A thin ring of radius `R` is made of a material of density `rho` and Young's modulus `Y`. If the ring is rotated about its centre in its own plane with angular velocity `omega` , find the small increase in its radius.

Explanation

To find the small increase in the radius of a thin ring of radius \( R \), made of a material with density \( \rho \) and Young's modulus \( Y \), when it is rotated about its center in its own plane with angular velocity \( \omega \), we can follow these steps:

### Step 1: Understand the Forces Acting on the Ring
When the ring rotates, each element of the ring experiences a centripetal force due to its circular motion. This force is provided by the tension in the ring.

### Step 2: Consider an Element of the Ring
Take a small element of the ring. The tension \( T \) acts tangentially to the ring. For a small angle \( \theta \), the horizontal components of the tension from both sides of the element will contribute to the net force acting towards the center of the ring.

### Step 3: Set Up the Equation for Centripetal Force
The net force acting towards the center due to the tension can be expressed as:
\[ F_{\text{net}} = 2T \sin(\theta) \]
For small angles, \( \sin(\theta) \approx \theta \). Thus, we can write:
\[ F_{\text{net}} \approx 2T \theta \]

### Step 4: Relate the Net Force to Centripetal Force
The centripetal force required for the mass \( dm \) of the element moving in a circle of radius \( R \) is given by:
\[ F_{\text{centripetal}} = dm \cdot \omega^2 R \]

### Step 5: Calculate the Mass of the Element
The mass \( dm \) of the small element can be expressed in terms of the density \( \rho \) and the geometry of the ring:
\[ dm = \rho \cdot dV = \rho \cdot (dL \cdot A) \]
where \( dL \) is the length of the element and \( A \) is the cross-sectional area. The length of the element can be approximated as \( R \cdot d\theta \), so:
\[ dm = \rho \cdot (R \cdot d\theta) \cdot A \]

### Step 6: Equate Forces
Setting the net force equal to the centripetal force gives:
\[ 2T \theta = \rho (R \cdot d\theta) A \cdot \omega^2 R \]

### Step 7: Solve for Tension \( T \)
From the equation above, we can isolate \( T \):
\[ T = \frac{\rho A \omega^2 R^2 \theta}{2\theta} = \frac{\rho A \omega^2 R^2}{2} \]

### Step 8: Relate Tension to Stress and Strain
The tension in the ring produces a stress, which is related to strain through Young's modulus \( Y \):
\[ \text{Stress} = \frac{T}{A} = Y \cdot \text{Strain} \]
The strain is defined as:
\[ \text{Strain} = \frac{\Delta R}{R} \]
Thus, we can write:
\[ \frac{T}{A} = Y \cdot \frac{\Delta R}{R} \]

### Step 9: Substitute for Tension
Substituting our expression for \( T \) into the equation gives:
\[ \frac{\frac{\rho A \omega^2 R^2}{2}}{A} = Y \cdot \frac{\Delta R}{R} \]
This simplifies to:
\[ \frac{\rho \omega^2 R^2}{2} = Y \cdot \frac{\Delta R}{R} \]

### Step 10: Solve for the Increase in Radius \( \Delta R \)
Rearranging the equation to solve for \( \Delta R \):
\[ \Delta R = \frac{\rho \omega^2 R^3}{2Y} \]

### Final Result
Thus, the small increase in the radius of the ring when it is rotated with angular velocity \( \omega \) is:
\[ \Delta R = \frac{\rho \omega^2 R^3}{2Y} \]