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Energy Lost in rolling spheres

  1. Sep 12, 2012 #1
    1. The problem statement, all variables and given/known data
    There is a tube attached to a board in a fashion that a ball can be dropped in the top and the tube curves to the right 90°. If a ball of mass 7.6g is dropped into the top of the tube, what is the minimum height the exit point of the tube needs to be in order for the ball to go at least 1 meter.

    Here is a diagram:
    https://dl.dropbox.com/u/14174746/diagram.png [Broken]

    Radius of the ball is unknown.

    A friction test was done, the board was put on an angle such that the tube made a half-pipe. The board was angled so one side was slightly higher than the other and when the ball was dropped in it would reach the other side of the tube perfectly (because normally friction would stop it from going all the way). One side had to be 1.1cm higher than the other for this to happen, however I don't know how I'm supposed to use this information in relevance to the energy lost due to friction.

    2. Relevant equations
    Ep = mgh
    Ek = 0.5mv2
    Basic Kinematic equations etc.

    I don't know what equation I need to use for energy of the rolling


    3. The attempt at a solution
    I found the maximum velocity the ball could have when leaving the end of the tube by determining the kinetic energy based on it's potential energy:
    Ep= mgh = 0.0417088 (not rounded)
    Ek = 0.5mv2
    Ek = Ep
    0.5mv2 = mgh
    0.5v2 = gh
    v2 = 2gh
    v = √(2gh)
    v = 3.313m/s

    That velocity is without any loss of energy due to friction, or due to rolling. Our teacher hinted that the energy lost due to these two factors is in fact significant and will affect the outcome.

    My problem is how to determine the energy lost from friction and from rolling, I've done some research into the formulas of a rolling sphere but they all require radius which I am not given. I was told that the radius of the sphere was irrelevant.

    Here's the rolling formula anyway:
    vcm = ωR

    That is the only formula I could make sense of without any outside instruction, that the velocity of the center of mass is equal to the frequency of rotation multiplied by the radius ( I think )

    Thanks in advance for any help.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 13, 2012 #2

    CWatters

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    Without friction both sides could be at same height and it would roll down one side and all the way up the other.

    The fact that a height difference was required gives you a way to calculate the additional energy required due to friction.

    EDITED: I changed "friction/rolling" to just "friction". See later posts.
     
    Last edited: Sep 13, 2012
  4. Sep 13, 2012 #3
    I know... I just don't know how
     
  5. Sep 13, 2012 #4

    CWatters

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    Ok..

    In the half pipe case it started of with PEstart=mghstart and ended up with PEend=mghend. So how much energy did it loose?

    Perhaps assume it would loose the same amount of energy when set up as per the diagram. My only hesitation with this approach is that the velocity is different when it's set up as a half pipe and when set up as per the diagram. It might be worth asking your teacher if it's ok to assume they are approximately the same OR does he want you to correct for the difference somehow. If you assume it's the same then perhaps state that in the answer.
     
  6. Sep 13, 2012 #5
    I believe we were okay to assume they lose the same amount of energy in both. The friction was tested in class and given to us that way because it was a 'less significant' factor. The true factor was the energy supposedly lost in rolling, in which all of the students who did the project last year failed because they didn't consider it.
    I think I understand how to reduce the energy based on what you said, how would I go about determining the loss of energy due to rolling?

    Just a side note: None of this has been taught to us (rolling), we were asked to research and figure it out on our own
     
  7. Sep 13, 2012 #6

    CWatters

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    Hmm. Ok try these three equations..

    The energy stored in a rotating object (eg the ball as it exit the chute) is..

    E = 0.5 * I * ω2................[1]

    where

    w is the angular velocity
    I is the moment of inertia

    The moment of inertia I of a sphere is

    I = 0.4 m * r2...................[2]

    where
    m is the mass
    r is the radius

    The rolling equation

    V=ωr
    so
    r=V/ω.................................................[3]

    See what happens if you substitute [2] and [3] into [1]. With luck r will dissapear.
     
  8. Sep 13, 2012 #7

    CWatters

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    I'd agree. In the diagram it falls about 50cm. In the half pipe the difference is 1cm. So the loss is order 2% which is small.

    Incidentally in the half pipe the ball starts rotating as it falls, so at the bottom it has stored energy. However it looses that on the way back up the other side so this is measuring just the loss due to friction not the energy stored in the ball.
     
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