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Work Energy Theorem Paradox

  1. Jan 20, 2014 #1
    Consider a solid sphere and a cube of equal mass, both on a frictionless table. Now, you apply a force to both objects at the point of contact between the object and the table. Then the linear accelerations of both objects will be the same (since the same force is applied to the two objects of equal mass). Therefore, at some point in time, the velocities of both objects will be the same so their translational kinetic energies will also be the same.

    Additionally, since you are applying a nonzero torque to the sphere, the sphere will rotate and therefore have a nonzero rotational kinetic energy at the same point in time. Therefore, the total kinetic energy of the sphere is greater than the total kinetic energy of the cube (since it does not rotate).

    However, by the Work-Energy Theorem, shouldn't both objects have the same total kinetic energy since they both traverse the same distance (same accelerations) and both have the same force applied to them? Can someone explain this apparent paradox?
     
  2. jcsd
  3. Jan 20, 2014 #2

    jbriggs444

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    The work done on the sphere by the force is equal to the product of the force times the distance moved by the point of contact. If the sphere is allowed to rotate as a result of the applied force, the distance moved by the point of contact will be greater than the distance moved by the sphere.

    Strictly speaking, it's not quite a product. It's an integral of the instantaneous force applied times the incremental distance moved.
     
  4. Jan 20, 2014 #3
    The work applied on the ball is in fact larger because even though the two objects move at the same speed the ball rotates while the cube doesn't which causes the point where the force is applied to move faster (compounded translation plus rotation). The formula for the power P = Fv uses the speed v of the point of contact, not the speed of the center of mass.
     
  5. Jan 20, 2014 #4
    There is no paradox.
    You started off with the erroneous premise that the force is the same. The force on the sphere is greater to obtain the results that you stated.

    A common junior physcs problem is as follows:
    Consider two identical spheres. One rolls down an inclined plane that has a typical coefficient of friction, say 0.3 so there is no slippage. The other slides down a frictionless inclined plane of the same angle.
    The rolling sphere will be slower.
     
  6. Jan 20, 2014 #5
    Thanks jbriggs444 and dauto! That makes sense. After some algebra I see that [itex]W=\frac{F^2 t^2}{2\beta m}[/itex] while translational kinetic energy is [itex]\frac{F^2 t^2}{2m}[/itex].

    I think you misunderstood my question. The point is that the force applied on both objects is the same; though, in order to obtain the same total kinetic energy in both the sphere and the cube, a smaller force must be applied to the sphere for the same amount of time (by a factor of [itex]\sqrt{\beta}[/itex]).
     
  7. Jan 20, 2014 #6
    OK
    More work was done on the sphere. Jbriggs and dauto are fine.
     
  8. Jan 21, 2014 #7
    Maybe I am missing something, but on a frictionless table there shouldn't be any force at the contact point with the object other than the vertical gravitational force (for a cube it would be a contact surface anyway).
     
  9. Jan 21, 2014 #8
    I mean the force is applied, not naturally there. For example, a string is attached to the bottom of the cube and you pull the string parallel to the table. You apply that force yourself in some way.
     
  10. Jan 21, 2014 #9
    It all depends how you apply the force at the sphere. If you apply it radially (vertically to the surface) it will just translate, if you apply it tangentially it will just rotate. In general you have a mixture between the two, with the total energy the same in all cases.
     
  11. Jan 21, 2014 #10

    jbriggs444

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    If you apply force and the sphere is not restrained, it will always translate. This remains true, even if the force is applied tangent to the sphere. Newton's second law still applies. F=ma

    The total energy is not the same in all cases, as has been explained already.
     
  12. Jan 22, 2014 #11

    sophiecentaur

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    There are no paradoxes in the ideal world of School Mechanics.
     
  13. Jan 22, 2014 #12

    A.T.

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    Related Puzzle:

    https://www.youtube.com/watch?v=vWVZ6APXM4w
     
  14. Jan 22, 2014 #13

    russ_watters

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    Note that the OP says nothing about the distances through which the force is applied, so it must be making the erroneous assumption that they are equal.
     
  15. Jan 22, 2014 #14
    The puzzle is not that interesting; it is interesting why this is a puzzle to begin with. The answer should matter a lot to teachers of physics.
     
  16. Jan 22, 2014 #15

    sophiecentaur

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    It may be interesting to you but it's no surprise to me that people who do not 'know' Physics come to the most amazing conclusions about the way the World works. Just read a few of the posts we get on PF - and they are from people who are, at least. a bit interested in the topic, even if they may not know a lot.
     
  17. Jan 22, 2014 #16

    sophiecentaur

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    They do not show you close-ups of the bullet holes.. . . . . .
     
  18. Jan 22, 2014 #17
    I understood the puzzle was being offered to some fellow maintainers of science related channels. What was remarkable is that they were easily going to consider conservation of energy, but not of momentum.
     
  19. Jan 22, 2014 #18
    If the force is applied to a fixed point it has to rotate with the sphere in order to stay tangential. So there can not be a net translation.

    As for F=ma, let me invert your argument: if there is no force (but only a torque), then there is no acceleration.
     
    Last edited: Jan 22, 2014
  20. Jan 22, 2014 #19

    Dale

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    If there is a net force then it will translate, regardless of whether the force is tangential or not and regardless of whether it is applied to a fixed point or not.
     
  21. Jan 22, 2014 #20

    Dale

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    Newtons second law is better stated as ∑F=ma. Otherwise you invite comments like Fantasist's.
     
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