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 of a Solid Sphere.
The moment of inertia for a solid sphere (mass \( M \), radius \( R \)) about its diameter is given by:\[I_{\text{sphere (diameter)}} = \frac{2}{5} M R^2.\]Step 2: Moment of Inertia about the Specified Axis.
Applying the Parallel Axis Theorem, the moment of inertia about an axis parallel to the diameter and \( d \) distance from the center is:\[I_{\text{parallel}} = I_{\text{center}} + M d^2.\]Given:- \( d = 10 \, \text{cm} = 0.1 \, \text{m} \) (distance from the center of each sphere to the axis).For each individual sphere, the moment of inertia about the specified axis is:\[I_{\text{sphere (axis)}} = \frac{2}{5} M R^2 + M d^2.\]Substituting \( R = 0.1 \, \text{m} \) 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.
Considering two spheres, and that the axis passes symmetrically through their 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 Result.
The total moment of inertia for the system is:\[\boxed{\frac{14}{5} M R^2}.\]