Question

A bob of mass \( m \) is suspended by a light string of length \( L \). It is imparted a minimum horizontal velocity at the lowest point \( A \) such that it just completes a half circle, reaching the topmost position \( B \). The ratio of kinetic energies \( \frac{(\text{K.E.})_A}{(\text{K.E.})_B} \) is:

• A. 3 : 2
• B. 5 : 1
• C. 2 : 5
• D. 1 : 5

Answer 1

The Correct Option is B

To determine the ratio of kinetic energies \( \frac{(\text{K.E.})_A}{(\text{K.E.})_B} \), an analysis of a simple pendulum undergoing circular motion is required.

Key considerations include:

1. At the lowest point \( A \), the bob possesses maximum kinetic energy and zero potential energy.
2. At the highest point \( B \), the bob has maximum potential energy and the minimum kinetic energy necessary to sustain circular motion.

The solution proceeds as follows:

1. The condition for the bob to just reach point \( B \) necessitates that its velocity \( v_B \) at point \( B \) maintains string tension. This minimum required velocity can be determined using the centripetal force equation:
2. Kinetic energy at point \( B \) is expressed as:
3. Conservation of energy between point \( A \) and point \( B \) dictates that total energy at \( A \) equals total energy at \( B \):
4. Substitute the derived expression for \( v_B^2 \):
5. Simplify the equation to solve for \( v_A^2 \):
6. Consequently, the kinetic energy at point \( A \) is:
7. Calculate the ratio of kinetic energies:

The resultant ratio is \(\frac{5}{1}\).