Here is a question that I am sure was probably answered in the heaps of documents I have read but just didn't catch it. I am mainly concerned about the rotation of a rigid body so I am going to say that the body is just fixed with its center of mass at (0,0,0). If a force, F(x,y,z), is applied to a point p(x,y,z) then I can find the torque with this: t = p x F The toque is a vector so that must be the axis that the body rotates around (correct so far?). I am also assuming that the norm of the cross product must be the amount of torque that is being applied to rotate the body around the axis. But, to then find the angular acceleration of the body I need to know the moment of inertia about that axis ... this is where my "math loving" and "physics hating" halves of my brain collide. Are my assumptions correct so far? How does one calculate the moment of inertia about an axis? Thanks Edited by - jag_oes on December 17, 2001 5:34:44 PM

The moment of inertia of a point particle mass m and distance r from the axis of rotation is defined as I = m(r^2). The degree of difficulty encountered in calculating I for a rigid body depends on its symmetry. In general if one has an equation for the bodies shape, by the principle of superposition, it is possible to calculate I by integration, as this sums the moments of inertia of elements of the solid where dI = (r^2)dm (assuming the body is of uniform density). For example to find I for a sphere, first consider a solid ring of width dx and radius x, its moment of inertia is given by the above as all elements are equidistant from the centre. Now express its mass in terms of its volume as a function of x and its density. Integrate with respect to x from 0 to r where r is the maximum radius, this gives you I for a disc. Now consider the sphere to be composed of thin discs on top of each other the I for each being a function of its radius x expressed in terms of z on the axis of rotation; integrate again from z = -r to +r and you get I for the sphere. I found a good site at: http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html which also goes into the perpendicular and parallel axis theorems.

Ok, your post, although nicely put, was everything I already knew. Let me say that I have found the inertia tensor for my object (it is just a box anyway):

[ Ixx  0   0   ][  0  Iyy  0   ][  0    0  Izz ] 

However, that only helps me for rotating around the x-, y- and z-axes. What if I wanted to rotate around the vector (1, 1, 1) ? See what I am saying? Of course the axis of rotation would be difined by the cross product of the torque and point that the force is being applied to but you get the idea ...

Have a look at this paper:

http://www-2.cs.cmu.edu/~baraff/sigcourse/notesd1.pdf

It explains how the inertia tensor is related to angular momentum among other things.

Does that help?

That looks a lot better .. I will have a look and let you know. Thank you very much ...

Edited by - jag_oes on December 20, 2001 5:19:43 PM

I have another question about moment of inertia that maybe someone can help me out with. I want to make an application where the user can draw their own rigid body and then it will bounce around the screen of walls and I am having some trouble calculating the moment of inertia of an arbitrary shape. The rigid body is drawn with a bunch of straight lines so that makes it a little bit easier ... the density is also constant throughout the body. I know that I can calculate the moment of inertia about the x- and y-axis (and therefore the origin) with the following integrals:

Ix = -p/3 òcy3 dx Iy = -p/3 òcx3 dy 

But, when I try to generalize this for my purposes (for programming) it does not work ... maybe someone can check my algrebra.

First I set up a parameterization of my lines with the following equations:

parametric equation for a line connecting (x0, y0) with (x1, y1)x = x0 (1 - t) + x1 ty = y0 (1 - t) + y1 t

Finding the derivatives of the parametric equations gives me this:

dx = (x1 - x0) dtdy = (y1 - y0) dt

Plugging back into my line integrals I get this:

Ix = -p/3 ò01[y0(1 - t) + y1 t]3 (x1 - x0) dt Iy = -p/3 ò01[x0(1 - t) + x1 t]3 (y1 - y0) dt 

Doing a u and evaluating the integrals I get the final equations:

     -p (x1 - x0)(y14 - y04)Ix = ---------------------           12 (y1 - y0)     -p (y1 - y0)(x14 - x04)Iy = ---------------------           12 (x1 - x0)

Now, using the property of line integrals that you can just sum the individual
line integrals of a piecewise function, I made a few functions that would find
the moment of inertia for an object made up of a number of vertices. The functions
do not output the correct numbers even though they work properly. I am pretty
sure it must be an error in my work somewhere ... does anyone have any ideas?



Thanks.

P.S.

I am talking about a 2D rigid body now ...

P.S. x 2

Also, the reason the equations I derived did not work was because of when I tried it on a square ... sooner or later when you start plugging in numbers you get a divide by zero ...

Ok, forget all the nonsense above ... but can anyone tell me what the moment of inertia about the origin of a square with side lengths of two would be? Its density is 1 too ...

Edited by - jag_oes on December 21, 2001 2:19:17 PM

Teehee... you wrote almost your entire post in Greek. Poorly-spelled Greek, too