A uniform disc of mass m and radius r is suspended through a wire attached to its Centre. If the time period of the torsional oscillations be T, what is the torsional constant of the wire?

Explanation

To find the torsional constant \( K \) of the wire when a uniform disc of mass \( m \) and radius \( r \) is suspended through it, we can follow these steps:

### Step 1: Understand the formula for the time period of torsional oscillations
The time period \( T \) of torsional oscillations is given by the formula:
\[ T = 2\pi \sqrt{\frac{I}{K}} \]
where \( I \) is the moment of inertia and \( K \) is the torsional constant.

### Step 2: Determine the moment of inertia of the disc
For a uniform disc of mass \( m \) and radius \( r \), the moment of inertia \( I \) about its center is given by:
\[ I = \frac{1}{2} m r^2 \]

### Step 3: Substitute the moment of inertia into the time period formula
Now, substitute the expression for \( I \) into the time period formula:
\[ T = 2\pi \sqrt{\frac{\frac{1}{2} m r^2}{K}} \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides gives:
\[ T^2 = (2\pi)^2 \cdot \frac{\frac{1}{2} m r^2}{K} \]
This simplifies to:
\[ T^2 = \frac{4\pi^2 \cdot \frac{1}{2} m r^2}{K} \]

### Step 5: Rearrange to solve for \( K \)
Rearranging the equation to solve for \( K \) gives:
\[ K = \frac{4\pi^2 \cdot \frac{1}{2} m r^2}{T^2} \]
This simplifies to:
\[ K = \frac{2\pi^2 m r^2}{T^2} \]

### Final Result
Thus, the torsional constant \( K \) of the wire is:
\[ K = \frac{2\pi^2 m r^2}{T^2} \]

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