Question

A car of mass \( m \) moves on a banked road having radius \( r \) and banking angle \( \theta \). To avoid slipping from the banked road, the maximum permissible speed of the car is \( v_0 \). The coefficient of friction \( \mu \) between the wheels of the car and the banked road is:

• A. \( \mu = \frac{v_0^2 + rg \tan \theta}{rg - v_0^2 \tan \theta} \)
• B. \( \mu = \frac{v_0^2 + rg \tan \theta}{rg + v_0^2 \tan \theta} \)
• C. \( \mu = \frac{v_0^2 - rg \tan \theta}{rg + v_0^2 \tan \theta} \)
• D. \( \mu = \frac{v_0^2 - rg \tan \theta}{rg - v_0^2 \tan \theta} \)

Hint:

In banked road problems, the frictional force helps to balance the centripetal force required to keep the car on the curved path. The coefficient of friction can be derived from the force balance equations.

Answer 1

The Correct Option is C

In this problem, we analyze the forces acting on a car moving on a banked road. The forces include the gravitational force \( mg \), the normal force \( N \), and the frictional force \( f = \mu N \). 
Step 1: Decompose the forces acting on the car. 
- The vertical force balance gives: \[ N \cos \theta + \mu N \sin \theta = mg \] 
- The horizontal force balance gives: \[ N \sin \theta - \mu N \cos \theta = \frac{mv_0^2}{r} \] 
Step 2: Solve the system of equations for \( \mu \). From the vertical force balance: \[ N = \frac{mg}{\cos \theta + \mu \sin \theta} \] Substitute this into the horizontal force balance equation: \[ \frac{mg}{\cos \theta + \mu \sin \theta} \sin \theta - \mu \frac{mg}{\cos \theta + \mu \sin \theta} \cos \theta = \frac{mv_0^2}{r} \] Simplifying the equation, we get the final expression for \( \mu \): \[ \mu = \frac{v_0^2 - rg \tan \theta}{rg + v_0^2 \tan \theta} \] Thus, the correct expression for the coefficient of friction is option (3).