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

Explanation

To find the torque acting on a square plate of mass 120 g and edge 5.00 cm that rotates about one of its edges with a uniform angular acceleration of 0.2 rad/s², we can follow these steps:

### Step 1: Convert mass and dimensions to SI units
- The mass of the plate is given as 120 g. To convert this to kilograms:
\[ m = 120 \, \text{g} = 0.120 \, \text{kg} \]
- The edge length of the plate is given as 5.00 cm. To convert this to meters:
\[ a = 5.00 \, \text{cm} = 0.0500 \, \text{m} \]

### Step 2: Calculate the moment of inertia (I)
The moment of inertia \( I \) of a square plate rotating about one of its edges can be calculated using the formula:
\[ I = \frac{m a^2}{3} \]
Substituting the values we have:
\[ I = \frac{0.120 \, \text{kg} \cdot (0.0500 \, \text{m})^2}{3} \]
Calculating \( (0.0500 \, \text{m})^2 \):
\[ (0.0500)^2 = 0.0025 \, \text{m}^2 \]
Now substituting this back into the moment of inertia formula:
\[ I = \frac{0.120 \cdot 0.0025}{3} = \frac{0.0003}{3} = 0.0001 \, \text{kg m}^2 \]

### Step 3: Calculate the torque (T)
The torque \( T \) can be calculated using the formula:
\[ T = I \alpha \]
Where \( \alpha \) is the angular acceleration, given as \( 0.2 \, \text{rad/s}^2 \). Substituting the values we have:
\[ T = 0.0001 \, \text{kg m}^2 \cdot 0.2 \, \text{rad/s}^2 \]
Calculating the torque:
\[ T = 0.00002 \, \text{kg m}^2/\text{s}^2 = 2 \times 10^{-5} \, \text{N m} \]

### Final Answer
The torque acting on the plate is:
\[ T = 2 \times 10^{-5} \, \text{N m} \]