A disc of radius `R` is pivoted at its rim. The period for small oscillations about an axis perpendicular to the plane of disc is

Options

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

Explanation

To find the period of small oscillations of a disc pivoted at its rim, we can follow these steps:

### Step 1: Identify the parameters
- We have a disc of radius \( R \) pivoted at its rim.
- The center of mass of the disc is located at a distance \( R/2 \) from the pivot point.

### Step 2: Calculate the moment of inertia
The moment of inertia \( I_A \) about the pivot point can be calculated using the parallel axis theorem:
\[ I_A = I_{cm} + m \cdot d^2 \]
where:
- \( I_{cm} \) is the moment of inertia about the center of mass,
- \( m \) is the mass of the disc,
- \( d \) is the distance from the center of mass to the pivot point, which is equal to \( R \).

For a disc, the moment of inertia about its center of mass is given by:
\[ I_{cm} = \frac{1}{2} m R^2 \]

Substituting \( d = R \):
\[ I_A = \frac{1}{2} m R^2 + m R^2 = \frac{1}{2} m R^2 + \frac{2}{2} m R^2 = \frac{3}{2} m R^2 \]

### Step 3: Use the formula for the period of a physical pendulum
The period \( T \) of small oscillations for a physical pendulum is given by:
\[ T = 2\pi \sqrt{\frac{I_A}{m g d}} \]
where:
- \( g \) is the acceleration due to gravity,
- \( d \) is the distance from the pivot to the center of mass, which is \( R/2 \).

Substituting \( I_A \) and \( d \):
\[ T = 2\pi \sqrt{\frac{\frac{3}{2} m R^2}{m g \cdot \frac{R}{2}}} \]

### Step 4: Simplify the expression
Cancelling \( m \) and simplifying:
\[ T = 2\pi \sqrt{\frac{\frac{3}{2} R^2}{g \cdot \frac{R}{2}}} \]
\[ = 2\pi \sqrt{\frac{3R}{g}} \]

### Final Result
Thus, the period for small oscillations about an axis perpendicular to the plane of the disc is:
\[ T = 2\pi \sqrt{\frac{3R}{g}} \]