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 the 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: We are asked to find the maximum speed with which the car can move in the circular path. The force that provides circular motion is the centripetal force. In this case, this centripetal force required for moving in a circular path is provided by the friction. The centripetal petal force is given by the formula,
${F_c} = \dfrac{{m{v^2}}}{R}$
Where m is the mass, v is the velocity and R is the radius of the circular path.
By equating this centripetal force with friction and solving for the velocity we will reach the final answer.

Complete step by step answer:
It is given that the car has a mass m.
The radius of the circular path in which it is moving is given as R.
The static friction between the road and the tyre is given as ${\mu _s}$.
 We are asked to find the maximum speed with which the car can move in the circular path.
We know that centripetal petal force is the force that provides circular motion. It always acts towards the center of the circular path. In our case, this centripetal force required for moving in a circular path is provided by the friction. The centripetal petal force is given by the formula,
${F_c} = \dfrac{{m{v^2}}}{R}$
Where m is the mass, v is the velocity and R is the radius of the circular path.
So, we can write force of friction, ${F_r}$ is equal to the centripetal force ${F_c}$
That is, ${F_r} = {F_c}$
Now let us substitute the value of centripetal force.
Thus,
${F_r} = \dfrac{{m{v^2}}}{R}$ …………………..(1)
We need to find the maximum value of v.
The value of velocity will be maximum when friction will be maximum. We know that the force of friction is given by the formula,
${F_r} = \mu {\rm N}$ where N is the normal reaction force.
When the car is moving on the road the weight of the car is balanced by the normal reaction force. So we can equate the normal force $N = mg$.
Therefore,
${F_r} = \mu mg$
This is the maximum frictional force.
Thus, we can write equation 1 as,
$\Rightarrow \mu mg = \dfrac{{m{v^2}}}{R}$
On solving this for v, we get
$\Rightarrow v = \sqrt {R\mu g} $
In the question it is given static friction is represented as ${\mu _s}$.
$\therefore v = \sqrt {R{\mu _s}g} $

Hence the correct answer is option C.

Note:
The maximum value of friction that we took is known as the limiting friction. The static friction is the friction when a body is placed on the surface of another. For a body to start moving it should overcome the maximum value of this static friction which is called the limiting friction.