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 solve this problem, we apply energy conservation between points \( A \) and \( B \).

Step 1: Energy conservation between \( A \) and \( B \):  

\(\frac{1}{2} m V_L^2 = \frac{1}{2} m V_B^2 + mg(2L)\)

where:  
- \( V_L \) is the velocity at point \( A \) (the lowest point),  
- \( V_B \) is the velocity at point \( B \) (the highest point).  

Rearranging the equation:  

\(V_L^2 = V_B^2 + 4gL\)
 

Step 2: Calculate \( V_L \) (Minimum Velocity at \( A \)):  
Since the bob must just complete the circular path, the minimum velocity at \( A \) must be such that:  

\(V_L = \sqrt{5gL}\)
 

Step 3: Calculate \( V_B \):  
Applying energy conservation:  

\(\frac{1}{2} m V_L^2 = \frac{1}{2} m V_B^2 + mg(2L)\)

Substitute \(V_L = \sqrt{5gL}\):  

\(\frac{1}{2} m (5gL) = \frac{1}{2} m V_B^2 + 2mgL\)

Simplifying:  

\(\frac{5}{2} mgL = \frac{1}{2} m V_B^2 + 2mgL\)

\(\frac{1}{2} m V_B^2 = \frac{5}{2} mgL - 2mgL\)

\(V_B^2 = gL\)

\(V_B = \sqrt{gL}\)
 

Step 4: Calculate the ratio of kinetic energies:  

\(\left(\frac{\text{(K.E.)}_A}{\text{(K.E.)}_B}\right) = \frac{\frac{1}{2} m V_L^2}{\frac{1}{2} m V_B^2} = \frac{V_L^2}{V_B^2} = \frac{5gL}{gL} = 5\)

Thus, the ratio of kinetic energies \( \left(\frac{\text{(K.E.)}_A}{\text{(K.E.)}_B}\right) \) is 5 : 1.

The Correct Answer is: 5 : 1