Puzzled about circular motion (whirling a bung on a string)

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Dear forums

Hello. I teach physics, and whilst I think that my knowledge is fairly sound, once every now and I then I encounter something which baffles me like so....

In a simple demo of circular motion, a rubber bung is whirled around the head in a horizontal plane. If the bung is allowed to wrap around the finger APPARENTLY it should speed up.

However, if we consider the equation for centripetal force: F = MV^2/r

The mass is constant, and presumably the force remains fairly constant as I am providing that. The radius has decreased, so surely the velocity should also decrease to some extent to keep the fraction constant.

How do I resolve this please ! I am starting to think I should just take an early retirement !

Cheers

Matt
 

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  • #2
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Dear forums

Hello. I teach physics, and whilst I think that my knowledge is fairly sound, once every now and I then I encounter something which baffles me like so....

In a simple demo of circular motion, a rubber bung is whirled around the head in a horizontal plane. If the bung is allowed to wrap around the finger APPARENTLY it should speed up.

However, if we consider the equation for centripetal force: F = MV^2/r

The mass is constant, and presumably the force remains fairly constant as I am providing that. The radius has decreased, so surely the velocity should also decrease to some extent to keep the fraction constant.

How do I resolve this please ! I am starting to think I should just take an early retirement !

Cheers

Matt
I think you are confusing V with ω, linear velocity with angular velocity. The speedup you notice is not V but angular velocity ω.
The equation can also be written as

F = mv2/R = mω2R , using v = ωR. This makes

ω = √(F/mR).

As R decreases, angular velocity ω increases.

Not sure if this explanation is enough??
 
  • #3
Simon Bridge
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That was my thought too.

[itex]v=\sqrt{rF/m}[/itex]

[itex]\omega = \sqrt{F/mr}[/itex]

@mklein: how do you mean "APPARENTLY"? How could you tell?
 
  • #4
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  • #5
Simon Bridge
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Step b is a model of a satellite about 200 km from the Earth where there is still some air resistance and so energy is transferred to the atmosphere and the satellite descends. As that happens, its time to orbit the Earth grows smaller and the satellite speeds up.
... not actually wrong as such, but they are certainly guilty of sloppy writing.

Leaving aside the applicability of the model to the situation ... the time to orbit does get shorter ... but they don't say which speed they are talking about. [itex]v=2\pi r/T[/itex] or [itex]\omega = 2\pi/T[/itex]
 
  • #6
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Hi Simon

I used the word "apparently" because the websites describing the demo state that we should be able to notice an increase in speed.

Here is somebody with an MIT phd also stating that the speed should increase:

http://www.physlink.com/education/askexperts/ae154.cfm

I am still unsure !
 
  • #7
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Just in case you didn't notice it, on the practicalphysics.org website it was actually the reference to speed in the procedure part b which I was initially confused about
 
  • #8
Simon Bridge
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I used the word "apparently" because the websites describing the demo state that we should be able to notice an increase in speed.
Thanks - sooo... it wasn't because you'd done an experiment and noticed this then?
As it happens you'd cleared it up in a post subsequent to my question.
Just in case you didn't notice it, on the practicalphysics.org website it was actually the reference to speed in the procedure part b which I was initially confused about
I think we are posting past each other - that was the bit I quoted. Glad I got it right.
 
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  • #9
Simon Bridge
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Since L = mvr is then a constant of the motion, the product of v and r is L/m, hence r and v would be inversely proportional.
Oooh - sneaky.
In this case the force is not a constant of the motion? [itex]F=Lv/r^2[/itex]

In the first post - the argument was that the force is constant. Which would make for [itex]v=(F/L)r^2[/itex]... I can see how someone could get confused.
(Now I think I'm forgetting something ... I'll sleep on it.) [edit] got it - in order for the force to also be a constant, the speed and radius have to increase or decrease together. If they are left to do their own thing, they have an inverse relation.

In the previous example - the writing was sloppy because the author failed to make clear which speed he was referring to.

From this example, I can see that your confusion is actually over the actual conditions when twirling a string around your finger. Either that or you are having fun with us?

Perhaps a variation on the glass-tube and weight experiment could be done to decide the matter?
Try it with a string+bung twirled around a pole (get it going and let go vs keep it going).
 
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  • #10
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Your quote is from the "teaching notes" section, I am refering to the "procedure" section. Sorry to be pedantic !

"a Whirl the bung fairly gently in a horizontal plane above the head.

b Allow the string to wrap round a finger. If the bung is whirling slowly at first, the consequent increase in speed will be apparent."


Anyway, there are two references to "speed" on the page.

Thanks very much for your help, I do appreciate it.
 
  • #11
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Definitely not doing this for fun !

I am not entirely sure what the force condition is when a rubber bung on a string wraps itself around a finger !

I also admit that I am very rusty on the angular momentum it's equations - simply because it is not something I teach and I haven't looked at it in many years. I was hoping to be able to explain things using the centripetal force equation.
 
  • #12
Simon Bridge
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For fun - describe the problem to a year 11 class (16-17yo) and see what they come up with.
I was hoping to be able to explain things using the centripetal force equation.
... well that didn't work.

Turned out more subtle than it initially looked.

Angular momentum is constant in the absence of an unbalanced torque.
But when you let the string wrap around your finger you are going to give it some extra tugs just from the idiomotor effect.
You'd have to do a careful experiment to see if you have been tugging on the string.

Also look at it the other way - let the speed decrease by L=mvr and then see what happens to the force (halve r, doubles v, what happens to the force?)
 
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  • #13
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Indeed !

Thanks again

Matt

It would be fun to pose the question to my a-level class - but then not so fun when they give up and ask for the answer ! Just looking up the idiomotor effect - not familiar with that !
 
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  • #14
rcgldr
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The speed is constant because in this case the force is perpendicular to the path of the spiraling (involute of circle) bung. Since speed is constant, tension increases relative to 1/r. Take a look at post #4 in this thread:

https://www.physicsforums.com/showthread.php?t=328121
 
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  • #15
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Thanks rcgldr

So how does this link with the equation: F=mw^2R (w=angular frequency)

Suppose we halve R. Therefore, as you say, F doubles. So does this mean that W^2 must quadruple, so therefore w doubles?
 
  • #16
rcgldr
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So how does this link with the equation: F=mw^2R (w=angular frequency).
Indirectly, in this case s = wR = constant. If R is halved, then w is doubled.
 
  • #17
Simon Bridge
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Considering we have two examples of people missing this (four if you count above) perhaps someone should write this up for a common reference?

That is a really great description btw.
The first example (nuffield #4) seems to try to treat the "wind around the finger" as being the same as just shortening the string ... but then, it was not intended as a rigorous treatment.
The second example (physlink #6) was explicitly central and the next response proposes the "pull the string through a hole" experiment. The physics would be that of post #2 in the PF link (rclgdr #14).

Which would appear to cover everything - thanks.
I'm thinking that entire thread needs to be written up for teachers.
 
  • #18
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Yes please - without too much (any !) calculus !
 
  • #19
Simon Bridge
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In NZ, y11 physics (16-17yo) only just involves some calculus.
It's a tad tricky teaching moment of inertia without integration.

@rcldgr: you up for it - it's your work - or you happy for others to use it?
 
  • #20
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One way of looking at this is in terms of energy changes.Imagine that the finger is replaced by a rod which is held in position and that initially the bung is set rotating in a horizontal circle,the string,by necessity,being at a downward angle to the vertical.As the string starts to wrap around the post the potential energy will decrease due mainly to the string and bung following a downward spiral.This decrease of potential energy can results in an increase of kinetic energy and therefore speed.
What really happens is more complicated but for A level quite generous simplifying assumptions are allowed
 
  • #21
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I think I'm looking at this problem in a different way to others on this thread.In my case the mass is spiralling downwards but I think others are assuming that the string is being constrained to wrap itself at a fixed height on the rod.
 
  • #22
rcgldr
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@rcldgr: you up for it - it's your work - or you happy for others to use it?
I'm happy to have others use it, and it's not really my work, since there are various web articles that cover bits and pieces of these problems. Generally when someone asks about this issue, I refer to that previous thread, but it would be nice to have some of this stuff in a more accessable location.

I think I'm looking at this problem in a different way to others on this thread.In my case the mass is spiralling downwards but I think others are assuming that the string is being constrained to wrap itself at a fixed height on the rod.
The scenario previously being considered, is a puck sliding on a frictionless surface, so that there is no vertical movement, just the spiraling movement due to an attached string wrapping around a post or being pulled or released through a hole.
 

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