Moment of inertia of a cube

Classical Mechanics Level 2

What is the moment of inertia of a cube with uniform mass distribution, if the axis of rotation is its body diagonal? The cube has an edge of length $a$ and mass $M$ . The result can be written as $I= \frac{\alpha}{\beta}Ma^2$ . Select the value of $\alpha+\beta$ as the answer.

Notes

The moment of inertia of a thin square around an axis that is perpendicular to the square and goes through its center is $\frac{1}{6}Ma^2$ .
Try to solve this problem without doing complex calculations.

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1 solution

Laszlo Mihaly
Oct 13, 2017

The moment of inertia of a cube around any axis that goes through its center is $\frac{1}{6}Ma^2$ .

First, if the axis goes through two opposing face centers, the moment of inertia is the same as the moment of inertia of a square. Just imagine that you start with a thin square and stack more of them to the axis until they make a cube.

Second, by symmetry, the moment of inertia around the 3 perpendicular axes that we can have this way are all equal to the same value. That means the moment of inertia tensor is diagonal and all of the diagonal elements are equal.

Third, if the moment of inertia tensor is diagonal and all of the diagonal elements are equal then the moment of inertia is totally isotropic (as long as the axis goes through the center).

Moment of inertia of a cube

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