Torque with attraction and friction

neufneufneuf

M is a mass (point) which attract the disk (disk has a mass too). Walls (black lines) don't have mass. No gravity. No friction on walls. In the image, it's like "sum" force increase rotationnal speed. I imagine the adjust the rotationnal speed and velocity and contact with left wall only for have the N1 force, if it's not possible I imagine an external system to do that, it's only just before the top wall.

1/ What's wrong with forces ?

2/ Before the contact with top wall, if there is friction on left wall, the disk turn ?

2a/ If the disk is turning, if there is friction added to left wall that cancel N1+FM exactly, just before the contact with up wall, no force on up walls and the disk continue to turn ?

dlouth15

Just to get this clear, do you believe the disk should turn in the real world? Or do you think that the forces as they are drawn in the diagram suggest that the disk should turn?

neufneufneuf

No, the disk don't turn in the real world ! it's like I see forces (and to find error of Algodoo too). I think I done error in my diagram and I don't drawn forces like they should. I explain like I do, like I done details maybe you can find what's wrong.

dlouth15

OK. I would not expect turning in the real world either but in addition, I don't think your diagram says that the disk should turn. The forces drawn in seem OK. The reason I don't think the diagram suggests turning is that all the lines of force pass through the centre of the disk. You need off-centre forces to cause torque and therefore turning.

neufneufneuf

But I drawn force at the CG, friction is at outer circle.

Take a wheel (bike) and attached a rope in the center, the force is 45° with vertical this attact wheel in one direction. Wheel turn, no ? The energy is kinetics and rotationnal. I understand where come from thekinetics energy but for rotationnal ?

dlouth15

Yes but you said there was no friction on the walls. I took this to mean that there was no friction between any object and the walls.

If it is the case that there is friction between the walls and the disk then, yes, there would be turning as it rolls towards the top wall. But this turning would stop when the overall motion of the disk stops when it hits the top wall. Your diagram shows a normal force downwards from the top wall which means that the disk has already hit the top wall and is no longer turning.

In any case, the forces as you have drawn them all pass through the centre of the disk and therefore in theory there should not be any torque.

neufneufneuf

yes, no friction at start and after a while I give friction (only at left wall). The disk stop turning when there is contact with top wall, even with top wall there is no friction ?

dlouth15

These things should have been specified at the beginning. Anyway the diagram as it stands with the forces as drawn does not predict that the wheel turns after it has hit the top wall. Which is also what we would expect in the real world. So there is no problem. The diagram is consistent with reality.

neufneufneuf

What force cancel the rotation after contact with up wall ?

dlouth15

There is still friction with the side wall. This causes the rotation to stop when it hits the top.

neufneufneuf

Ah ! ok. I would like to know the point of application of sum, how can I do ? I know only the point of application for FM, N1, N2 and friction. It's the center of mass of forces ?

dlouth15

In general all the forces are added (including any off-centre forces) and the resultant force is applied to the centre of mass and this combined with the mass of the object tells you how the object accelerates linearly (i.e. how the centre of mass accelerates). But additionally if you have off-centre forces then you need to work out the resultant moment of force or torque. This you combine with the moment of inertia to obtain the angular acceleration. Unfortunately it can't be done with a single simple vector diagram.

neufneufneuf

and for this example, you think the point of application of sum is in the center of disk ?

dlouth15

neufneufneuf wrote: »

and for this example, you think the point of application of sum is in the center of disk ?

I think friction is unnecessarily complicating matters. The object has come to rest against the two walls. Friction has come into play in stopping the rotation but once the rotation has stopped there are no more forces due to friction.

There are really only three forces at work. There is the force of attraction to the fixed point and then there are the normals from the two walls. These three forces act through the centre of the disk which happens also to be the centre of mass. The three forces balance each other out. There is no further movement and no further rotation.

This is what the physics would say and it is also what our intuition would tell us would happen in the real world.

Vectors N1, N2 and FM sum to zero.

neufneufneuf

Sure, but have you a link where I can learn to compute the point of application of sum ?

dlouth15

I would suggest a first year university physics textbook, the chapters on mechanics.

dlouth15

Here's a diagram of the physical setup described by neufneufneuf. I've drawn in all the forces that effect large red disk.



The purpose of this is to show that with a clear diagram the problem is a lot simpler than it might otherwise appear.

The main force [latex]\mathbf{M}[/latex] is the attraction between the the disk and the black dot.

The pull of the disk towards the dot produces two normal forces, [latex]\mathbf{N}_1[/latex], and [latex]\mathbf{N}_2[/latex] due to the the left and top walls respectively. The line of the force acts through the centre of the disk and the centre of the dot and so there's no torque due to this.

These normal forces act perpendicular to the plane of the wall but because the object is a disk, the line of the forces passes through the centre of the disk and therefore there's no torque due to them.

In the bottom there's a diagram showing how, when you arrange the vectors head to tail, they form a closed figure. This means that there's no resultant force on the object causing it to accelerate in any direction. I've done this in a separate area to avoid cluttering up the main diagram.

There's no distances or angles indicated on the diagram but if there were you could use them to work out the magnitude of the two normal forces, [latex]\mathbf{N}_1[/latex], and [latex]\mathbf{N}_2[/latex].

Because none of the forces act off-centre there's no overall torque causing the object to rotate. If there's no friction, however, and the object is already rotating, it will continue to rotate indefinitely at the same rate. Any friction will cause the object to stop rotating.

neufneufneuf

Yes like you drawn forces there is not torque. I gave 3 images with 3 cases in last message, one case is the same than this. I'm agree with that image, but if ball turn (with friction or not) on left wall and move towards top wall: the ball move towards top wall, the force is only towards up wall. N1 is not the same value than this image. So when the ball reached top wall, if N1 is different, N2 must be different too and force "sum" exist. Maybe it exists only a fraction of second and after all forces are like this image.