Question 1186952
Here's how to calculate the moment of inertia for each scenario:

**Understanding Moment of Inertia**

Moment of inertia (I) is a measure of an object's resistance to rotational acceleration. It depends on the mass distribution of the object and the axis of rotation.  For a point mass *m* rotating at a distance *r* from the axis, I = mr².  For extended objects, we often use the parallel axis theorem: I = I_cm + md², where I_cm is the moment of inertia about the center of mass, *m* is the mass, and *d* is the distance between the axis of rotation and the center of mass.

**1. Moment of Inertia of a Single Sphere**

The moment of inertia of a solid sphere about an axis through its center is (2/5)mr².  In our case:

I_sphere = (2/5) * 2.0 kg * (0.25 m)² = 0.05 kg*m²

**2. Calculations**

**(a) Axis through the center of the square and perpendicular to its plane:**

*   Distance from each sphere to the axis:  Half the diagonal of the square. The diagonal is √2 * side, so half the diagonal is (√2 * 3.00 m) / 2 = 2.12 m (approximately).
*   Using the parallel axis theorem for each sphere: I = I_sphere + m * d² = 0.05 kg*m² + 2.0 kg * (2.12 m)² = 9.03 kg*m²
*   Since there are four spheres: I_total = 4 * 9.03 kg*m² = 36.12 kg*m²

**(b) Axis through one of the masses and perpendicular to its plane:**

*   One sphere is on the axis, so its contribution is just I_sphere.
*   The other three spheres are at distances L, L, and √2 * L from the axis.
*   I_total = I_sphere + m*(0)² + m*(3)² + m*(3)² + m*(3√2)² = 0.05 kg*m² + 2*9 kg*m² + 2*18 kg*m² = 72.05 kg*m² (approximately)

**(c) Axis through two adjacent masses parallel to the plane:**

*   The two spheres on the axis contribute only their I_sphere terms.
*   The other two spheres are at a distance of L from the axis.
*   I_total = 2 * I_sphere + 2 * m * L² = 2 * 0.05 kg*m² + 2 * 2.0 kg * (3.00 m)² = 36.1 kg*m²

**(d) Axis in the plane running diagonally through two masses:**

*   The two spheres on the axis contribute only their I_sphere terms.
*   The other two spheres are at a distance of L/2 from the axis.
*   I_total = 2 * I_sphere + 2 * m * (L/2)² = 2 * 0.05 kg*m² + 2 * 2.0 kg * (1.5 m)² = 9.1 kg*m²

**Summary of Answers (Approximate):**

*   (a) 36.12 kg*m²
*   (b) 72.05 kg*m²
*   (c) 36.1 kg*m²
*   (d) 9.1 kg*m²