A uniform density mass $M$ cube has side length $a$ and moment of inertia through its center of mass and perpendicular to two sides $I_{c m}=\frac{1}{6} M a^{2}$. What is the moment of inertia for an axis along one side, midway between two edges as labeled with the rotation arrow in the diagram?

Step 1 ：First, we need to understand the Parallel Axis Theorem. The Parallel Axis Theorem states that the moment of inertia about an axis parallel to and a distance d away from an axis through the center of mass is given by \(I = I_{cm} + Md^2\), where \(I_{cm}\) is the moment of inertia about the center of mass, M is the mass of the object, and d is the distance between the two axes.

Step 2 ：In this problem, we are given that the moment of inertia through the center of mass \(I_{cm} = \frac{1}{6}Ma^2\), and we are asked to find the moment of inertia for an axis along one side, midway between two edges. This axis is a distance of \(\frac{a}{2}\) away from the center of mass.

Step 3 ：Substituting the given values into the Parallel Axis Theorem, we get \(I = \frac{1}{6}Ma^2 + M(\frac{a}{2})^2\).

Step 4 ：Solving this equation, we get \(I = \frac{1}{6}Ma^2 + \frac{1}{4}Ma^2\).

Step 5 ：Adding these two terms together, we get \(I = \frac{5}{12}Ma^2\).

Step 6 ：So, the moment of inertia for an axis along one side, midway between two edges is \(\boxed{\frac{5}{12}Ma^2}\).