A uniform ring of radius `R` and made up of a wire of cross - sectional radius `r` is rotated about its axis with a frequency `f`. If density of the wire is `rho ` and young's modulus is `Y` . Find the fractional change in radius of the ring .

Explanation

To solve the problem of finding the fractional change in the radius of a rotating uniform ring, we will follow these steps:

### Step 1: Understand the Problem
We have a uniform ring of radius \( R \) made from a wire with a cross-sectional radius \( r \). It rotates about its axis with a frequency \( f \). We need to find the fractional change in the radius of the ring when it is subjected to this rotation.

### Step 2: Define Relevant Variables
- Let \( \rho \) be the density of the wire.
- Let \( Y \) be the Young's modulus of the wire.
- The tension \( T \) in the wire will be influenced by the centripetal force required for the rotation.

### Step 3: Consider a Small Element of the Ring
Consider a small element \( dL \) of the ring. The angle subtended by this element at the center is \( d\theta \). The tension in the wire will provide the necessary centripetal force for this element.

### Step 4: Write the Expression for Centripetal Force
The centripetal force required for the mass \( dm \) of the element rotating at radius \( R \) with angular velocity \( \omega \) is given by:
\[ F_c = dm \cdot R \cdot \omega^2 \]
where \( \omega = 2\pi f \).

### Step 5: Relate Tension and Centripetal Force
The tension in the wire contributes to the centripetal force. The vertical components of the tension from two sides of the element will add up:
\[ 2T \sin(d\theta) \approx 2T d\theta \quad (\text{for small } d\theta) \]
Thus, we equate the centripetal force to the tension:
\[ 2T d\theta = dm \cdot R \cdot (2\pi f)^2 \]

### Step 6: Express Mass Element \( dm \)
The mass element \( dm \) can be expressed in terms of the density \( \rho \) and the volume:
\[ dm = \text{Volume} \times \text{Density} = A \cdot dL \cdot \rho \]
where \( A = \pi r^2 \) is the cross-sectional area of the wire.

### Step 7: Substitute \( dm \) into the Force Equation
Substituting \( dm \) into the equation gives:
\[ 2T d\theta = \pi r^2 \cdot dL \cdot \rho \cdot R \cdot (2\pi f)^2 \]

### Step 8: Relate Change in Length to Tension
The change in length \( \Delta L \) due to tension can be expressed using Young's modulus:
\[ \Delta L = \frac{T \cdot L}{A \cdot Y} \]
where \( L = 2\pi R \) is the circumference of the ring.

### Step 9: Calculate the Fractional Change in Radius
The fractional change in radius \( \frac{\Delta R}{R} \) can be related to the change in length:
\[ \frac{\Delta L}{L} = \frac{\Delta R}{R} \]

### Step 10: Combine All Relationships
From the earlier steps, we can derive:
\[ \frac{\Delta R}{R} = \frac{T \cdot 2\pi R}{\pi r^2 \cdot Y} \]
Substituting the expression for \( T \) derived from the centripetal force equation will yield the final result.

### Final Expression
After substituting and simplifying, we arrive at:
\[ \frac{\Delta R}{R} = \frac{4\pi^3 f^2 r^2 \rho}{Y} \]

### Conclusion
The fractional change in the radius of the ring is given by:
\[ \frac{\Delta R}{R} = \frac{4\pi^3 f^2 r^2 \rho}{Y} \]