A ring of diameter 0.4 m and of mass 10 kg is rotating about its axis at the rate of 1200 rpm. The angular momentum of the ring is

Options

A. `60.28 "kgm"^(2) s^(-1)`
B. `55.26 "kgm"^(2)s^(-1)`
C. `40.28 "kgm"^(2) s^(-1)`
D. `50.28 "kgm"^(2)s^(-1)`

Explanation

To find the angular momentum of a ring rotating about its axis, we can use the formula:

\[ L = I \cdot \omega \]

where:
- \( L \) is the angular momentum,
- \( I \) is the moment of inertia,
- \( \omega \) is the angular velocity in radians per second.

### Step 1: Convert the angular velocity from RPM to radians per second

Given the rotation rate is 1200 RPM (revolutions per minute), we can convert this to radians per second using the following conversion:

\[ \omega = \text{RPM} \times \frac{2\pi \text{ radians}}{1 \text{ revolution}} \times \frac{1 \text{ minute}}{60 \text{ seconds}} \]

Substituting the values:

\[ \omega = 1200 \times \frac{2\pi}{60} \]

Calculating this:

\[ \omega = 1200 \times \frac{2\pi}{60} = 1200 \times \frac{\pi}{30} = 40\pi \text{ radians/second} \]

### Step 2: Calculate the moment of inertia \( I \) of the ring

The moment of inertia \( I \) for a ring is given by the formula:

\[ I = M \cdot R^2 \]

where:
- \( M \) is the mass of the ring,
- \( R \) is the radius of the ring.

Given:
- Diameter of the ring = 0.4 m, hence the radius \( R = \frac{0.4}{2} = 0.2 \) m,
- Mass \( M = 10 \) kg.

Substituting the values into the moment of inertia formula:

\[ I = 10 \cdot (0.2)^2 = 10 \cdot 0.04 = 0.4 \text{ kg m}^2 \]

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

Now, substituting the values of \( I \) and \( \omega \) into the angular momentum formula:

\[ L = I \cdot \omega = 0.4 \cdot (40\pi) \]

Calculating this:

\[ L = 0.4 \cdot 40 \cdot \pi = 16\pi \text{ kg m}^2/\text{s} \]

Using \( \pi \approx \frac{22}{7} \):

\[ L \approx 16 \cdot \frac{22}{7} \approx \frac{352}{7} \approx 50.2857 \text{ kg m}^2/\text{s} \]

Rounding to two decimal places, we find:

\[ L \approx 50.28 \text{ kg m}^2/\text{s} \]

### Conclusion

The angular momentum of the ring is approximately \( 50.28 \text{ kg m}^2/\text{s} \).

### Final Answer

The correct option is option 4: \( 50.28 \text{ kg m}^2/\text{s} \).

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