Question

A car of mass 800 kg moves on a circular track of radius 40 m. If the coefficient of friction is 0.5, then maximum velocity with which the car can move is:
A. 7 m/s
B. 14 m/s
C. 8 m/s
D. 12 m/s

Answer 1

Hint: If a car goes round in a curved path or a circular track then it requires some centripetal force which is directed towards the centre of the circular path. While a car rounds the curve the wheels of the car have a tendency to leave the curved path and regain the straight-line path. This force of friction between the wheels and the path opposes this tendency of the wheels. Thus, this force of friction moves towards the centre of the circular path and provides the centripetal force.

Formula used:
In a circular motion, a frictional force acts towards the centre of the circular track and provides the necessary centripetal force as,
\[\dfrac{{m{v^2}}}{r} \le \mu mg\]
Where m is the mass of a body, v is the speed, r is the radius of the circular track, \[\mu \] is the coefficient of friction and G is the acceleration due to gravity.

Complete step by step solution:
Given mass of a car=800 kg
The radius of a car=40 m
Coefficient of friction, \[\mu \] = 0.5
As we know that
\[\dfrac{{m{v^2}}}{r} = \mu mg\]
\[{v^2} = \mu rg\]
\[\Rightarrow v = \sqrt {\mu rg} \]
By substituting the values, we get
\[v = \sqrt {0.5 \times 40 \times 9.8} \]
\[\therefore v = 14\,{\rm{ m/s}}\]
Therefore, the maximum velocity with which the car can move is 14 m/s.

Hence option B is the correct answer.

Note: If an object is moving in a circular path, then it is constantly changing its direction. In this instance, the object is moving tangent to the circle and the direction of the velocity vector is the same as the direction of the motion of an object.