A mass is whirled in a circular path with a constant angular velocity and its angular momentum is L. If the string is now halved keeping the angular velocity same, the angular momentum is

Options

A. `L/4`
B. `L/2`
C. L
D. 2L

Answer

To solve the problem, we need to analyze the situation step by step.

### Step 1: Understand the initial conditions
We have a mass \( m \) whirling in a circular path with a radius \( R \) and a constant angular velocity \( \omega \). The angular momentum \( L \) of the mass can be expressed as:
\[ L = m R^2 \omega \]

### Step 2: Determine the new conditions after halving the string
When the string is halved, the new radius \( R' \) becomes:
\[ R' = \frac{R}{2} \]
The angular velocity \( \omega \) remains the same.

### Step 3: Calculate the new angular momentum
The angular momentum \( L' \) for the new radius can be expressed as:
\[ L' = m (R')^2 \omega \]
Substituting \( R' \):
\[ L' = m \left(\frac{R}{2}\right)^2 \omega \]
This simplifies to:
\[ L' = m \left(\frac{R^2}{4}\right) \omega \]
\[ L' = \frac{m R^2 \omega}{4} \]

### Step 4: Relate the new angular momentum to the original angular momentum
From the initial condition, we know:
\[ L = m R^2 \omega \]
Thus, we can express \( L' \) in terms of \( L \):
\[ L' = \frac{L}{4} \]

### Conclusion
The new angular momentum \( L' \) when the string is halved while keeping the angular velocity the same is:
\[ L' = \frac{L}{4} \]