Question

A turn of radius $20m$ 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?

• A. $14.7km/h$ and $54km/h$
• B. $14.2km/h$ and $50km$
• C. $11.7km/h$ and $44km/h$
• D. $17.7km/h$ and
  $34km/h$

Answer 1

The correct option is A $14.7km/h$ and $54km/h$

Given speed of the vehicle $v=36$ km/hr$=36\times \frac{5}{18}=10m/s$

turning radius $r=20m$ and static friction ${\mu }_s=0.4$

The bend angle of the road $\theta ={tan}^{-1}\left(\frac{v^2}{rg}\right)={tan}^{-1}0.5$

When the car travels at maximum speed so that it slips upwards and the component $\mu R$ acts downwards. So,$R_1-mg\cos \theta -\frac{m{v}^2}{r}\sin \theta =0$-------(A)

and $\mu R_1+mgsin\theta -\frac{m{v}^2}{r}\cos \theta =0$----------(B)

Solving the equation we get,

$v_1=\sqrt{rg\frac{\tan \theta -\mu }{1+\mu \tan \theta }}=\sqrt{20\times 10(\frac{0.1}{1.2}})=4.082$m/s=14.7 km/h

Hence its possible speed will be between 14.7 and 54 km/h