# Would equations for a spinning top be an (x,y,z) vector [duplicate]

I am following the equations on this page, and for torque it is $mgr\sin\theta$, but I am curious about $r$.

I am working on a game and I want to correctly model the top, and am curious if $r$ should be the distance from the tip of the top to the center of mass as an $(x,y,z)$ vector, or should it just be the scalar version?

• What is your level of understanding of vector mechanics and dynamics? – John Alexiou Sep 1 '14 at 21:33
• See physics.stackexchange.com/a/74484/392 for how to apply the equations of motion. – John Alexiou Sep 1 '14 at 21:34
• possible duplicate of Derivation of Newton-Euler equations of motion – John Alexiou Sep 1 '14 at 21:35
• @ja72 - I thought my understanding was pretty good, but it has been years since I studied physics, but when I look at how to solve these equations I realized I will have some questions. – James Black Sep 1 '14 at 23:27
• Please post your questions in Physics as you encounter them. – John Alexiou Sep 2 '14 at 11:14

If you are just interested in the rate of precession, then $r$ just has to be the distance from the support (tip of the top) to the center of mass - this is the distance that gives rise to the torque through $mgr\sin\phi$
However, if you want to do this as a vector equation, then you have the instantaneous angular momentum (vector) of the top $I\vec\omega$ which points along the axis of the top, and the torque vector which is at right angles to this (namely, $\vec {r}\times\vec F_g$ - possibly with the sign flipped...)
I think for your purpose, if you assume that the rate of precession is constant you don't need to delve down into the details of the vector math / calculus - you will get the right answer just using the distance to the COM and applying the correct value of $\phi$. Do pay attention to the direction of the precession (from the link, you can see it is in the same direction that the top is spinning - so looking from above, if the top spins clockwise, then it also precesses clockwise).