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Write the kinetic energy of the sphere

  1. Apr 29, 2009 #1
    1. The problem statement, all variables and given/known data
    I have a string(length= r) and negligible mass to which a uniform sphere of radius r is attached on one end. I fix upper end of the string to the ceiling and release the sphere when it made angle θ0 with the vertical. I have to write the kinetic energy of the sphere when it makes an angle θ with the vertical.


    3. The attempt at a solution
    Say, the angular velocity of the center of mass about the point of suspension be ω when it makes an angle θ.
    I have a doubt:
    Way 1:
    KE=0.5*Iω2
    Using parallel axis theorem;
    'I' about the point of suspension= (2/5)mr2+m(2r)2
    =4.4mr2
    KE=2.2mr2ω2

    Way 2:
    KE=0.5*mvcm2
    v=2rω
    KE=2mr2ω2

    Which of these is correct. My book gives the method 1 as a correct solution. But I cannot understand whats wrong with method 2?

    What I think is that method 1 gives kinetic energy when the sphere itself is rotating about its own axis with the same angular velocity 'ω' as the center of mass has about the point of suspension. Am I right?
    But in the case specified does the sphere rotate about its axis??? If yes, from where does the torque come from? The tension and the mass both pass from the center of mass, so they will have zero torque.
    Please help me. I am confused!!
     
  2. jcsd
  3. Apr 30, 2009 #2

    tiny-tim

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    Hi Ritwik! :smile:

    Method 2 is the one we usually use for a sphere, because we usually treat the sphere as a point mass (so that the whole mass moves with the same speed, rω).

    When the string is a lot longer than the radius of the sphere, that's accurate enough.

    But here, the string equals the radius, and the approximation isn't good enough …

    different parts of the sphere are moving with different speeds, and the average speed isn't:wink:

    so we can't treat the sphere as a point mass, and we have to do it "properly". :smile:
     
  4. Apr 30, 2009 #3
    Re: Sphere+Pendulum

    Further complication:
    Since the sphere has angular acceleration, the string is not
    in line with the centre of the sphere.

    To make it agree with the expected answer, the string should
    be replaced by a light stiff rod rigidly attached to the sphere.

    The original problem is close to an example of chaos theory!
    Although I would see it as an example of 2 "normal modes" of
    oscillation. The 2 variables are the angles of rotation of the sphere,
    and the angle of the string. In one normal mode these are in phase,
    and in the other antiphase.




    David
     
    Last edited: Apr 30, 2009
  5. Apr 30, 2009 #4
    Re: Sphere+Pendulum

    I understood that exactly! the sphere will also rotate about its center, right? But from where did the torque to cause this roatation came from?
     
  6. Apr 30, 2009 #5
    Re: Sphere+Pendulum

    That was exactly the point I was making.
    In fact the sphere's angle of rotation about its centre is slightly greater
    than the string's angle. This enables the string to provide the torque required.


    David
     
  7. Apr 30, 2009 #6
    Re: Sphere+Pendulum

    First please tell me, the angular velocity about the sphere about its axis should be equal to the angular velocity of the center of mass about the suspension point for method 1 to be correct, right?
    That means the there is rotation in the sphere about its center of mass! Now to cause this rotation, there must be a torque on the body. From where did this torque come from?
     
  8. Apr 30, 2009 #7

    tiny-tim

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    No!!!

    This has nothing to do with rotation!

    (and there's no torque!)

    The bits of the sphere at height h above the centre are moving at speed (r+h)ω,

    so the moment of inertia of those bits has a factor of (r+h)2ω, and the average of that is not r2ω … the lower bits have more contribution than the upper bits …

    to put it another way, the centre of mass is not at the same place as the centre of moment of inertia.
     
  9. Apr 30, 2009 #8
    Re: Sphere+Pendulum

    Oh thanks!! I understood. And if I replace the string with a stiff rod, will there be any rotation about the center? Then I presume method two will work there. Right?
     
  10. Apr 30, 2009 #9
    Re: Sphere+Pendulum

    I see through that, but the thing is that it can be possible only when the sphere rotates. I dont wish to be rude, I just want to clarify my concepts.
    Please see through my viewpoint: suppose the sphere does not undergo rotation about its centre. Then can I say that all the particles on the sphere will have the same velocity as the centre of mass? Without rotation about its own axis, is there any other way that the velocities of the particles in/on the sphere are different? I dont think so.
     
  11. Apr 30, 2009 #10
    Re: Sphere+Pendulum

    No. You still need to take into account the sphere's rotation about its
    centre of mass. But now the "rigid" connection between rod and sphere
    ensures that they have the same angular velocity, and the rigid connection
    provides the necessary torque, which string can't do.
     
  12. Apr 30, 2009 #11

    tiny-tim

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    The sphere is rotating about the top of the string.

    That is why the moment of inertia needs to be calculated relative to an axis through the top of the string. :wink:

    Each point on the sphere is moving in an arc whose centre is the top of the string.

    That is all you need for Way 1 in the book. :smile:
     
  13. Apr 30, 2009 #12
    Re: Sphere+Pendulum

    Thanks a lot. I get it now.
     
  14. Apr 30, 2009 #13
    Re: Sphere+Pendulum

    Thank you! I see through it! Can you find a practical example where all the points on the sphere do have the same velocity?
    Will this do? -----> Say a rigid rod is attached to the center of the sphere and the attachment is such that the sphere does not rotate about it centre. Now will the way 1 be applicable?
     
    Last edited: Apr 30, 2009
  15. Apr 30, 2009 #14
    Re: Sphere+Pendulum

    ?????
     
  16. Apr 30, 2009 #15
    Re: Sphere+Pendulum

    Yes. Cut the sphere almost in two so that it is like a yoyo,
    and attach the string at its centre.
     
  17. Apr 30, 2009 #16
    Re: Sphere+Pendulum

    I get it completely now, i think a rod will be better than a string to avoid rotations about the centre?
    A string might still allow rotation about the center. What do you say?
     
  18. Apr 30, 2009 #17
    Re: Sphere+Pendulum

    I am sorry. A string would work too if attached to the centre as even if there are forces there on centre of mass their net torque woul be zero. Am I right?
    Thanks a lot!!
     
  19. Apr 30, 2009 #18
    Re: Sphere+Pendulum

    It might "allow" it, but with the only forces on the sphere now being
    gravity and tension applied to the centre, there is now nothing to cause it.

    The sphere could be spinning with constant angular velocity, (string loop
    round the axle slips rather than winds up) but the sphere still moves as if
    it were a point mass at its centre.
     
  20. Apr 30, 2009 #19
    Re: Sphere+Pendulum

    exactly

    OTOH a rod welded to the axle would constrain the sphere to
    rotate with it, and such a joint can transmit the required torque.
     
    Last edited: Apr 30, 2009
  21. Apr 30, 2009 #20
    Re: Sphere+Pendulum

    Thank you very very much!!!
     
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