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Axle of Rotation

  1. Jul 29, 2010 #1
    I was wondering, if say a rocket in space receives a force perpendicular to its body (on some point on the body), how would one find the axle in which the rotation occurs? Would it be along its center of mass?

    Thanks.
     
  2. jcsd
  3. Jul 30, 2010 #2

    Doc Al

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    Staff: Mentor

    That force does two things:
    - it produces an acceleration of the center of mass
    - it produces an angular acceleration about the center of mass

    Combine those two accelerations to find the instantaneous axis of rotation, which will generally not be about its center of mass.
     
  4. Jul 30, 2010 #3
    Why would the angular acceleration be about the center of mass?

    How does the acceleration of the center of mass shift the axis of rotation?
     
  5. Jul 30, 2010 #4
    I would have stopped right there.
     
  6. Jul 31, 2010 #5

    Doc Al

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    Staff: Mentor

    You can treat the effect of the force as being the combination of a rotation about the center of mass plus a translation of the center of mass.
    To find the translational acceleration of each portion of the body, add the acceleration due to rotation to the acceleration of the center of mass. You'll find a point where the total acceleration is zero. That's the axis of rotation.
     
  7. Aug 3, 2010 #6
    Sorry but I'm still quite confused. I felt that the center of mass would translate in space but ultimately the rotation will always be around the center of mass no matter where it is, is this correct?

    Also, if the above is true, is it possible to mathematically prove why the center of mass is always the axle of rotation?
     
  8. Aug 4, 2010 #7
    You can logically prove why the center of mass is always the "axle of rotation".

    When an object spins on ice, all the forces on the object must balance. If the object rotates about any point other than CM then there will be an acceleration of the CM which is not possible because the net force on the object is zero.

    p.s. we're not talking about rotation, we're talking about an instant where the accelerations due to the forces add up to zero at a point.
     
  9. Aug 4, 2010 #8
    It depends on what you mean by "axe o rotation". If you mean the points that are left fixed by the rotation/translation, then, as Doc Al said, you can find this line by intersecting the planes orthogonal to the velocities of two points of the body. In general, not only it's not true that this line goes through the center of mass, but it can even lie outside the body.
    For instance, the instantaneous axe of rotation of a wheel af a car is the point the wheel touches the ground, and the instantaneous axe for the moon is the center of the earth.
    Moreover the IA of rotation depends on the reference frame you are in, so it has no real physical significance.

    What has relevance is the direction of this axe, not it's origin. So, don't really bother on finding such axe: you can say that the body rotates about a direction, and you can put the origin of the axe wherever you like, the rotation doesn't change.
     
  10. Aug 4, 2010 #9
    Sorry about that I wasn't very clear; I just meant the axis of rotation relative to the object itself, which would be the center of mass of the object but the axis of rotation with respect to a ground observer would be different, is this correct?

    I'm kind of confused on how to add angular acceleration and translational acceleration of a certain point.

    I don't see why there would be a net force on the spinning object if we assume the pivot is not the CM and what if a net force is exerted on an object in space as in the example of my first post?


    I also found:

    http://en.allexperts.com/q/Physics-1358/only-rotate-around-center.htm [Broken]

    The above post almost answers my question but can someone expand on the above explanation as to why rotating on another axis is not stable and why the object would automatically want to go into a stable axis?

    Thanks.
     
    Last edited by a moderator: May 4, 2017
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