Question

A turn of radius 20 m is banked for the vehicles going at a speed of 36 km/h. If the coefficient of static friction between the road and the tyre is 0.4, what are the possible speeds of a vehicle so that it neither slips down nor skids up?

Answer 1

Given:
Speed of vehicles = v = 36 km/hr = 10 m/s
Radius = r = 20 m
Coefficient of static friction = μ = 0.4
Let the road be banked with an angle $\theta$. We have:
$$\begin{gathered} \theta ={\tan }^{-1}\frac{v^2}{rg} \\ ={\tan }^{-1}\frac{100}{20\times 10} \\ ={\tan }^{-1}\left(\frac{1}{2}\right) \\ \Rightarrow \mathrm{tan\theta }=0.5 \end{gathered}$$



When the car travels at the maximum speed, it slips upward and μN₁ acts downward.
Therefore we have:
$$\begin{gathered} N_1-mg\cos \theta -\frac{m{v}_1^2}{r}\sin \theta =0...\left(\mathrm{i}\right) \\ \mu N_1+mg\sin \theta \mathit{-}\frac{m{v}_{\mathit{1}}^{\mathit{2}}}{r}\cos \theta =0...\left(\mathrm{ii}\right) \end{gathered}$$

On solving the above equations, we get:
$$\begin{gathered} v_1=\sqrt{rg\frac{\mu +\tan \theta }{1-\mu \tan \theta }} \\ =\sqrt{20\times 10\times \frac{0.9}{0.8}} \\ =15\mathrm{m}/\mathrm{s}=54\mathrm{km}/\mathrm{hr} \end{gathered}$$



Similarly, for the other case, it can be proved that:
$$\begin{gathered} v_2=\sqrt{rg\frac{\mathrm{tan\theta }-\mathrm{\mu }}{\sqrt{1-\mathrm{\mu }\mathrm{tan\theta }}}} \\ =\sqrt{20\times 10\times \frac{0.1}{1.2}} \\ =4.08\mathrm{m}/\mathrm{s}=14.7\mathrm{km}/\mathrm{hr} \end{gathered}$$

Thus, the possible speeds are between 14.7 km/hr and 54 km/hr so that the car neither slips down nor skids up.