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Confusing paradox

  1. Sep 15, 2010 #1
    I've read that the acceleration produced in the center of mass of a rigid body is equal to:

    AccelerationCenter of mass = Forceexternal / Mass​


    However, that leads to a counter-intuitive conclusion.
    The reason that seems unintuitive to me is that I would expect some rotation to happen.
    Where am I going wrong?
     
  2. jcsd
  3. Sep 15, 2010 #2

    gabbagabbahey

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    The hammer can (and probably will, depending on where the string is attached) rotate about its center of mass (CoM), even if its CoM travels in a straight line.

    Try spinning a basketball on your finger and slowing move your finger/arm up and down. The ball will continue to rotate about its CoM, even while its CoM moves up and down.
     
  4. Sep 15, 2010 #3
    Thanks, but won't that violate the energy conservation principle then, since we managed to achieve rotational kinetic energy AND the linear kinetic energy (that treating the object as a point mass under external force F will give)
     
  5. Sep 16, 2010 #4
    This kind of reasoning is wrong. You can, for example, describe circular motion using vectors.

    There is no paradox, the mass point model is not applicable here. If you want to include rotation, use the rigid body model.
     
  6. Sep 16, 2010 #5

    Doc Al

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    No, energy is still conserved. The work done is force X displacement of the contact point; when the object rotates, the displacement is greater and you end up doing more work. That additional work provides the rotational energy.
     
  7. Sep 16, 2010 #6

    Doc Al

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    The reasoning is correct and follows Newton's 2nd law. As long as the net force on the object remains constant, the object's center of mass will accelerate in a straight line.
     
  8. Sep 16, 2010 #7
    OK. Correct me if I'm wrong.

    The work done by the force is the sum of:
    1. Force X Displacement of the CoM
    2. Torque of the force about the angle of rotation X Angle by which the body rotates
    The first part is the linear kinetic energy and the second the rotational.

    Right?
     
  9. Sep 16, 2010 #8

    Doc Al

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    Essentially correct.

    Those work-like terms are really examples of pseudo-work. They are consequences of Newton's 2nd law.

    The most basic definition of the work done by a force acting on a system is Force X Displacement of the point of contact of the force. That includes all the work done on the system by that force.

    In this case, there's no big issue since the work done will equal the sum of linear and rotational KE. But you can easily have cases where the two give different answers, due to some change in internal energy.
     
  10. Sep 16, 2010 #9
    Qute from OP:
    I assumed causal `since´ (since = because).
    I also assumed `in a straight line´ to refer to the motion, not only the acceleration.
    Then it reads: "... the center of mass (of some object) accelerates moving in a straight line, because the involved qantities are vectors."
    I still think the because-clause is wrong. (Test: some object = the moon, involved vectorial qantities are gravitational force of the earth and momentum of the moon.)
    So I consider the quote from the OP to be at least misleading. If I missed some finer nuances in his post, please tell me which.

    btw. As long as the acceleration lasts, there will probably be oscillatory motion of the hammer in addition to the (CoM) acceleration. When acceleration stops rotation sets in, depending on angular momentum at that time.
     
  11. Sep 16, 2010 #10

    Doc Al

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    We assume that the hammer is initially at rest and that the applied force remains constant in magnitude and direction. In that simple case, the center of mass of the body will accelerate (and move) in a straight line.

    No matter what the direction of the force is, the acceleration of the center of mass will be in the same direction. So if the force changes direction, so will the acceleration. And of course if there were an initial velocity not in line with the force, the motion of the center of mass (as seen in the original frame of reference) will not be a straight line.
     
  12. Sep 16, 2010 #11
    Did you really read what I wrote?
    Would it move in a curve if you choose to describe it without vectors?
     
  13. Sep 16, 2010 #12

    Doc Al

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    Yes, but I can't say I understood your point. (Sorry.) Did you read what I wrote? Do you disagree with any of it?
    I don't really know what you're getting at. As long as the force acts in a single direction, the motion of the center of mass will be in a straight line (assuming it starts from rest).
     
  14. Sep 16, 2010 #13
    I try to say that there is no causal relationship between moving in a straight line and vectors being involved in the description of the motion (as the OP seems to do). I even offered a counterexample.
    Otherwise, I don´t disagree.
     
  15. Sep 16, 2010 #14
    @maimonides
    My reasoning behind the line in question was.

    F=ma

    F and a being vectors, must then have the same direction.
    So the acceleration must have the same direction as the force, and if the force is constant in direction, so is the acc., resulting in one dimensional motion (assuming zero initial velocity)

    I do agree it wasn't well worded though
     
  16. Sep 16, 2010 #15

    rcgldr

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    Rotation will occur. Note that the "resistance" (linear and angular inertia) by the hammer to acceleration of the string is reduced when the string is attached away from the center of mass. In order to generate the same force when attached at the lighter end, the string must accelerate much faster in order to generate the same amount of force. The hammer's center of mass will accelerate based on the amount of force / (mass of hammer) as mentioned above and it will also experience angular acceleration. In order to keep the force constant, the string's rate of acceleration would have to decrease as the hammer's rotation caused the strings line of force to get closer to the hammer's center of mass.

    Once the hammer is moving, if the force applied by the string has any component perpendicular to the direction of movement of the center of mass of the hammer, then that perpendicular component will result in a change in direction (which is also an acceleration), but not magnitude of speed of the hammer. Unless the line of force at the other end of the string went through the center of mass of the hammer (assuming a massless string), you'd get a pendulum like motion of the center of mass of the hammer. Even in this case, the total acceleration of the center of mass, including linear and centripetal acceleration, will equal force / (mass of hammer).
     
    Last edited: Sep 16, 2010
  17. Sep 16, 2010 #16

    DaveC426913

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    Conventionally one removes this complicating component by assuming the force is short and sharp enough to elimiate any issues of changes in vectors.
     
  18. Sep 16, 2010 #17

    rcgldr

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    I was updating my previous post to note that total acceleration of the hammer's center of mass remains equal to force / (mass of hammer), even in the pendulum motion case.
     
  19. Sep 16, 2010 #18

    DaveC426913

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