A car moves on a horizontal circular road of radius R, the speed increases at a rate dv/dt = a.

Question

A car moves on a horizontal circular road of radius R, the speed increases at a rate \(\frac {dv}{dt} = a\). The frictional coefficient between the road and the tire is µ . Now, find the speed at which the car will skid.

Answer 1

The net acceleration of the car is the vector sum of centripetal acceleration and tangential acceleration. By Newton’s second law the net force of friction acting on the car is equal to mass multiplied by net acceleration.

Here, at any time t, the speed of the car becomes V,

the net acceleration in the plane road is \(\sqrt{\left(\frac {v^2}R\right)^2 + a^2}\).

This acceleration is provided by the frictional force. At the moment car will slide