Rotational mechanics string wrap around twig

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  • #1
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A mass m is tied to a string which is wrapped to a twig of a tree. How many minimum times should we wrap the string so that it doesn't fall?

Here is what i got....
1.There is friction between the twig (or fixed cylinder) and the string.
2. if the total friction between the string wrapped around in 1 circle around the cylinder it will be a piece of cake.
3. The thing is that i am unable to find that because the operating friction force at every point from the free body diagram will be different
 

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  • #4
A mass m is tied to a string which is wrapped to a twig of a tree. How many minimum times should we wrap the string so that it doesn't fall?

Here is what i got....
1.There is friction between the twig (or fixed cylinder) and the string.
2. if the total friction between the string wrapped around in 1 circle around the cylinder it will be a piece of cake.
3. The thing is that i am unable to find that because the operating friction force at every point from the free body diagram will be different
Well, I've seen your image, I got a question now: Do you actually mean "twig" as the "twig" of trees and "strings" as strings?
If so, you are now facing an extremely difficult question, here's why:

You need data of both surfaces to calculate the friction between them,right? Now:

1.Twigs have an extremely complicated surface, composed of dead cells,lichen and other stuff.
2.Strings have an extremely complcated surface too, different materials,weaving techniques affects it's surface.

So you are now dealing with a problem involving so many different topics which every single one of them is so ridiculously complex that I DON'T believe that's what the question, or you originally meant.

So, the problem must come with some sort of "assume the friction between the twig and the string is xxxxx", or else this will be solveless.
 
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  • #5
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Well, I've seen your image, I got a question now: Do you actually mean "twig" as the "twig" of trees and "strings" as strings?
If so, you are now facing an extremely difficult question, here's why:

You need data of both surfaces to calculate the friction between them,right? Now:

1.Twigs have an extremely complicated surface, composed of dead cells,lichen and other stuff.
2.Strings have an extremely complcated surface too, different materials,weaving techniques affects it's surface.

So you are now dealing with a problem involving so many different topics which every single one of them is so ridiculously complex that I DON'T believe that's what the question, or you originally meant.

So, the problem must come with some sort of "assume the friction between the twig and the string is xxxxx", or else this will be solveless.
Yeah i know. The twig is supposed to be considered as a cylinder. So the surface is of a plain cylinder! Even the string is a plain and cylindrical
 
  • #6
Yeah i know. The twig is supposed to be considered as a cylinder. So the surface is of a plain cylinder! Even the string is a plain and cylindrical
Yeah, I know it should be, but you still can't know the friction between them.
Think about this: If you drew a right triangle with its two legs being 3 and 4, you know the last side is 5, because 3 and 4 are precise values.
(Pythagorean theorem)
But a "flat surface" isn't precise at all for calculating friction.

Or do you just need a solution with the friction considered as x?(Answer will be a polynomial of x)
 
  • #7
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I get it! I do know it's complications. I need a value with respect to coefficient of friction and mass m. I got why it's so hard..... That's why I asked this doubt
 
  • #8
haruspex
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This is a standard topic (given the simplifying assumption of a cylindrical twig) but there are two pieces of information missing:
  • The coefficient of static friction
  • The force applied at the other end of the string. If there is no force at all at the other end then, in principle, the string will always slip.
The solution is quite interesting. It involves some calculus.

By the way, your thread title is rather inapt. Nothing is rotating here.
 
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  • #9
This is a standard topic (given the simplifying assumption of a cylindrical twig) but there are two pieces of information missing:
  • The coefficient of static friction
  • The force applied at the other end of the string. If there is no force at all at the other end then, in principle, the string will always slip.
The solution is quite interesting. It involves some calculus.
Woah. calculus? I haven't reach math topics that hard yet, but I am now really interested about the solution since it has something to do it. Can you somehow post it after the two informaton missing shown above are fulfilled?

ps: i originally thought this will not be a tricky question as long as the friction between is being defined
 
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  • #10
haruspex
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Woah. calculus? I haven't reach math topics that hard yet, but I am now really interested about the solution since it has something to do it. Can you somehow post it after the two informaton missing shown above are fulfilled?

ps: i originally thought this will not be a tricky question as long as the friction between is being defined
Take a look at https://en.m.wikipedia.org/wiki/Capstan_equation
 
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  • #11
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This is a standard topic (given the simplifying assumption of a cylindrical twig) but there are two pieces of information missing:
  • The coefficient of static friction
  • The force applied at the other end of the string. If there is no force at all at the other end then, in principle, the string will always slip.
The solution is quite interesting. It involves some calculus.

By the way, your thread title is rather inapt. Nothing is rotating here.
Yeah i know it isnti about rotation, actually we were given this question just after rotation, anyway sorry about that..... And the coefficient of friction is to be taken n. And the other end will require no force at the other end if there is friction because friction will be opposing the weight of the block
 
  • #12
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Oh ohk i got this.... Ah let me go through it once though. Thanks.
 
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  • #13
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My question has nothing to do with a capstan.... Take a rope and a cylinder.... If you wind the rope around it and attach a mass to it, yoyou' see that it will slide but if the no. of windings is enough it won't slide due to the friction holding the rope acting between the cylinder and the rope!!! Yes it will require integration.... That's no problem, my concern is to calculate friction acting on one element!
 
  • #15
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@haruspex he doesn't actually mean that. What he is trying to do is to take a piece of string, wrap it onto it with only the friction between the string and cylinder holding it.
Yes exactly!
 
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  • #16
haruspex
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@haruspex he doesn't actually mean that. What he is trying to do is to take a piece of string, wrap it onto it with only the friction between the string and cylinder holding it.
But why would there be any? There would have to be a normal force, which must come from tension at the far end of the string or from the weight of the string.
If you put a few turns on then you will find (from the capstan solution) that you need very little tension. The weight of a dangling end of string might well be enough. But it cannot be zero, and in order to solve the question you must hypothesise a nonzero value.
 
  • #17
which must come from tension at the far end of the string or from the weight of the string.
Yeah, and that tension causing the string to wrap tightly on the cylinder which then creates friction which then stops the mass M from moving.

Are me and @Priyanka88 thinking this the wrong way?
 
  • #18
haruspex
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Yeah, and that tension causing the string to wrap tightly on the cylinder which then creates friction which then stops the mass M from moving.

Are me and @Priyanka88 thinking this the wrong way?
My point is that you have to supply a value, either for a tension applied at the far end of the string or for the density of the string itself. With the first option you can then apply the capstan equation; with the second it is going to be a different equation (and might be rather tricky).

If you take the string as massless and do not apply sufficient tension at the other end then it will slip.
 
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  • #19
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My point is that you have to supply a value, either for a tension applied at the far end of the string or for the density of the string itself. With the first option you can then apply the capstan equation; with the second it is going to be a different equation (and might be rather tricky).

If you take the string as massless and do not apply sufficient tension at the other end then it will slip.
Yes! And my question is concerned with your 2nd case! Obviously the string is massed here! I am unable to make that tricky equation! That's where i need help
 
  • #21
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Just asking,didn't your question come with that?o_O
Well i said there is friction.... Which means it is massed! If anyone is really solving it take it M1 if you want!
 
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  • #22
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Just asking,didn't your question come with that?o_O
Or just consider a value d as its linear mass density! That'll be easier
 
  • #24
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Yes! And my question is concerned with your 2nd case! Obviously the string is massed here! I am unable to make that tricky equation! That's where i need help
Let the string have linear density ρ and the cylinder radius r.
If we consider a string element length rdθ at angle θ above the horizontal (starting where the string contacts the twig above the mass), it has weight ρgrdθ.
If the tension there is T(θ) then by force balance we can obtain the differential equation
##\frac{dT}{d\theta}=gr\rho (\cos(\theta)-\mu\sin(\theta))-\mu T##
This much you might be able to find for yourself. Since you are unfamiliar with calculus I will lay out the solution I found.

When the smoke clears:
##T=ce^{-\mu\theta}+\frac{gr\rho}{1+\mu^2}[2\mu(\cos(\theta)-1)-(\mu^2-1)\sin(\theta)]##
At θ=0 we have T=Mg, so c=Mg; at the other end, after n+½ turns, θ=(2n+1)π and T≤0.
##2n+1≥\frac 1{\mu\pi}\ln(\frac{M(1+\mu^2)}{4r\rho\mu})##.
 
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  • #25
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@haruspex please explain how that differential equation came.... Cause i tried and mine is different.... The rest of all i understood!
 

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