College Physics (4th Edition)

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

Chapter 5 - Problems - Page 189: 64


The radial acceleration is $0.0257~m/s^2$

Work Step by Step

We can find the angular speed of the Earth as it rotates: $\omega = \frac{2\pi~rad}{(24)(3600~s)}$ $\omega = 7.27\times 10^{-5}~rad/s$ We can find the radius of rotation for an object at a latitude of $40.2^{\circ}~N$ of the equator: $r = (6.38\times 10^6~m)~cos~40.2^{\circ}$ $r = 4.87\times 10^6~m$ We can find the radial acceleration: $a_r = \omega^2~r$ $a_r = (7.27\times 10^{-5}~rad/s)^2~(4.87\times 10^6~m)$ $a_r = 0.0257~m/s^2$ The radial acceleration is $0.0257~m/s^2$
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