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 ratio of kinetic energy to potential energy is

Options

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

Answer

To find the ratio of kinetic energy (KE) to potential energy (PE) for a rotating ring, we can follow these steps:

### Step 1: Calculate the Kinetic Energy (KE)
The kinetic energy of the ring can be calculated using the formula:
\[ KE = \frac{1}{2} I \omega^2 \]
where \(I\) is the moment of inertia of the ring and \(\omega\) is the angular frequency.

For a thin ring of radius \(R\), the moment of inertia \(I\) is given by:
\[ I = mR^2 \]
where \(m\) is the mass of the ring. The mass can be calculated using the density \(\rho\), cross-sectional area \(a\), and the circumference of the ring:
\[ m = \rho \cdot a \cdot L = \rho \cdot a \cdot (2\pi R) \]
Thus, substituting for \(m\):
\[ I = (\rho \cdot a \cdot 2\pi R) R^2 = 2\pi \rho a R^3 \]
Now substituting \(I\) back into the kinetic energy formula:
\[ KE = \frac{1}{2} (2\pi \rho a R^3) \omega^2 = \pi \rho a R^3 \omega^2 \]

### Step 2: Calculate the Potential Energy (PE)
The potential energy stored in the material of the ring due to its elongation can be calculated using the formula:
\[ PE = \frac{1}{2} \frac{T \cdot \Delta L}{L} \]
where \(T\) is the tension, \(\Delta L\) is the change in length, and \(L\) is the original length.

The tension \(T\) can be related to the Young's modulus \(Y\) as:
\[ T = Y \cdot \frac{A}{L} \cdot \Delta L \]
The elongation \(\Delta L\) can be expressed as:
\[ \Delta L = \frac{T \cdot L}{Y \cdot A} \]
Substituting for \(T\):
\[ \Delta L = \frac{(Y \cdot A \cdot \Delta L) \cdot L}{Y \cdot A} = \frac{T \cdot L}{Y \cdot A} \]
Now, substituting this back into the potential energy formula:
\[ PE = \frac{1}{2} Y \cdot A \cdot \frac{T \cdot L}{Y \cdot A} = \frac{1}{2} \cdot \frac{T^2 \cdot L}{Y \cdot A} \]
Using the earlier expression for \(T\):
\[ PE = \frac{1}{2} \cdot \frac{(Y \cdot A \cdot \Delta L)^2 \cdot L}{Y \cdot A} \]
After simplification, we find:
\[ PE = \frac{1}{2} \cdot \frac{Y \cdot A \cdot (2\pi R)^2 \cdot \rho^2 \cdot \omega^2}{Y} \]
This leads to:
\[ PE = \frac{1}{2} \cdot \frac{2\pi^2 R^2 \rho^2 \omega^2 A}{Y} \]

### Step 3: Calculate the Ratio of KE to PE
Now, we can find the ratio of kinetic energy to potential energy:
\[ \frac{KE}{PE} = \frac{\pi \rho a R^3 \omega^2}{\frac{1}{2} \cdot \frac{2\pi^2 R^2 \rho^2 \omega^2 A}{Y}} \]
This simplifies to:
\[ \frac{KE}{PE} = \frac{2\pi \rho a R^3 \omega^2 Y}{2\pi^2 R^2 \rho^2 \omega^2 A} \]
Canceling out common terms:
\[ \frac{KE}{PE} = \frac{Y}{\rho R^2} \]

### Final Answer
Thus, the ratio of kinetic energy to potential energy is:
\[ \frac{KE}{PE} = \frac{Y}{\rho R^2} \]