Question

Two spheres are given with radius \( r = 10 \, \text{cm} \), and the distance between them is \( 20 \, \text{cm} \). The axis passes through the midpoint of the distance between the two spheres. What is the moment of inertia of the system?

Hint:

Use the Parallel Axis Theorem to compute the moment of inertia about axes other than the center or diameter. Always account for symmetry when dealing with multiple objects.

Answer 1

Step 1: Moment of Inertia for a Solid Sphere.
The moment of inertia of a solid sphere of mass \( M \) and radius \( R \) about its diameter is: \[ I_{\text{sphere (diameter)}} = \frac{2}{5} M R^2. \] Step 2: Moment of Inertia about the Given Axis (Parallel Axis Theorem).
Using the Parallel Axis Theorem, the moment of inertia about an axis parallel to the diameter and at a distance \( d \) from the center is: \[ I_{\text{parallel}} = I_{\text{center}} + M d^2. \] Here: - \( d = 10 \, \text{cm} = 0.1 \, \text{m} \) (distance from the center of each sphere to the axis). For each sphere, the moment of inertia about the given axis is: \[ I_{\text{sphere (axis)}} = \frac{2}{5} M R^2 + M d^2. \] Substitute \( R = 0.1 \, \text{m} \) and \( d = 0.1 \, \text{m} \): \[ I_{\text{sphere (axis)}} = \frac{2}{5} M (0.1)^2 + M (0.1)^2. \] \[ I_{\text{sphere (axis)}} = \frac{2}{5} M (0.01) + M (0.01) = \frac{2}{5} M (0.01) + \frac{5}{5} M (0.01). \] \[ I_{\text{sphere (axis)}} = \frac{7}{5} M (0.01). \] Step 3: Total Moment of Inertia for Two Spheres.
Since there are two spheres and the axis passes symmetrically through the midpoint, the total moment of inertia is: \[ I_{\text{total}} = 2 \times I_{\text{sphere (axis)}} = 2 \times \frac{7}{5} M R^2. \] \[ I_{\text{total}} = \frac{14}{5} M R^2. \] Step 4: Final Answer.
The total moment of inertia of the system is: \[ \boxed{\frac{14}{5} M R^2}. \]