A particle attached to a string is rotating with a constant angular velocity and its angular momentum is L. If the string is halved and angular velocity is kept constant, the angular momentum will be

Options

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

Answer

To solve the problem, we need to understand how angular momentum is affected when the radius of rotation is changed while keeping the angular velocity constant.

### Step-by-Step Solution:

1. **Understand Angular Momentum**:
The angular momentum \( L \) of a particle rotating about an axis is given by the formula:
\[ L = I \omega \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

2. **Moment of Inertia**:
For a particle of mass \( m \) rotating at a radius \( r \), the moment of inertia \( I \) is:
\[ I = m r^2 \]
Therefore, the angular momentum can be expressed as:
\[ L = m r^2 \omega \]

3. **Initial Conditions**:
Initially, the particle is rotating at radius \( r \) with angular momentum \( L \):
\[ L = m r^2 \omega \]

4. **Halving the String Length**:
When the string is halved, the new radius becomes \( r' = \frac{r}{2} \).

5. **New Moment of Inertia**:
The new moment of inertia \( I' \) when the radius is halved is:
\[ I' = m (r')^2 = m \left(\frac{r}{2}\right)^2 = m \frac{r^2}{4} \]

6. **Angular Momentum with New Radius**:
The angular momentum with the new radius while keeping the angular velocity \( \omega \) constant is:
\[ L' = I' \omega = \left(m \frac{r^2}{4}\right) \omega = \frac{m r^2 \omega}{4} \]

7. **Relating New Angular Momentum to Original**:
Since \( L = m r^2 \omega \), we can substitute:
\[ L' = \frac{L}{4} \]

8. **Final Answer**:
Therefore, the new angular momentum \( L' \) when the string is halved and angular velocity is kept constant is:
\[ L' = \frac{L}{4} \]

### Conclusion:
The angular momentum after halving the string while keeping the angular velocity constant is \( \frac{L}{4} \).

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