A ring of radius R is made of a thin wire of material of density `rho`, having cross-section area a and Young's modulus y. The ring rotates about an axis perpendicular to its plane and through its centre. Angular frequency of rotation is `omega`.
The tension in the ring will be

Options

A. `(arhoR^(2)omega^(2))/(2)`
B. `a rho R^(2) omega^(2)`
C. `2a rho R^(2) omega^(2)`
D. `(arhoR^(2) omega^(2))/(4)`

Answer

To find the tension in the rotating ring, we can follow these steps:

### Step 1: Understanding the System
We have a ring of radius \( R \) made of a thin wire with density \( \rho \), cross-sectional area \( a \), and Young's modulus \( y \). The ring rotates about an axis perpendicular to its plane with an angular frequency \( \omega \).

### Step 2: Consider a Small Segment of the Ring
Let's take a small segment of the ring that subtends an angle \( \theta \) at the center. The length of this small segment is given by:
\[ \text{Length} = R \theta \]

### Step 3: Calculate the Mass of the Small Segment
The volume of this small segment can be calculated as:
\[ \text{Volume} = \text{Area} \times \text{Length} = a \cdot (R \theta) = aR\theta \]
Thus, the mass \( dm \) of this small segment is:
\[ dm = \rho \cdot \text{Volume} = \rho \cdot (aR\theta) = \rho a R \theta \]

### Step 4: Forces Acting on the Segment
The tension \( T \) in the wire acts on both ends of the small segment. The net force acting towards the center due to the tension can be derived from the components of the tension:
\[ F_{\text{net}} = 2T \cos\left(\frac{\theta}{2}\right) \]
For small angles, we can approximate \( \cos\left(\frac{\theta}{2}\right) \approx 1 \), so:
\[ F_{\text{net}} \approx 2T \]

### Step 5: Relate the Net Force to Centripetal Force
The net force acting towards the center must equal the centripetal force required to keep the mass moving in a circle:
\[ F_{\text{net}} = \frac{dm \cdot v^2}{R} \]
where \( v \) is the linear velocity. Since \( v = R \omega \), we can write:
\[ F_{\text{net}} = \frac{dm \cdot (R \omega)^2}{R} = dm \cdot R \omega^2 \]

### Step 6: Substitute \( dm \) into the Equation
Substituting \( dm = \rho a R \theta \) into the centripetal force equation gives:
\[ 2T = \left(\rho a R \theta\right) R \omega^2 \]

### Step 7: Solve for Tension \( T \)
Now, we can simplify this equation:
\[ 2T = \rho a R^2 \omega^2 \theta \]
Dividing both sides by \( 2\theta \):
\[ T = \frac{\rho a R^2 \omega^2 \theta}{2\theta} = \frac{\rho a R^2 \omega^2}{2} \]

### Final Answer
Thus, the tension in the ring is:
\[ T = \rho a R \omega^2 \]