# Rate of change of angular acceleration?

1. Mar 24, 2013

### risecolt

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 $\alpha$/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*$\alpha$ = 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:

$\alpha$ = $\alpha$

ω = $\alpha$t + ωo

S = 1/2 $\alpha$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 $\alpha$t^2 + ωot + So
$\alpha$ 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]

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I have discovered a new method that I might be able to use.
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 = $\alpha$t + ωo
ω = $\alpha$t
ω = $\alpha$t
sqrt(gsinθ) = $\alpha$t
t = sqrt(gsinθ)/$\alpha$

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

WARNING:
ψ = $\alpha$/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. Mar 24, 2013

### Staff: Mentor

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

What is S?
α depends on θ, how does your integration work?

3. Mar 24, 2013

### risecolt

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 $\alpha$.
ψ = ψ
$\alpha$ = ψt + $\alpha$
etc.

4. Mar 24, 2013

### Staff: Mentor

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

ψ depends on S, so your approach does not work.

5. Mar 24, 2013

### risecolt

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.

6. Mar 24, 2013

### Staff: Mentor

Degrees are evil. And I think you are describing α/ω Or dα/dθ.

7. Mar 24, 2013

### risecolt

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.

8. Mar 24, 2013

### Staff: Mentor

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