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Distance of a ball in a circle

  1. Jul 15, 2013 #1
    1. The problem statement, all variables and given/known data
    In the figure on the right, a 4.0 kg ball is attached to the end of a 1.6 m rope, which is fixed at O.
    The ball is held at A, with the rope horizontal, and is given an initial downward velocity. The ball moves through three quarters of a circle and arrives at B, with the rope barely under tension.

    r80SUSD.jpg


    2. Relevant equations
    K1 + U1 = K2 + U2


    3. The attempt at a solution
    I tried with the following:

    R = 1.6m
    m = 4kg

    K1 = [itex]\frac{1}{2}[/itex]m[itex]v_{a}^{2}[/itex]
    U1 = mgR
    K2 = [itex]\frac{1}{2}[/itex]m[itex]v_{b}^{2}[/itex]
    U2 = mgR

    But in a solution I found it was used U2 = mg2R

    Why it was used 2R and not R? The distance to use in a circle is from the center to the particle or is from the bottom to the particle?
    Because as A is in a distance R from center and B too, I tried with R for both but it seems that it needs to be done using a distance from bottom
     
  2. jcsd
  3. Jul 15, 2013 #2
    That U equation is for Potential energy. The equation is mass x g x distance. The distance you would use isn't too important. U1 and U2 just have to be measured from the same spot so that the change in potential energy would be mgR.
     
  4. Jul 15, 2013 #3
    Ok, so from center A will be -R and B will be R?
     
    Last edited: Jul 15, 2013
  5. Jul 15, 2013 #4
    Sorry I guess i wasn't too clear. Its the height that you would be measuring so if you did use the center you would just need to look at how high each point is from the center.
     
  6. Jul 16, 2013 #5

    haruspex

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    If it were R, how fast would the ball be travelling at the top of the circle?
     
  7. Jul 16, 2013 #6

    PeterO

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    I keep reading the problem, and don't find a question??? Are you trying to calculate vo perhaps?
     
  8. Jul 16, 2013 #7

    haruspex

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    Forget that - I misinterpreted your equations.
    You have U = mgR at A. How could it also be mgR at B, given that B is height R above A?
     
  9. Jul 17, 2013 #8

    rude man

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    Assuming the question is, 'what is the initial imparted velocity', use energy conservation. Note that the tension on the rope imparts force to the mass but since it's always orthogonal to the directin of motion, that force does no work.

    So, simply, initial kinetic energy = change in potential energy.
     
  10. Jul 17, 2013 #9
    When the ball arrives at B the tension in the rope becomes zero. This does not necessarily mean that the ball has come to rest.
    It is not 100% clear what the question is asking but, in my experience of questions of this sort, I would say the ball completes a circle.
    If that is the case then the ball has KE at point B, as well as PE.
     
  11. Jul 17, 2013 #10

    haruspex

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    I think you've made the same presumption I made in my first post on this thread, that the poster has made the usual mistake of forgetting about centripetal acceleration. But the question posed concerns the potential energy at B. duplaimp appears to have measured the PE at A relative to the bottom of the loop but that at B relative to A.
     
  12. Jul 17, 2013 #11
    Haruspex: I agree, some information missing. My inclination is that it would be a mistake to assume that the ball comes to rest at B.
     
  13. Jul 17, 2013 #12

    PeterO

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    This is the problem with the Original Post! There is no question posed !!!!
     
  14. Jul 17, 2013 #13

    haruspex

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    It doesn't say what question the poster was trying to answer, but it does pose this question from the poster:
    Not wonderfully clear, I agree, but I think there's enough circumstantial evidence to interpret U1 and U2 as representing the PEs at A and B respectively. If so, the error is evident.
     
  15. Jul 17, 2013 #14

    PeterO

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    Understood. Seems OP was unaware that the standard formula U = mgh had had the h value replaced by "the height above lowest point" value - R and 2R respectively.
     
  16. Jul 17, 2013 #15

    rude man

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    Let
    vi = initial velocity at A
    vf = final velocity at B

    EDIT:

    Technician is right, the velocity at B is not zero. Matter of fact, it's v = vf = sqrt(gR). That result should be obvious: at the top, the centripetal force still has to be mv^2/R and that has to be equal to mg + T, but T(B) = 0.

    So we have the complete picture: 1/2 mvi^2 = 1/2 mvf^2 + mgR. With vf known, vi is immediately determined.

    There is no information missing.
     
    Last edited: Jul 17, 2013
  17. Jul 17, 2013 #16

    PeterO

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    It is not clear that OP had any difficulty coping with the centripetal acceleration etc - he just wondered why U was mgR in one position and mg2R in the other. Presumably he had forgotten that the R and 2R referred to the height above a reference point, not the distance from the centre.
     
  18. Jul 17, 2013 #17

    rude man

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    So it's agreed there is no "missing information"?

    The only possible questions are what are the initial & final velocities & I have answered both.
     
  19. Jul 17, 2013 #18

    PeterO

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    I eventually found the OP's question.

    It is at the bottom

    "But in a solution I found it was used U2 = mg2R

    Why it was used 2R and not R? The distance to use in a circle is from the center to the particle or is from the bottom to the particle?
    Because as A is in a distance R from center and B too, I tried with R for both but it seems that it needs to be done using a distance from bottom"
     
  20. Jul 17, 2013 #19

    rude man

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    OK, so that's easy to find when you know the velocity at B.
     
  21. Jul 17, 2013 #20

    PeterO

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    You don't need to find the velocity anywhere - you just have to realise that the potential energy U is calculated using the vertical displacement above some reference point/level - and the the solution OP had seen that reference level was the bottom of the circle.
    Point A is One Radius above the reference point, so the usual mgh had become mgR
    Point B is two Radii above the reference point, so the usual mgh had become mg2R.
     
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