Question

A car is moving along a circular path having coefficient of friction 0.5 and radius of curvature 16.2 m. Then the maximum velocity of the car that can travel without falling outwards is (Acceleration due to gravity = 10 ms\(^{-2} \))

• A. \( 18 \) ms\(^{-1} \)
• B. \( 32.4 \) kmh\(^{-1} \)
• C. \( 18 \) kmh\(^{-1} \)
• D. \( 32.4 \) ms\(^{-1} \)

Hint:

The maximum velocity without skidding occurs when the centripetal force is equal to the maximum static frictional force. Use the formula \( v_{max} = \sqrt{\mu gr} \), where \( \mu \) is the coefficient of friction, \( g \) is the acceleration due to gravity, and \( r \) is the radius of curvature. Ensure consistent units when performing the calculation and conversion if required.

Answer 1

The Correct Option is B

For a car moving along a circular path without skidding outwards, the centripetal force required is provided by the frictional force between the tires and the road.
The maximum frictional force is given by \( f_{max} = \mu N \), where \( \mu \) is the coefficient of friction and \( N \) is the normal force.
On a level road, the normal force \( N \) is equal to the gravitational force \( mg \), where \( m \) is the mass of the car and \( g \) is the acceleration due to gravity.
So, \( f_{max} = \mu mg \).
The centripetal force required for circular motion with velocity \( v \) and radius \( r \) is \( F_c = \frac{mv^2}{r} \).
For the car not to skid, the centripetal force must be less than or equal to the maximum frictional force: \( \frac{mv^2}{r} \le \mu mg \) \( v^2 \le \mu gr \) The maximum velocity \( v_{max} \) is given by: \( v_{max} = \sqrt{\mu gr} \) Given: Coefficient of friction \( \mu = 0.
5 \) Radius of curvature \( r = 16.
2 \) m Acceleration due to gravity \( g = 10 \) ms\(^{-2} \) Substitute the values: \( v_{max} = \sqrt{0.
5 \times 10 \times 16.
2} \) \( v_{max} = \sqrt{5 \times 16.
2} \) \( v_{max} = \sqrt{81} \) \( v_{max} = 9 \) ms\(^{-1} \) Now, convert the velocity from ms\(^{-1} \) to kmh\(^{-1} \): \( v_{max} = 9 \frac{\text{m}}{\text{s}} \times \frac{3600 \text{ s}}{1 \text{ hour}} \times \frac{1 \text{ km}}{1000 \text{ m}} \) \( v_{max} = 9 \times 3.
6 \) kmh\(^{-1} \) \( v_{max} = 32.
4 \) kmh\(^{-1} \)