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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:

\[
I = \rho \int_0^L \int_0^L \int_0^L (x^2 + y^2) \, dx \, dy \, dz
= \rho \int_0^L \int_0^L \left( \int_0^L x^2 \, dx + \int_0^L y^2 \, dy \right) dz
\]

Since the integrals over \( x \) and \( y \) are independent of \( z \), they can be simplified:

\[
I = \rho \int_0^L dz \left( \int_0^L x^2 \, dx + \int_0^L y^2 \, dy \right)
\]

Evaluating these integrals:

\[
\int_0^L x^2 \, dx = \frac{L^3}{3}, \quad \int_0^L y^2 \, dy = \frac{L^3}{3}
\]

So,

\[
I = \rho \int_0^L dz \left( \frac{L^3}{3} + \frac{L^3}{3} \right) = \rho \int_0^L dz \left( \frac{2L^3}{3} \right)
= \frac{2L^3}{3} \cdot \rho \cdot L
= \frac{2L^4}{3} \cdot \frac{M}{L^3} = \frac{2ML^2}{3}
\]

Thus, the moment of inertia of the cube rotating about an axis through one of its edges is:

\[
I = \frac{2ML^2}{3}
\]

---

For **The Cube II**, the moment of inertia is calculated about an axis through its center (and the center of one of its faces). Using the parallel axis theorem, we would expect the moment of inertia for the cube rotating about its center to be smaller compared to that along its edge, due to the mass being distributed closer to the axis.

Would you like further details on the second part, or have any questions so far?

Here are 5 questions to expand on this information:
1. What is the physical meaning of moment of inertia in rotational dynamics?
2. How does the choice of axis affect the moment of inertia of an object?
3. Can the parallel axis theorem be applied to any arbitrary shape?
4. How would the moment of inertia change if the mass distribution of the cube was non-uniform?
5. What are other common shapes (like a sphere or cylinder) where the moment of inertia plays a key role in rotational dynamics?

**Tip:** When calculating moments of inertia, always ensure you're clear about the axis of rotation since it can drastically change the result.

A wooden cube with side length L and total mass M is rotating about an axis passing through one of its edges. Use an integral to calculate the moment of inertia.

Solve the moment of inertia for a wooden cube rotating about an axis along one of its edges using integral calculus. Detailed step-by-step solution for physics students at the university level.

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