Angular velocity of falling box

[GeorgeM]

Hi,

I'm preparing a computer software to simulate the fall of an object for an academic project. The object is rotating (not rolling) over the circular bottom point. I know that the angular velocity at the horizontal point is ω = √(3g/L). I would like to calculate the angular velocity at every angle (from 90° to 180°). Can you please provide a simple formula for this?

Thanks

 

[mfb]

It is your project.
Energy conservation should work. For every rotation angle, you can calculate the potential energy, the difference to the original value got converted to kinetic energy.

 

[GeorgeM]

So, you are proposing the following:

The difference in PE is equal to the KE, that is ΔPE = KE.
Ex: PE₁ - PE₂ = ½ m⋅v²
Then I solve for v, but this v is the vector from the COM to the ground. Will then have to calculate ω.
Or else, use the angular KE?
Is that right?

 

[mfb]

The kinetic energy of a rotating object is not 1/2 m v^2. You'll have to find the center of rotation and the moment of inertia around it, or sum rotational energy (around a different point) plus lateral motion.

 

[GeorgeM]

Yes, my mistake, KE = ½⋅I⋅ω²
May I ask what you mean by the centre of rotation? Is this the pivot point on the ground?

 

[mfb]

Yes, my mistake, KE = ½⋅I⋅ω²

If you measure I around a point that does not move, yes. But then I depends on the angle if you want to be precise.

May I ask what you mean by the centre of rotation? Is this the pivot point on the ground?

Yes.

 

[GeorgeM]

But then I depends on the angle if you want to be precise

Is this because of the opposite forces of the rotating body towards the rotation point? Can you clarify this please?

 

[mfb]

The point that has contact to the surface changes over time.

 

[GeorgeM]

Hi again,
I did the calculations on my software, but the graph of ω is not what I expect. See series 1 on attached image (x axis =frame no, y-axis = ω).
It should show a slower increase in acceleration at the beginning of the fall. Any ideas?

 

[mfb]

Where is series 1?
Yes, acceleration should increase over time.

 

[GeorgeM]

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...

 

[mfb]

Yes, series 1

I don't see series 1 there. Just 2, 3 and 4.

 

[GeorgeM]

Is only series 1 in the graph at post #11

 

[davenn]

Is only series 1 in the graph at post #11

you are still not understanding mfb's comment

there is NO series 1 plotted in your graph, only series 2, 3, and 4

so where is the series 1 data ?D

 

[GeorgeM]

There IS one series in ejs-graph-jpg

 

[mfb]

Yes, but that graph shows the expected result. So where is the problem? The attachment of post 9, where you discussed the problem with "series 1", does not have a series 1.