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. (a)`2pisqrt((2r)/(3g))`
B. (b)`4pisqrt((r)/(2g))`
C. (c)`2pisqrt((3r)/(2g))`
D. (d)None of these

Explanation

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

### Step 1: Understand the Problem
We have a uniform disc that is oscillating in a plane parallel to its plane. The point of suspension is located at a distance of \( \frac{r}{2} \) from the center of the disc.

### Step 2: Moment of Inertia Calculation
Using the parallel axis theorem, we can find the moment of inertia \( I \) of the disc about the point of suspension.

1. The moment of inertia of the disc about its center is given by:
\[ I_{\text{cm}} = \frac{1}{2} m r^2 \]

2. The distance from the center to the point of suspension is \( d = \frac{r}{2} \).

3. According to the parallel axis theorem:
\[ I = I_{\text{cm}} + m d^2 \]
Substituting the values:
\[ I = \frac{1}{2} m r^2 + m \left(\frac{r}{2}\right)^2 \]
\[ I = \frac{1}{2} m r^2 + m \cdot \frac{r^2}{4} \]
\[ I = \frac{1}{2} m r^2 + \frac{1}{4} m r^2 = \frac{2}{4} m r^2 + \frac{1}{4} m r^2 = \frac{3}{4} m r^2 \]

### Step 3: Time Period Formula
The time period \( T \) for small oscillations is given by:
\[ T = 2\pi \sqrt{\frac{I}{M g d}} \]
where \( M \) is the mass of the disc, \( g \) is the acceleration due to gravity, and \( d \) is the distance from the center of mass to the point of suspension.

Substituting the values:
\[ T = 2\pi \sqrt{\frac{\frac{3}{4} m r^2}{m g \cdot \frac{r}{2}}} \]

### Step 4: Simplifying the Expression
1. Cancel \( m \) from the numerator and denominator:
\[ T = 2\pi \sqrt{\frac{\frac{3}{4} r^2}{g \cdot \frac{r}{2}}} \]

2. Simplifying further:
\[ T = 2\pi \sqrt{\frac{\frac{3}{4} r^2}{\frac{g r}{2}}} \]
\[ T = 2\pi \sqrt{\frac{3 r^2}{4} \cdot \frac{2}{g r}} = 2\pi \sqrt{\frac{3 r}{2g}} \]

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