Question

A car is moving on a circular track of radius $R$. The road is banked at an angle $\theta$. ${\mu }_s$ is the coefficient of static friction between the car and the track. Find the maximum speed with which the car can move safely on the track.

• A. ${\left[\frac{rg(\sin \theta +{\mu }_s\cos \theta )}{\cos \theta +{\mu }_s\sin \theta }\right]}^{\frac{1}{2}}$
• B. ${\left[\frac{rg(\cos \theta +{\mu }_s\sin \theta )}{\cos \theta -{\mu }_s\sin \theta }\right]}^{\frac{1}{2}}$
• C. ${\left[\frac{rg(\sin \theta +{\mu }_s\cos \theta )}{\cos \theta -{\mu }_s\sin \theta }\right]}^{\frac{1}{2}}$
• D. None

Answer 1

The correct option is C ${\left[\frac{rg(\sin \theta +{\mu }_s\cos \theta )}{\cos \theta -{\mu }_s\sin \theta }\right]}^{\frac{1}{2}}$

In this case, frictional force is also used to support the requirement of centripetal force along with that provided by normal reaction.
For maximum safe speed, friction will also act at its limiting value i.e $f_{max}$


Applying the equation of dynamics along horizontal and vertical direction:
$N\sin \theta +f_{max}cos\theta =\frac{m{v}_{max}^2}{r}$
The maximum value of friction:
$f_{max}={\mu }_{s}N$
$\therefore N\sin \theta +{\mu }_{s}Ncos\theta =\frac{m{v}_{max}^2}{r}..\left(i\right)$

In vertical direction, the system is in a state of equilibrium:
$N\cos \theta =f_{max}\sin \theta +mg$
$\therefore N\cos \theta -{\mu }_{s}N\sin \theta =mg...\left(ii\right)$

Dividing Eq.$\left(i\right)$ by Eq.$\left(ii\right)$, we get:
$\frac{\sin \theta +{\mu }_s\cos \theta }{\cos \theta -{\mu }_s\sin \theta }=\frac{v_{max}^2}{rg}$

$\therefore v_{max}=\sqrt{\frac{rg(\sin \theta +{\mu }_s\cos \theta )}{(\cos \theta -{\mu }_s\sin \theta )}}$