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Speed of a bead turned through X degrees

  1. Sep 15, 2016 #1
    1. The problem statement, all variables and given/known data
    A small bead is threaded on a smooth circular wire of radius r which is fixed in a vertical plane. The bead is projected from the lowest point of the wire with speed √6gr. FInd the speed of the bead when it has turned through: (a) 60°, (b) 90°, (c) 180°, (d) 300°.

    Answers: (a) √5gr, (b) √4gr, (c) √2gr, (d) √5gr

    2. The attempt at a solution
    While this question is explained here, I do not understand one part. Why PE = m * g * h = m * g * (r - r * cos(θ)). Specifically h = r - r * cos(θ)? Why isn't it equal to just h = r cos 60°?

    Taking this image (as I see the problem):
    serie4-7v.png
    We have φ = 60°. R is h and equals to r cos 60°, where r is the adjacent line to the angle. Why the correct height h is r - r cos 60° and not just r cos 60°?
     
  2. jcsd
  3. Sep 15, 2016 #2

    BvU

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    The potential energy is taken as zero when ##\phi=0## So you want h to be the height above the reference height zero. ##h - h_0 = ...##
     
  4. Sep 15, 2016 #3

    TSny

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    h is the vertical height of the bead above the bottom of the wire loop, as explained by BvU.
    upload_2016-9-15_10-21-7.png
    Your picture indicates that the loop is rotating about the vertical axis with angular velocity ##\Omega##. But, apparently the problem wants you to assume the loop is not rotating.
     
  5. Sep 15, 2016 #4
    What would be the correct graphical representation of the problem? Could you please give an example?

    And with your image I think I got it. If we take two positions h = 0 at α = 0 and r = 10 cm (for example). When the bead moves to α = 60° the height is not 0, since the bead is in the air (if we take h = 0 as the moment when the bead is on the ground). In order to find h we need to subtract 10 cos 60° from the original radius, which is equal to 10. So h = 10 - 10 cos 60 = 5 cm. Is this the right logic?
     
  6. Sep 15, 2016 #5

    TSny

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    Yes, that's right.
     
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