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,

$v=36km/hr$

$=10m/s$

$r=20m,\mu =0.4$

The road is banked with an angle.

$\theta =ta{n}^{-1}\frac{v^2}{rg}$

$=ta{n}^{-1}\frac{100}{20\times 10}$

$=ta{n}^{-1}\left(\frac{1}{2}\right)$

$ortan\theta =0.5$

When the car travels at maximum speed, so that it slips upward $\mu R_1$ acts downward. As shown in Fig. 1,

$R_1-mgcos\theta -\frac{m{v}_1^2}{r}sin\theta =0...\left(i\right)$

$and\mu R_1+vmgsin\theta -\frac{m{v}_1^2}{r}cos\theta =\mathrm{0....}\left(ii\right)$

Solving equations we get,

$v_1=\sqrt{rg\frac{\mu +tan\theta }{1-\mu tan\theta }}$

$=\sqrt{20\times 10\times \frac{0.9}{0.8}}$

$=15m/s=54km/hr$

Similarly it can be proved that,

$v_2=\sqrt{rg\frac{tan\theta -\mu }{\sqrt{1-\mu tan\theta }}}$

$=\sqrt{20\times 10\times \frac{0.1}{1.2}}$

$=4.08m/s=14.7km/hr$