Question

A turn of radius $20$ m is banked for the vehicles going at a speed of $36km/h$. 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? $(g=9.8m/s^2)$

• A. $2.2m{s}^{-1}\le v\le 4.2m{s}^{-1}$
• B. $4.2m{s}^{-1}\le v\le 15m{s}^{-1}$
• C. $15m{s}^{-1}\le v\le 18.5m{s}^{-1}$
• D. $15m{s}^{-1}\le v\le 22.5m{s}^{-1}$

Answer 1

The correct option is B $4.2m{s}^{-1}\le v\le 15m{s}^{-1}$

Angle of banking is given by:

$\tan \theta =\frac{v^2}{rg}$

Given: $r=20m,v=36km/h=10m/s$

Hence, $\tan \theta =\frac{{10}^2}{20\times 9.8}=\frac{5}{9.8}$

$\sin \theta =\frac{5}{11},\cos \theta =\frac{9.8}{11}$

The maximum speed for a vehicle not to skid up is given by:

$v_{max}\le \sqrt{\frac{gr(\sin \theta +{\mu }_s\cos \theta )}{\cos \theta -{\mu }_s\sin \theta }}$

$v_{max}\le \sqrt{\frac{9.8\times 0.4(\frac{5}{11}+0.4\times \frac{9.8}{11})}{\frac{9.8}{11}-0.4\times \frac{5}{11}}}=15m/s$

The minimum speed for a vehicle not to slip down is given by:

$v_{min}\ge\sqrt{\dfrac{gr(\sin\theta-\mu_{s}\cos\theta)}{\cos\theta

+\mu_{s}\sin\theta}}$

$v_{min}\ge \sqrt{\frac{9.8\times 0.4(\frac{5}{11}-0.4\times \frac{9.8}{11})}{\frac{9.8}{11}+0.4\times \frac{5}{11}}}=4.2m/s$