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 will 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 (conversion from cm to m)
- Frequency of rotation (f) = 10 rotations/sec

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

### Step 3: Calculate the angular velocity (ω)
Angular velocity (ω) is related to the frequency (f) by the formula:
\[ \omega = 2 \pi f \]
Substituting the value of f:
\[ \omega = 2 \pi \times 10 \]
\[ \omega = 20 \pi \, \text{rad/sec} \]

### Step 4: Calculate the angular momentum (L)
Angular momentum (L) is given by the formula:
\[ L = I \omega \]
Substituting the values of I and ω:
\[ L = 0.04 \, \text{kg m}^2 \times 20 \pi \, \text{rad/sec} \]
\[ L = 0.8 \pi \, \text{kg m}^2/\text{s} \]

### Step 5: Calculate the numerical value of L
Using the approximation \( \pi \approx 3.14 \):
\[ L = 0.8 \times 3.14 \]
\[ L \approx 2.512 \, \text{kg m}^2/\text{s} \]

Thus, the angular momentum of the ring is approximately:
\[ L \approx 2.51 \, \text{Joule sec} \]

### Final Answer:
The value of angular momentum with respect to the axis of rotation is approximately **2.51 Joule sec**.

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