Question

A circular race course track has a radius of 500$$$$m and is banked to${10^ \circ }$ . If the coefficient of friction between tyres of vehicles and the road surface is $0.25$. Compute:
(A) The maximum speed to avoid slipping.
(B) The optimum speed to avoid wear and tear.

Answer 1

A vehicle requires a centripetal force while taking a turn on the road; this force is produced by sloping down the roads inwards at the turn. This force on the banked road is produced by the horizontal component of the normal reaction force.

Complete step by step answer:
At the optimum speed the normal reaction is enough to provide the necessary centripetal force to avoid slipping. But if we move at speed higher than the permissible speed, the vehicle tends to slip outwards. To avoid such slipping, the frictional force acts inwards and provides the additional centripetal force. Now this gives us a maximum speed to avoid slipping.
The formula for the maximum speed is given by,
${V_{\max }} = \sqrt {Rg\tan \left( {\theta + \varphi } \right)}$
R = 500$$$$m(given)
g = 10$$$$m/{s^2} (acceleration due to gravity)
$\theta = {10^ \circ }$ (given)
$\varphi = {\tan ^{ - 1}}\mu$ ($\mu$ is coefficient of friction)
${\tan ^{ - 1}}\mu = {\tan ^{ - 1}}\left( {0.25} \right) = {14^ \circ }$
Putting all the values in the above equation, we get,
${V_{\max }} = \sqrt {500 \times 10 \times \tan \left( {{{24}^ \circ }} \right)}$
{V_{\max }} = 47.18$$$$m/s
Hence, the maximum speed to avoid slipping on a banked road is $47.18$ $m/s$
For the optimum speed we know that the horizontal component of normal reaction is enough to provide the required centripetal force so the formula is given by,
${V_o} = \sqrt {Rg\tan \theta }$
${V_o} = \sqrt {500 \times 10 \times \tan {{10}^ \circ }}$
{V_o} = 29.69$$$$m/s

Hence, The optimum speed to avoid wear and tear is 29.69$$$$m/s

Note: The vehicles moving on the road are heavy and while taking a turn the required centripetal force may not be provided by the friction alone which may lead to wear and tear of wheels. To avoid such things, roads are banked inwards at the turn. Thus the necessary centripetal force is provided by the horizontal component of normal reaction and its vertical component balances the weight of the vehicle.