Robittybob1

Some good points there. If it was extremely low friction the jar might just not move at all and it would transcribe a circle drawn in the opposing direction to the rotation. If there was extreme friction no matter how fast I turn it there will be no slippage. So definitely the coefficient of friction will be a factor. Weight of the jar??? Mass is involved with friction. Friction force is proportional to mass but also the centrifugal force will be proportional to mass as well. So mass might not be a factor as long as it didn't exceed my ability to give it angular momentum. Is the speed of rotation a factor? Keeping the other factors the same there will be a speed at which the jar will slip at a certain radius. You won't be able to change that. Will the speed that is needed to make it slip be the same speed no matter where it is put? I am assuming there will be no slipping back from the radial line as the acceleration I'll be able to give it will not be great enough. So at all times the speed the jar is going is the speed of the surface at that particular radius. If it started at 1m and progresses to a radius of 1.300m at the same angular rotation of 1 Hz the force required to hold it in circular motion is greater as the radius increases so once it starts to slip it shouldn't stop, for at all times the centrifugal force is greater than the force of friction (being constant). So the idea of slowing the merry go round down once it begins to slip is rather impossible as the slipping off will happen faster than my ability to slow it down again. I'll have to have an edge so the jar doesn't smash.

OK as I've conceded above that there is only one force on the ball and that is the centripetal force, let's cast our minds back to the sliding issue discussed during the week, will the coin slide along a radial line, with only slight curve back from the radial line as I expect or as you said a transverse or transverse component. I had asked you about how much forward or backward from the radial line you'd expect after slipping 300 mm. If you answered that I missed it sorry, for today is good day to run the test. Would it be forward or backward of the radial line? The above statement sounds a bit like radial line movement. Do you still stick to the transverse concept? Remember in the video the slip line was drawn perpendicular and forward wrt to the radial line. I'll have a flat bottomed pint sized glass jar filled with water, and spin the merry go round by hand. I'll draw a line out from the center, place the jar on the line 300 mm from the edge and then "attempt to spin the whole system fast enough to generate enough centrifugal force to throw the jar off the edge". (I'll slow it down when it is near the edge as I don't want broken glass.) I'll then slow it to a stop and see what path the jar took. Predictions please.

If something is related by 2 Pi/T does that mean it was in radians originally and is now turned back to rotations/unit time? I'll get it soon. Hey - I'm tired now and I had a great day. Thanks.

I was wondering if it was angular velocity. Thanks for the clarification. I can't accept that the whole system is only the ball. It must be all that is within the circle that the ball turns. Who decides such matters? If you look at the ball only how much of the circle do you analyse? Just a small portion to be sure it is travelling in a circle! OK if that was the case there is only one force acting on the ball, but ....

Great, I see what you mean now. Did you see my post on the other thread where I had a strange result can you give me a little help please http://www.scienceforums.net/topic/88420-centrifugal-forces-appear-to-act-opposite-to-gravity-how-is-this-possible/page-24#entry868731 That bit.

That helped a lot. Was this a typo?

I would like to answer the question but first I've got to understand the question. You are the first to describe the "centrifugal force acting on the back". Where is the back? Since there is tension in the string I can only conceive of a force at each end of the string, so in your view which end of the string is the centripetal force acting? It surely must originate at the pivot point and that force is transmitted to the object via the tension in the string. So that leaves only one place where the centrifugal force can originate and that is at the other end of the string, in other words within and by the mass itself.

does m = mass? T-1 = /second or per unit time. L what is this variable? I have seen L being used for angular momentum - what is it in your equation? Finally found what I trying to say: http://en.wikipedia.org/wiki/Angular_momentum#/media/File:Ang_mom_2d.png

It is the tangential component of the momentum. Tangential being the direction perpendicular to the radial lines of a circle (curve) The radial line is the shortest distance between them or even more basic just the distance between the two points ( a point on the circumference has a constant distance to the center point in a circle. (they are radii or radial lines maybe.) If an object is travelling directly along the radius it is not going to have any angular momentum wrt the center point but it will still have momentum.

Thanks, but can you use the word "momentum" in a sentence scientifically explaining how the string gets tension? Does the tension vary proportionally to momentum? The centripetal force is going to be Fc = mv^2/r where as momentum is just mv, so just looking at it the difference is a factor of v/r. v/r is that something like a frequency? v = 2 Pi r/T, t= period, divide both sides by r and we get v/r = 2 Pi/T, v/r = constant /T seconds, so can we call that a frequency?

OK I have told you how the ball achieves its angular momentum so where does the tension in the string come from?

Exactly. That is how the experiment starts off. It isn't a diversion, I am trying to get Swansont to tell me where the centripetal force comes from; progressing from that situation. What caused the tension in the string that he mentioned #462? That is where that angle of motion comes into the equation for if you throw the ball just in a radial direction that won't produce circular motion either. So it is more than just giving the ball initial momentum but a momentum component perpendicular to the radials, and that is the basis of angular momentum the "mv" part and then multiply that by the length of the radius or in total L = mvr where v = tangential velocity component of the initial momentum vector.

something missing there. You have to have the ball in transverse motion as well. The pivot point is an immovable object (a fixed point). It isn't capable of pulling the string or making the ball move.

OK you tie a string to a ball and the other end to a pivot point. Where does the centripetal force come from?

Thanks. "what is providing that alleged force. ...... just an object that's moving in a circle (a ball on a string was the scenario)." Can you put tension on that string without two forces, one being applied at each end? If one force is called the centripetal force what is the other called?

Are you using the theory of the Roche Limit? http://en.wikipedia.org/wiki/Roche_limit in most cases the object falling toward the Sun is attempting to orbit the Sun's mass.

It was part of a whole paragraph which I had this idea about but I was too tired to check the idea with formulas. But it was still looking promising so I deleted the majority of the paragraph and just left a bit even though I knew it wasn't making a whole lot of sense. I was just too tired to complete it, sorry. But I was looking at any two point masses and calculating the angular momentum between them even though they were not attached in any way but just passing each other in space. The angular momentum is based on the vector component that was perpendicular to the distance between them but the momentum is dependent on the relative velocity component between them. The ratio between the linear and angular has possibly a trig function relationship but I couldn't find what I was looking for just yet. Can you see what I am trying to say? Can you help me on this please? Below is what I had written originally.

I haven't got that text any longer but you will just about guarantee that it is drawn with an immoveable center of rotation. It is this central point that the object's centrifugal force acts on the same center that is providing the centripetal forces. If there is allowed to be unbalanced forces in our discussion, what happens if the central point loses its immovable character? Then the centrifugal force over balances the centripetal part. I don't like the unbalanced concept for didn't Newton talk of equal and opposite forces, so you should see a balance at all times. With gravitational forces the objects move wrt each other in binary patterns orbiting around a barycenter. Even in the best set-up the immoveable central point will "move" to some degree unbelievable as that sounds (even I struggle with that).

Someone else mentioned a centrifuge! Could centrifuge be thought of as the ratio of momentum to angular momentum?

Good point. Ball bearings reduce friction not because they are frictionless but because they roll. http://www.scienceforums.net/topic/88420-centrifugal-forces-appear-to-act-opposite-to-gravity-how-is-this-possible/page-19#entry867736 I remember that post now. A rolling ball will have an extra angular momentum to consider. Is this what you used to see?

Yes I was wrong, very wrong. Is friction present at all times even when there is no lateral forces? But we can't call that friction a force can we? I might have to start a thread on friction to begin to understand friction. I had the idea the force will build up till it exceeded the frictional force. So if friction is the centripetal force what force overcomes the frictional force? Stuff in a centrifuge tube would be similar to the experiments of objects accelerated in a tube performed by Mike and Robittybob. The masses definitely move along the tube. This is my understanding of a centrifuge: With the centrifuge the denser material displaces any lighter (less dense) material, as happens slowly in a gravitational field but in a centrifuge the G-forces are much greater hence denser material settles out more rapidly.

I'll look back for it, but every situation will be slightly different. Rolling balls are rather frictionless so won't have a significant friction force compared to sliding object on a rotating table. It is Friday night here so with a bit of luck we'll be able to test it within 24 hours.

Tell me do you agree with Swansont "(Momentum of an object traveling in a circle is not conserved; whatever is exerting the centripetal force is changing it)"? I see his point, but when it is thought of as angular momentum it is conserved. It would be a bit unusual to say "(The Linear Momentum of an object traveling in a circle is not conserved; whatever is exerting the centripetal force is changing it) Either you are traveling in circle or a straight line you can't be both.

Yes but in another way an object rotating about the point of rotation that is not moving laterally might not have any linear momentum at all.

That is a surprise, The momentum of an object traveling in a circle has angular momentum and that is conserved. I listened to a lecture on that the day before yesterday. If you work out its instantaneous momentum the same amount of momentum is present the next moment but at different direction, true, but that is allowed under angular momentum rules. We allow friction to be part of the centripetal force but that only appears if the object is pushed over another surface. So what is the push force that makes friction into a centripetal force?