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

Answer: (B)

5 : 1

Answer 2

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