Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 5 - Circular Motion; Gravitation - Problems - Page 132: 4

Answer

$a = 3.9 ~m/s^2$

Work Step by Step

We can find the velocity of the speck as it completes 45 rpm on the wheel through the formula: $v = \frac{(45)(2\pi r)}{T}$ $v = \frac{(45)(2\pi)(0.175 ~m)}{60 ~s}$ $v = 0.825~m/s$ We can then use the velocity of the speck to find the acceleration: $a = \frac{v^2}{r}$ $a = \frac{(0.825 ~m/s)^2}{(0.175 ~m)}$ $a = 3.9 ~m/s^2$
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