A uniform stick of length l and mass m lies on a smooth table. It rotates with angular velocity `omega` about an axis perpendicular to the table and through one end of the stick. The angular momentum of the stick about the end is

Options

A. `ml^(2)omega`
B. `(ml^(2)omega)/(3)`
C. `(ml^(2)omega)/(12)`
D. `(ml^(2)omega)/(6)`

Answer

To find the angular momentum of a uniform stick of length \( l \) and mass \( m \) rotating about an axis perpendicular to the table through one end of the stick, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We have a uniform stick of length \( l \) and mass \( m \) that rotates about one end with an angular velocity \( \omega \). We need to find the angular momentum \( L \) of the stick about the end.

2. **Use the Angular Momentum Formula**: The angular momentum \( L \) of a rotating object can be calculated using the formula:
\[ L = I \cdot \omega \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

3. **Determine the Moment of Inertia**: For a uniform stick rotating about one end, the moment of inertia \( I \) is given by:
\[ I = \frac{1}{3} m l^2 \]
This formula is derived from the standard moment of inertia for a rod rotating about an end.

4. **Substitute the Values**: Now, substitute the expression for \( I \) into the angular momentum formula:
\[ L = \left(\frac{1}{3} m l^2\right) \cdot \omega \]

5. **Simplify the Expression**: This gives us the final expression for the angular momentum:
\[ L = \frac{1}{3} m l^2 \omega \]

### Final Answer:
The angular momentum of the stick about the end is:
\[ L = \frac{1}{3} m l^2 \omega \]