A uniform square plate of mass m and edge a initially at rest startas rotating about one of the edge under the action of a constant torque `tau`. Then at the end of the `5^(th)` sec after start

Options

A. angular momentum is equal to `5tau`
B. kinetic energy is equal to `(75tau^(2))/(ma^(2))`
C. angular momentum is equal to `2.5 tau`
D. kinetic energy is equal to `(75tau^(2))/(2ma)`

Answer

To solve the problem step by step, we will follow the outlined approach based on the information provided in the video transcript.

### Step 1: Calculate the Moment of Inertia (I)
The moment of inertia of a uniform square plate about an edge can be treated as that of a rod rotating about one of its ends. The formula for the moment of inertia of a rod about one end is given by:

\[ I = \frac{1}{3} m a^2 \]

where \( m \) is the mass of the plate and \( a \) is the length of the edge.

### Step 2: Calculate the Angular Acceleration (α)
The angular acceleration can be calculated using the formula:

\[ \alpha = \frac{\tau}{I} \]

Substituting the expression for \( I \):

\[ \alpha = \frac{\tau}{\frac{1}{3} m a^2} = \frac{3\tau}{m a^2} \]

### Step 3: Calculate the Angular Velocity (ω) at the end of 5 seconds
The angular velocity after time \( t \) can be calculated using the formula:

\[ \omega = \alpha t \]

Substituting the expression for \( \alpha \) and \( t = 5 \) seconds:

\[ \omega = \left(\frac{3\tau}{m a^2}\right) \times 5 = \frac{15\tau}{m a^2} \]

### Step 4: Calculate the Angular Momentum (L)
The angular momentum can be calculated using the formula:

\[ L = I \omega \]

Substituting the expressions for \( I \) and \( \omega \):

\[ L = \left(\frac{1}{3} m a^2\right) \left(\frac{15\tau}{m a^2}\right) = \frac{15\tau}{3} = 5\tau \]

### Step 5: Calculate the Kinetic Energy (KE)
The kinetic energy of the rotating plate can be calculated using the formula:

\[ KE = \frac{1}{2} I \omega^2 \]

Substituting the expressions for \( I \) and \( \omega \):

\[ KE = \frac{1}{2} \left(\frac{1}{3} m a^2\right) \left(\frac{15\tau}{m a^2}\right)^2 \]

Calculating \( \omega^2 \):

\[ \omega^2 = \left(\frac{15\tau}{m a^2}\right)^2 = \frac{225\tau^2}{m^2 a^4} \]

Now substituting this back into the KE equation:

\[ KE = \frac{1}{2} \left(\frac{1}{3} m a^2\right) \left(\frac{225\tau^2}{m^2 a^4}\right) = \frac{75\tau^2}{2 m a^2} \]

### Final Answers
1. **Angular Momentum (L)** at the end of the 5th second is \( 5\tau \).
2. **Kinetic Energy (KE)** at the end of the 5th second is \( \frac{75\tau^2}{2 m a^2} \).