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Moment of inertia and force needed to tilt/change axis of rotation

  1. Sep 2, 2012 #1
    Consider a freely rotating body. Let the axis of rotation be the z-axis. For simplicity assume all the mass of the body is concentrated in the x-y-plane, i.e. the plane in which the body rotates.

    I have read about the moment of inertia tensor on wikipedia, but I don't see how I would combine it with a torque to tilt the axis of rotation.

    Suppose the above rotating body indeed has a solid axis, albeit of zero mass, sticking out at one end with length [itex]\gt l[/itex]. At [itex]z=l[/itex] we apply a force perpendicular to the axis for a distance of [itex]\Delta s[/itex] in the direction of [itex]-x[/itex].

    Code (Text):

      |<- apply force
      |
      |
    =====  <- x-y plane of rotation
    What will happen to the to the overall rotation.

    a) Will the axis tilt only a certain amount or does the force applied induce a rotation that keeps going and combines with the previous rotation.

    b) What is the formula to get the tilt angle or the angular speed? I assume it somehow combines the inertia tensor and the force F or torque [itex]l\times F[/itex]?

    Thanks,
    Harald.
     
  2. jcsd
  3. Sep 2, 2012 #2
    Found it myself. In general it is

    [tex]\tau = I\cdot\dot{\vec{\omega}}[/tex]

    where [itex]\tau[/itex] is the torque, the equivalent of force for linear motion, [itex]I[/itex] is the moment of inertia tensor (i.e. 3x3 matrix) and [itex]\dot{\vec{\omega}}[/itex] is the three-vector of angular acceleration. The rest seems to be to put in the special case values. And I reckon that applying a torque that that is not just parallel to [itex]\dot{\vec{\omega}}[/itex] will result in an angular velocity component, not just in a tilt of the rotational axis.
     
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