Calculate the time period of a uniform square plate of side `'a'` if it is suspended through a corner.

Explanation

To calculate the time period of a uniform square plate of side 'a' suspended through a corner, we can follow these steps:

### Step 1: Determine the Moment of Inertia
The moment of inertia (I) of a square plate about its center of mass (CM) is given by the formula:
\[ I_{CM} = \frac{m a^2}{6} \]
where \( m \) is the mass of the plate and \( a \) is the side length.

Since the plate is suspended from a corner, we need to use the parallel axis theorem to find the moment of inertia about the corner. The distance from the center of mass to the corner is \( \frac{a}{\sqrt{2}} \).

Using the parallel axis theorem:
\[ I = I_{CM} + m d^2 \]
where \( d = \frac{a}{\sqrt{2}} \).

Substituting the values:
\[ I = \frac{m a^2}{6} + m \left(\frac{a}{\sqrt{2}}\right)^2 \]
\[ I = \frac{m a^2}{6} + m \frac{a^2}{2} \]
\[ I = \frac{m a^2}{6} + \frac{3m a^2}{6} \]
\[ I = \frac{4m a^2}{6} = \frac{2m a^2}{3} \]

### Step 2: Calculate the Torque
When the plate is displaced by an angle \( \theta \), the torque (\( \tau \)) due to gravity acting on the center of mass is given by:
\[ \tau = m g \cdot \text{(perpendicular distance)} \]
The perpendicular distance from the line of action of the weight to the pivot point (corner) is:
\[ \text{Perpendicular distance} = \frac{a}{\sqrt{2}} \cdot \theta \]

Thus, the torque can be expressed as:
\[ \tau = m g \cdot \left(\frac{a}{\sqrt{2}} \cdot \theta\right) \]

### Step 3: Relate Torque to Angular Acceleration
According to Newton's second law for rotation:
\[ \tau = I \alpha \]
where \( \alpha \) is the angular acceleration. Since \( \alpha = \frac{d^2\theta}{dt^2} \), we can write:
\[ m g \cdot \left(\frac{a}{\sqrt{2}} \cdot \theta\right) = I \frac{d^2\theta}{dt^2} \]

### Step 4: Substitute Moment of Inertia
Substituting the moment of inertia we found earlier:
\[ m g \cdot \left(\frac{a}{\sqrt{2}} \cdot \theta\right) = \frac{2m a^2}{3} \frac{d^2\theta}{dt^2} \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ \frac{d^2\theta}{dt^2} + \frac{3g}{2a} \theta = 0 \]

This is a simple harmonic motion equation of the form:
\[ \frac{d^2\theta}{dt^2} + \omega^2 \theta = 0 \]
where \( \omega^2 = \frac{3g}{2a} \).

### Step 6: Calculate the Time Period
The angular frequency \( \omega \) is related to the time period \( T \) by:
\[ T = 2\pi \sqrt{\frac{I}{\tau}} \]
Substituting \( \omega \):
\[ T = 2\pi \sqrt{\frac{2a}{3g}} \]

### Final Result
Thus, the time period \( T \) of the uniform square plate suspended through a corner is:
\[ T = 2\pi \sqrt{\frac{2a}{3g}} \]

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