Chapter 5, Problem 93GP

To determine

To find: The formula for maximum and minimum velocity.

Answer to Problem 93GP

  $V_{\max }=\sqrt{Rg\left(\frac{\mu Rg+V_0^2}{Rg-\mu V_0^2}\right)}$

  $V_{\min }=\sqrt{Rg\left(\frac{V_0^2-\mu Rg}{Rg+\mu V_0^2}\right)}$

Explanation of Solution

Given Data:

  $\tan \theta =\frac{V_0^2}{rg}$

Formula Used:

  $F_C=\frac{m{V}^2}{r}$

and $F=ma$

Calculation:

  

  $N\cos \theta =F\sin \theta +mg$

  $\Rightarrow N\cos \theta =\mu N\sin \theta +mg$

  $\Rightarrow N\left(\cos \theta -\mu \sin \theta \right)=mg$

  $N=\frac{mg}{\cos \theta -\mu \sin \theta }$

again

  $N\sin \theta +F\cos \theta =\frac{m{v}^2}{R}$

i.e., $N\sin \theta +\mu N\cos \theta =\frac{m{v}^2}{R}$

  $N\left(\sin \theta +\mu \cos \theta \right)=\frac{m{v}^2}{R}$

  $\frac{mg}{\cos \theta -\mu \sin \theta }\left(\sin \theta +\mu \cos \theta \right)=\frac{m{v}^2}{R}$

  $V=\sqrt{Rg\left(\frac{\mu +\tan \theta }{1-\mu \tan \theta }\right)}$

Substituting the values

  $V_{\max }=\sqrt{Rg\left(\frac{\mu Rg+V_0^2}{Rg-\mu V_0^3}\right)}$

Similarly for minimum velocity

  

  $N\cos \theta +F\sin \theta =mg$

Or $N\cos \theta +\mu N\sin \theta =mg$

  $\Rightarrow N=\frac{mg}{\cos \theta +\mu \sin \theta }$

Again

  $N\sin \theta -F\cos \theta =\frac{m{V}_{\min }^2}{R}$

  $N\left(\sin \theta -\mu \cos \theta \right)=\frac{m{V}_{\min }^2}{R}$

  $\frac{mg}{\cos \theta +\mu \sin \theta }\left(\sin \theta -\mu \cos \theta \right)=\frac{m{V}_{\min }^2}{R}$

Or $V_{\min }=\sqrt{Rg\left(\frac{\sin \theta -\mu \cos \theta }{\cos \theta +\mu \sin \theta }\right)}$

i.e., $V_{\min }=\sqrt{Rg\left(\frac{\tan \theta -\mu }{1+\mu \tan \theta }\right)}$

Substituting $\tan \theta =\frac{V_0^2}{Rg}$

  $V_{\min }=\sqrt{Rg\left(\frac{V_0^2/Rg-\mu }{1+\mu V_0^2/Rg}\right)}$

  $\Rightarrow V_{\min }=\sqrt{Rg\left(\frac{V_0^2-\mu Rg}{Rg+\mu V_0^2}\right)}$