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Step-by-Step Solution
Step 1: Identify the components of the system
The system has two identical solid spheres, each of:
• Mass $M = 1.5\,\text{kg}$
• Radius $r = 0.5\,\text{m}$
They are attached to the ends of a light rod such that the distance between their centers is $5\,\text{m}$. We are to find the moment of inertia about an axis perpendicular to the rod through its midpoint.
Step 2: Determine the distance from the axis of rotation to each sphere’s center
Since the total distance between the centers of the spheres is $5\,\text{m}$, the midpoint of the rod is equidistant from each sphere. Thus, each sphere’s center is at a distance
$$
d = \frac{5}{2} = 2.5\;\text{m}
$$
from the axis of rotation.
Step 3: Use the formula for the moment of inertia of one sphere including the parallel axis shift
The moment of inertia of a solid sphere of mass $M$ and radius $r$ about its own center is
$$
I_{\text{center}} = \frac{2}{5} \, M \, r^2.
$$
However, in this problem, we need the moment of inertia about an axis that is $d$ away from the center (using the Parallel Axis Theorem). Thus, for one sphere,
$$
I_{\text{sphere about midpoint axis}}
= \frac{2}{5} \, M \, r^2 + M \, d^2.
$$
Step 4: Calculate the moment of inertia for one sphere
Plug in the given values:
$$
I_{\text{sphere about midpoint axis}}
= \frac{2}{5} \times 1.5 \times (0.5)^2 \;+\; 1.5 \times (2.5)^2.
$$
First part (about the center):
$$
\frac{2}{5} \times 1.5 \times (0.5)^2
= \frac{2}{5} \times 1.5 \times 0.25
= 0.6 \times 0.25
= 0.15\;\text{kg·m}^2.
$$
Second part (parallel axis shift):
$$
1.5 \times (2.5)^2
= 1.5 \times 6.25
= 9.375\;\text{kg·m}^2.
$$
Thus, for one sphere:
$$
I_{\text{sphere}} = 0.15 + 9.375 = 9.525\;\text{kg·m}^2.
$$
Step 5: Calculate the total moment of inertia for both spheres
There are two identical spheres, so the total moment of inertia is:
$$
I_{\text{total}} = 2 \times 9.525 = 19.05\;\text{kg·m}^2.
$$
Step 6: State the final answer
The moment of inertia of the system about the given axis is:
$$
\boxed{19.05\;\text{kg·m}^2.}
$$
