A particle performing uniform circular motion gas angular momentum `L`. If its angular frequency is double and its kinetic energy halved, then the new angular momentum is :

Options

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

Explanation

To solve the problem, we need to analyze the relationship between angular momentum, angular frequency, and kinetic energy. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the relationship between angular momentum and angular frequency
The angular momentum \( L \) of a particle in uniform circular motion is given by the formula:
\[ L = I \omega \]
where \( I \) is the moment of inertia and \( \omega \) is the angular frequency.

### Step 2: Understand the relationship between kinetic energy and angular frequency
The kinetic energy \( K \) of a particle in circular motion is given by:
\[ K = \frac{1}{2} I \omega^2 \]

### Step 3: Relate the initial conditions to the new conditions
Let the initial angular frequency be \( \omega \) and the initial kinetic energy be \( K \). According to the problem:
- The new angular frequency becomes \( 2\omega \).
- The new kinetic energy becomes \( \frac{K}{2} \).

### Step 4: Express the moment of inertia in terms of kinetic energy and angular frequency
From the kinetic energy formula, we can express the moment of inertia \( I \):
\[ K = \frac{1}{2} I \omega^2 \implies I = \frac{2K}{\omega^2} \]

### Step 5: Calculate the new moment of inertia with the new kinetic energy and angular frequency
Using the new kinetic energy \( \frac{K}{2} \) and the new angular frequency \( 2\omega \):
\[ K' = \frac{1}{2} I' (2\omega)^2 \]
Substituting \( K' = \frac{K}{2} \):
\[ \frac{K}{2} = \frac{1}{2} I' (4\omega^2) \implies K = 4 I' \omega^2 \implies I' = \frac{K}{4\omega^2} \]

### Step 6: Calculate the new angular momentum
Now, substituting \( I' \) into the angular momentum formula for the new conditions:
\[ L' = I' (2\omega) = \left(\frac{K}{4\omega^2}\right)(2\omega) = \frac{2K}{4\omega} = \frac{K}{2\omega} \]

### Step 7: Relate the new angular momentum to the initial angular momentum
Now, we can relate \( L' \) to the initial angular momentum \( L \):
\[ L = I \omega = \frac{2K}{\omega^2} \cdot \omega = \frac{2K}{\omega} \]
Thus,
\[ L' = \frac{K}{2\omega} = \frac{L}{4} \]

### Final Answer
The new angular momentum \( L' \) is:
\[ L' = \frac{L}{4} \]

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