Let’s go over the calculation for Part C carefully, where we need to find the moment of inertia of the combination about an axis parallel to the bar through both balls. Since the axis is parallel to the bar and passes through each of the balls at either end, this setup means:

1. For Part C, the bar itself does not contribute to the moment of inertia because the axis is parallel to the bar and coincides with the line containing both balls.
2. The moment of inertia will therefore only include the contributions from the two point masses at the ends of the bar.

Solution

Step 1: Calculate the Moment of Inertia for Each Ball

For a point mass, the moment of inertia $I$ about an axis at a perpendicular distance $r$ is: $I = m r^2$ However, in this case, each ball is located directly on the axis, meaning the distance $r = 0$.

Thus, the moment of inertia for each ball about this axis is: $I_{\text{ball}} = 0$

Step 2: Total Moment of Inertia

Since both balls are positioned on the axis of rotation, the moment of inertia contributed by each ball is zero. Consequently, the total moment of inertia about this axis is: $I_{\text{total}} = 0$

Final Answer

$I = 0 \, \text{kg} \cdot \text{m}^2$

Would you like a more detailed breakdown, or do you have further questions?

Here are five follow-up questions to deepen understanding:

1. Why does positioning masses directly on the axis of rotation result in zero