A circular ring of mass 1kg and radius 0.2m executes 10 revolutions per sec. Its angular momentum would be - (kg-`m^2`/sec)

Options

A. 0.025
B. 0.25
C. 2.5
D. 25

Answer

To find the angular momentum of a circular ring, we can use the formula:

\[ J = I \omega \]

where:
- \( J \) is the angular momentum,
- \( I \) is the moment of inertia,
- \( \omega \) is the angular velocity.

### Step 1: Calculate the Moment of Inertia (I)

The moment of inertia \( I \) of a circular ring about its axis is given by the formula:

\[ I = m r^2 \]

where:
- \( m \) is the mass of the ring,
- \( r \) is the radius of the ring.

Given:
- Mass \( m = 1 \, \text{kg} \)
- Radius \( r = 0.2 \, \text{m} \)

Substituting the values:

\[ I = 1 \times (0.2)^2 \]
\[ I = 1 \times 0.04 \]
\[ I = 0.04 \, \text{kg} \cdot \text{m}^2 \]

### Step 2: Calculate the Angular Velocity (ω)

The angular velocity \( \omega \) can be calculated from the number of revolutions per second \( N \):

\[ \omega = 2 \pi N \]

Given:
- \( N = 10 \, \text{revolutions per second} \)

Substituting the value:

\[ \omega = 2 \pi \times 10 \]
\[ \omega = 20 \pi \, \text{rad/s} \]

### Step 3: Calculate Angular Momentum (J)

Now we can substitute \( I \) and \( \omega \) into the angular momentum formula:

\[ J = I \omega \]

Substituting the values we calculated:

\[ J = 0.04 \times (20 \pi) \]
\[ J = 0.04 \times 20 \times 3.14 \]
\[ J = 0.04 \times 62.8 \]
\[ J = 2.512 \, \text{kg} \cdot \text{m}^2/\text{s} \]

### Final Answer

The angular momentum of the circular ring is:

\[ J = 2.512 \, \text{kg} \cdot \text{m}^2/\text{s} \]

---