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 determine the coefficient of friction \( \mu \) for a car moving on a banked road, we need to analyze the forces acting on the car and use the given conditions.

The forces involved are: 

• Centripetal force required to keep the car moving in a circle of radius \( r \).
• The gravitational force acting downward.
• The normal force perpendicular to the surface of the road.
• The frictional force, which can act either up or down the slope depending on the speed of the car relative to the design speed.

Let's consider the car at the verge of slipping, where friction is maximized. The centripetal force required is given by:

\(F_c = \frac{mv_0^2}{r}\)

Let's resolve the forces parallel and perpendicular to the inclined plane.

Balance of forces perpendicular to the incline gives:

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

Balance of forces parallel to the incline (where friction opposes slip):

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

Solving these equations will yield the expression for the coefficient of friction \( \mu \).

First, express \( N \) from the perpendicular balance:

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

Substitute \( N \) in the parallel balance equation:

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

Simplifying and solving for \( \mu \), you arrive at:

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

This matches the correct answer:

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

Answer 2

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).