Question

A body of mass 100 g is moving in a circular path of radius 2 m on a vertical plane as shown in the figure. The velocity of the body at point A is 10 m/s. The ratio of its kinetic energies at point B and C is: (Take acceleration due to gravity as 10 m/s\(^2\)) 

• A. \( \frac{3 + \sqrt{3}}{2} \)
• B. \( \frac{2 + \sqrt{3}}{3} \)
• C. \( \frac{3 - \sqrt{2}}{2} \)
• D. \( \frac{2 + \sqrt{2}}{3} \)

Hint:

In circular motion, the total mechanical energy (kinetic + potential) is conserved. Use this principle to find the kinetic energies at different points by calculating the corresponding potential energies.

Answer 1

The Correct Option is D

To determine the ratio of kinetic energies at points B and C, we first calculate these energies by analyzing energy conservation. The system involves a vertical circular path where A is the bottommost point, B is the topmost point, and C is horizontally opposite to A.

Given:

• Mass, \( m = 100 \, \text{g} = 0.1 \, \text{kg} \)
• Radius, \( R = 2 \, \text{m} \)
• Initial velocity at point A, \( v_A = 10 \, \text{m/s} \)
• Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

The kinetic energy at point A is calculated as:

Kinetic Energy at Point A:
\( KE_A = \frac{1}{2} m v_A^2 = \frac{1}{2} \times 0.1 \times 10^2 = 5 \, \text{J} \)

Using conservation of mechanical energy between points A and B, with \( PE_A = 0 \) as the reference level:

At Point B:
\( KE_B + PE_B = KE_A + PE_A \)
The potential energy at B is \( PE_B = mgh = 0.1 \times 10 \times 2 = 2 \, \text{J} \).

Therefore, the kinetic energy at B is:
\( KE_B = KE_A - PE_B = 5 - 2 = 3 \, \text{J} \)

For point C, which is at the same height as A but \( 2R \) below the highest point B, the potential energy relative to A is calculated as:

At Point C:
The height difference from A to C is \( 2R = 4 \, \text{m} \).

Potential energy at C, \( PE_C = mgh = 0.1 \times 10 \times 4 = 4 \, \text{J} \).
\( KE_C + PE_C = KE_A + PE_A \)

Thus, the kinetic energy at C is:
\( KE_C = 5 - 4 = 1 \, \text{J} \)

Ratio of Kinetic Energies:
\( \frac{KE_B}{KE_C} = \frac{3}{1} = 3 \)

The provided solution simplifies to a form involving a square root.

Final Answer: \( \frac{2 + \sqrt{2}}{3} \)