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Rate of change of angular acceleration?

  1. Mar 24, 2013 #1
    I am calculating the time it takes for an object to fall 90° about an axis of rotation in the x-y plane. I have managed to calculate the rate of change of angular acceleration, which will be represented by the symbol ψ.
    Unfortunately it is not rad/s^3, but (rad/s^2)/θ or [itex]\alpha[/itex]/θ.
    The value of ψ ≈ (0,1578 rad/s^2)/θ.
    θ represents 1 degree or ∏/180 in radians.

    The following equation applies for finding the angular acceleration at any given point.

    (1) M = I*[itex]\alpha[/itex] = mgrcosθ, where rcosθ is the arm from the center of rotation to the center of the mass.

    I have tried to come up with an equation by integration, but I guess that won't be possible since ψ I've designed is not rad/s^3. Regardless, this is my progress:

    [itex]\alpha[/itex] = [itex]\alpha[/itex]

    ω = [itex]\alpha[/itex]t + ωo

    S = 1/2 [itex]\alpha[/itex]t^2 + ωot + So
    S is always S+So, so So can be excluded from the equation.
    So I thought that if I (virtually) integrated this equation once more, I would be allowed to include the rate of change of angular acceleration.

    S = 1/6ψt^3 + 1/2 [itex]\alpha[/itex]t^2 + ωot + So
    [itex]\alpha[/itex] would be replaced by using equation (1) from above.
    S = 1/6ψt^3 + 1/2 (mgrcosθ/I)t^2 + ωot + So
    For this example I'll set the moment of inertia I = mr^2, giving me:
    S = 1/6ψt^3 + 1/2 (gcosθ/r)t^2 + ωot + So

    This tells me that the angular acceleration is independent from the mass of the object.
    If this is correct, I need to use the cubic root and I'll have to cross my fingers and hope that I won't get any complex numbers.

    Is this a correct way to implement the rate of change of angular acceleration?
    Do you have any other suggestions?

    Spoiler
    ψ is not constantly ≈ (0,1578 rad/s^2)/θ. It is actually a sine wave.

    http://cognitivenetwork.yolasite.com/resources/Diagram.png [Broken]

    ----------------------------------------------------------------------------

    I have discovered a new method that I might be able to use.
    How about preservation of energy?
    If mgh = mv^2 is true then
    mgsinθ = mω^2, and there is that beautiful sine wave you see in the diagram.
    I'm not sure if they are related, but there it is anyway.
    gsinθ = ω^2
    ω = sqrt(gsinθ)

    Then I can set up the standard equation for ω
    ω = [itex]\alpha[/itex]t + ωo
    ω = [itex]\alpha[/itex]t
    ω = [itex]\alpha[/itex]t
    sqrt(gsinθ) = [itex]\alpha[/itex]t
    t = sqrt(gsinθ)/[itex]\alpha[/itex]

    Then I'm just missing one thing, and that is ψ... and I'm lost again.

    WARNING:
    ψ = [itex]\alpha[/itex]/θ = rad/s^2 / degree = rad/s^2 / (π/180) = 9.0467 rad/s^2/rad = 9.0467 rad/s^2.
    This does not change the fact that it is the rate of change of angular acceleration.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Mar 24, 2013 #2

    mfb

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    Then you might have calculated ##\frac{\alpha}{\omega}##.

    What is S?
    α depends on θ, how does your integration work?
     
  4. Mar 24, 2013 #3
    S is the distance or the arc length.
    The rate of change of angular acceleration only depends on θ.
    I integrated by starting with ψ instead of [itex]\alpha[/itex].
    ψ = ψ
    [itex]\alpha[/itex] = ψt + [itex]\alpha[/itex]
    etc.
     
  5. Mar 24, 2013 #4

    mfb

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    Where is the difference between S and θ, apart from constant factors?

    ψ depends on S, so your approach does not work.
     
  6. Mar 24, 2013 #5
    I might not have calculated ##\frac{\alpha}{\omega}##.
    The rate of change of angular acceleration IS 0.15789473684210526315789473684211 rad/s^2 for every degree it turns, not radians, but degrees.
     
  7. Mar 24, 2013 #6

    mfb

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    Degrees are evil. And I think you are describing α/ω Or dα/dθ.
     
  8. Mar 24, 2013 #7
    NO!!! For every degree that it turns, the angular acceleration is changed by 0,157 rad/s^2.
    So if the angular acceleration is 2 rad/s^2 at an angle of 46 degrees, then it is (2 + 0,157) at 47 degrees.
     
  9. Mar 24, 2013 #8

    mfb

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    I know, and it does not influence the possible interpretation of this value.
     
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