# Rotational motion

1. Sep 15, 2005

### Dhl

hi!

i've been pondering about rotational motion (of a rigid body). i've been doing some internet research on it, but i have some trouble putting things together...
i want to apply forces on the particle parts of a body. one website told me that when you know the axis of rotation, you compute r (the shortest distance between the particle and the axis), then the torque is t = r X F (F being the force). i also learned that motion of a rigid body can be expressed as linear motion + rotation about the center of mass (CM), so i assume that any rotational axis will go through CM.(correct me if i'm wrong so far...)
can i just set r=CMP (vector from CM to the particle)? if this is not correct, i wonder how i could know the axis of rotation *before* i know the torque...

2. Sep 15, 2005

### mukundpa

Correct, the rigid body on application of force translate and rotate about an axis passing through its center of mass, but if the body (say a door panel) is not free, may only rotate about an axis not passing through ats center of mass. Any how, a motion both translational and rotational simultaneously may be considered as purely rotational instataneously, about some axis called instatinious axis of rotational.

3. Sep 17, 2005

### Dhl

thanks a lot. :)
so, if i have a body of n particles (which can rotate freely), and two forces F1 and F2 that act on two of those particles, i could calculate rotational acceleration by:

$r_{1}$ = vector from CM to particle 1
$r_{2}$ = vector from CM to particle 2

$r_{1} \times F_{1} + r_{2} \times F_{2} = \sum_{i=0}^n \tau$ (sum of torques)

Inertia $I = \left(\begin{array}{ccc}\sum_{i=0}^n m_{i}( r_{iy}^2+r_{iz}^2)&\sum_{i=0}^n m_{i} r_{ix}r_{iy}&\sum_{i=0}^n m_{i} r_{ix}r_{iz}\\\sum_{i=0}^n m_{i} r_{ix}r_{iy}&\sum_{i=0}^n m_{i}( r_{ix}^2+r_{iz}^2)&\sum_{i=0}^n m_{i} r_{iz}r_{iy}\\\sum_{i=0}^n m_{i} r_{ix}r_{iz}&\sum_{i=0}^n m_{i} r_{iz}r_{iy}&\sum_{i=0}^n m_{i}( r_{ix}^2+r_{iy}^2)\end{array}\right)$

rotational acceleration $a=I^{-1} \sum_{i=0}^n \tau$
am i still right?

4. Sep 18, 2005

### mukundpa

Yes, I think so.

5. Sep 18, 2005

### neurocomp2003

search for chris hecker tutorial notes on game physics...it has a somewhat awkward wording but relatively easy tutorial on what your lookign for.

6. Sep 21, 2005

### Dhl

The Chris Hecker tutorial helped me a lot... articles explaining physics for use in games seem more understandable to me than physics papers. However, there was a thought that embarassed me: Assuming an object of two particles with the same mass, these will have the following vectors from the CM: (a,b,c), (-a,-b,-c) because they must be on the opposite side of CM. The resulting inertia matrix, then, is:

$$2m \left(\begin{array}{ccc} b^2+c^2&-ab&-ac\\-ab&a^2+c^2&-bc\\-ac&-bc&a^2+b^2\end{array}\right)$$

which has a determinant of 0 and can't be inverted, so the motion can't be computed, too

If nature used particles to compute rotational motion, the universe would have crashed long ago :(

Last edited: Sep 21, 2005
7. Sep 21, 2005

### mukundpa

some problem in computing of Moment of Inertia. r^2 can not be negative.

8. Sep 21, 2005

### Dhl

The inertia matrix I posted was wrong, it should be:

$$I = \left(\begin{array}{ccc}\sum_{i=0}^n m_{i}( r_{iy}^2+r_{iz}^2)&-\sum_{i=0}^n m_{i} r_{ix}r_{iy}&-\sum_{i=0}^n m_{i} r_{ix}r_{iz}\\-\sum_{i=0}^n m_{i} r_{ix}r_{iy}&\sum_{i=0}^n m_{i}( r_{ix}^2+r_{iz}^2)&-\sum_{i=0}^n m_{i} r_{iz}r_{iy}\\-\sum_{i=0}^n m_{i} r_{ix}r_{iz}&-\sum_{i=0}^n m_{i} r_{iz}r_{iy}&\sum_{i=0}^n m_{i}( r_{ix}^2+r_{iy}^2)\end{array}\right)$$

thus the negative items in the matrix for 2 particles.