A circular track of radius 600 m is tio be designed for cars at an average speed of 180 km/hr. What should be the angle of banking of the track?

Explanation

Let the angle of banking be `theta`. The forces on the car are ure.
a. weight of the car Mg downward and
b. normal force N.

For proper bnking sttic frictioN/Al force is not needed.
For vertical direction the acceleration is zero. So,
`N costheta=Mg` ..........i
for horizontal direection the acceleration is `v^2/r` towards the centre, so that `
`N sintheta=Mv^2/r`..........ii
From i and ii
`tantheta=v^2/rg`
Putting the values `tantheta=((180 km/hr)^2)/((600 m)(10 m/s^2))=0.4167`
or, `theta=22.6^0`