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

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 equation: \[ N = \frac{mg}{\cos \theta + \mu \sin \theta} \] 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 \( \mu \): \[ \mu = \frac{v_0^2 - rg \tan \theta}{rg + v_0^2 \tan \theta} \]

Final Answer:

The correct expression for the coefficient of friction is \( \boxed{\frac{v_0^2 - rg \tan \theta}{rg + v_0^2 \tan \theta}} \), which corresponds to option (3).