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 analyze the situation step by step.

### Step 1: Understand the Initial Conditions
We have a particle of mass \( m \) attached to a string of length \( r \), rotating about an axis with a constant angular velocity \( \omega \). The angular momentum \( L \) of the particle can be expressed as:
\[ L = I \omega \]
where \( I \) is the moment of inertia.

### Step 2: Calculate the Initial Moment of Inertia
The moment of inertia \( I \) for a point mass at a distance \( r \) from the axis of rotation is given by:
\[ I = m r^2 \]
Thus, the initial angular momentum can be rewritten as:
\[ L = m r^2 \omega \]

### Step 3: Halve the Length of the String
When the string is halved, the new radius \( r' \) becomes:
\[ r' = \frac{r}{2} \]

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

### Step 5: Calculate the New Angular Momentum
Since the angular velocity \( \omega \) remains constant, the new angular momentum \( L' \) can be calculated as:
\[ L' = I' \omega = \left(m \frac{r^2}{4}\right) \omega \]

### Step 6: Relate New Angular Momentum to Initial Angular Momentum
Substituting the expression for \( L \):
\[ L' = \frac{1}{4} (m r^2 \omega) = \frac{L}{4} \]

### Conclusion
Thus, the new angular momentum after halving the string length while keeping the angular velocity constant is:
\[ L' = \frac{L}{4} \]

### Final Answer
The angular momentum will be \( \frac{L}{4} \).

---