Hint 5

(a) For a flat plate the normal ${\bf e}_3$ is always a principal axis. In this case the corresponding principal moment is

$$I_3={5Ma^2\over 3}.$$

To find the other two principal moments and axes we have to solve the $2\times 2$ eigenvalue problem:

$$\pmatrix{2-\lambda&-3\cr -3&8-\lambda}\pmatrix{x_1\cr x_2}={\bf0}.$$

This gives $\lambda=5\pm 3\sqrt 2$ and hence

$$I_1={Ma^2\over 6}(5-3\sqrt 2),\qquad I_2={Ma^2\over 6}(5+3\sqrt 2).$$

The corresponding principal axes lie along the directions

$${\bf x}_1=(1,\,\sqrt2-1,\,0),\qquad {\bf x}_2=(1,\,-\sqrt2-1,\,0).$$

(b) Use the symmetry of the brick to find the principal axes. The corresponding principal moments are

$$I_1={M\over 12}(b^2+c^2),\qquad I_2={M\over 12}(c^2+a^2),\qquad I_3={M\over 12}(a^2+b^2).$$

(c) Taking our coordinate axes to lie parallel to the edges of the cube, with the ${\bf e}_3$ axis lying along the edge on which the fixed point lies, a possible set of principal axes is given by the directions:

$${\bf x}_1=(1,\,1,\,0),\qquad {\bf x}_2=(1,\,-1,\,0),\qquad {\bf x}_3=(0,\,0,\,1).$$

(If you can't deduce these from symmetry considerations, construct the moment of inertia tensor and solve the eigenvalue problem.) The corresponding principal moments are

$$I_1={1\over 6}Ma^2, \qquad I_2={2\over 3}Ma^2, \qquad I_3={2\over 3}Ma^2.$$