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Dynamics of circular motion. Particle on the top of a circle

  1. Jan 12, 2013 #1
    1. The problem statement, all variables and given/known data
    A particle of mass M is on the top of a vertical circle without initial velocity. It starts to fall clockwise.

    Find the angle with respect to the origin, where the particle leaves the circle.

    bc8d1bbbf5.gif

    2. Relevant equations
    v=ωXr

    3. The attempt at a solution

    I used two unitary vectors..
    εr : which direction is the same as the line described by the radio in the separation point.
    εθ: perpendicular to εr (the direction of the tangential velocity in the separation point)

    r=R er

    v=dr/dt = Rder/dt
    v=ωXr (counterclockwise, so I changedthe sign to negative)
    v=-R(ωXer)

    v = -R(-ω) = ωR
    α=dω/dt

    a = dv/dt = R(α + ω d/dt)

    d/dt = -(ωX) = -ωer

    a = α R - ω^2 R er

    Forces:
    Weight, Normal force
    Centripetal is a resultant?

    F(eθ)==> mgcosθ = αRm
    F(er)==> N-mgsenθ= -ω^2Rm

    So:

    α=dω/dt= (dω/dθ)(dθ/dt)=ωdω/dθ

    θ
    ∫αdθ =
    ∏/2

    ω
    ∫ωdω
    ω0

    From F(er) : α= (gcosθ)/R

    θ
    ∫ (gcosθ)/Rdθ = (gsenθ)/R
    ∏/2

    ω
    ∫ωdω= (ω^2)/2
    ω0


    ω^2 = (2gsenθ)/R

    I come back to the force equations, (with zero normal force) and replace ω^2

    And I get nonsenses like 2=1. Did I do something wrong?

    Solution: 48,19º (by the teacher)
     
  2. jcsd
  3. Jan 12, 2013 #2

    haruspex

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    Not sure exactly where, but you have a sign error. The first integral will be negative, but the second positive.
    You've not filled in the 'lower' bound correctly after integrating.
    Could have got this much more quickly and reliably by conservation of energy. Note that the RHS should be 1-sin, not sin.
     
  4. Feb 13, 2014 #3
    I found it solved.

    rbxrgn.jpg
     
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