To solve the first part of the problem (The Cube I), we need to calculate the moment of inertia of a wooden cube with side length $L$ and mass $M$, rotating about an axis passing through one of its edges.

Step 1: General formula for moment of inertia

For a body rotating about a specific axis, the moment of inertia $I$ is given by:

$I = \int r^2 \, dm$

Where:

• $r$ is the distance from the axis of rotation,
• $dm$ is a small mass element.

Step 2: Setup the problem

We know the cube has side length $L$, and we want to rotate it around an axis along one of its edges.

1. Coordinate system: Let's place the cube in a 3D Cartesian coordinate system. The axis of rotation can be chosen along the $z$-axis, running through the edge of the cube at $x = 0$ and $y = 0$.

2. Mass element $dm$: The mass density $\rho$ is given by:

$\rho = \frac{M}{L^3}$

A small volume element $dV$ is $dx \, dy \, dz$, and the corresponding mass element is $dm = \rho \, dV = \rho \, dx \, dy \, dz$.

Step 3: Expression for the distance from the axis

The distance $r$ from the axis of rotation (along one edge) to the mass element at coordinates $(x, y, z)$ is given by:

$r^2 = x^2 + y^2$

Since the axis is along the $z$-axis, we ignore the $z$-coordinate for this calculation.

Step 4: Integral setup

Now, we integrate over the entire volume of the cube:

$I = \rho \int_0^L \int_0^L \int_0^L (x^2 + y^2) \, dx \, dy \, dz$

Step 5: Simplify and solve

This integral breaks down as: