Physics (10th Edition)

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

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


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: $$\theta=\cos^{-1}\Big(\frac{r_A}{r_E}\Big)=70.53^o$$
Small 1582072870
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.