A uniform disc of radius R is pivoted at point O on its circumstances. The time period of small oscillations about an axis passing through O and perpendicular to plane of disc will be

Options

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

Answer

To solve the problem of finding the time period of small oscillations of a uniform disc pivoted at a point on its circumference, we can follow these steps:

### Step 1: Understand the System
We have a uniform disc of radius \( R \) and mass \( M \) that is pivoted at point \( O \) on its circumference. We need to find the time period of small oscillations about an axis passing through \( O \) and perpendicular to the plane of the disc.

### Step 2: Use the Formula for Time Period of a Physical Pendulum
The time period \( T \) of a physical pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{I}{mgd}} \]
where:
- \( I \) is the moment of inertia about the pivot point,
- \( m \) is the mass of the disc,
- \( g \) is the acceleration due to gravity,
- \( d \) is the distance from the pivot point to the center of mass of the disc.

### Step 3: Calculate the Moment of Inertia
To find the moment of inertia \( I \) about point \( O \), we can use the parallel axis theorem:
\[ I = I_{cm} + Md^2 \]
where \( I_{cm} \) is the moment of inertia about the center of mass (which is at the center of the disc) and \( d \) is the distance from the center of mass to the pivot point.

For a disc, the moment of inertia about the center is:
\[ I_{cm} = \frac{1}{2} MR^2 \]
The distance \( d \) from the pivot point \( O \) to the center of the disc (which is at a distance \( R \) from \( O \)) is \( R \).

Thus, we can substitute into the parallel axis theorem:
\[ I = \frac{1}{2} MR^2 + M(R^2) = \frac{1}{2} MR^2 + MR^2 = \frac{3}{2} MR^2 \]

### Step 4: Calculate the Distance \( d \)
The distance \( d \) from the pivot point \( O \) to the center of mass of the disc is simply the radius \( R \).

### Step 5: Substitute into the Time Period Formula
Now we can substitute \( I \) and \( d \) into the time period formula:
\[ T = 2\pi \sqrt{\frac{\frac{3}{2} MR^2}{MgR}} \]
Here, the mass \( M \) cancels out:
\[ T = 2\pi \sqrt{\frac{\frac{3}{2} R^2}{gR}} = 2\pi \sqrt{\frac{3R}{2g}} \]

### Step 6: Final Result
Thus, the time period of small oscillations is:
\[ T = 2\pi \sqrt{\frac{3R}{2g}} \]