## Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson

# 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: $\frac{F_nsin\theta\times r}{v^2} = \frac {F_ncos\theta}{g} \\ gF_nsin\theta\times r=v^2F_ncos\theta \\ tan\theta =\frac{v^2}{gr}\\ \theta = \frac{25^2}{9.81 \times 350} \approx 10^{\circ}$

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