b) The Earth finishes 1 revolution, which is $\Delta \theta=2\pi rad$, around the sun in $\Delta t=365.25 days=3.156\times10^7s$. Therefore, $$\overline\omega=\frac{\Delta\theta}{\Delta t}=+1.99\times10^{-7}rad/s$$
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