Question

Four spheres each of mass \( M \) and radius \( R \) are placed with their centres on the corners of a square of side \( L \). The moment of inertia of the system about any side of the square is

• A. \( \frac{4}{3} MR^2 + ML^2 \)
• B. \( \frac{3}{5} MR^2 + 2ML^2 \)
• C. \( \frac{8}{5} MR^2 + 2ML^2 \)
• D. \( \frac{6}{5} MR^2 + ML^2 \)

Hint:

To calculate the moment of inertia of objects arranged in a specific configuration, consider both the individual moments and use the parallel axis theorem for objects not centered on the axis of rotation.

Answer 1

The Correct Option is C

Step 1: Moment of inertia of a single sphere.
The moment of inertia of a single sphere about an axis passing through its center is given by: \[ I_{\text{sphere}} = \frac{2}{5} m R^2 \] where \( m \) is the mass and \( R \) is the radius of the sphere. Step 2: Moment of inertia of spheres at the corners.
Each sphere is placed at a corner of the square, and the axis of rotation is along one side of the square. For each sphere, we use the parallel axis theorem to calculate the moment of inertia about the axis passing through the side of the square: \[ I_{\text{total}} = I_{\text{sphere}} + md^2 \] where \( d \) is the distance from the center of the sphere to the axis. In this case, \( d = \frac{L}{\sqrt{2}} \). Step 3: Calculate the total moment of inertia.
There are four spheres, and we calculate the total moment of inertia by summing the contributions from each sphere: \[ I_{\text{total}} = 4 \left( \frac{2}{5} MR^2 + M \left( \frac{L}{\sqrt{2}} \right)^2 \right) \] Simplifying: \[ I_{\text{total}} = 4 \left( \frac{2}{5} MR^2 + \frac{ML^2}{2} \right) \] \[ I_{\text{total}} = \frac{8}{5} MR^2 + 2ML^2 \] Step 4: Conclusion.
Thus, the moment of inertia of the system about any side of the square is \( \frac{8}{5} MR^2 + 2ML^2 \), which corresponds to option (C).