Question

A large parking place has uniform slope of angle $\theta$ with the horizontal. A driver wishes to drive his car in a circle of radius $R$, at constant speed. Coefficient of static friction between the tyres and the ground is $\mu$. What greatest speed can the driver achieve without slipping? Assume entire load of the car on the front wheels.

• A. $\sqrt{gRtan\theta}$
• B. $\sqrt{gRcot\theta}$
• C. $\sqrt{gR(sin\theta + \mu cos\theta)}$
• D. $\sqrt{gR(\mu cos\theta - sin\theta)}$

Answer 1

Answer: (C)

$\sqrt{gR(\sin\theta + \mu \cos\theta)}$

Answer 2

To determine the greatest speed the driver can achieve without slipping, we need to analyze the forces acting on the car. The car is moving in a circle of radius R on the inclined plane of angle θ.

1. Forces Acting on the Car: Let m be the mass of the car and v be its speed.

   • Gravitational Force (mg): Acts vertically downwards.
   • Normal Force (N): Acts perpendicular to the inclined plane.
   • Static Friction Force (f_s): For the greatest speed, the car tends to slip outwards and upwards along the incline. Therefore, the static friction force f_s acts inwards and downwards along the incline, opposing this tendency to slip.

2. Coordinate System: Choose axes parallel and perpendicular to the inclined plane.

   • x-axis: Parallel to the incline, pointing towards the center of the circle (which is along the incline).
   • y-axis: Perpendicular to the incline, upwards.

3. Resolving Forces:

   • Normal Force (N): Acts along the positive y-axis.
   • Gravitational Force (mg):
     • Component perpendicular to incline (downwards along y-axis): mg cosθ
     • Component parallel to incline (downwards along x-axis): mg sinθ
   • Static Friction Force (f_s): For maximum speed, the car tends to slip outwards and upwards on the incline. So, friction acts inwards and downwards along the incline (along the positive x-axis).
     • f_s = μN (for maximum speed).

4. Equations of Motion:

   • Perpendicular to incline (ΣFy = 0): The car is not accelerating perpendicular to the incline. N - mg cosθ = 0 N = mg cosθ

   • Parallel to incline (ΣFx = mv²/R): The net force along the incline provides the centripetal acceleration mv²/R. Both the component of gravity and friction act in the direction of the centripetal force for maximum speed. mg sinθ + f_s = mv²/R

5. Substitute f_s = μN and N = mg cosθ: mg sinθ + μ(mg cosθ) = mv²/R

6. Solve for v: Divide by m: g sinθ + μg cosθ = v²/R g(sinθ + μ cosθ) = v²/R v² = gR(sinθ + μ cosθ) v = \sqrt{gR(\sin\theta + \mu \cos\theta)}

Thus, the greatest speed is: v = \sqrt{gR(\sin\theta + \mu \cos\theta)}