Calculating Angular Velocity for Falling Object Simulation

[GeorgeM]

This is actually ω = √(6Δ(gh)/l²) l square is in the root.

Another note re the formula in post #18: velP = velP_pre + sqrt(((3*gA)*(1-sin((angle) * PI / 180.0) ))/length ) (this is eq2 attached image)
This equation will give different ω from the one in post #1, see attached. ω based on post #18 at 90° is 180 rad/sec! This is given a rod of 2m.

 

[BvU]

Could you post a list of symbols, their dimension, and what they stand for ? You have way too many different symbols for a simple 1 degree of freedom simulation. One moment vertical is 90 degrees, the other horizontal. Intractable.
There is no way to check your dimensions are correct. $\omega$ is pretty unambiguous, but vel is confusing and now I see an A pop up, etc. etc.
The pictures are meaningless, you don't say what is plotted.

ω based on post #18 at 90° is 180 rad/sec

This is so easy to check with the formulas in post #1 that I am surprised you post it (that takes longer, and it's clearly nonsense)

 

[GeorgeM]

I have the graphs annotated, x-axis = degrees, y-axis = omega.
If formulas in post #1 are nonsense, then why you couldn't say this in the first place? The other thing, is that none else did any commitment for this question, or said that my answer is wrong. I'm not saying it is right, I know is not right and that's why I'm here!

 

[BvU]

I'm sorry, I still can't make out what these pictures represent. I know I see omega as a function of some angle. I suppose the angle is in degrees and omega is in radians/second ? But you do not give me anything that I can use to reproduce these results.

The formulas in post #1 are fine. Your subsequent processing in the algorithm is not.
That's why you end up with something like

ω based on post #18 at 90° is 180 rad/sec

when it is so easy to see that

ω = √(6Δ(gh)/l²)

gives $\omega^2 = 6 * 10 \, m/s^2 * 1 m / (2m)^2 = 15 /s$

This is actually ω = √(6Δ(gh)/l²) l square is in the root.

Another note re the formula in post #18: velP = velP_pre + sqrt(((3*gA)*(1-sin((angle) * PI / 180.0) ))/length )

There I do not see L²in the root. So this is the one that is wrong. It is also confusing to use a symbol vel that is not explained. We are now at post #39 and not making much progress.

The other thing, is that none else did any commitment for this question, or said that my answer is wrong

The simple reason is that it has become hard to follow for others than you. I don't know how I can help you get on a better track. It does not help if I do the exercise for you. I have interspersed some sensible (imho) advice. Can you do something with that ? How would you want to be assisted better in this ?

 

[GeorgeM]

There I do not see L2in the root.

There isn't a L² since

$$\omega^2 = {3g\over l} (1-\sin\theta)\quad$$ so
$$\omega = \sqrt{{3g\over l} (1-\sin\theta)\quad} \ \ \ (1)$$
Given by the equation in post #18

As I said, I also tried the formula form post #1 $$\omega = \sqrt{\frac{6\Delta(gh)} { l^2}} \ \ \ (2)$$

Then I have two different equations for ω. And, I did a comparison to see on whether they return the same values, but they don't.

In code those are:
Note velP is the ω

for(float angle=90;angle>0;angle--) // loop from 90 → 0
{
hyp = length/2; // half of rod's height
cenX = sin((angle) * PI / 180.0)*hyp; // height of COM

gh = gA*cenX; // g*height of COM

velP = velP_pre + sqrt((6*(gh_pre-gh))/pow(length, 2)); //--- > Equation 2
velP = velP_pre + sqrt(((3*gA)*(1-sin((angle) * PI / 180.0) ))/length ); // --> Equation 1

omegaList.push_back(velP); // store omegas

velP_pre = velP;
gh_pre = gh;
}

omegaList will store the 90 different omegas. I used the list to draw the graphs for post #38

Is this clear so far please?

Thank you!

 

[GeorgeM]

I'll have to ask again in the future then about this. It is not a coursework, or an assignment. I have worked out formula (1) post #40, but I don't see an option to iterate the angles (90-0) without a loop (starting from 90->0). Then again, I cannot draw the rod without using another loop for each resulted ω from the previous step.
Anyhow, thank you very much for committing to this.