The system consists of a uniform bar and two point masses (balls) at its ends. The bar has a length \( L = 2.00 \, \text{m} \) and mass \( M = 4.50 \, \text{kg} \). Each ball has a mass \( m = 0.300 \, \text{kg} \).
The moment of inertia of a uniform bar about an axis parallel to the bar and at a distance \( d \) from its center is given by the parallel axis theorem:
\[
I_{\text{bar}} = I_{\text{center}} + Md^2
\]
where \( I_{\text{center}} = \frac{1}{12}ML^2 \) is the moment of inertia about the center of the bar, and \( d = 0.500 \, \text{m} \) is the distance from the axis to the center of the bar.
Substituting the values:
\[
I_{\text{center}} = \frac{1}{12} \times 4.50 \, \text{kg} \times (2.00 \, \text{m})^2 = 1.500 \, \text{kg} \cdot \text{m}^2
\]
\[
I_{\text{bar}} = 1.500 \, \text{kg} \cdot \text{m}^2 + 4.50 \, \text{kg} \times (0.500 \, \text{m})^2 = 2.625 \, \text{kg} \cdot \text{m}^2
\]
Each ball is treated as a point mass, and its moment of inertia about the axis is given by:
\[
I_{\text{ball}} = m \times (d + \frac{L}{2})^2
\]
For each ball, the distance from the axis is \( d + \frac{L}{2} = 0.500 \, \text{m} + 1.00 \, \text{m} = 1.50 \, \text{m} \).
Substituting the values:
\[
I_{\text{ball}} = 0.300 \, \text{kg} \times (1.50 \, \text{m})^2 = 0.675 \, \text{kg} \cdot \text{m}^2
\]
Since there are two balls, the total moment of inertia for the balls is:
\[
I_{\text{balls}} = 2 \times 0.675 \, \text{kg} \cdot \text{m}^2 = 1.350 \, \text{kg} \cdot \text{m}^2
\]
The total moment of inertia of the system is the sum of the moments of inertia of the bar and the balls:
\[
I_{\text{total}} = I_{\text{bar}} + I_{\text{balls}} = 2.625 \, \text{kg} \cdot \text{m}^2 + 1.350 \, \text{kg} \cdot \text{m}^2 = 3.975 \, \text{kg} \cdot \text{m}^2
\]
\[
\boxed{3.975 \, \text{kg} \cdot \text{m}^2}
\]
Answered
