Question

A circular racecourse track has a radius of $500m$ and is banked to ${10^\circ }$ . if the coefficient of friction between tyres of the vehicle 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

Try to illustrate the situation which may help you in the account for all the components of forces contributing to the equilibrium of the vehicle. In other words, resolve the normal reaction, frictional force, and weight into their respective components and deduce which of the components act in the vertical and horizontal directions. Thus obtained an expression for velocity.
Formula Used:
Maximum speed to avoid slipping: ${v_{\max }} = \sqrt {\dfrac{{rg\left( {\tan \theta + \mu } \right)}}{{1 - \mu \tan \theta }}}$
Optimum speed to avoid wear and tear: ${v_{optimum}} = \sqrt {rg\tan \theta }$
Here, $r$ is the track radius, $g$ is the acceleration due to gravity, $\theta$ is the banking angle, and $\mu$ is the coefficient of friction.

Complete answer:

Let us reconstruct the question to derive an expression for a vehicle moving on this blanked racetrack in general.
Let that mass of the vehicle be $m$ , and let it move with a velocity $v$ . Since it's moving in a circular racetrack with radius $r$ , it experiences a centrifugal force of magnitude ${F_c} = \dfrac{{m{v^2}}}{r}$ directed outwards.
Let $\theta$ be the angle of banking and let $f$ be the frictional force acting between the road and the tyres.
Note that we divide the forces in play into their corresponding components to ease our understanding and make straightforward calculations. These are shown in the diagram.
Now, for a balanced system, the
Total upward force $=$ total downward force
$\Rightarrow N\cos \theta = mg + f\sin \theta$
$\Rightarrow mg = N\cos \theta - f\sin \theta$
In the same way, we look at the lateral forces, which for a balanced system:
$\dfrac{{m{v^2}}}{r} = N\sin \theta + f\cos \theta$
On dividing the above two equations:
$\dfrac{{\dfrac{{m{v^2}}}{r}}}{{mg}} = \dfrac{{N\sin \theta + f\cos \theta }}{{N\cos \theta - f\sin \theta }}$
$\Rightarrow \dfrac{{{v^2}}}{{rg}} = \dfrac{{N\sin \theta + f\cos \theta }}{{N\cos \theta - f\sin \theta }}$
Now, substituting frictional force $f = \mu N$ in the above equation we get:
$\dfrac{{{v^2}}}{{rg}} = \dfrac{{N\sin \theta + \mu N\cos \theta }}{{N\cos \theta - \mu N\sin \theta }}$
$\Rightarrow \dfrac{{{v^2}}}{{rg}} = \dfrac{{\sin \theta + \mu \cos \theta }}{{\cos \theta - \mu \sin \theta }}$
Divide both the numerator and the denominator by $\cos \theta$ :
$\dfrac{{{v^2}}}{{rg}} = \dfrac{{\tan \theta + \mu }}{{1 - \mu \tan \theta }}$
$\Rightarrow v = \sqrt {\dfrac{{rg\left( {\tan \theta + \mu } \right)}}{{1 - \mu \tan \theta }}}$
a) Now, we can calculate the maximum speed to avoid slipping:
This is the maximum speed that the vehicle can travel under the influence of friction. This is known by the expression we derived above:
$\left( {\tan {{10}^ \circ } = 0.1763} \right)$
${v_{\max }} = \sqrt {\dfrac{{rg\left( {\tan \theta + \mu } \right)}}{{1 - \mu \tan \theta }}}$
$\Rightarrow {v_{\max }} = \sqrt {\dfrac{{500 \times 9.8\left( {0.1763 + 0.23} \right)}}{{1 - \left( {0.25 \times 0.1763} \right)}}} = \sqrt {2185.182}$
On further solving we get,
${v_{\max }} = 46.75m{s^{ - 1}}$
b) The optimum speed to avoid wear and tear can be given by disregarding friction and taking $\mu = 0$ , in which case:
${v_{optimum}} = \sqrt {rg\tan \theta }$
$\Rightarrow {v_{optimum}} = \sqrt {500 \times 9.8 \times 0.1763} = \sqrt {863.87}$
On further solving we get,
${v_{optimum}} = 29.39m{s^{ - 1}}$
Therefore, the maximum possible speed for the vehicle to stop slipping is $46.75m{s^{ - 1}}$ and the optimum speed for the vehicle to avoid wear and tear is $29.39m{s^{ - 1}}$ .

Note:
Always remember that for the first part we considered an expression for velocity when friction was prevalent. This is because to dodge slipping the vehicle must get a grip, and this grip is provided by the friction among the road and the tyres. For the second part, the major source of wear and tear of the tyres is due to frictional forces that act between the tyres and the road.