Question

A car of mass m is taking a circular turn of radius 'r' on a frictional level road with a speed v. In order that the car does not skid
A. $\dfrac{mv^2}{r} \geq \mu mg$
B. $\dfrac{mv^2}{r} \leq \mu mg$
C. $\dfrac{mv^2}{r} = \mu mg$
D. $\dfrac{v}{r} = \mu mg$

Answer 1

Hint: Friction is a necessary requirement for the car to be able to describe a circle. More the friction more can be the velocity of the car in the circular path and it will do so without slipping (or skidding).

Complete answer:
Let us first get a complete comprehension of the forces involved:
1. Weight of the car mg acting downwards.
2. Normal force outward from the surface of the road (due to the road) which is equal to the weight simply when the road is level.
3. Centripetal force $mv^2/r$ acting towards the centre of the circle.
4. Frictional force that also points towards the centre.

When an object like our car describes circular motion, centripetal force pointing towards the centre of the circle acts on the car. A pseudo force called centrifugal force tends to act outward from the circle. Centrifugal force is the reason why the person sitting in a car experiences a shift away from the centre of the circle. The car has a tendency to move along this centrifuge and to start skidding.
The frictional force opposes this centrifugal force. More the amount of friction between the road and the car more will be its ability to oppose the centrifugal force. Centrifugal force is equal in magnitude to centripetal force which means it is just $mv^2/r$. And we know that frictional force is given by:
$F = \mu N$;
in our case we get the magnitude of the frictional force as:
$F = \mu mg$.
Therefore in order to avoid skid, this frictional force has to be more than the centrifugal force.
So,
$\dfrac{mv^2}{r} \leq \mu mg$

Therefore the correct answer is option (B).

Note:
One might think that the option with the equal sign must be correct. But the equal sign will only give us a condition when there will be a perfect balance. We are asked the condition when the car does not skid so less than or equal sign option is the one we should prefer.