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Rotational motion

  1. Sep 15, 2005 #1

    Dhl

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    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. jcsd
  3. Sep 15, 2005 #2

    mukundpa

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    Homework Helper

    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.
     
  4. Sep 17, 2005 #3

    Dhl

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    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:

    [itex]r_{1}[/itex] = vector from CM to particle 1
    [itex]r_{2}[/itex] = vector from CM to particle 2

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

    Inertia [itex]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)[/itex]

    rotational acceleration [itex]a=I^{-1} \sum_{i=0}^n \tau[/itex]
    am i still right?
     
  5. Sep 18, 2005 #4

    mukundpa

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    Yes, I think so.
     
  6. Sep 18, 2005 #5
    search for chris hecker tutorial notes on game physics...it has a somewhat awkward wording but relatively easy tutorial on what your lookign for.
     
  7. Sep 21, 2005 #6

    Dhl

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    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:

    [tex]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)[/tex]

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

    If nature used particles to compute rotational motion, the universe would have crashed long ago :(
     
    Last edited: Sep 21, 2005
  8. Sep 21, 2005 #7

    mukundpa

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    some problem in computing of Moment of Inertia. r^2 can not be negative.
     
  9. Sep 21, 2005 #8

    Dhl

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

    [tex]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)[/tex]

    thus the negative items in the matrix for 2 particles.
     
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