Pendulum and rotation

neufneufneuf

Thanks, it's better sure.

neufneufneuf

When you computed the position of the center of gravity, it's in 2d (disk) or 3d (sphere) for ball ? Algodoo it's a 2d software and when I drawn forces I think in 2d not in 3d. Maybe the problem come only in 2d.

dlouth15

neufneufneuf wrote: »

Thanks, it's better sure.

Ok. So lets start with the concept itself of centre of gravity. If we imagine a arbitrary body of mass [latex]m_1[/latex] and a point outside at position [latex]\mathbf{r}_{2}[/latex], then an object at [latex]\mathbf{r}_{2}[/latex] will be attracted gravitationally to all parts of the large body.

Each of the parts of the large body will generate a small force, but all these forces will sum up to a single resultant force (here labeled [latex]\mathbf{W}[/latex]) in the general direction of the large body.



Do you agree with that so far?

neufneufneuf

Yes, I'm agree. But I see a "line" where there is no torque not a point. You're thinking in 2d ?

dlouth15

neufneufneuf wrote: »

Yes, I'm agree. But I see a "line" where there is no torque not a point. You're thinking in 2d ?

I didn't mention torque; I just wanted to know if you agreed with that statement that the small forces from the different parts of the large body added up to effectively a single large force.

neufneufneuf

yes I'm agree

dlouth15

neufneufneuf wrote: »

yes I'm agree

Ok, so we know the force by adding up all the bits of the object, calculating the individual small forces and then adding them together to get the big force F. We know its magnitude and direction.

Now imagine we were to concentrate all the mass, [latex]m_1[/latex], to a single point. We must find the position where this point is located such that it produces the same force F on the point mass located at [latex]\mathbf{r}_2[/latex].

This position that we need to find is called the centre of gravity with respect to [latex]r_2[/latex].

Do you understand so far?

neufneufneuf

yes, great !

dlouth15

neufneufneuf wrote: »

yes, great !

Ok, now please write down a formula which takes two masses a distance between them and tells you the magnitude of the gravitational force.

Then, rearrange it so that you supply the two masses, the magnitude of the gravitational force and the new formula gives you the distance between the objects.

neufneufneuf

For me it's an integrate 2D (if the object is 2d), I need to integrate all the surface of disk, it's that ?

dlouth15

neufneufneuf wrote: »

For me it's an integrate 2D (if the object is 2d), I need to integrate all the surface of disk, it's that ?

Yes. You need to sum up all the small forces on the disk to get a single force F. Then you rearrange the standard gravitational formula so that given a force and two masses you can calculate the distance. Subtract this distance from the distance between the centre of the disk and the attracting point and you have the deviation of the centre of gravity from the centre of mass.

neufneufneuf

I put the formula for a disk in the image, it's that ?

d is the distance from mass (attraction) to the center of disk
x= angle

dlouth15

Can you explain your reasoning behind it?

neufneufneuf

k is the "force" of attraction it is mMG with gravity, denominator is the distance from mass to a precise point

I lost the sqrt !

dlouth15

neufneufneuf wrote: »

k is the "force" of attraction it is mMG with gravity, denominator is the distance from mass to a precise point

Do you mean the denominator is the distance from the point (r, x) in polar coordinates to the point M?

Also you seem to be adding up magnitudes of all the forces rather than adding the forces as vector quantities though I could be misunderstanding what you are trying to do.

neufneufneuf

I forgot the sqrt, look at my last message. And true I need to multiply by cos(x)

dlouth15

neufneufneuf wrote: »

I forgot the sqrt, look at my last message. And true I need to multiply by cos(x)

I think you are on the right track but could you explain it in a bit more detail perhaps with the help of a diagram showing how the expression within the integral is constructed. That way we can know if your reasoning is correct.

Also you should say what the integral as a whole means (is it the magnitude of a force? is it a distance? etc). What are we getting after we do this integral?

neufneufneuf

A disk is a lot of circle, so it's easier to calculate only a circle. For have the center of gravity of this circle, I need to resolve the equation :

integrate(from -a to a of F) - integrate(from a to -a of F) = 0

a is an angle

F is k*cos(x)/(sqrt((r*sin(x))²+(d-r*cos(x))²)

Like that it's easier, a is the angle to find, after center of gravity is r*cos(a)

do you agree ?

Image is the integrate and solution

dlouth15

I understand and agree with the approach of treating the disk as a lot of rings giving a double integral.


But please explain this using a diagram: k*cos(x)/(sqrt((r*sin(x))²+(d-r*cos(x))²). I think you may be incorrectly calculating the x-component of the force by multiplying by cos(x). It is not cos(x) but the cos of the angle at the point M you need to use I think.

neufneufneuf

yes it's not cos(x) it's cos(w) with w = atan ( r*sin(x) / ( d - r*cos(x) ) ?

dlouth15

By "diagram" I meant a picture with lines, angles, labels etc. illustrating the angle w and the angle x and and so forth.

You should be able to find a simpler formula for cos(w) that doesn't involve atan. Remember your school trigonometry. Cos = adjacent / hypothenuse.

neufneufneuf

w = (d-r*cos(x)) / ( sqrt( sin²(x) + (d-r*cos(x))² ) )


I found this link, it's like that you compute ?

http://physics.stackexchange.com/questions/23868/center-of-gravity-vs-center-of-mass

dlouth15

I think the reason you are going wrong is that you are skipping important steps in the derivation of the formulas.

Anyway it is taking to long to explain so here it is

[latex]$$F_{x}=\frac{GM_{1}M_{2}}{\pi}\int_{r=0}^{1}\int_{\theta=0}^{2\pi}\frac{\left(d-r\cos\theta\right)}{\left(\left(d-r\cos\theta\right)^{2}+r^{2}\sin^{2}\theta\right)^{3/2}}r\mathrm{\, d\theta\,\mathrm{d}}r$$[/latex].

This is then used to calculate the position of the centre of gravity.

neufneufneuf

Ok, thanks I will try to find my error.

Do you know if force from wall is applied to center of mass or center of gravity. When formula says the perpendiculary force of the mouvement don't works, the force must be applied to the center of mass or the center of gravity ? I don't find links about that in Internet

dlouth15

neufneufneuf wrote: »

Ok, thanks I will try to find my error.

Do you know if force from wall is applied to center of mass or center of gravity. When formula says the perpendiculary force of the mouvement don't works, the force must be applied to the center of mass or the center of gravity ? I don't find links about that in Internet

Centre of mass I think.

neufneufneuf

Why it's the center of mass ? For me it's possible to think with the sum of all forces. In a linear study, the force from wall has always the same direction for each point of a solid.

dlouth15

The normal to a surface (the outer wall) always acts perpendicular to that surface. Since the object is circular, that means that the force acts through the centre of the object which in this case is also the centre of mass.

neufneufneuf

I understood the perpendicularity, it's ok. And I understood why Algodoo lost energy, it's because there are chocks and the restitution was not to 1, all is fine.

What's happen if the circle turn before start, the angular velocity is conserved and the force from wall don't work too ? or some point works in positive and other in negative and the sum is 0 ?

Another case: if wall is part of circle, the force from wall is perpendicular to all particules of the circle but if it's a part of an ellipse, the force from wall works ? I thought not an ellipse with same change of slope but very asymmetrical slope, a slope with a function of d^3 for example. Maybe it's necessary to take not a circle for the object to move but an ellipse with non homogeneous density, take density for have no torque when mass attract it.

neufneufneuf

I drawn a circle on an ellipse, the angle of force from wall and the movement of ball is not perpendicular so N works for some forces in positive and negative values I think.

neufneufneuf

I tested on Algodoo, and a circle don't change its energy but an ellipse on a circle can ...