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

To determine the maximum safe velocity of a car negotiating a curved road, we need to consider both the bank angle and the frictional force acting on the car. Let's derive the expression for this scenario step by step:

1. When a car moves on a banked curve, the forces acting on the car are:

   • The gravitational force mg acting vertically downward.
   • The normal force N acting perpendicular to the surface of the road.
   • The frictional force f that acts parallel to the road surface.

2. For the car to move safely along the curve without slipping, the net force towards the center must provide the necessary centripetal force:

   m v^2/R = N \sin \theta + f \cos \theta

3. Decomposing the gravitational force into components and applying Newton's second law in the vertical direction, we get:

   • Vertical force balance: N \cos \theta = mg + f \sin \theta

   By substituting f = \mu_s N, we get:

   N \cos \theta = mg + \mu_s N \sin \theta

4. Rearrange to solve for N:

   N = \frac{mg}{\cos \theta - \mu_s \sin \theta}

5. Substitute the expression for N into the centripetal force equation:

   m v^2/R = \frac{mg \sin \theta + \mu_s mg \cos \theta}{\cos \theta - \mu_s \sin \theta}

   Cancel out m and solve for v^2:

   v^2 = gR \left(\frac{\sin \theta + \mu_s \cos \theta}{\cos \theta - \mu_s \sin \theta}\right)

6. Simplify using trigonometric identities:

   v^2 = gR \frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta}

   Thus, the maximum safe velocity v is:

   v = \sqrt{g R \frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta}}

Hence, the correct answer is:

v = \sqrt{g R \frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta}}

This option accounts for the combination of friction and banking angle to determine the maximum safe velocity.