A particle of mass 2 kg is on a smooth horizontal table and moves in a circular path of radius 0.6 m. The height of the table from the ground is 0.8 m. If the angular speed of the particle is `"12 rad s"^(-1)` , the magnitude of its angular momentum about a point on the ground right under the centre of the circle is

Options

A. `14.4 kg-m^(2)s^(-1)`
B. `8.64kg-m^(2)s^(-1)`
C. `20.16kg-m^(2)s^(-1)`
D. `11.52 kg-m^(2)s^(-1)`

Explanation

To find the magnitude of the angular momentum of a particle moving in a circular path about a point on the ground directly beneath the center of the circle, we can follow these steps:

### Step 1: Identify the Given Data
- Mass of the particle, \( m = 2 \, \text{kg} \)
- Radius of the circular path, \( r = 0.6 \, \text{m} \)
- Height of the table from the ground, \( h = 0.8 \, \text{m} \)
- Angular speed, \( \omega = 12 \, \text{rad/s} \)

### Step 2: Calculate the Distance from the Point on the Ground to the Particle
The distance \( R \) from the point on the ground (directly beneath the center of the circular path) to the particle can be calculated using the Pythagorean theorem. The horizontal distance is the radius of the circle, and the vertical distance is the height of the table.

\[ R = \sqrt{(r)^2 + (h)^2} = \sqrt{(0.6)^2 + (0.8)^2} \]

Calculating the squares:

\[ R = \sqrt{0.36 + 0.64} = \sqrt{1} = 1 \, \text{m} \]

### Step 3: Calculate the Linear Momentum of the Particle
The linear momentum \( P \) of the particle is given by:

\[ P = m \cdot v \]

Where \( v \) is the linear velocity of the particle. The linear velocity can be calculated using the formula:

\[ v = r \cdot \omega \]

Substituting the values:

\[ v = 0.6 \cdot 12 = 7.2 \, \text{m/s} \]

Now, calculate the linear momentum:

\[ P = 2 \cdot 7.2 = 14.4 \, \text{kg m/s} \]

### Step 4: Calculate the Angular Momentum
The angular momentum \( L \) about the point on the ground is given by:

\[ L = R \cdot P \]

Substituting the values we found:

\[ L = 1 \cdot 14.4 = 14.4 \, \text{kg m}^2/\text{s} \]

### Final Answer
The magnitude of the angular momentum about the point on the ground right under the center of the circle is:

\[ \boxed{14.4 \, \text{kg m}^2/\text{s}} \]

---