Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 8 - Dynamics II: Motion in a Plane - Exercises and Problems - Page 199: 24


$v = 5.5~m/s$

Work Step by Step

We can use the equation for centripetal force to find the speed at the bottom. $\sum F = \frac{mv^2}{r}$ $T-mg = \frac{mv^2}{r}$ $v^2 = \frac{r~(T-mg)}{m}$ $v = \sqrt{\frac{r~(T-mg)}{m}}$ $v = \sqrt{\frac{(1.5~m)[15~N-(0.500~kg)(9.80~m/s^2)]}{0.500~kg}}$ $v = 5.5~m/s$
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.