​ A car moves on a horizontal circular road of radius R. The speed of the car is increasing at a rate dv/dt = a . ​

Question

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

Answer 1

The net acceleration of the car is the vector sum of the centripetal acceleration and the tangential acceleration. By Newton’s second law the friction force on the car is (mass)×(net acceleration).

Here, at any time t, the speed of the car becomes V; therefore, the net acceleration in the plane of the road is 

This acceleration is provided by the frictional force. At the moment, the car will slide if it reaches the speed as given by M \(\sqrt{\left(\frac{V}{R}\right)^2 + (a^2)}\) = \(\mu Mg \Rightarrow\) v = [R²(\(\mu\)g² - a²)]^1/4