Question

A car goes on a banked track with angle of banking 37º. It comes across a turn of radius 100m. If coefficient of
friction between the tyres and the surface is 1/√3, then the maximum velocity with which the car can take the turn,
assuming g= 10 m/s2, is -
please answer fast and correct​

Answer 1

Given:

The angle of banking  i.e. θ = 37º

radius r = 100m

coefficient of friction = $\mu_s$ = 1/√3 = 0.5773

g = 10 m/$s^2$

To find:

v =?

Explanation:

The velocity of a car at an inclined angle is given by the formula

Vmax = $\displaystyle \sqrt{\frac{rg(sin\theta + \mu_scos\theta)}{(cos\theta-\mu_s sin\theta)} }$

where

r= radius of curvature of the curve

g = acceleration due to gravity

$\mu_s$ =  coefficient of friction between the tires and the surface

Put given values in the above equation, we get

         = $\displaystyle \sqrt{\frac{100 \times 10 (sin37 + 0.5773cos37) }{cos37 - 0.5773\ sin37 } }$

         =  $\displaystyle \sqrt{ \frac{1059.67}{0.45123}}$

Vmax = 48.46 m/s