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 μ. Find the speed at which the car will skid.

Answer 1

Let v be the speed of the car.
Since the motion is non-uniform, the acceleration has both radial (a_r) and tangential (a_t) components.
$$\begin{gathered} a_r=\frac{v^2}{R} \\ {a}_t=\frac{dv}{dt}=a \\ \mathrm{Resultant}\mathrm{magnitude}=\sqrt{{\left(\frac{v^2}{R}\right)}^2+a^2} \end{gathered}$$



From free body diagram, we have:
$$\begin{gathered} mN=m\sqrt{{\left(\frac{v^2}{R}\right)}^2+a^2} \\ \Rightarrow \mu mg=m\sqrt{{\left(\frac{v^2}{R}\right)}^2+a^2} \\ \Rightarrow {\mu }^2{g}^2=\frac{v^4}{R^2}+a^2 \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{v^4}{R^2}=({\mathrm{\mu }}^2{g}^2-a^2) \\ \Rightarrow v^4=({\mathrm{\mu }}^2{g}^2-a^2)R^2 \\ \Rightarrow v=\left[\right({\mathrm{\mu }}^2{\mathrm{g}}^2-{\mathrm{a}}^2)R^2{]}^{1/4} \end{gathered}$$