Physics (10th Edition)

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

Chapter 8 - Rotational Kinematics - Problems - Page 214: 40

Answer

a) The tangential speed of the person in Ecuador is $465.74m/s$ b) $\theta=70.53^o$

Work Step by Step

a) The Earth finishes 1 revolution, or $2\pi$ radians, in $23.9h=8.604\times10^4s$. Its angular speed is $$\omega=\frac{2\pi}{8.604\times10^4}=7.3\times10^{-5}rad/s$$ The Earth's radius $r_E=6.38\times10^6m$. The tangential speed of the person in Ecuador, therefore, is $$v_E=\omega r_E=465.74m/s$$ b) Take the tangential speed of a person in place A to be $v_A$, where $v_A=1/3v_E$. We have $$\frac{v_A}{v_E}=\frac{\omega r_A}{\omega r_E}=\frac{r_A}{r_E}=\frac{1}{3}$$ As the figure below shows, we can find the angle $\theta$ by using its cosine: $$\theta=\cos^{-1}\Big(\frac{r_A}{r_E}\Big)=70.53^o$$
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