A particle moving in a circular path has an angular momentum of L. If the frequency of rotation is halved, then its angular momentum becomes

Options

A. `(L)/(2)`
B. L
C. `(L)/(3)`
D. `(L)/(4)`

Explanation

To solve the problem, we need to understand the relationship between angular momentum and frequency of rotation. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the formula for angular momentum
The angular momentum \( L \) of a rotating body is given by the formula:
\[ L = I \omega \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

### Step 2: Relate angular velocity to frequency
Angular velocity \( \omega \) can be expressed in terms of frequency \( f \):
\[ \omega = 2 \pi f \]
Thus, we can rewrite the angular momentum as:
\[ L = I (2 \pi f) = 2 \pi I f \]

### Step 3: Identify the effect of halving the frequency
If the frequency of rotation is halved, the new frequency \( f' \) is:
\[ f' = \frac{f}{2} \]

### Step 4: Substitute the new frequency into the angular momentum formula
Now, substituting \( f' \) into the angular momentum formula:
\[ L' = 2 \pi I f' = 2 \pi I \left(\frac{f}{2}\right) = 2 \pi I \frac{f}{2} = \frac{1}{2} (2 \pi I f) = \frac{L}{2} \]

### Step 5: Conclusion
Thus, when the frequency is halved, the new angular momentum \( L' \) becomes:
\[ L' = \frac{L}{2} \]

### Final Answer
The angular momentum becomes \( \frac{L}{2} \).

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