Question

State the formula for the safety speed of a vehicle on a curved banked road.

• A. \(v = \sqrt{rg\sin\theta}\)
• B. \(v = \sqrt{rg\cos\theta}\)
• C. \(v = \sqrt{rg\tan\theta}\)
• D. \(v = \sqrt{\frac{g}{r\tan\theta}}\)

Hint:

Banked roads help reduce dependence on friction by allowing the normal reaction to provide centripetal force. The safe speed depends on radius \(r\), gravitational acceleration \(g\), and banking angle \(\theta\).

Answer 1

The Correct Option is C

Concept: When a vehicle moves along a curved road, it requires a centripetal force directed towards the center of the circular path. On a banked road, the road is inclined at an angle \(\theta\) with respect to the horizontal. This inclination helps provide the necessary centripetal force through the components of the normal reaction. A banked road allows vehicles to move safely around curves even without relying entirely on friction. The horizontal component of the normal reaction provides the centripetal force required for circular motion.

Step 1: Forces acting on the vehicle.
Two main forces act on the vehicle:
• Weight of the vehicle \(mg\) acting vertically downward
• Normal reaction \(N\) perpendicular to the road surface

Step 2: Resolving forces.
The components of the normal reaction are:
• \(N\cos\theta\) balancing the weight
• \(N\sin\theta\) providing the centripetal force

Step 3: Applying circular motion condition.
Centripetal force required: \[ \frac{mv^2}{r} \] Equating the horizontal component: \[ N\sin\theta = \frac{mv^2}{r} \] From vertical balance: \[ N\cos\theta = mg \] Dividing the two equations: \[ \tan\theta = \frac{v^2}{rg} \] Therefore, \[ v = \sqrt{rg\tan\theta} \] Thus, the safe speed of a vehicle on a banked curve is given by: \[ v = \sqrt{rg\tan\theta} \]