A circular ring of mass 1kg and radius 20 cms rotating on its axis with 10 rotations/sec. The value of angular momentum in joule sec with respect to the axis of rotation will be?

Options

A. 5
B. 2.51
C. 1.25
D. 2500

Answer

To find the angular momentum of a circular ring rotating about its axis, we can follow these steps:

### Step 1: Identify the given values
- Mass of the ring (m) = 1 kg
- Radius of the ring (r) = 20 cm = 0.2 m (convert cm to m)
- Frequency of rotation (f) = 10 rotations per second

### Step 2: Calculate the angular speed (ω)
The angular speed (ω) in radians per second can be calculated using the formula:
\[ \omega = 2\pi f \]
Substituting the given frequency:
\[ \omega = 2 \pi \times 10 = 20\pi \text{ rad/s} \]

### Step 3: Calculate the moment of inertia (I) of the ring
The moment of inertia (I) for a circular ring rotating about its axis is given by:
\[ I = m r^2 \]
Substituting the values of mass and radius:
\[ I = 1 \times (0.2)^2 = 1 \times 0.04 = 0.04 \text{ kg m}^2 \]

### Step 4: Calculate the angular momentum (L)
The angular momentum (L) can be calculated using the formula:
\[ L = I \omega \]
Substituting the values of moment of inertia and angular speed:
\[ L = 0.04 \times 20\pi \]
Calculating this:
\[ L = 0.04 \times 20 \times 3.14 = 0.04 \times 62.8 = 2.512 \text{ kg m}^2/\text{s} \]
Since 1 kg m²/s is equivalent to 1 Joule second (J·s), we can say:
\[ L = 2.512 \text{ J·s} \]

### Final Answer
The value of angular momentum with respect to the axis of rotation is:
\[ L = 2.512 \text{ J·s} \]

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