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Velocity of Center of Mass of Disk

  1. Oct 7, 2015 #1
    1. The problem statement, all variables and given/known data
    ?temp_hash=c75b6228ea0a231ab93864c726808e60.png

    2. Relevant equations


    3. The attempt at a solution

    I think if I can locate the instantaneous axis of rotation ,then may be motion/velocity of CM could be determined . In the attached figure I have marked the two points A and B which are instantaneously at rest .The black arrows depict their respective velocities w.r.t CM .

    Now consider point A . If it is instantaneously at rest ,then ##V_{CM}## has to be along the green vector .Similarly If we consider point B , then ##V_{CM}## should be along the red vector .

    But that is a contradiction . I am not very sure if this is the right way to approach this problem .

    I would be grateful if somebody could help me with the problem .
     

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  2. jcsd
  3. Oct 7, 2015 #2

    TSny

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    Locating the instantaneous axis of rotation sounds like a very good idea!

    Are the points A and B instantaneously at rest? Hint: Are the strings changing their angles with respect to the vertical?
     
  4. Oct 7, 2015 #3
    Yes . I have been thinking on this line . No , the points are not at rest . I am guessing that the left string is rotating clockwise whereas right is rotating anticlockwise w.r.t the support . Is that correct ?

    So what should be the next line of thinking ?
     
  5. Oct 7, 2015 #4

    TSny

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    Yes.

    Can you see the direction that point A (on the disc) must be moving?
     
  6. Oct 7, 2015 #5
    Before I answer your question , I would like to clarify ,whether it is correct to say that point A is instantaneously at rest w.r.t the string ?
     
  7. Oct 7, 2015 #6

    TSny

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    Yes. It is like a ball rolling without slipping on a board that is being tilted.
     
  8. Oct 7, 2015 #7
    Fine .

    Are you getting ##V_{CM} = \frac{\omega R}{cos\frac{\alpha}{2}}##

    In the attached figure ,the green dot represents the IAOR and red vector represents the direction of velocity of CM .

    Does that make sense ?
     

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  9. Oct 7, 2015 #8

    TSny

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    I believe your answer is correct.
     
  10. Oct 7, 2015 #9
    Did you have a look at the attached figure as well ?
     
  11. Oct 7, 2015 #10

    TSny

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    Yes, it looks good to me. Something to think about: does the answer make sense for ##\alpha## approaching zero?
     
  12. Oct 7, 2015 #11
    Thank you very much :smile:

    Yes ,it makes sense. I am assuming this situation corresponds to when the two strings are aligned one on top of the other vertically (i.e both the strings are vertical) . In this case ##V_{CM} = \omega R##

    Do you see any issue ?
     
    Last edited: Oct 7, 2015
  13. Oct 7, 2015 #12
    A side question -

    If a disk is rolling without slipping(accelerated/non-accelerated) on a stationary horizontal surface ,the bottommost point is at rest . Is it correct to say that this point does not have any kind of acceleration , neither linear nor centripetal acceleration ? I am concerned more about centripetal acceleration .
     
  14. Oct 7, 2015 #13

    TSny

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    I think it makes sense, as you say. You can also think about ##\alpha## going to 180o, with the two strings horizontal.
     
  15. Oct 7, 2015 #14

    TSny

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    The point will have a centripetal acceleration (i.e., upward) at the instant the point is at rest.
     
    Last edited: Oct 7, 2015
  16. Oct 7, 2015 #15
    The point is at rest i.e v=0 .Shouldn't the centripetal acc. v2/R be zero as well? Please explain this apparent contradiction .
     
  17. Oct 7, 2015 #16

    TSny

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    Look at the motion https://en.wikipedia.org/wiki/Cycloid. Is the vertical component of velocity changing when the point is in contact with the surface?

    What if you switch to the inertial reference frame moving with the center of the disk? In this frame would the point have a vertical component of acceleration at the lowest position? Do accelerations change when switching between inertial reference frames?
     
  18. Oct 8, 2015 #17
    Okay .

    There is another problem involving similar concepts of rolling and use of Instantaneous axis . Please see the attached picture .

    Do you agree that in the inertial frame moving with the center of the sphere

    1) speed of the lower tip is 'v' towards left .
    2) speed of the topmost point is 'v' vertically upwards .
    3 rod is undergoing pure rotation about the center of the sphere .
    4) the center of the sphere acts as the instantaneous axis .
    5) the CM of the rod is moving with speed v/√2 .
     

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    Last edited: Oct 8, 2015
  19. Oct 8, 2015 #18

    TSny

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    I agree with all 5.
     
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