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risecolt
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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
----------------------------------------------------------------------------
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.
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
----------------------------------------------------------------------------
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.
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