Well I wonder if there's an "easy" (or at least solveable) way to convert the second moment of inertia around any arbitrary angle.
The area moment of inertia to an axis B is defined by $r$ being the distance to axis B:
$$ I_B = \int_A r^2 dA$$
Or:
$$I_x = \iint_A y^2 dy dx $$
$$I_y = \iint_A x^2 dx dy $$
I know of the perpendicular axis theorem and the parallel axis theorem (and the stretch rule). But I wonder if using these I can actually convert any area moment of inertia to any axis.
More to the point, say I have an irregular shape, I know that $I_x = C_0$ and $I_y = C_1$. Now using this can I get the area moment of inertia around an axis which makes an angle of $\alpha$ with the $x$ axis?
Best Answer
If the shape is located and oriented in a "good" way relative to the $x$- and $y$-axes, the answer is yes, you can derive the second moment around any axis in the plane given only $I_x$ and $I_y.$
More generally, for an arbitrary shape at an arbitrary location and orientation, you will also need the product moment of inertia, $$I_{xy} = -\iint_A xy \, dx dy.$$ It's also conventional in this context to write $I_{xx}$ and $I_{yy}$ instead of $I_x$ and $I_y.$ Using this notation, then for a shape lying entirely in the $x,y$-plane with the centroid of the shape at $(0,0),$ the moment of inertia around an axis given by a unit vector $\hat n$ can be computed (using matrix notation) by the use of the inertia tensor, $I,$ as follows: $$I_{\hat n} = \hat n^T I \hat n = \hat n^T \begin{pmatrix} I_{xx} & I_{xy} & 0 \\ I_{xy} & I_{yy} & 0 \\ 0 & 0 & I_{xx} + I_{yy} \end{pmatrix} \hat n. $$ Note that $\hat n$ can be, but is not required to be, a vector in the $x,y$-plane. You can use the parallel axis theorem to get the moment of inertia around an axis parallel to $\hat n$ that does not pass through the shape's centroid.
This is a special case of a more general formula for the moment of inertia of a three-dimensional object around the axis through the centroid of the object in the direction of the vector $\hat n$: $$I_{\hat n} = \hat n^T \begin{pmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{xy} & I_{yy} & I_{yz} \\ I_{xz} & I_{yz} & I_{zz} \end{pmatrix} \hat n. $$ (These notes offer this notation; also see this page and the page that follows it, or these lecture notes and the notes for the following lecture.)
For an object entirely in the $x,y$-plane, the $z$-coordinate of any point is zero, hence $I_{xz} = I_{yz} = 0,$ and we can use the perpendicular axis theorem to substitute for $I_{zz}.$
You can simplify the calculation further, however, by finding the principal axes of your object. This is a set of orthogonal axes such that when the axis of rotation is described by a vector $\hat n$ relative to those axes, the moment of inertia is simply $$I_{\hat n} = \hat n^T \begin{pmatrix} I_{xx} & 0 & 0 \\ 0 & I_{yy} & 0 \\ 0 & 0 & I_{zz} \end{pmatrix} \hat n $$ when the principal axes are taken as the $x$-, $y$-, and $z$-axes. That is, if you place your object on your coordinate plane so that its principal axes coincide with your $x$- and $y$-axes, or conversely choose $x$- and $y$-axes that correspond with the object's principal axes, then you can derive the moment of inertia around any axis merely by knowing its moments of inertia around the two axes. But that assumes you are able to find the principal axes in the first place, which (for an irregular body) may involve computing $I_{xy}$ in some initially chosen coordinate system and then finding the eigenvectors of the inertia tensor in that coordinate system, which indicate the principal axes.
A possibly useful fact: if your planar object has any bilateral symmetry, the axis of symmetry is one of your principal axes. That's easy to confirm by setting one of your coordinate axes to the axis of symmetry and computing $I_{xy}$ in that coordinate system.