Rate of change of angular acceleration?

In summary, the conversation discusses calculating the time it takes for an object to fall 90° about an axis of rotation in the x-y plane and the rate of change of angular acceleration represented by the symbol ψ. The value of ψ is approximately (0,1578 rad/s^2)/θ, where θ represents 1 degree or ∏/180 in radians. The equation M = I*\alpha = mgrcosθ is used to find the angular acceleration at any given point, and the moment of inertia is assumed to be I = mr^2. The possibility of using integration and the preservation of energy method to calculate ψ is also mentioned. However, it is noted that ψ is not constantly ≈ (
  • #1
risecolt
38
0
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.
 
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  • #2
risecolt said:
Unfortunately it is not rad/s^3, but (rad/s^2)/θ or [itex]\alpha[/itex]/θ.
Then you might have calculated ##\frac{\alpha}{\omega}##.

What is S?
α depends on θ, how does your integration work?
 
  • #3
mfb said:
Then you might have calculated ##\frac{\alpha}{\omega}##.

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

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.
 
  • #4
Where is the difference between S and θ, apart from constant factors?

ψ depends on S, so your approach does not work.
 
  • #5
mfb said:
Then you might have calculated ##\frac{\alpha}{\omega}##.

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

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
Degrees are evil. And I think you are describing α/ω Or dα/dθ.
 
  • #7
mfb said:
Degrees are evil. And I think you are describing α/ω Or dα/dθ.

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
risecolt said:
NO! For every degree that it turns, the angular acceleration is changed by 0,157 rad/s^2.
I know, and it does not influence the possible interpretation of this value.
 

Related to Rate of change of angular acceleration?

1. What is the definition of rate of change of angular acceleration?

The rate of change of angular acceleration is the measure of how quickly the angular acceleration of an object is changing over a certain period of time. It is represented by the symbol alpha and is measured in radians per second squared.

2. How is the rate of change of angular acceleration related to angular velocity?

The rate of change of angular acceleration is directly related to angular velocity. This means that if the angular velocity of an object is increasing, the rate of change of angular acceleration is positive, and if the angular velocity is decreasing, the rate of change of angular acceleration is negative.

3. What factors can affect the rate of change of angular acceleration?

The rate of change of angular acceleration can be affected by a few different factors. These include the mass and shape of the object, the applied torque, and the moment of inertia. These factors can either increase or decrease the rate of change of angular acceleration.

4. How is the rate of change of angular acceleration calculated?

The rate of change of angular acceleration can be calculated by taking the derivative of the angular acceleration function with respect to time. This can be done using calculus or by using sensors to measure the change in angular velocity over a specific time interval.

5. Why is understanding the rate of change of angular acceleration important in physics?

The rate of change of angular acceleration is an important concept in physics because it helps us understand the motion and behavior of rotating objects. It allows us to predict how fast an object will rotate and how quickly it will change its rotation, which is crucial in many real-world applications such as engineering and robotics.

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