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Moment of inertia and force needed to tilt/change axis of rotation |
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| Sep2-12, 10:07 AM | #1 |
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Moment of inertia and force needed to tilt/change axis of rotation
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:
|<- apply force | | ===== <- x-y plane of 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. |
| Sep2-12, 02:57 PM | #2 |
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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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