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Conceptual question about torque

  1. Sep 16, 2009 #1
    Imagine an object in space with no other forces acting on it. Why is it that if you apply a torque on the object, it will rotate around its center of mass? That is, why does it "choose" to rotate around the center of mass? I cant come up with an answer and this question has been bugging me for a long time. Any sort of insight would be greatly appreciated.
     
  2. jcsd
  3. Sep 16, 2009 #2
    The minimal moment of inertia is about an axis that passes trough the center of mass.
     
  4. Sep 16, 2009 #3
    In addition to what bp_psy said - the minimal moment of inertia allows for the most effective rotation (an object would most likely rotate around an axis that is the easiest to rotate around).
     
  5. Sep 17, 2009 #4

    A.T.

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    It is conservation of the linear momentum. Imagine the object as a bunch of point masses, let's say just two identical ones. Rotation around any other point than their midpoint would mean that the sum of their linear momentums varies over time.

    Why linear momentum is conserved, we don't know.
     
    Last edited: Sep 17, 2009
  6. Sep 17, 2009 #5

    Andy Resnick

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    Since a torque is defined as a force applied via a lever arm, there is a change to the *angular* momentum. Although the general motion of the (rigid) object will allow for movement of the center of mass, the total motion can be decomposed into linear motion of the center of mass (given by the component of the impressed force parallel to the lever arm) and a rotation given by the component perpendicular to the lever arm.
     
  7. Sep 17, 2009 #6

    A.T.

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    1) I don't understand you decomposition. The force component perpendicular to the lever arm, doesn't only cause rotation it also causes linear motion.

    2) How do you define the lever arm anyway without already assuming the center of mass as the center of rotation?

    3) All that doesn't answer the OP question, why for a freely rotating object in space it is the center of mass that moves inertially (or is at rest). The answer to that in terms of a more fundamental law is given in post #4.
     
  8. Sep 17, 2009 #7
    Such a decomposition follows form the Newton equations. As soon as you have at least two particles, one can introduce the notion of the center of inertia R. Its equations are free ones in absence of an external force. The relative coordinate motion is determined with inter-particle forces; normally it is oscillations and rotations. But the notion of CI is valid even for non interacting particles. It is just such a variable that depends on external force solely.
     
    Last edited: Sep 17, 2009
  9. Sep 17, 2009 #8
    Although looks trivial, I would like to let you know that equations for CI coordiante R contain the total mass M. There is no such a particular particle with such a mass actually, that is why the CI is called a quasi-particle. Its equations are particle-like (Newton ones) but for non existing particle. Still, it is very practical: we often assocoate R with the body coordinates if we look from far away.

    Similarly, the relative distances like r = r1 - r2 are determined with reduced masses so they also correspond to quasi-particles. It is important to understant that often (if not always) we observe quasi-particle properties - proper oscillation frequencies, average motion, etc., not particle ones.
     
  10. Sep 17, 2009 #9

    Andy Resnick

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    We may not 'know' in a philosophical sense, but we do know that linear momentum is conserved because the result of an experiment does depend on *where* in the universe it was perfomed, and angular momentum is conserved becasue the results of an experiment do not depend on the direction we are looking when we perform the experiment.
    Conservation of energy occurs becasue the results of an experiment do not depend on *when* we perform the experiment.

    (Noether's theroem).
     
  11. Sep 17, 2009 #10
    In replacing a flat tire, I tighten the lug nuts (on an 8-cm radius) by positioning the lug wrench handle directly over the wheel's axis of rotation. If my hand is closer than 8-cm from the lug, the wheel rotates one way when I apply force. If my hand is further than 8 cm, the wheel rotates the other way when I apply force. There is a point at 8 cm (when my hand is directly at the axis of rotation) where there is no rotation about the wheel's center of mass at all, even if the wheel is floating in free space. So your statement "Imagine an object in space with no other forces acting on it. Why is it that if you apply a torque on the object, it will rotate around its center of mass" is not always true.
     
  12. Sep 17, 2009 #11

    DaveC426913

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    Right. In this special case, the force is applied in a line that passes through the centre of mass, thus zero torque.

    However, correct me if I'm wrong, this state can only exist instantaneously. As soon as the wrench tightens the nut through any angle, the distance changes, causing torque to rise from zero. It works OK when there's friction on the wheel, but would not work in free-floating space.
     
  13. Sep 17, 2009 #12
    In replacing a flat tire, I tighten the lug nuts (on an 8-cm radius) by positioning the lug wrench handle directly over the wheel's axis of rotation. If my hand is closer than 8-cm from the lug, the wheel rotates one way when I apply force. If my hand is further than 8 cm, the wheel rotates the other way when I apply force. There is a point at 8 cm (when my hand is directly at the axis of rotation) where there is no rotation about the wheel's center of mass at all, even if the wheel is floating in free space. So your statement "Imagine an object in space with no other forces acting on it. Why is it that if you apply a torque on the object, it will rotate around its center of mass" is not always true.
    In this case, I would use a ratchet torque wrench.
    [Edit] In every case, there will be a linear momentum transfer to the object. When I tighten the wheel lugs, the axle is on a jack, and the wheel is free to rotate, so my hand position has to be in line with the axle.
     
    Last edited: Sep 17, 2009
  14. Sep 17, 2009 #13

    DaveC426913

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    Yes. My point was merely that it only starts off in line with the axle. The geometry changes as soon as the nut turns (which is relevant to the OP).
     
  15. Sep 17, 2009 #14

    rcgldr

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    It won't, the center of rotation will depend on where the torque is applied. Imagine a rocket with a side nozzle at one end and an opposing side nozzle in the middle, as opposed to the other end. If the opposing nozzle is at the other end (and the thrust is the same), the rocket's center of mass doesn't move. If the opposing nozzle is not at the other end, then the rocket's center of mass moves in a circle.
     
  16. Sep 18, 2009 #15
    Even when the opposing nozzle is not placed at the other end and as long as the net force(due to the releasing gases from the nozzle) on the rocket is zero, COM wouldn't accelerate.
    So i think we can sum it up as,
    Non-zero force;non-zero torque *may* have a different center of rotation.
    Zero force; non-zero torque *should* have the COR at the COM.
     
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