A ring of radius 'r' and mass per unit length 'm' rotates with an angular velocity `'omega'` in free space then tension will be :

Options

A. zero
B. `(1)/(2)momega^(2)r^(2)`
C. `momega^(2)r^(2)`
D. `mromega^(2)`

Answer

To solve the problem of finding the tension in a rotating ring, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the System:**
- We have a ring of radius \( r \) and mass per unit length \( m \) rotating with an angular velocity \( \omega \).
- The ring is in free space, meaning there are no external forces acting on it.

2. **Identifying Forces:**
- Consider a small segment of the ring subtending an angle \( d\theta \) at the center. The length of this segment is \( r d\theta \).
- The mass of this small segment is given by \( \text{mass} = \text{mass per unit length} \times \text{length} = m \cdot (r d\theta) \).

3. **Centripetal Force Requirement:**
- For the segment to maintain circular motion, a net centripetal force must act towards the center of the ring. This force can be provided by the tension in the ring.
- The tension \( T \) in the ring creates components that contribute to the centripetal force.

4. **Components of Tension:**
- The tension \( T \) acts along the ring, and we can resolve it into components. Each side of the segment contributes a component of tension towards the center.
- The vertical component of tension towards the center is \( T \sin\left(\frac{d\theta}{2}\right) \) from each side, giving a total contribution of \( 2T \sin\left(\frac{d\theta}{2}\right) \).

5. **Setting Up the Equation:**
- The net centripetal force required for the mass \( m \cdot (r d\theta) \) moving with velocity \( v \) is given by:
\[ F_{\text{centripetal}} = \frac{m \cdot (r d\theta) \cdot v^2}{r} \]
- Since \( v = \omega r \), we can substitute to get:
\[ F_{\text{centripetal}} = \frac{m \cdot (r d\theta) \cdot (\omega r)^2}{r} = m \cdot (r d\theta) \cdot \omega^2 r \]

6. **Equating Forces:**
- Set the centripetal force equal to the net tension force:
\[ 2T \sin\left(\frac{d\theta}{2}\right) = m \cdot (r d\theta) \cdot \omega^2 r \]

7. **Simplifying the Equation:**
- For small angles, \( \sin\left(\frac{d\theta}{2}\right) \approx \frac{d\theta}{2} \). Thus, we can rewrite the equation as:
\[ 2T \cdot \frac{d\theta}{2} = m \cdot (r d\theta) \cdot \omega^2 r \]
- This simplifies to:
\[ T d\theta = m \cdot r^2 \cdot \omega^2 d\theta \]

8. **Final Result:**
- Canceling \( d\theta \) from both sides (assuming \( d\theta \neq 0 \)):
\[ T = m \cdot r^2 \cdot \omega^2 \]

### Conclusion:
The tension \( T \) in the ring is given by:
\[ T = m \cdot r^2 \cdot \omega^2 \]