Question

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

Answer 1

Since the motion is non-uniform, the acceleration has both radial and tangential components.

$a=\frac{v^2}{r}$

$a_t=\frac{dv}{dt}=a$

Resultant magnitude $=\sqrt{{\left(\frac{v^2}{r}\right)}^2+a^2}$

Now, $mN=m\sqrt{{\left(\frac{v^2}{r}\right)}^2+a^2}$

$\Rightarrow \mu mg=m\sqrt{{\left(\frac{v^2}{r}\right)}^2+a^2}$

$\Rightarrow {\mu }^2{g}^2=\frac{v^4}{r^2}+a^2$

$\Rightarrow \frac{v^4}{r^2}=({\mu }^2{g}^2-a^2)$

$\Rightarrow v^4=({\mu }^2{g}^2-a^2)r^2$

$\Rightarrow v=\left[\right({\mu }^2{g}^2-a^2\left)r^2\right]1/4$