Solve the following :

Question:

A turn of radius $20 \mathrm{~m}$ is banked for the vehicles going at a speed of $36 \mathrm{~km} / \mathrm{h}$. 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?

Solution:

For banking angle,

$\tan \theta=^{\frac{V^{2}}{R g}}$

$=\frac{\left(36 \times \frac{5}{48}\right)^{2}}{(20)(10)}$

$\tan \theta=0.5$

$V \max =\sqrt{\operatorname{Rg} \frac{(\mu+\tan \theta)}{(1-\mu \tan \theta)}}$

$=\sqrt{\frac{20 \times 10 \times(0.4+0.5)}{(1-0.4 \times 5)}}$

$=15 \mathrm{~m} / \mathrm{s}$

$15 \times \frac{18}{5} \mathrm{kmph}$

$=54 \mathrm{kmph}$

$V_{\min }=\sqrt{\operatorname{Rg} \frac{(\tan \theta-\mu)}{(1+\mu \tan \theta)}}$

$=\sqrt{\frac{20 \times 10 \times(0.5-0.4)}{(1+0.4 \times 0.5)}}$

$=4 \mathrm{~m} / \mathrm{s}$

$\mathrm{V}_{\mathrm{min}}=14.7 \mathrm{kmph}$

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