Rotational Mechanics

The correct answer is: Body diagonal.

We can answer this question without actually calculating the moment of inertia of a cube, just from the general properties of the moment of inertia.

• The moment of inertia in the most general case is a $3 \times 3$ symmetric tensor. If we pick the right frame of reference the tensor can be diagonalized. For a cube the frame of reference that reflects the symmetries of the cube is trivial, it is an $x,y,z$ Descartian system, with the origin in the center of the cube and the 3 axes going through the 6 face centers. The moment of inertia $I$ is the same for rotation around the $x$ or the $y$ or the $z$ axis. (It happens to be $I=\frac{Ma^2}{6}$ , but that is not important).

• If all three diagonal elements of the tensor are the same, than the moment of inertia is isotropic, i.e. it is the same for any axis (that goes through the center of mass). That includes the axis that is along the body diagonal.

• If we parallel-shift the axis of rotation from the center by a distance $d$ the new moment of inertia is $I'=I+d^2M$ , where $M$ is the mass of the object. We can generate the moment of inertia around an edge by shifting the axis of rotation from the center by a distance of $\sqrt{2} (a/2)$ yielding $I_{edge}=I+(a^2/2)M$ . We can generate the moment of inertia around a face diagonal by shifting the axis of rotation by a distance of $a/2$ yielding $I_{face diag.}=I+(a^2/4)M$ . Both of these are larger than $I$ and therefore we have to pick the body diagonal out of the options provided for answers.

In summary, for a cube the moment of inertia is the same for all of the axis that go through the center (similar to a sphere). For any other axis that does not go through the center the moment of inertia is larger.

@Laszlo Mihaly , @Beatriz Sampaio , could u pls post ur solution for this problem ?