Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 5 - Section 5.3 - Circular Motion - Example - Page 80: 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}}$
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.