A particle of mass 0.2 kg is moving in a circle of radius 1 m with `f=(2//pi)sec^(-1)`, then its angular momentum is :

Options

A. `0.8kg-m^(2)//s`
B. `2kg-m^(2)//s`
C. `8kg-m^(2)//s`
D. `16kg-m^(2)//s`

Answer

To find the angular momentum of a particle moving in a circle, we can use the formula:

\[ L = mvr \]

where:
- \( L \) is the angular momentum,
- \( m \) is the mass of the particle,
- \( v \) is the linear velocity,
- \( r \) is the radius of the circle.

However, we can also express the linear velocity \( v \) in terms of the angular frequency \( \omega \) and the radius \( r \):

\[ v = \omega r \]

The angular frequency \( \omega \) can be calculated from the frequency \( f \) using the relationship:

\[ \omega = 2\pi f \]

Given:
- Mass \( m = 0.2 \, \text{kg} \)
- Radius \( r = 1 \, \text{m} \)
- Frequency \( f = \frac{2}{\pi} \, \text{s}^{-1} \)

### Step 1: Calculate the angular frequency \( \omega \)

\[ \omega = 2\pi f = 2\pi \left(\frac{2}{\pi}\right) = 4 \, \text{rad/s} \]

### Step 2: Calculate the linear velocity \( v \)

Using \( v = \omega r \):

\[ v = 4 \times 1 = 4 \, \text{m/s} \]

### Step 3: Calculate the angular momentum \( L \)

Now, substituting \( m \), \( v \), and \( r \) into the angular momentum formula:

\[ L = mvr = 0.2 \times 4 \times 1 = 0.8 \, \text{kg m}^2/\text{s} \]

### Final Answer

The angular momentum of the particle is:

\[ L = 0.8 \, \text{kg m}^2/\text{s} \]

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