A ring of mass `m` and radius `r` rotates about an axis passing through its centre and perpendicular to its plane with angular velocity `omega`. Its kinetic energy is

Options

A. `1/2 mr^(2) omega^(2)`
B. `mromega^(2)`
C. `mr^(2)omega^(2)`
D. `1/3 mr^(2) omega^(2)`

Answer

To find the kinetic energy of a ring rotating about an axis passing through its center and perpendicular to its plane, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the parameters**: We have a ring with mass \( m \) and radius \( r \) rotating with an angular velocity \( \omega \).

2. **Understand the formula for rotational kinetic energy**: The kinetic energy \( K \) of a rotating object is given by the formula:
\[ K = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia of the object.

3. **Determine the moment of inertia for the ring**: The moment of inertia \( I \) of a ring about an axis passing through its center and perpendicular to its plane is given by:
\[ I = m r^2 \]

4. **Substitute the moment of inertia into the kinetic energy formula**: Now, we can substitute \( I \) into the kinetic energy formula:
\[ K = \frac{1}{2} (m r^2) \omega^2 \]

5. **Simplify the expression**: This simplifies to:
\[ K = \frac{1}{2} m r^2 \omega^2 \]

6. **Conclusion**: Therefore, the kinetic energy of the ring is:
\[ K = \frac{1}{2} m r^2 \omega^2 \]

### Final Answer:
The kinetic energy of the ring is \( \frac{1}{2} m r^2 \omega^2 \).

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