Question

A car goes on a horizontal circular road of radius R, the speed increasing at a constant rate dv/dt=a. The friction coefficient between the road and the tyre is µ. Find the speed at which the car will skid.

Answer 1

If the speed of the body is increasing in case of the circular motion. This means that the magnitude of the Velocity is changing at the Constant rate.

This also tell that there is a tangential acceleration which is produced due to the change in magnitude of the velocity.

∴  tangential acceleration = dv/dt = a

Now, Centripetal Acceleration = v²/r

∴ Net Acceleration = $\sqrt{  (\frac{v^{2} }{r})^2 + a^2}$

∴ m × accele.(net) =  $m\sqrt{  (\frac{v^{2} }{r})^2 + a^2}$

⇒ There will be Frictional Force which will balances the Force produced due to the tangential acceleration.

∴ μmg = $m\sqrt{  (\frac{v^{2} }{r})^2 + a^2}$

⇒  μ²g² = v⁴/r² + a²

⇒ v⁴/r² =  μ²g² - a²

∴ v = $[r^2(u^2g^2 - a^2)]^{\frac{1}{4} }$

Hope it helps.