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 - General Problems: 90

Answer

See work below.

Work Step by Step

We want to obtain the dimensions of acceleration, length divided by time squared, $\frac{L}{T^2}$. To work with, we have speed v, with dimensions of length over time $\frac{L}{T}$, and the radius r, with dimensions of length, L. To get units of time squared in the denominator, we must square the speed. However, the dimensions of speed squared are $\frac{L^2}{T^2}$. This has an extra power of length compared to what we want, so divide by the radius. Our final answer is $\frac{v^2}{r}$. Dimensionless factors such as 2 or $\pi$ cannot be determined using dimensional analysis.
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