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Step-by-Step Solution
Step 1: Identify the Relevant Physical Principle
We want to calculate the moment of inertia of two identical solid spheres about an axis perpendicular to the rod and passing through its midpoint. The key ideas are:
The moment of inertia of a solid sphere of mass m and radius r about an axis through its center is
\frac{2}{5} m r^{2} .
The Parallel Axis Theorem: I_{\text{new axis}} = I_{\text{center}} + m d^{2}, where d is the distance between the new axis and the original axis passing through the center of the sphere.
Step 2: List the Given Information
Mass of each sphere, m = 2\,\text{kg} .
Radius of each sphere, r = 0.10\,\text{m} (10 cm).
Distance between centers of the two spheres = 0.40\,\text{m} .
The rotation axis passes through the midpoint of the rod connecting the centers, so each sphere’s center is 0.20\,\text{m} from the axis.
Step 3: Moment of Inertia of One Sphere About the Midpoint Axis
Moment of inertia about its own center:
\[
I_{\text{center}} = \frac{2}{5}\,m\,r^{2}
= \frac{2}{5} \times 2 \,\text{kg} \times (0.10\,\text{m})^{2}
= \frac{4}{5} \times 0.01 \,\text{kg}\,\text{m}^{2}
= 0.008 \,\text{kg}\,\text{m}^{2}.
\]
Using the Parallel Axis Theorem to shift by d = 0.20\,\text{m} :
\[
I_{\text{shifted}} = I_{\text{center}} + m\,d^{2}
= 0.008 \,\text{kg}\,\text{m}^{2} + 2\,\text{kg} \times (0.20\,\text{m})^{2}
= 0.008 + 0.08
= 0.088 \,\text{kg}\,\text{m}^{2}.
\]
Step 4: Total Moment of Inertia of Both Spheres
Since there are two identical spheres, the total moment of inertia is:
\[
I_{\text{total}} = 2 \times 0.088 \,\text{kg}\,\text{m}^{2}
= 0.176 \,\text{kg}\,\text{m}^{2}.
\]
Step 5: Convert to the Desired Form
The problem expresses the answer in multiples of 10^{-3} \,\text{kg}\,\text{m}^{2} :
\[
0.176 \,\text{kg}\,\text{m}^{2} = 176 \times 10^{-3} \,\text{kg}\,\text{m}^{2}.
\]
Final Answer
The moment of inertia of the system about the stated axis is
176 \times 10^{-3}\,\text{kg}\,\text{m}^{2}.
