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Rotating ball and string Rotational Kinetic Energy?

  1. Dec 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Hello,
    I am a bit confused on when rotational kinetic energy exists and when linear kinetic energy exists. For example when we spin a string with a ball attached to the end it has kinetic energy of 0.5mv^2 of the ball when we were solving for velocity of the ball at certain points in the vertical circle of the ball. But know that i learned rotational motion, why don't we consider rotational energy in that example instead of just mgh and 0.5mv^2? There was a question where there was a ball attached to a bar and the system of bar and ball attached to the bar are moving and we used just rotational kinetic energy to solve the question and didn't account for 0.5mv^2.

    2. Relevant equations



    3. The attempt at a solution
    I was thinking maybe it has to do with something about the string maybe being negligible or something but still not sure. I would like some clarification on this matter please.
     
  2. jcsd
  3. Dec 9, 2013 #2

    tiny-tim

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    Hi Vontox7! :smile:
    If the string has length R and the ball has radius r, then the KE is 1/2 mR2ω2 + 1/2 2/5 mr2ω2

    = 1/2 m(R2 + 2/5 r22

    r2 is usually so much smaller than R2 that the extra energy can be ignored. :wink:
     
  4. Dec 9, 2013 #3
    Okay so just simplifying KE = 1/2m(R^2+0)w^2 =1/2mR^2w^2 = 1/2mR^2(v^2/R^2) = 1/2mv^2 ?
    Btw thank you very much for help it is greatly appreciated !
     
  5. Dec 9, 2013 #4

    haruspex

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    Strictly speaking, since the string is attached to the outside of the ball:
    1/2 m(R+r)2ω2 + 1/2 2/5 mr2ω2
    For the motion of the mass centre about the axis, you can handle it in either of two ways:
    - as linear motion, mv2/2
    - as rotational motion, mr2ω2/2
    Since v = rω, these are the same.
    If you also want to account for the rotation of the object about its mass centre then you add Iω2/2.
     
  6. Dec 9, 2013 #5
    The problem is I know doubt the method we used back in high school to solve questions of objects rotating in a circle like a pendulum problem and using energy to find the balls maximum height and so forth because back then we didn't account for rotation kinetic energy why? Maybe I am not understanding what exactly is rotational kinetic energy.
     
  7. Dec 10, 2013 #6

    tiny-tim

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    Hi Vontox7! :smile:
    There's nothing special about rotational kinetic energy.

    It's just the sum (strictly, integral) of the ordinary 1/2 mv2 kinetic energy of the individual parts.
    The school pendulum solution has a simple L in it.

    That L is √(R2 + 2/5 r2).

    Since you can't measure L very precisely (and since the string or rod isn't totally negligible either), the school solution is correct, it just uses a slightly different L. :wink:
     
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