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 - Page 77: 5.6


10 degrees

Work Step by Step

We find the x and y components of the forces: x: $\frac{mv^2}{r} = F_nsin\theta$ y: $mg = F_ncos\theta$ We set each equation equal to m, and then we set the equations equal to each other: $\frac{F_nsin\theta\times r}{v^2} = \frac {F_ncos\theta}{g} \\ gF_nsin\theta\times r=v^2F_ncos\theta \\ tan\theta =\frac{v^2}{gr}\\ \theta = \frac{25^2}{9.81 \times 350} \approx 10^{\circ}$
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.