A uniform square plate of mass 2.0 kg and edge 10 cm rotates about one of its diagonals under the action of a constant torque of 0.10 Nm. Calculate the angular momentum and the kinetic energy of the plate at the end of the fifth second after the start.

Explanation

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

### Step 1: Calculate the Moment of Inertia (I)
The moment of inertia for a square plate rotating about one of its diagonals is given by the formula:

\[ I = \frac{m l^2}{12} \]

Where:
- \( m = 2.0 \, \text{kg} \) (mass of the plate)
- \( l = 10 \, \text{cm} = 0.1 \, \text{m} \) (edge length of the plate)

Substituting the values:

\[ I = \frac{2.0 \times (0.1)^2}{12} = \frac{2.0 \times 0.01}{12} = \frac{0.02}{12} = \frac{1}{600} \, \text{kg m}^2 \approx 0.00167 \, \text{kg m}^2 \]

### Step 2: Calculate Angular Acceleration (α)
Using the torque equation:

\[ \tau = I \alpha \]

Where:
- \( \tau = 0.1 \, \text{Nm} \) (torque)

Rearranging the equation to find angular acceleration (\( \alpha \)):

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

Substituting the values:

\[ \alpha = \frac{0.1}{\frac{1}{600}} = 0.1 \times 600 = 60 \, \text{rad/s}^2 \]

### Step 3: Calculate Angular Velocity (ω) at the End of the Fifth Second
Using the kinematic equation for rotational motion:

\[ \omega = \omega_0 + \alpha t \]

Where:
- \( \omega_0 = 0 \, \text{rad/s} \) (initial angular velocity)
- \( t = 5 \, \text{s} \)

Substituting the values:

\[ \omega = 0 + 60 \times 5 = 300 \, \text{rad/s} \]

### Step 4: Calculate Angular Momentum (L)
The angular momentum is given by:

\[ L = I \omega \]

Substituting the values:

\[ L = \frac{1}{600} \times 300 = \frac{300}{600} = 0.5 \, \text{kg m}^2/\text{s} \]

### Step 5: Calculate Kinetic Energy (KE)
The kinetic energy in rotational motion is given by:

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

Substituting the values:

\[ KE = \frac{1}{2} \times \frac{1}{600} \times (300)^2 \]

Calculating:

\[ KE = \frac{1}{2} \times \frac{1}{600} \times 90000 = \frac{45000}{600} = 75 \, \text{J} \]

### Final Answers
- Angular Momentum \( L = 0.5 \, \text{kg m}^2/\text{s} \)
- Kinetic Energy \( KE = 75 \, \text{J} \)