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 half a circle, reaching the topmost position B. The ratio of kinetic energies \(\left(\frac{K.E.}{K.E.}\right)_A \), \(\text{to}\), \(\left(\frac{K.E.}{K.E.}\right)_B\) is:

• A. 3:2
• B. 5:1
• C. 2:5
• D. 1:5

Hint:

In vertical circular motion, use \textbf{energy conservation} to relate velocities at different points. At the topmost point, potential energy is maximum, and kinetic energy is minimum.

Answer 1

The Correct Option is B

Step 1: Applying Energy Conservation between Points A and B
According to the principle of conservation of energy: \[ \text{Total Energy at A} = \text{Total Energy at B} \] The total energy is the sum of kinetic energy (K.E.) and potential energy (P.E.).
Step 2: Energy at the Lowest Point (A)
At the lowest point A: - The bob has only kinetic energy. - Potential energy is taken as zero at this point. \[ K.E._A = \frac{1}{2} m v_H^2 \] where \( v_H \) is the velocity at point A.
Step 3: Energy at the Highest Point (B)
At the highest point B: - The bob has both kinetic energy and potential energy. - The height at point B is 2L, so the potential energy is: \[ P.E._B = mg(2L) \] - The velocity at B is given by: \[ v_L = \sqrt{5gL} \] So, the kinetic energy at point B is: \[ K.E._B = \frac{1}{2} m v_L^2 = \frac{1}{2} m (5gL) = \frac{5}{2} m g L \]
Step 4: Finding the Ratio of Kinetic Energies
Using energy conservation: \[ K.E._A + P.E._A = K.E._B + P.E._B \] Since \( P.E._A = 0 \), \[ K.E._A = K.E._B + mg(2L) \] Substituting \( K.E._B = \frac{5}{2} m g L \), \[ K.E._A = \frac{5}{2} m g L + 2 mg L \] \[ K.E._A = \frac{5}{2} mgL + \frac{4}{2} mgL = \frac{9}{2} mgL \] \[ K.E._B = \frac{5}{2} mgL \] So, the ratio is: \[ \frac{K.E._A}{K.E._B} = \frac{\frac{9}{2} mgL}{\frac{5}{2} mgL} = \frac{9}{5} \] \[ = 5:1 \] Final Answer: The ratio of kinetic energies is 5:1.