A square plate of mass 120g and edge 5.00 cm rotates about one of the edges. If it has a unifrom angular acceleration of `0.2 rad/s^2`, what torque acts on the plate?

Explanation

To find the torque acting on the square plate, we can use the relationship between torque, moment of inertia, and angular acceleration. The formula for torque (τ) is given by:

\[ \tau = I \cdot \alpha \]

where:
- \( \tau \) is the torque,
- \( I \) is the moment of inertia,
- \( \alpha \) is the angular acceleration.

### Step 1: Calculate the Moment of Inertia (I)

For a square plate rotating about one of its edges, the moment of inertia can be calculated using the formula:

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

where:
- \( m \) is the mass of the plate,
- \( L \) is the length of one side of the square plate.

Given:
- Mass \( m = 120 \, \text{g} = 0.120 \, \text{kg} \) (since we need to convert grams to kilograms),
- Edge length \( L = 5.00 \, \text{cm} = 0.0500 \, \text{m} \) (since we need to convert centimeters to meters).

Now, substituting the values into the moment of inertia formula:

\[ I = \frac{1}{3} (0.120 \, \text{kg}) (0.0500 \, \text{m})^2 \]

Calculating \( I \):

\[ I = \frac{1}{3} (0.120) (0.0025) = \frac{1}{3} (0.0003) = 0.0001 \, \text{kg m}^2 \]

### Step 2: Calculate the Torque (τ)

Now that we have the moment of inertia, we can calculate the torque using the angular acceleration:

Given:
- Angular acceleration \( \alpha = 0.2 \, \text{rad/s}^2 \).

Now substituting \( I \) and \( \alpha \) into the torque formula:

\[ \tau = I \cdot \alpha = 0.0001 \, \text{kg m}^2 \cdot 0.2 \, \text{rad/s}^2 \]

Calculating \( \tau \):

\[ \tau = 0.00002 \, \text{N m} \]

### Final Answer

The torque acting on the plate is:

\[ \tau = 0.00002 \, \text{N m} \, \text{or} \, 20 \, \mu \text{N m} \]

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