# How do I convert cardinal axis inertias to an arbitrary axis inertia? I've tried to reduce this problem to the most basic form. I know the equation for a spring-mass system is generally:

$$F_{\mbox{applied}} = kx + m\ddot{x} \\$$

but, in my case, I have a spring that's not aligned with any particular axis. The load is a generic rectangle with a mass $m$ and a moment of inertia $J$. I am attempting to control the mass by applying a force in line with the spring. I know the vector that defines the orientation of the spring, and I know the vector that defines the orientation of the spring's attachment to the center of mass of the load.

My trouble is that I know the load is going to translate along $x$, translate along $y$, and rotate about $z$ (giving some rotation angle $\psi$). I'm pretty sure I can convert from the cardinal axes to the spring axis to find spring position $s$ and spring acceleration $\ddot{s}$, but I'm not sure how I convert the mass and inertia to an equivalent inertia $J_{\mbox{eq}}$ for the spring equation. What I'd like to have is:

$$F_{\mbox{applied}} = ks + J_{\mbox{eq}}\ddot{x} \\$$

I think I convert to $s$ coordinates with the dot product. First I find the tangential velocity of the spring's attachment point by:

$$v_{\mbox{from rotation}} = \mbox{cross}(\omega, r) \\ v_{\mbox{from rotation}} = \mbox{cross}(\dot{\psi}, r) \\$$

then I add that to the translational information:

$$v_{\mbox{from translation}} = <\dot{x}, \dot{y}, 0> \\$$

to get:

$$v_{\mbox{total}} = v_{\mbox{from translation}} + v_{\mbox{from rotation}} \\$$

and then finally convert to the spring coordinates with:

$$\dot{s} = \mbox{dot}(v_{\mbox{total}}, \hat{s}) \\$$

Position and acceleration are the same to convert cardinal to spring coordinates.

My question is, do I do the same with inertias? What would that look like? My guess is to setup the moment of inertia as $<0, 0, J>$, cross that with the moment arm $r$ to get a $J_{\mbox{rot eq}} = \mbox{cross}(J, r)$, then try to do the same steps with $J_{\mbox{translational}} = <m, m, 0>$, then dot product again to convert to spring coordinates, but this doesn't feel like the right answer.

I'm a little lost, and I'm not sure what to call this kind of a conversion either, so I'm having a hard time searching for results online. I'm not looking for a worked out solution, just a couple tips/pointers in the right direction would be a tremendous help.

• I'll comment too that I've been looking at the moment of inertia tensor, but my understanding of that is that it's meant for entirely rotational inertias. That is, there is still no "conversion" between rotational and translational inertia via the inertia tensor. – Chuck Jun 29 '18 at 12:58
• The block is pinned to the center and your problem is two-dimensional, yes? – aghostinthefigures Jul 1 '18 at 10:53
• @aghostinthefigures - assume the center is not pinned. If the center were pinned, there could be no x- and y- translation and the whole problem would reduce to the tangential component of the applied/spring forces. – Chuck Jul 1 '18 at 10:57
• But yes, the problem is two dimensional, horizontal with no gravity. – Chuck Jul 1 '18 at 11:12
• Last question before I take a crack at it; does the spring have a non-zero relaxation length? – aghostinthefigures Jul 1 '18 at 14:48