Question

Find the maximum velocity for skidding for a car moved on a circular track of radius 100 m. The coefficient of friction between the road and tyre is 0.2.
A. 0.14 m/s
B. 140 m/s
C. 1.4 m/s
D. 14 m/s

Answer 1

Hint: When the car is moving on a circular track then an outward centrifugal force acts on the body which is proportional to the square of the linear velocity of the car. To keep the car on the circular track without skidding there is a balancing force inward which is the static frictional force.

Formula used:
\[{F_0} = \dfrac{{m{v^2}}}{r}\]
here \[{F_0}\] is the outward force acting on the body of mass m in a circular path of radius r moving with linear velocity v.
\[{f_s} = \mu N\]
here \[{f_s}\] is the static friction between the tyre of the car and the road with a coefficient of friction \[\mu \] and N is the normal force acting.

Complete step by step solution:

Image: Car moving without skidding

The normal force acting on the car due to the surface of the road is equal to the weight of the car,
\[N = mg\]
The outward force acting on the car in the circular track is,
\[{F_o} = \dfrac{{mv_{\max }^2}}{r}\]
Here, \[{v_{\max }}\] is the maximum velocity for the skidding of the car.
The static friction between the car and the road is,
\[{f_s} = \mu mg\]

At the equilibrium condition for the car along the line joining the center of the circular track and the car, the net force along the line must be zero, i.e. the inward static frictional force must balance the outward centrifugal force. Hence, the magnitude of these two forces must be equal.
\[{F_o} = {f_s}\]
\[\Rightarrow \dfrac{{mv_{\max }^2}}{r} = \mu mg\]
\[\Rightarrow v_{\max }^2 = \mu gr\]
\[\Rightarrow {v_{\max }} = \sqrt {\mu gr} \]

Here,
\[\mu = 0.2\]
\[\Rightarrow g = 10\,m/{s^2}\]
\[\Rightarrow r = 100\,m\]
Putting the values, we get
\[{v_{\max }} = \sqrt {0.2 \times 10 \times 100} m/s\]
\[\Rightarrow {v_{\max }} = \sqrt {200} m/s\]
\[\therefore {v_{\max }} \approx 14\,m/s\]

Therefore, the correct option is D.

Note: If we increase the speed of the car more than the maximum velocity for the skidding then the outward force on the car will be greater than the static friction acting on it, and its result in the skidding of the car outside the circular track.