Answer: A) \( 2ma^2 \)
Step 1: Understanding the problem
We are calculating the moment of inertia \( I \) of four particles, each of mass \( m \), located at the corners of a square of edge \( a \). The axis of rotation is perpendicular to the plane of the square and passes through the center of the square.
Step 2: Distance of each particle from the center of the square
The diagonal of the square has a length given by:
\[
\text{Diagonal} = \sqrt{a^2 + a^2} = a\sqrt{2}
\]
Each particle is located at half the diagonal distance from the center, so the distance \( r \) of each particle from the center is:
\[
r = \frac{a\sqrt{2}}{2}
\]
Step 3: Moment of inertia of each particle
The moment of inertia of a single particle about the axis perpendicular to the square and passing through its center is:
\[
I_{\text{single}} = m r^2
\]
Substitute \( r = \frac{a\sqrt{2}}{2} \):
\[
I_{\text{single}} = m \left( \frac{a\sqrt{2}}{2} \right)^2
\]
Simplify the square of the distance:
\[
I_{\text{single}} = m \left( \frac{a^2 \cdot 2}{4} \right) = m \frac{2a^2}{4} = \frac{ma^2}{2}
\]
Step 4: Total moment of inertia
Since there are 4 particles, the total moment of inertia \( I \) is:
\[
I = 4 \times I_{\text{single}}
\]
Substitute \( I_{\text{single}} = \frac{ma^2}{2} \):
\[
I = 4 \times \frac{ma^2}{2} = 2ma^2
\]
Final Answer:
The moment of inertia of the system is:
\[
I = 2ma^2
\]
.png)
0 Comments