Question

A car is moving on a circular track banked at an angle of 45°. If the maximum permissible speed of the car to avoid slipping is twice the optimum speed of the car to avoid the wear and tear of the tyres, then the coefficient of static friction between the wheels of the car and the road is

• A. \(0.3\)
• B. \(0.5\)
• C. \(0.4\)
• D. \(0.6\)

Hint:

Remember that the relationship between maximum and optimum speeds on a banked curve involves both gravitational and frictional forces.

Answer 1

The Correct Option is D

Let's analyze the situation. Let the radius of the circular track be $r$.
The angle of banking is $\theta = 45^\circ$.
The maximum permissible speed to avoid slipping is $v_{max}$.
The optimum speed to avoid wear and tear is $v_{opt}$. Given that $v_{max} = 2 v_{opt}$.
The optimum speed $v_{opt}$ is the speed at which no frictional force is required.
$v_{opt} = \sqrt{rg \tan \theta}$
Since $\theta = 45^\circ$, $\tan \theta = \tan 45^\circ = 1$.
$v_{opt} = \sqrt{rg}$
The maximum permissible speed to avoid slipping is given by:
$v_{max} = \sqrt{rg \left( \frac{\tan \theta + \mu_s}{1 - \mu_s \tan \theta} \right)}$
where $\mu_s$ is the coefficient of static friction.
Given $v_{max} = 2 v_{opt}$, we have:
$2 \sqrt{rg} = \sqrt{rg \left( \frac{\tan \theta + \mu_s}{1 - \mu_s \tan \theta} \right)}$
Squaring both sides, we get:
$4 rg = rg \left( \frac{\tan \theta + \mu_s}{1 - \mu_s \tan \theta} \right)$
$4 = \frac{\tan \theta + \mu_s}{1 - \mu_s \tan \theta}$
Since $\theta = 45^\circ$, $\tan \theta = 1$.
$4 = \frac{1 + \mu_s}{1 - \mu_s}$
$4(1 - \mu_s) = 1 + \mu_s$
$4 - 4\mu_s = 1 + \mu_s$
$3 = 5 \mu_s$
$\mu_s = \frac{3}{5} = 0.6$
Therefore, the coefficient of static friction between the wheels of the car and the road is 0.6. Final Answer: The final answer is $\boxed{(4)}$