What should be the angle of banking of a circular track of radius $600 m$ which is designed for cars at an average speed of $180 k m / h$ ?

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Solution

Let the angle of banking be $θ$ . The forces on the car are as shown in the figure.
(a) weight of the car Mg downward and
(b) normal force N.
For proper banking, static frictional force is not needed.
For vertical direction the acceleration is zero. So,
N cos $θ$ = Mg .......... (i)
For horizontal direction, the acceleration is $v^{2} / r$ towards the centre, so that
N sin $θ = M v^{2} / r$ .......... (ii)
From (i) and (ii), $t a n θ = v^{2} / r g$
Putting the values, $t a n θ = \frac{180 (k m / h)^{2}}{(600 m) (10 m / s^{2})} = 0.4167 \Rightarrow θ = {22.6}^{\circ}$

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