Rotating sheet

$I_{z}=\int{(x^{2}+y^{2})dm}=\int{x^{2}dm}+\int{y^{2}dm}=I_{y} +I_{x}=10$ (1)

for the moment of inertia with respect to the diagonal line.

The axis satisfies the equation $\frac{y}{5} +\frac{x}{2} =0$ thus the distance from a point (x,y) to the axis is $r= \frac{ |\frac{y}{5} + \frac{x}{2}|}{sqrt{(1/5)^{2}+(1/2)^{2}}}$ .

$I=\int{r^{2}dm}=\int{dm(\frac{y}{5} +\frac{x}{2})^{2}\frac{100}{29}}$

$=\frac{100}{29}(\int{y^{2}dm/25}+\int{x^{2}dm/4}+\int{xydm/5}) =\frac{100}{29}(I_{x}/25+I_{y}/4)$

( $\int{xydm/5}=0$ because of symmetry).

So $I_{x}/25+I_{y}/4=\frac{29}{20}$ (2).

Solve (1) and (2) we obtain $I_{y}=5$ .

Choose another 3D coordinate Ox'y'z with Oy' is one of the diagonal lines of the rectangle sheet, and the Oxy plane is the same with Ox'y' plane. Using the theorem of perpendicular axes, we get $I_z=I_o=I_{x'}+I_{y'}=I_{x'}+I \Rightarrow I_{x'}=I_o-I=5~\mbox{kg}\cdot\mbox{m}^2=I$ .

In detail: the distance between a point $i$ in the sheet and the Oz axis is $r_i=\sqrt{{{x'}_i}^2+{{y'}_i}^2}$ , and the equation for the inertial angular momentum is:

$I_z=\sum m_i r_i^2=\sum m_i {{x'}_i}^2 + \sum m_i {{y'}_i}^2=I_{y'} + I_{x'}$

OK, good. We get $\sum m_i {{x'}_i}^2 =\sum m_i {{y'}_i}^2=I_{y'}=I_{x'}=I$

Now let's consider an axis in Ox'y' that passes through O and make an angle $c$ with Ox' axis; the equation for the inertial momentum of that axis $I'$ is:

$I'=\sum m_i {r'}_i^2=\sum m_i {{x'}_i \cos(c) + {y'}_i \sin(c)}^2$

$\Rightarrow I'=\cos^2(c) \sum m_i {{x'}_i}^2 + \sin^2(c) \sum m_i {{y'}_i}^2+2\sin(c)\cos(c) \sum m_i {{x'}_i}{{y'}_i}$

$\Rightarrow I'=I+2\sin(c)\cos(c) \sum m_i {{x'}_i}{{y'}_i}$

So if there is one more axis (that also passes through O and in the Õ'y' plane) with the same inertial angular momentum $I$ , it means $\sum m_i {{x'}_i}{{y'}_i}=0$ and all axis in Õ'y' plane that passes through O have the same inertial angular momentum $I$ .

It means the inertial angular momentum along Oy is just equal to $I=5~\mbox{kg}\cdot\mbox{m}^2$ .