Lines in the XY Plane
Table of Contents
- Preamble
- General
- At the Origin
- Endpoints on the x Axis and y Axis
- Parallel to x Axis and Perpendicular to y Axis
Preamble
In the previous post, we found the general expression for the moment of inertia of a line. In this post, we are going to take the special case where the line is restricted to the $xy$ plane and derive some simplified formulas. This will be help in the next post on triangles. This is meant to be more of a reference, which is why I have provided a table of contents.
One interesting thing to notice in the inertia tensors is that $I_{zz} = I_{xx} + I_{yy}$. This is true for any object restricted to the $xy$ plane (not just lines). You can easily prove this using the definitions of these values.
General
Recall the general inertia tensor of a line from the previous post. Since we are restricted to the $xy$ plane, $\alpha_z = \pi/2$. Therefore $\sin \alpha_z = 1$ and $\cos \alpha_z = 0$. Furthermore, there is now a much more direct relationship between $\alpha_x$ and $\alpha_y$. It is a bit redundant to use both. Therefore, we use the standard convention from cylindrical and polar coordinate and denote $\phi_L = \alpha_x = \pi/2 - \alpha_y$. We do likewise with $\beta_x$ and $\beta_y$ and denote $\phi_{\text{cm}} = \beta_x = \pi/2 - \beta_y$.
Therefore, the inertia tensor is the following. Recall that $D = \abs{\b{r}_{\text{cm}}}$.
\[\m{I} = \tfrac{1}{12} M L^2 \begin{bmatrix} \sin^2 \phi_L & - \cos \phi_L \sin \phi_L & 0 \\ - \sin \phi_L \cos \phi_L & \cos^2 \phi_L & 0 \\ 0 & 0 & 1 \end{bmatrix} + M D^2 \begin{bmatrix} \sin^2 \phi_{\text{cm}} & - \cos \phi_{\text{cm}} \sin \phi_{\text{cm}} & 0 \\ - \sin \phi_{\text{cm}} \cos \phi_{\text{cm}} & \cos^2 \phi_{\text{cm}} & 0 \\ 0 & 0 & 1 \end{bmatrix}\]Note For the signs of the product of inertia values to be correct, we need to measure $\phi_L$ and $\phi_{\text{cm}}$ using the standard conventions (positive side of the $x$-axis, counterclockwise).
Of course we can also write this in terms of lengths instead of angles.
\[\m{I} = \tfrac{1}{12} M \begin{bmatrix} L_y^2 & - L_x L_y & 0 \\ - L_y L_x & L_x^2 & 0 \\ 0 & 0 & L^2 \end{bmatrix} + M \begin{bmatrix} y_{\text{cm}}^2 & - x_{\text{cm}} y_{\text{cm}} & 0 \\ - y_{\text{cm}} x_{\text{cm}} & x_{\text{cm}}^2 & 0 \\ 0 & 0 & D^2 \end{bmatrix}\]Note for the product of inertia values to be correct, we need to measure pairs $(L_x, L_y)$ and $(x_{\text{cm}}, y_{\text{cm}})$ using the standard conventions (right and up are positive).
Coordinate Formula
Let $\b{r}_1 = (x_1, y_1, 0)$ and $\b{r}_2 = (x_2, y_2, 0)$ be the endpoints of the line.
Recall the formula for the moments of inertia and product of inertia from the previous post. Applying these to the special case of a triangle on the $xy$ plane gives the following.
\[\m{I} = \tfrac{1}{3} M \begin{bmatrix} y_1^2 + y_2^2 + y_1y_2 & - \tfrac{1}{2}[ (x_2 + x_1) (y_2 + y_1) + x_1y_1 + x_2y_2 ] & 0 \\ - \tfrac{1}{2}[ (x_2 + x_1) (y_2 + y_1) + x_1y_1 + x_2y_2 ] & x_1^2 + x_2^2 + x_1x_2 & 0 \\ 0 & 0 & \abs{\b{r}_1}^2 + \abs{\b{r}_2}^2 + \b{r}_1 \cdot \b{r}_2 \end{bmatrix}\]This is nice because it’s a pure coordiante definition of a line (which will come in handy when we analyze general triangles in the next post). Using the law of cosines \(\abs{\b{r}_2 - \b{r}_1}^2 = \abs{\b{r}_1}^2 + \abs{\b{r}_2}^2 - 2(\b{r}_1 \cdot \b{r}_2) \\[10pt] \b{r}_1 \cdot \b{r}_2 = \tfrac{1}{2} \left ( \abs{\b{r}_1}^2 + \abs{\b{r}_2}^2 - \abs{\b{r}_2 - \b{r}_1}^2 \right )\) , then we can get rid of that dot product.
\[\m{I} = \tfrac{1}{3} M \begin{bmatrix} 3y_1^2 + 3y_2^2 - (y_2 - y_1)^2 & - \tfrac{1}{2}[ (x_2 + x_1) (y_2 + y_1) + x_1y_1 + x_2y_2 ] & 0 \\ - \tfrac{1}{2}[ (x_2 + x_1) (y_2 + y_1) + x_1y_1 + x_2y_2 ] & 3x_1^2 + 3x_2^2 - (x_2 - x_2)^2 & 0 \\ 0 & 0 & 3\abs{\b{r}_1}^2 + 3\abs{\b{r}_2}^2 - \abs{\b{r}_2 - \b{r}_1}^2 \end{bmatrix}\] \[I_{zz} = \tfrac{1}{6} M \left ( 3\abs{\b{r}_1}^2 + 3\abs{\b{r}_2}^2 - L^2 \right ) = \lambda \cdot \tfrac{1}{6} \abs{\b{r}_2 - \b{r}_1} \left ( 3\abs{\b{r}_1}^2 + 3\abs{\b{r}_2}^2 - \abs{\b{r}_2 - \b{r}_1}^2 \right )\]This is cool because we will see a very similar looking for the moment of inertia of a triangular surface. If you can find a similarly nice way to expression $I_{xx}$, $I_{yy}$, and $I_{xy}$ then please let me know.
At the Origin
Center of Mass at the Origin
In this case, $\phi_{\text{cm}} = \phi_L$ and $D = 0$.
Simplifying from the general case on the $xy$ plane we just solved for in the previous section.
\[\begin{align} \m{I} &= \tfrac{1}{12} M L^2 \begin{bmatrix} \sin^2 \phi_L & - \cos \phi_L \sin \phi_L & 0 \\ - \sin \phi_L \cos \phi_L & \cos^2 \phi_L & 0 \\ 0 & 0 & 1 \end{bmatrix} = \tfrac{1}{12} M \begin{bmatrix} L_y^2 & - L_x L_y & 0 \\ - L_y L_x & L_x^2 & 0 \\ 0 & 0 & L^2 \end{bmatrix} \end{align}\]Endpoint at the Origin
In this case, $\phi_{\text{cm}} = \phi_L$ and $D = L/2$.
Simplifying from the general case on the $xy$ plane we just solved for in the previous section.
\[\begin{align} \m{I} &= \tfrac{1}{12} M L^2 \begin{bmatrix} \sin^2 \phi_L & - \cos \phi_L \sin \phi_L & 0 \\ - \sin \phi_L \cos \phi_L & \cos^2 \phi_L & 0 \\ 0 & 0 & 1 \end{bmatrix} + \tfrac{1}{4} M L^2 \begin{bmatrix} \sin^2 \phi_L & - \cos \phi_L \sin \phi_L & 0 \\ - \sin \phi_L \cos \phi_L & \cos^2 \phi_L & 0 \\ 0 & 0 & 1 \end{bmatrix} \\[10pt] &= \tfrac{1}{3} M L^2 \begin{bmatrix} \sin^2 \phi_L & - \cos \phi_L \sin \phi_L & 0 \\ - \sin \phi_L \cos \phi_L & \cos^2 \phi_L & 0 \\ 0 & 0 & 1 \end{bmatrix} \\[10pt] &= \tfrac{1}{3} M \begin{bmatrix} L_y^2 & - L_x L_y & 0 \\ - L_y L_x & L_x^2 & 0 \\ 0 & 0 & L^2 \end{bmatrix} \end{align}\]Endpoints on the $x$ Axis and $y$ Axis
In this case, $\phi_{\text{cm}} = \pi-\phi_L$ and $D = L/2$. To convince yourself that is true requires a bit of geometry.
Simplifying from the general case on the $xy$ plane we just solved for in the previous section.
\[\begin{align} \m{I} &= \tfrac{1}{12} M L^2 \begin{bmatrix} \sin^2 \phi_L & - \cos \phi_L \sin \phi_L & 0 \\ - \sin \phi_L \cos \phi_L & \cos^2 \phi_L & 0 \\ 0 & 0 & 1 \end{bmatrix} + \tfrac{1}{4} M L^2 \begin{bmatrix} \sin^2 \phi_L & \cos \phi_L \sin \phi_L & 0 \\ \sin \phi_L \cos \phi_L & \cos^2 \phi_L & 0 \\ 0 & 0 & 1 \end{bmatrix} \\[10pt] &= \tfrac{1}{3} M L^2 \begin{bmatrix} \sin^2 \phi_L & \tfrac{1}{2} \cos \phi_L \sin \phi_L & 0 \\ \tfrac{1}{2} \sin \phi_L \cos \phi_L & \cos^2 \phi_L & 0 \\ 0 & 0 & 1 \end{bmatrix} \\[10pt] &= \tfrac{1}{3} M \begin{bmatrix} L_y^2 & \tfrac{1}{2} L_x L_y & 0 \\ \tfrac{1}{2} L_y L_x & L_x^2 & 0 \\ 0 & 0 & L^2 \end{bmatrix} \end{align}\]Be careful with the sign of $L_x$ and $L_y$ when applying the formula. For instance, in the above diagram $L_x$ is negative while $L_y$ is positive.
Parallel to $x$ Axis and Perpendicular to $y$ Axis
In these special cases, we assume $\phi_L = 0$, i.e. the line is parallel to the $x$-axis. Of course, if we instead assume that $\phi_L = \pi/2$ (the line is parallel to the $y$-axis), then the inertia tensors can be derived symmetrically.
General
First, we consider the general case.
Simplifying from the general case on the $xy$ plane we just solved for in the previous section. Recall that $D = \abs{\b{r}_{\text{cm}}}$.
\[\m{I} = \tfrac{1}{12} M L^2 \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} + M D^2 \begin{bmatrix} \sin^2 \phi_{\text{cm}} & - \cos \phi_{\text{cm}} \sin \phi_{\text{cm}} & 0 \\ - \sin \phi_{\text{cm}} \cos \phi_{\text{cm}} & \cos^2 \phi_{\text{cm}} & 0 \\ 0 & 0 & 1 \end{bmatrix}\]Again, we can instead write this in terms of lengths.
\[\m{I} = \tfrac{1}{12} M L^2 \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} + M \begin{bmatrix} y_{\text{cm}}^2 & - x_{\text{cm}} y_{\text{cm}} & 0 \\ - y_{\text{cm}} x_{\text{cm}} & x_{\text{cm}}^2 & 0 \\ 0 & 0 & 1 \end{bmatrix}\]Centered at a Distance
Now we assume the line intersects the axis at its center of mass. Therefore $\phi_{\text{cm}} = \pi/2$.
Simplifying from the previous result. Recall that $D = \abs{\b{r}_{\text{cm}}}$.
\[\m{I} = \tfrac{1}{12} M L^2 \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} + M D^2 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} = M \begin{bmatrix} D^2 & 0 & 0 \\ 0 & \tfrac{1}{12} L^2 & 0 \\ 0 & 0 & \tfrac{1}{12} L^2 + D^2 \end{bmatrix}\]Center on an Axis
Finally, we assume the center of mass is at the origin ($D = 0$).
We can either simplify from the previous result, or from the Center of Mass at the Origin result.
\[\m{I} = \tfrac{1}{12} M L^2 \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\]This gives us the standard formula you will see for the moment of inertia of a line.