Question

A car is moving with velocity V at the top of a semi-circular hill of radius 40 m such that the normal force on it is zero. Find the velocity (V) of the car. [use \( g = 10 \, \text{ms}^{-2} \)]

• A. \( 10 \, \text{m/s} \)
• B. \( 15 \, \text{m/s} \)
• C. \( 20 \, \text{m/s} \)
• D. \( 25 \, \text{m/s} \)

Hint:

At the top of a vertical circle, \( N = m(g - v^2/R) \). Setting \( N = 0 \) gives \( v = \sqrt{gR} \), the critical velocity.

Answer 1

The Correct Option is C

Concept: At the top of a circular hill, the centripetal force required for circular motion is provided by the difference between weight (mg) and normal force (N). When normal force is zero, weight alone provides the centripetal force.

Step 1: Apply Newton's second law at the top. Net downward force = centripetal force: \[ mg - N = \frac{mV^2}{R} \] Given \( N = 0 \): \[ mg = \frac{mV^2}{R} \]

Step 2: Cancel mass and solve for V. \[ g = \frac{V^2}{R} \quad \Rightarrow \quad V^2 = gR \] \[ V = \sqrt{gR} \]

Step 3: Substitute given values. \( g = 10 \, \text{m/s}^2 \), \( R = 40 \, \text{m} \): \[ V = \sqrt{10 \times 40} = \sqrt{400} = 20 \, \text{m/s}. \]