Question

A car of mass m is moving on a level circular track of radius, R. If \[{\mu _s}\]represents the static friction between the road and tyres of the car, the maximum speed of the car in circular motion is given by:
A. \[\sqrt {Rg/{\mu _S}} \]
B. \[\sqrt {mRg/{\mu _S}} \]
C. \[\sqrt {{\mu _S}Rg} \]
D. \[\sqrt {{\mu _S}mRg} \]

Answer 1

Hint:Frictional force is a special type of force which is usually against the direction of motion. In the above problem, the frictional force is tangentially inwards directed towards the centre of the circular path given by the tangent at that point on the curve while the direction of the centripetal force is tangentially outwards from the curvature of the circular path. For a car to stay on the track, the Frictional force should be equal to the Centripetal force.

Formula used:
Frictional force, \[f = {\mu _S}mg\]
where \[{\mu _s}\]= coefficient of static friction, m = mass of the object and g = acceleration due to gravity = 9.8 \[m{s^{ - 2}}\]
Centripetal force, \[F = m{v^2}/R\]
where m = mass of the object, v = velocity of the object and R = radius of the track.

Complete step by step solution:
Given: mass of the car = m, radius of level circular track = R, \[{\mu _s}\]= coefficient of static friction between road and tyres of the car.
Let v be the velocity of the car. The condition for the car wheels to remain in contact with the track is:
\[\text{Centripetal force} \le \text{frictional force}\]
So, \[m{v^2}/R \le {\mu _S}mg\]- (1)
And \[{v^2}/R \le {\mu _S}g\]
\[v \le \sqrt{{\mu _S}Rg}\]
So, the maximum velocity is,
\[{v_{\max }} = \sqrt{{\mu _S}Rg}\]

Hence option (C) is the correct answer.

Note: If the centripetal force is greater than the force of friction, then the car will not remain on the circular track and sway outwards. If the frictional force is considerably greater than the centripetal force, the car will not move at all along the circular path. In order that the motion occurs as desired along the path, the two opposing forces should balance out each other.