Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 5 - Circular Motion; Gravitation - Search and Learn - Page 137: 3

Answer

$v_{max}=\sqrt{\frac{gr(v_0^2-\mu_s gr)}{gr+\mu_s v_0^2}}$ $v_{min}=\sqrt{\frac{gr(v_0^2+\mu_s gr)}{gr-\mu_s v_0^2}}$

Work Step by Step

When no friction is required, $F_N\sin(\theta)=m\frac{v_0^2}{r}$ $v_0^2=\frac{F_N\sin(\theta)r}{m}$ Also, $F_N\cos(\theta)=mg$ When there is friction, for maximum velocity, the force of friction will act in the direction down the slope. The forces acting on the car in the vertical direction are $F_N\cos(\theta)$, vertical component of the normal force, and $F_G$, the force of gravity. $\sum F_y=F_N\cos(\theta)-mg-\mu F_N\sin(\theta)=0$ $F_N\cos(\theta)-\mu F_N\sin(\theta)=mg$ $mg-\mu m\frac{v_0^2}{r}=mg$ In the horizontal direction, the forces acting on the car are $F_f\cos(\theta)=\mu_s F_N\cos(\theta)$ and $F_N\sin(\theta)$ $\sum F_x=F_N\sin(\theta)-\mu_s F_N\cos(\theta)=\frac{mv_{max}^2}{r}$ $v_{max}^2=\frac{F_N\sin(\theta)r}{m}-\frac{\mu_s F_N\cos(\theta)r}{m}=v_0^2-\mu gr$ $\frac{gr}{v_{max}^2}=\frac{F_N\cos(\theta)-\mu F_N\sin(\theta)}{F_N\sin(\theta)-\mu_s F_N\cos(\theta)}=\frac{mg+\mu_s m\frac{v_0^2}{r}}{m\frac{v_0^2}{r}-\mu_s mg}$ $\frac{gr}{v_{max}^2}=\frac{gr+\mu_s v_0^2}{v_0^2-\mu_s gr}$ $v_{max}=\sqrt{\frac{gr(v_0^2-\mu_s gr)}{gr+\mu_s v_0^2}}$ For minimum velocity, the force of friction will act in the direction up the slope. Therefore, $v_{min}=\sqrt{\frac{gr(v_0^2+\mu_s gr)}{gr-\mu_s v_0^2}}$
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.