A disc is hinged such that it can freely rotate in a vertical plane about a point on its radius. If radius of disc is `'R'` , then what will be minimum time period of its simple harmonic motion ?

Options

A. `2pisqrt((R)/(g))`
B. `2pisqrt((3R)/(2g))`
C. `2pi sqrt((sqrt(2)R)/(g))`
D. `2pisqrt((R)/(2g))`

Explanation

To find the minimum time period of the simple harmonic motion of a disc hinged at a point on its radius, we will follow these steps:

### Step 1: Understand the system
The disc can be treated as a physical pendulum. The time period \( T \) of a physical pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{I}{mgx}} \]
where:
- \( I \) is the moment of inertia of the disc about the hinge point,
- \( m \) is the mass of the disc,
- \( g \) is the acceleration due to gravity,
- \( x \) is the distance from the hinge to the center of mass of the disc.

### Step 2: Calculate the moment of inertia
For a disc of radius \( R \) and mass \( m \), the moment of inertia about its center is:
\[ I_{center} = \frac{1}{2} m R^2 \]
When the disc is hinged at a point on its radius, we can use the parallel axis theorem to find the moment of inertia about the hinge point:
\[ I = I_{center} + m d^2 \]
where \( d \) is the distance from the center of the disc to the hinge point. Since the hinge is at a distance \( \frac{R}{2} \) from the center (as the hinge is on the radius), we have:
\[ d = \frac{R}{2} \]
Thus,
\[ I = \frac{1}{2} m R^2 + m \left(\frac{R}{2}\right)^2 = \frac{1}{2} m R^2 + \frac{1}{4} m R^2 = \frac{3}{4} m R^2 \]

### Step 3: Determine the distance to the center of mass
The center of mass of the disc is at a distance of \( \frac{R}{2} \) from the hinge point (since the hinge is at the radius). Therefore, we have:
\[ x = \frac{R}{2} \]

### Step 4: Substitute values into the time period formula
Now, substituting \( I \) and \( x \) into the time period formula:
\[ T = 2\pi \sqrt{\frac{\frac{3}{4} m R^2}{mg \cdot \frac{R}{2}}} \]
This simplifies to:
\[ T = 2\pi \sqrt{\frac{\frac{3}{4} R^2}{g \cdot \frac{R}{2}}} \]
\[ = 2\pi \sqrt{\frac{3R}{2g}} \]

### Step 5: Find the minimum time period
To find the minimum time period, we need to find the condition for \( x \). The minimum time period occurs when \( x = \frac{R}{\sqrt{2}} \). Thus, substituting this back into the time period formula:
\[ T_{min} = 2\pi \sqrt{\frac{\frac{3}{4} m R^2}{mg \cdot \frac{R}{\sqrt{2}}}} \]
This simplifies to:
\[ T_{min} = 2\pi \sqrt{\frac{3R\sqrt{2}}{2g}} \]

### Final Result
Thus, the minimum time period of the simple harmonic motion of the disc is:
\[ T_{min} = 2\pi \sqrt{\frac{R\sqrt{2}}{g}} \]