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? $\frac{1}{6} M a^{2}$ $\frac{5}{12} M a^{2}$ $\frac{17}{36} M a^{2}$ $\frac{1}{3} M a^{2}$

Step 1 ：The moment of inertia of a body about an axis is given by the sum of the products of the masses of its particles and the squares of their distances from the axis.

Step 2 ：The moment of inertia of a body about an axis parallel to and a distance d away from an axis through its center of mass is given by the parallel axis theorem, which states that \(I = I_{cm} + Md^2\), where \(I_{cm}\) is the moment of inertia of the body about an axis through its center of mass, M is the mass of the body, and d is the distance between the two axes.

Step 3 ：In this case, the axis is along one side of the cube, midway between two edges, so the distance d is \(\frac{a}{2}\).

Step 4 ：Given that \(I_{cm} = \frac{1}{6} M a^{2}\) and d = \(\frac{a}{2}\), we can substitute these values into the parallel axis theorem to find the moment of inertia for an axis along one side, midway between two edges.

Step 5 ：Substituting the given values, we get \(I = \frac{1}{6} M a^{2} + M (\frac{a}{2})^{2} = \frac{5}{12} M a^{2}\).

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