Find time period of uniform disc of mass m and radius r suspended through point `r//2` away from centre, oscillating in a plane parallel to its plane.

Options

A. `2pisqrt((2r)/(3g))`
B. `4pisqrt((r)/(2g))`
C. `2pisqrt((3r)/(2g))`
D. None of these

Explanation

To find the time period of a uniform disc of mass \( m \) and radius \( r \) suspended at a point \( \frac{r}{2} \) away from its center, we can follow these steps:

### Step 1: Identify the pivot point and the center of mass
The disc is pivoted at a point that is \( \frac{r}{2} \) from its center. The center of mass of the disc is at its geometric center.

### Step 2: Calculate the distance from the pivot to the center of mass
The distance \( d \) from the pivot point to the center of mass is given as \( \frac{r}{2} \).

### Step 3: Use the formula for the 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 of the disc 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 to the center of mass.

### Step 4: Calculate the moment of inertia \( I \)
The moment of inertia of a uniform disc about its center is:
\[ I_{cm} = \frac{1}{2} m r^2 \]
To find the moment of inertia about the pivot point, we use the parallel axis theorem:
\[ I = I_{cm} + md^2 \]
Substituting \( d = \frac{r}{2} \):
\[ I = \frac{1}{2} m r^2 + m \left(\frac{r}{2}\right)^2 = \frac{1}{2} m r^2 + m \frac{r^2}{4} = \frac{1}{2} m r^2 + \frac{1}{4} m r^2 = \frac{3}{4} m r^2 \]

### Step 5: Substitute \( I \) into the time period formula
Now substituting \( I \) into the time period formula:
\[ T = 2\pi \sqrt{\frac{\frac{3}{4} m r^2}{mg \cdot \frac{r}{2}}} \]

### Step 6: Simplify the expression
\[ T = 2\pi \sqrt{\frac{\frac{3}{4} r^2}{g \cdot \frac{r}{2}}} \]
\[ = 2\pi \sqrt{\frac{3r}{2g}} \]

### Final Answer
Thus, the time period \( T \) of the oscillation of the disc is:
\[ T = 2\pi \sqrt{\frac{3r}{2g}} \]