Recent content by 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...

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...

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...

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...

So my units for ω are in (√m)/s ? I shouldn't then use this formula if this what you are saying.

There is also another reason: my frame rate is not stable at 25 fps, so multiplying by 0.04 is not always right. You are saying that (sqrt(3*(gh_pre-gh)/length)) is incorrect, is this because of the mixed units?

So, I'll need to have an assumption i.e. how many pixels per metre (pixPerM = 100) and then multiply all my x, y variables with this? What do you mean by look up the omega that goes with theta? Say theta(0) = 90°, I draw this, then I look an omega that goes with that theta(0)? Give me one...

I know, I'm wrong here, I just cannot visualise it as an example...

Height is in pixels, say 100 pixels long rod. Is this what you ask?

There IS one series in ejs-graph-jpg

Is only series 1 in the graph at post #11

The final bit is to get the simulation running. Here I'm trying to implement what you wrote at post #6 I iterate the list of ω (there are 90 different values there) as follows: loop list of ω { thetaF0 = (theta_pre - 0.04*list) theta_pre = thetaF0 ///then draw the animation } It...

Great! And now the diagram looks like that... I did one run with height of rod = 100 and ω seems the same. I would expect that due to (sqrt(3*(gh_0 - gh)/length)) , when length goes up, ω goes down...

Yes, series 1, but it shouldn't accelerate at this step. It should look like the attached image. Notice that for the first frames/angles the velocity is not as accelerated as on the later frames/angles. I think something is missing here...

Sure. BTW, there is only one series (see attach). In this graph (x axis = frame no, y-axis = ω) I have simplified the problem with a rod falling, not an object as in post #1. So, here are the parameters and steps: length_rod = 400 comDistance = length_rod/2 gA = 9.8 loop angle 90 - >0...