Results of the centrifugal force experiment

[Robittybob1]

I placed glass jar on a playground merry go round which had a very shiny plastic surface. There was minimal friction. If it was placed near the edge it didn't take much angular speed to make it slide off and the initial sliding motion was radial but the further out it went the more it slid backwards to the direction of spin. If I started it off close in it needed a much higher angular velocity to get it started and the degree of swing to a transverse direction was much more accentuated.

I understand this to be due to the very low friction meaning it was unable to accelerate to the linear velocity of the surface at the ever increasing radius. When it fell off it was always going slower than the tangential speed of the outer edge.

The speed at which it approaches the outer edge increases as it slides. Once it starts sliding you don't need to increase the rotational velocity any further.

[swansont]

So your prediction was falsified and physics still works.

[studiot]

However all credit for making the experiment and posting the results. +1

[DrP]

However all credit for making the experiment and posting the results. +1

Funny - I was thinking -1 for starting a new thread about it instead of just putting it in the existing thread that SwansonT made... which was split from at least one other. Also, I'd like to see what conclusions he draws from this as I bet he still thinks there is a centrifugal force that is pushing the glass rather than it just going in a straight line... or he is just playing a game of 'how long can we go round in circles about the same thing'. I'll remain neutral for now. lol.

 

Rob - what conclusions do you draw from your experiment regarding the forces on the glass?

 

 

PS - sorry - a little grumpy this morning... not been awake that long. x

Edited May 24, 2015 by DrP

[J.C.MacSwell]

I placed glass jar on a playground merry go round which had a very shiny plastic surface. There was minimal friction. If it was placed near the edge it didn't take much angular speed to make it slide off and the initial sliding motion was radial but the further out it went the more it slid backwards to the direction of spin. If I started it off close in it needed a much higher angular velocity to get it started and the degree of swing to a transverse direction was much more accentuated.

I understand this to be due to the very low friction meaning it was unable to accelerate to the linear velocity of the surface at the ever increasing radius. When it fell off it was always going slower than the tangential speed of the outer edge.

The speed at which it approaches the outer edge increases as it slides. Once it starts sliding you don't need to increase the rotational velocity any further.

One caution is that statement is somewhat ambiguous. It is sliding backwards relative to the non inertial frame of the merry go round, not relative to the ground, the ground being what is referenced in the "direction of spin" of the merry go round.

 

You have of course made it clear with the remainder of the description...as long as you don't go back to using the non inertial frame and conclude that there must be a real force pushing the jar outward.

[dimreepr]

Imatfaal, if you close this thread I’m afraid I’ll be forced to declare the bet void (as it’s obviously the same thread).

If this was in the lounge I’d agree with studiot but as it’s here I agree with DrP.

[Robittybob1]

So your prediction was falsified and physics still works.

It was more or less what I predicted. It was a very low coefficient of friction. Whether this had a bearing on the result will need further investigation. It certainly made me think about how friction works and what happens if the surface providing the normal force is accelerated in an ever changing direction from underneath an object. http://www.scienceforums.net/topic/88420-centrifugal-forces-appear-to-act-opposite-to-gravity-how-is-this-possible/page-22#entry868431

 

I could accept a very slight slippage back from the radial line, for the coin has to accelerate at the larger radii and that could mean a slight difference in the coin's tangential speed and the speed of the merry go round

One caution is that statement is somewhat ambiguous. It is sliding backwards relative to the non inertial frame of the merry go round, not relative to the ground, the ground being what is referenced in the "direction of spin" of the merry go round.

I had the radial line on the platform as a reference. If it had moved away from the line in the same direction as the spin I would have said "forward".

I haven't said anything about forces but it was evident as well that the jar didn't go off at a tangent to the perimeter but at an angle which was a combination of the rate it increased radius and the angular speed it had gained from being carried by the surface.

Funny - I was thinking -1 for starting a new thread about it instead of just putting it in the existing thread that SwansonT made... which was split from at least one other. Also, I'd like to see what conclusions he draws from this as I bet he still thinks there is a centrifugal force that is pushing the glass rather than it just going in a straight line... or he is just playing a game of 'how long can we go round in circles about the same thing'. I'll remain neutral for now. lol.

 

Rob - what conclusions do you draw from your experiment regarding the forces on the glass?

 

 

PS - sorry - a little grumpy this morning... not been awake that long. x

I would have posted in the other thread but I was not able to post into it (some sort of permission restriction had been imposed, so I was left without an option but to start another discussing the results that I had observed). It has taught me plenty.

As far as the reason the mass starts sliding in the first place, it is the very bit that is radial at the start of the motion that is in question. At all times the jar has inertia. Does inertia increase with velocity? There is some reason the frictional forces were not enough to hold it in circular motion. I am looking into how inertia works in conjunction with friction. I don't have sufficient knowledge to explain it yet.

The static friction is less than dynamic friction, so once it starts moving there is no easy way to stop it.

Can we say "inertia overcomes the frictional force"? Would that mean inertia increases as the rotational velocity rises? I'll have to find out if this is the case.

Do you know how inertia works?

Edited May 25, 2015 by Robittybob1

[imatfaal]

Rob - how are you measuring any of this?

 

I would have hoped that the comments in previous threads would have led you to understand that just watching a moving object and estimating its movement is completely useless - in some cases worse than useless. In some cases you merely get an inaccurate result and in others what you see is warped to meet your preconceptions; this isn't a dig at you - it is just what happens. Experiments must be measured and conclusions drawn from data not from fallible human perception.

 

Use your smart phone to take video of the experiment, upload it to you-tube/vimeo, then you can use it to prove your ideas; until then any report of corroboration is as insubstantial as hand-waving.

[swansont]

 

The static friction is less than dynamic friction, so once it starts moving there is no easy way to stop it.

 

That's backwards. The static friction is greater. Once it starts sliding, the friction decreases.

[Robittybob1]

Rob - how are you measuring any of this?

 

I would have hoped that the comments in previous threads would have led you to understand that just watching a moving object and estimating its movement is completely useless - in some cases worse than useless. In some cases you merely get an inaccurate result and in others what you see is warped to meet your preconceptions; this isn't a dig at you - it is just what happens. Experiments must be measured and conclusions drawn from data not from fallible human perception.

 

Use your smart phone to take video of the experiment, upload it to you-tube/vimeo, then you can use it to prove your ideas; until then any report of corroboration is as insubstantial as hand-waving.

I'll see if I can borrow a smart phone. As long as you don't ask me to film it from the rotating frame. In the weekend I hoped on the merry go round and got spun around. I hate that. Maybe I could mount the smart phone and film it in the rotating frame, for it all happens within a quadrant.

 

That's backwards. The static friction is greater. Once it starts sliding, the friction decreases.

Thanks for the correction, "the static friction is greater than the dynamic friction, so once it starts moving there is no easy way to stop it". That was what I had intended to say.

Edited May 25, 2015 by Robittybob1

[J.C.MacSwell]

 

I had the radial line on the platform as a reference. If it had moved away from the line in the same direction as the spin I would have said "forward".

I haven't said anything about forces but it was evident as well that the jar didn't go off at a tangent to the perimeter but at an angle which was a combination of the rate it increased radius and the angular speed it had gained from being carried by the surface.

Right. Backward in the reference frame of the platform.

[Robittybob1]

Right. Backward in the reference frame of the platform.

The equipment I used had a total radius of about 1 meter, so lets assume in one of my many runs I placed the object at 0.5 m to begin with. That being what I would call quite close to the center for it took quite a fast spin rate to get it sliding.

From the maths while it is not sliding the linear speed of any part of the platform is proportional to its radius, so the speed of the surface when the object started to slip at the 0.5 m mark is half the speed at the outer edge. If the object was to leave at a true tangent by the time it got to the outer edge it would have had to double its linear speed.

 

It must be achieving part of this requirement as if begins to slip backward a little, implying that there is a frictional force accelerating the object in the tangential direction WRT to any point the object is at on the platform.

 

But any increase in linear speed only makes the rate of slipping increase. I found you could make it start off slowly if I gradually brought the rotational rate up. To stop a slow slip it might be possible to slow the turntable. That may have happened but it wasn't done deliberately. You can hear the jar sliding over the surface so that would be indicated by a sound followed by no sound. The sound is the same regardless of the direction of slip. (I just tested that rubbing my had in different directions across my computer table, but the frequency changes, so sound could be used to measure the speed of slipping.

So if I do the video I'll see if we can record the sound as well.

Now what I was getting to was that if it started slipping at 0.5 meter the maximum transverse component (assuming the acceleration from the dynamic friction to be zero) would it have a linear speed equal to 0.5 Vt (Vt being the tangential velocity of the outer rim at 1 meter?

Somehow I have to account for the time it took to slip as well, That is where the video would assist for timing is rather awkward having to look after the object and spinning the turntable and then timing it as well. I'd need an able bodied assistant.

It is really complicated to think of all the motions involved.

 

PS: Thinking about it, this velocity vector would meet the outer circumference at 45 degrees, so would you accept that the object would appear to leave at a 45 degree angle to the tangent in this hypothetical situation?

In the physical situation there will some dynamical friction so the velocity of the object will be increasing as it slides across the surface so the angle it leaves at will be less than 45 degrees to the tangent, but as it is slipping it will not be at the true tangent.

Well that is the logic and now to see if I can record this on a smart phone.

Edited May 25, 2015 by Robittybob1

[swansont]

Now what I was getting to was that if it started slipping at 0.5 meter the maximum transverse component (assuming the acceleration from the dynamic friction to be zero) would it have a linear speed equal to 0.5 Vt (Vt being the tangential velocity of the outer rim at 1 meter?

Why would you assume this?

[Robittybob1]

Why would you assume this?

I am making some very unlikely conditions especial of having zero dynamic friction, yet sufficient friction to keep the mass rotating at 0.5 r with a velocity of 0.5 v (v being the linear velocity at the outer edge).

Inertia will then keep the mass going in the direction and speed it was at the time the static friction was exceeded. If my line diagram is right it will hit the circumference at about (but less than) 60 degrees rather than the 45 degrees I estimated before.

But this would be so unlikely as when I try and sense the difference between static and dynamic friction is not as great as that.

I'll see if there are any records of these measurements on the net.

If I get the smartphone and something up loaded I could draw the diagram and you could check to see if my reasoning is right.

Static and Kinetic Coefficient of Friction Reference Table

http://blog.mechguru.com/machine-design/typical-coefficient-of-friction-values-for-common-materials/

Edited May 26, 2015 by Robittybob1