Pendulum and rotation

neufneufneuf

No gravity here. No friction. It's a pendulum, but the ball is attract by fixed mass M. The ball can turn around itself. For me, a part of ball is more attract from M than other, the so ball turn around itselft ? If ball turn around itself, the rotation win energy so the ball move towards M slowly ?

neufneufneuf

No, I think there is no rotation. It's logical. But Fb give more torque to ball than Ft, true ? I think the torque is different? It's logical too. Now if I look at the lenght for each force, the fact that the ball don't turn itself change the lenght of work. It's not the same work for Ft or Fb because it's not the same raidus but if I look for 90° of movement, the diameter of ball modifiy the distance too. Each point (look at image for black point) don't work with the same lenght. The ball is accelerating, and after decelerating. So the time during each point has the force from attraction change too. This must change the work, no ?

Second image: it's possible to imagine an easy balls composed with black and red point only. I drawn forces and a ball and the perpendicular force, it's not easy to understand the effect of this perpendicular force. It seems to reduce the attraction force.

neufneufneuf

I tested on Algodoo, sure the software seems not works in every case but here the energy lost is very important, it's not a simple error.

neufneufneuf

I attached the Algodoo scene. Note it's not a pendulum with gravity but with mass M attraction at the bottomo of the circle. Look at Algodoo scene.

dlouth15

It is sometimes hard to make out what you are trying to achieve. Can I suggest something like this:



What you can see here is based on your first diagram. You have the pendulum of length [latex]r_1[/latex]. It is at an angle [latex]\theta[/latex] from the vertical. The ball at the end of the pendulum is attracted to the black object at the bottom of the picture with force [latex]\mathbf{F}[/latex].

We know that there will be a normal force [latex]\mathbf{N}[/latex] acting along the direction of the pendulum as indicated.

We could if we wished, calculate the component of the force [latex]\mathbf{F}[/latex] along the tangent to the circle, labeled [latex]\mathbf{R}[/latex] in this picture, and then using the parallelogram of forces (see diagram at bottom) work out the magnitude of the normal force [latex]\mathbf{N}[/latex].

The important point to note is that there are only two forces acting directly on the ball: the force of attraction, [latex]\mathbf{F}[/latex], and the normal force, [latex]\mathbf{N}[/latex], and neither of these are off-centre to the ball. Therefore we would not expect any rotation to occur to the ball unless it was already rotating.

The other thing is that for the sake of clarity, I've separated any constructions based on the forces on the body to another diagram and labelled that clearly.

neufneufneuf

You're sure the center of gravity is always the same point ? In the center of the circle ? For me, the sum of forces change at each point, so the center of gravity change too, no ?

dlouth15

neufneufneuf wrote: »

For me, the sum of forces change at each point, so the center of gravity change too, no ?

Please elaborate on this.

neufneufneuf

cg3.png:

For me, Fb > Fa and at final the cg is not in the center of circle but like orange point in image. Like circle don't turn, if I look for each point, the movement of each point is not perpendicular to the wall's force.


perpen2.png:

angles are:

black point: -32.4° for Fa, -46.15° for movement
red point: -45.68° for Fb, -47.19° for movement

the force from wall is 41.68 °

Angle between force and wall is :

black point: 87.83 °
red point: 88.87 °

in each case the force from wall is not perpendicalar to the movement, this force works.

Do you see that ?

dlouth15

OK, so you have two points giving rise to two forces which we can call [latex]\mathbf{F_1,top}[/latex], and [latex]\mathbf{F_1,bottom}[/latex]. The top point is closer than the lower point and therefore [latex]\mathbf{F_1,top}[/latex] is bigger than [latex]\mathbf{F_1,bottom}[/latex]. See diagram below.



However for every two points we can choose another two (in red). These are chosen in such a way that they are mirror images of the first two as reflected in a vertical line through the centre. The corresponding forces are [latex]\mathbf{F_2,top}[/latex] and [latex]\mathbf{F_2,bottom}[/latex].



Now I hope you can see that the forces due to the top two points, [latex]\mathbf{F_1,top}[/latex] and [latex]\mathbf{F_2,top}[/latex] match the forces due to the bottom points, [latex]\mathbf{F_1,bottom}[/latex] and [latex]\mathbf{F_2,bottom}[/latex].

This means that there's no net torque due to these four points.

This can be done for any pair of pair of points. We can find another pair that cancel out any torque caused by the first pair.

Therefore in general for a disk there's no net torque.

neufneufneuf

No torque, I'm agree, the force from wall pass through the center of mass, so no torque. I'm interesting about the perdicularity of wall and attraction force. For me the CG move more and more the ball is attracted (point orange is CG). The movement of CG is not perpendicular to the force from wall, I think, or my image is false ?

dlouth15

If you go back to my earlier picture:



We have drawn in 4 points with associated forces pointing to the attracting body. But of course the disk has an infinite number of points, not merely on the edge but the interior also. Therefore there are an infinite number of forces. But it turns out that all these forces sum up to a single force that appears to act through a single point in the disk. This is true regardless of how the disk is orientated with respect to the attracting object. In fact every object has such a point, and the point is known as the centre of gravity (or today more commonly called the centre of mass). The location of the centre of mass is dependent on the mass distribution of the object and is independent of any external forces.

In the case of a disk of uniform density, this point called the centre of gravity (or centre of mass) is located at the geometrical centre.

Now in the case of your diagram, we can reduce the forces to two. The first is the force of attraction between the circle and the attracting body, [latex]F[/latex] and the second is the normal to the outer wall. The disk is forced to move in such a way that the centre of gravity always moves along the inner circle. So we can work out the component of F in the direction of the tangent to this circle to give [latex]R[/latex]. Then we can use the parallelogram of forces at the bottom of the diagram to calculate the normal force [latex]N[/latex]. See below:



You can see that it is basically the same set of forces as for the pendulum diagram I put up earlier.

I recommend you follow my example of only drawing the main forces (in this case only two forces) in the main diagram. It gets confusing if you also add in sums of forces etc. These can be worked out elsewhere.

The final diagram just shows the resultant force, [latex]R[/latex], which is the sum of the attractive force, [latex]F[/latex], and the normal force [latex]N[/latex].



You can see that the resultant force, [latex]R[/latex] always follows the tangent to the inner dotted circle, which is the line that the centre of gravity follows.

neufneufneuf

I'm agree that center of mass is in the center of circle (unifom density) but I don't understand why center of gravity can be at the center of circle, can you explain ? There are only 2 forces in this image.

dlouth15

neufneufneuf wrote: »

I'm agree that center of mass is in the center of circle (unifom density) but I don't understand why center of gravity can be at the center of circle, can you explain ? There are only 2 forces in this image.

The centre of mass is a more accurate and modern way of saying centre of gravity. They are the same thing. A disk acts under gravity as if it were a single point located at the centre of mass.

Can you explain how you place the orange dot in your picture?

neufneufneuf

Center of mass is the center of gravity if the gravitationnal field is uniform, here it's not the case, so for me center of gravity is like orange point.

a link:

http://ruina.tam.cornell.edu/Book/COMRuinaPratap.pdf

http://physics.stackexchange.com/questions/50107/what-is-the-difference-between-center-of-mass-and-center-of-gravity

http://answers.yahoo.com/question/index?qid=20070222032943AA3KZaB

The center of gravity is the average location of the weight of an object. In a uniform gravitational field, each mass element would weigh the same so the center of gravity is identical to the center of mass. In a non-uniform gravitational field--like in dilasluis's problem--the center of gravity is not the same as the center of mass.

dlouth15

neufneufneuf wrote: »

Center of mass is the center of gravity if the gravitationnal field is uniform, here it's not the case, so for me center of gravity is like orange point.

a link:

http://ruina.tam.cornell.edu/Book/COMRuinaPratap.pdf

http://physics.stackexchange.com/questions/50107/what-is-the-difference-between-center-of-mass-and-center-of-gravity

http://answers.yahoo.com/question/index?qid=20070222032943AA3KZaB

That is interesting. You are correct. In general the centre of mass and the centre of gravity are not the same thing in a non-uniform gravitational field. Thanks for providing those links.

I'm not sure, however that you are correct in placing the centre of gravity outside the centre of mass in the case of a disk or sphere. Can you explain your reason for this.

neufneufneuf

Center of gravity is where there is an equal weight on all sides.

maybe that ?

or:

http://www.google.fr/imgres?sa=X&espv=210&es_sm=93&biw=1920&bih=978&tbm=isch&tbnid=Ux-OclGc1kZbtM:&imgrefurl=http://scijinks.jpl.nasa.gov/tides&docid=36kaRNpS9FTq6M&imgurl=http://scijinks.jpl.nasa.gov/_media/en/site/tides/tides3.jpg&w=300&h=311&ei=EsW9UvCbPMqO0AWOwoCYDg&zoom=1&iact=rc&dur=390&page=1&tbnh=144&tbnw=138&start=0&ndsp=59&ved=1t:429,r:1,s:0,i:86&tx=91&ty=93

or , there is maths on several links on Internet ?

I added an image with 2 forces, I think it's the same, the force from wall is not perpendicular with attraction

dlouth15

neufneufneuf wrote: »

http://www.google.fr/imgres?sa=X&espv=210&es_sm=93&biw=1920&bih=978&tbm=isch&tbnid=Ux-OclGc1kZbtM:&imgrefurl=http://scijinks.jpl.nasa.gov/tides&docid=36kaRNpS9FTq6M&imgurl=http://scijinks.jpl.nasa.gov/_media/en/site/tides/tides3.jpg&w=300&h=311&ei=EsW9UvCbPMqO0AWOwoCYDg&zoom=1&iact=rc&dur=390&page=1&tbnh=144&tbnw=138&start=0&ndsp=59&ved=1t:429,r:1,s:0,i:86&tx=91&ty=93

This is actually the centre of mass of the Earth-Moon system not the centre of gravity of the Earth being shown here.

I will get back to you on the other points. I think the centre of gravity of a uniform density disk under the pull of an attracting object may be a point slightly ahead of the centre of mass. I'm looking into how to calculate it. For a sphere of uniform density, the centre of gravity and centre of mass are the same.

neufneufneuf

Please, show message #15 on this forum :

http://www.physicsforums.com/showthread.php?t=592784

I think it's possible to calculate with a point for fixed mass M and a disk for the object that move.

dlouth15

Basically the centre of gravity (not centre of mass) for a disk attracted to a point is very similar to what we've had before.

Recalling this image already posted:



From this image we have already agreed that there's no torque due to the attracting point. There is no tendency for the object to rotate if it is not already rotating. If it is rotating, it will continue to rotate at the same rate.

The effect of all the all the small forces from the bits of the disk add up to one large force which acts from the centre of the disk in the direction of the attracting point.

The only difference from what we've had before is that the force originates from a point slightly ahead of the centre of mass, but still in a straight line from the centre of the disk to the attracting point. See diagram below:



Note the small gap between the centre of the object and the start of the force vector.

Also, like before, the normal force also acts through the centre of the disk, therefore generating no torque.

So we have two forces, neither have torque. The parallelogram rule is applied like before and we get a resultant that points along the tangent to the inner dotted circle.

This is very much what we would expect in the real world. The Algodoo software also agrees.

neufneufneuf

For me, the orange point don't trace a circle, the center of gravity change more and more when it approach the fixed mass, show image. For you it is ?

dlouth15

neufneufneuf wrote: »

For me, the orange point don't trace a circle, the center of gravity change more and more when it approach the fixed mass, show image. For you it is ?

No, the centre of the disk (centre of mass) traces the dotted circle. The centre of gravity won't necessarily trace a circle but it will always be in a line from the centre of the disk to the attracting point; which is why there is no torque.

neufneufneuf

Imagine a big attraction and a big circle (where the ball move). The center of gravity move from center of circle of ball to outer circle, are you agree with that ?

dlouth15

neufneufneuf wrote: »

Imagine a big attraction and a big circle (where the ball move). The center of gravity move from center of circle of ball to outer circle, are you agree with that ?

I think the centre of gravity will move further from the centre of the disk when it is near the attracting point, yes.

dlouth15

I can't afford much more time on this. As far I can see, the disk doesn't rotate about itself. The centre of gravity (as opposed to the centre of mass) moves about the sphere. It moves further away from the centre of mass when it is close to the attracting point but because it is always in line with the centre of mass and the attracting point, its movement doesn't produce torque, which is what the thread is about.

neufneufneuf

I'm agree with you, the disk don't turn around itself. I drawn the center of gravity, for me it's logical the trajectory is not a circle.

dlouth15

neufneufneuf wrote: »

I'm agree with you, the disk don't turn around itself. I drawn the center of gravity, for me it's logical the trajectory is not a circle.

Yes, looks about right to me.

dlouth15

Here's a chart of how the centre of gravity deviates from the centre of the disk as the distance to the attracting point is varied. All units are in radii of the disk.

It starts with the attracting point on the edge of the disk which produces a large deviation of about 0.65 of the radius. It falls off rapidly after that.

neufneufneuf

So, it's possible to trace the trajectory of the center of gravity ? Have you a formula for compute the position of the CG ?

dlouth15

Yes it would be possible to trace it. I don't have a formula. I did it numerically.

Here is a smoother graph:

neufneufneuf

You have the program ? attraction, radius are parameters ? I would like to verify the perpendicularity of the trajectory. Like the trajectory is not the trajectory of a circle, for me, it's not perpendicular but maybe it's necessary to verify from start to end.

dlouth15

neufneufneuf wrote: »

You have the program ? attraction, radius are parameters ? I would like to verify the perpendicularity of the trajectory. Like the trajectory is not the trajectory of a circle, for me, it's not perpendicular but maybe it's necessary to verify from start to end.

It was done in a libraoffice spreadsheet. The only parameter is the distance of the attracting body from the centre of the disk in terms of the radius of the disk.

Rather than just supplying the spreadsheet I would rather explain the theory and show how you can use the theory to build your own spreadsheet. That would be a better use of this forum.

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) ) ?