I want to calculate the tensor of the moment of inertia. Consider this situation:


The dot represents a points mass, in size equal to $\frac{5}{4}m$. $m$ is the mass of the homogenous circle. I'm trying to calculate the tensor of inertia. Because this is in two dimensions, all components but $I_{xx}$, $I_{yy}$, $I_{zz}$ and $I_{xy}$ are zero. ($\bar{I}$ denotes the inertia around the mass center.)


But this is not the right answer. The right answer is supposed to be $\frac{10}{4}mR^2$, why?

$I_{zz}$ and $I_{yy}$ I can get for some reason, by the above method. For those it works.



These are correct. The last one is a bit easier because the axis passes through the mass center.

And then there's $I_{xy}$ which I cannot figure out how to calculate since I don't know the x and y position of the point mass, so I can't use the formula $I_{xy}=md_xd_y$. How would I calculate it? The answer is supposed to be $I_{xy}=-\frac{5mR^2}{4}$.


You don't need to apply Steiner's theorem onto the point mass. The point mass finds itself at a distance (apparently) $R$ of the x-axis. Since the moment of inertia is an extensive value, you can simply add all moments of inertia.

There's the moment of inertia of the solid disk with respect to it's diameter. You have to 'Steiner' that away from a distance $R$. Then you need to add $\frac{5}{4} mR^2$ which is the moment of inertia associated to a point-mass of mass $\frac{5}{4}m$ a distance $R$ away from the rotation axis

Therefore, $I_{xx} = \frac{mR^2}{4} + mR^2 + \frac{5}{4}mR^2 = \frac{10}{4}mR^2$

From the drawing, I strongly suppose the point mass finds itself at $(R,R)$, therefore lying on the diagonal. You should be able to conclude what the moment of Inertia is then. The minus sign you give for the supposed answer seems highly suspiciou, as the moment of inertia is a sum of positive quantities.


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