Question

A turn of radius 20 m is banked for a vehicle going at a speed of 36 Km/h. If the coefficient of static friction between road and tyre is 0.4, the maximum speed of a vehicale so that it does not skid up is 18 x Km/h then x is then

Answer 1

Given that,

Radius $r=20m$

Speed $v=36km/h$

Frictional coefficient $\mu =0.4$

Now, the road is banked with an angle

$\theta ={\tan }^{-1}\left(\frac{v^2}{rg}\right)$

$\theta ={\tan }^{-1}\left(\frac{100}{20\times 10}\right)$

$\theta ={\tan }^{-1}\left(\frac{1}{2}\right)$

$\tan \theta =0.5$

When the car travels at maximum speed so that it slips upward, $\mu R_1$ acts downward

So,

$R_1-mg\cos \theta -\frac{m{v}_1^2}{r}\sin \theta =\mathrm{0.....}\left(I\right)$

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

Now, after solving

$v_1^2=\sqrt{rg\frac{\tan \theta -\mu }{1+\mu \tan \theta }}$

$v_1^2=\sqrt{20\times 10\left(\frac{0.5-0.4}{1+0.4\times 0.5}\right)}$

$v_1^2=\sqrt{16.66}$

$v_1=4.08m/s$

$v_1=4.08\times \frac{18}{5}km/h$

$v_1=0.816\times 18$

$v_1=14.69km/h$

Hence, the value of $x$ is $0.816$