A uniform disc of radius r is to be suspended through a small hole made in te disc. Find the minimum possible time period of the disc for small oscillations. What should be the distance of the hole from the centre for it to have minimum time period?

Explanation

To solve the problem of finding the minimum possible time period of a uniform disc suspended through a small hole made in the disc, we will follow these steps:

### Step 1: Define the system
We have a uniform disc of radius \( r \) and mass \( m \). The disc is suspended through a small hole at a distance \( x \) from the center of the disc.

### Step 2: Moment of Inertia
The moment of inertia \( I \) of the disc about its center is given by:
\[ I_{cm} = \frac{1}{2} m r^2 \]
Using the parallel axis theorem, the moment of inertia about the suspension point (the hole) is:
\[ I = I_{cm} + m h^2 = \frac{1}{2} m r^2 + m x^2 \]

### Step 3: Time Period of Physical Pendulum
The time period \( T \) of a physical pendulum is given by:
\[ T = 2\pi \sqrt{\frac{I}{m g h}} \]
In our case, \( h = x \), so we can substitute:
\[ T = 2\pi \sqrt{\frac{\frac{1}{2} m r^2 + m x^2}{m g x}} \]
Cancelling \( m \) from the numerator and denominator, we have:
\[ T = 2\pi \sqrt{\frac{\frac{1}{2} r^2 + x^2}{g x}} \]

### Step 4: Finding Minimum Time Period
To find the minimum time period, we need to minimize \( T \). We can minimize \( T^2 \) instead:
\[ T^2 = 4\pi^2 \frac{\frac{1}{2} r^2 + x^2}{g x} \]
Let:
\[ f(x) = \frac{\frac{1}{2} r^2 + x^2}{x} \]
We will differentiate \( f(x) \) with respect to \( x \) and set the derivative to zero to find the critical points.

### Step 5: Differentiate and Set to Zero
Differentiating \( f(x) \):
\[ f'(x) = \frac{(2x)(x) - (\frac{1}{2} r^2 + x^2)}{x^2} \]
Setting \( f'(x) = 0 \):
\[ 2x^2 - \frac{1}{2} r^2 - x^2 = 0 \]
This simplifies to:
\[ x^2 = \frac{1}{2} r^2 \]
Taking the positive root, we have:
\[ x = \frac{r}{\sqrt{2}} \]

### Step 6: Minimum Time Period Calculation
Now substituting \( x = \frac{r}{\sqrt{2}} \) back into the expression for \( T \):
\[ T = 2\pi \sqrt{\frac{\frac{1}{2} r^2 + \left(\frac{r}{\sqrt{2}}\right)^2}{g \left(\frac{r}{\sqrt{2}}\right)}} \]
Calculating:
\[ T = 2\pi \sqrt{\frac{\frac{1}{2} r^2 + \frac{1}{2} r^2}{g \frac{r}{\sqrt{2}}}} = 2\pi \sqrt{\frac{r^2}{g \frac{r}{\sqrt{2}}}} = 2\pi \sqrt{\frac{r \sqrt{2}}{g}} \]

### Final Result
Thus, the minimum possible time period \( T \) for small oscillations of the disc is:
\[ T = 2\pi \sqrt{\frac{r \sqrt{2}}{g}} \]
And the distance of the hole from the center for it to have a minimum time period is:
\[ x = \frac{r}{\sqrt{2}} \]