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

To calculate the coefficient of friction \( \mu \) for a car on a banked road, we analyze the forces at play and the given conditions.

The forces include:

• The centripetal force, \(F_c = \frac{mv_0^2}{r}\), which is necessary for circular motion of radius \( r \).
• Gravity, acting downwards.
• The normal force, perpendicular to the road's surface.
• Friction, which can act up or down the incline depending on the car's speed relative to the design speed.

We consider the scenario where the car is on the verge of slipping, meaning friction is at its maximum. We then resolve the forces parallel and perpendicular to the inclined plane.

Perpendicular to the incline, the forces balance as:

\(N \cos \theta = mg + \frac{mv_0^2}{r} \sin \theta\)

Parallel to the incline, with friction opposing slip:

\(N \sin \theta - \frac{mv_0^2}{r}\cos \theta = \mu N \cos \theta\)

Solving these equations for \( \mu \) involves first expressing \( N \) from the perpendicular balance:

\(N = \frac{mg + \frac{mv_0^2}{r} \sin \theta}{\cos \theta}\)

Substituting this expression for \( N \) into the parallel balance equation yields:

\(\left( \frac{mg + \frac{mv_0^2}{r} \sin \theta}{\cos \theta} \right) \sin \theta - \frac{mv_0^2}{r}\cos \theta = \mu \left( \frac{mg + \frac{mv_0^2}{r} \sin \theta}{\cos \theta} \right) \cos \theta\)

Simplification and algebraic manipulation lead to the expression for the coefficient of friction:

\(\mu = \frac{v_0^2 - rg \tan \theta}{rg + v_0^2 \tan \theta}\)

This result is consistent with the correct formula:

\( \mu = \frac{v_0^2 - rg \tan \theta}{rg + v_0^2 \tan \theta} \)