Question

A car is negotiating a curved road of radius R. The road is banked at an angle $\theta$. The coefficient of friction between the types of the car and the road is $\mu_s$. The maximum safe velocity on this road is :

• A. $\sqrt{g R \frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta}}$
• B. $\sqrt{\frac{g}{R} \frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta}}$
• C. $\sqrt{\frac{g}{R^2} \frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta}}$
• D. $\sqrt{g R^2 \frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta}}$

Answer 1

The Correct Option is A

Speed Calculation 

We are given an equation involving velocity, the coefficient of static friction, and the angle \( \theta \). Let's break down the steps to find the velocity \( v \).

Step 1: Given Equation

The given equation is:

\(\frac{v^2}{Rg} = \frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta}\)

Where: - \( v \) is the velocity, - \( R \) is the radius of the curve, - \( g \) is the acceleration due to gravity, - \( \mu_s \) is the coefficient of static friction, - \( \theta \) is the angle of the curve.

Step 2: Solving for \( v \)

To solve for \( v \), multiply both sides of the equation by \( Rg \) and take the square root:

\(v^2 = Rg \left[ \frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta} \right]\)

Now, take the square root of both sides to get \( v \):

\(v = \sqrt{Rg \left[ \frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta} \right]}\)

Conclusion:

The velocity \( v \) is given by:

\(v = \sqrt{Rg \left[ \frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta} \right]}\)