Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 5 - Section 5.3 - Circular Motion - Example: 5.5

Answer

$ v= \sqrt{ \frac{gLcos^2\theta}{sin\theta}}$

Work Step by Step

We first find the two components: x: $ F_tcos\theta = \frac{mv^2}{Lcos\theta}$ y: $F_tsin\theta=mg \\ F_t = \frac{mg}{sin\theta}$ Using the y result for tension, we use the x-equation to find v: $ F_tcos\theta = \frac{mv^2}{Lcos\theta}$ $ \frac{mg}{sin\theta}cos\theta = \frac{mv^2}{Lcos\theta}$ $ v^2= \frac{gL}{sin\theta}cos^2\theta$ $ v= \sqrt{ \frac{gLcos^2\theta}{sin\theta}}$
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