Four identical solid spheres each of mass M and radius are fixed at four corners of a light square frame of side length `b` such that centres of spheres coincide with corners of square. Find out the moment of inertia of system about one side of square frame

Answer

To find the moment of inertia of the system of four identical solid spheres about one side of the square frame, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Configuration**:
We have four identical solid spheres, each of mass \( M \) and radius \( R \), fixed at the corners of a square frame of side length \( b \). The centers of the spheres coincide with the corners of the square.

2. **Identify the Axis of Rotation**:
We need to calculate the moment of inertia about one side of the square frame. Let's consider the side that connects two of the spheres (let's call them A and B).

3. **Moment of Inertia of Spheres A and B**:
For spheres A and B, the axis of rotation passes through their centers. The moment of inertia \( I_{AB} \) about this axis can be calculated using the formula for the moment of inertia of a solid sphere about an axis through its center:
\[ I = \frac{2}{5} M R^2 \]
Since we have two spheres (A and B), the total moment of inertia for A and B is:
\[ I_{AB} = I_A + I_B = \frac{2}{5} M R^2 + \frac{2}{5} M R^2 = \frac{4}{5} M R^2 \]

4. **Moment of Inertia of Spheres C and D**:
For spheres C and D, we need to use the parallel axis theorem since the axis of rotation does not pass through their centers. The distance from the center of each sphere to the axis is \( b \) (the side length of the square).
The moment of inertia about the center of each sphere is \( \frac{2}{5} M R^2 \). Using the parallel axis theorem:
\[ I_{C} = \frac{2}{5} M R^2 + M b^2 \]
\[ I_{D} = \frac{2}{5} M R^2 + M b^2 \]
Therefore, the total moment of inertia for spheres C and D is:
\[ I_{CD} = I_C + I_D = \left(\frac{2}{5} M R^2 + M b^2\right) + \left(\frac{2}{5} M R^2 + M b^2\right) = \frac{4}{5} M R^2 + 2 M b^2 \]

5. **Total Moment of Inertia**:
Now, we can find the total moment of inertia \( I \) about the chosen side of the square frame:
\[ I = I_{AB} + I_{CD} = \frac{4}{5} M R^2 + \left(\frac{4}{5} M R^2 + 2 M b^2\right) \]
Simplifying this gives:
\[ I = \frac{8}{5} M R^2 + 2 M b^2 \]

### Final Result:
The moment of inertia of the system about one side of the square frame is:
\[ I = \frac{8}{5} M R^2 + 2 M b^2 \]