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Lagrangian of a pulley system

  1. Apr 30, 2014 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    A uniform flexible chain of mass M and length L is hung under gravity over a frictionless pulley of radius a and moment of inertia I whose axle is at a fixed height above the ground. Write down the Lagrangian of this system in terms of a generalized coordinate l denoting the displacement below the axle of one end of the chain. Assume that L is sufficiently long that some part of the chain hangs freely from both sides of the pulley.

    2. Relevant equations
    L = T-V, V the potential energy term (which I should take to be of the form -mgz, where z is the distance of the CoM of the system from the fixed point and zero potential reference point there.


    3. The attempt at a solution
    Let ##l## be the distance of one end of the chain from the axle. Let ##l'## be the distance of the other end from the axle. Then ##l + l' + a\pi = L##. I don't see a way to incorporate the fixed height in (H), unless I introduce more parameters namely the height of the two ends above the ground ##z_1## and ##z_2## in which case ##z_1 + l = H## and ##z_2 + l' = H##.

    I am not really sure how to model the kinetic energy term. One point cannot have the entire mass of the string.

    Thanks.
     
  2. jcsd
  3. Apr 30, 2014 #2
    I don't think that the fixed height is crucial for solving the problem here, you are on the right track with just [itex]l[/itex] and [itex]l'[/itex] as the generalized coordinates since potential energy is defined up to an additive constant anyway. Since the chain is a continuous rather than a discrete body, perhaps you should consider differential kinetic and potential energy terms and then integrate to figure out the total Lagrangian.
     
  4. Apr 30, 2014 #3
    doesn't the chain move with speed dl/dt ?
     
  5. Apr 30, 2014 #4

    BvU

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    If l and l' have a constant sum, only one of them is enough to describe the state of the whole system....
     
  6. May 1, 2014 #5

    CAF123

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    Hi everyone,
    Ok, it turns out that the kinetic energy term was easy and the potential energy term not so. I went with your suggestion of integrating over the length of the strings to get the total potential energy contribution of each segment relative to the fixed axle.

    Yes, so since the string is inextensible and therefore of fixed length, if one end moves up the other end moves down with the same speed. So the kinetic energy of the string is simply ##(1/2) M \dot{l}^2## since all points on the string have the same speed.

    Yes, ##l + a\pi + l' = L \Rightarrow \dot{l} = - \dot{l'}## so they are related by this constraint and they cannot both be generlaised velocities.

    Put the zero of a coordinate system at the axle. Then the potential energy of a small piece of the string in segment from [-l,0] is ##dU = \rho d\ell g \ell## where ##\ell## is the distance of a small segment from the axle. So integrate over the whole segment gives ##U = \rho g l^2/2##

    I think this is the right idea but I can't get the other segments (the portion wrapped around the pulley and the portion of length l') in terms of known quantities.

    Edit: Alternatively, if I could find the centre of mass of the system (as a function of ##l##) which lies some distance ##h## below the axle then I could simply write the potential energy as -Mgh(l).

    Thanks.
     
    Last edited: May 1, 2014
  7. May 1, 2014 #6

    BruceW

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    yep. nice and simple :)

    Strictly speaking, ##\ell## is not the same as the distance from the axle, because the axle has finite width. But, anyway for this part of the calculation it doesn't matter, since x translation does not affect the potential energy of a vertical piece of string.

    Also, there should be a minus sign I think. You integrated from -l to zero right? surely there will be a minus sign..

    keep going. You talked before about l' being the length of the other section. So, write out l' in terms of l, and figure out the limits of integration, then integrate!

    Also, do you really need to integrate the segment wrapped around the pulley? Think about it for a bit.

    edit: sorry if my last line sounds condescending. It's not meant to be, I'm just being cautious about following the rules of PF, not giving away too much about the answer.
     
    Last edited: May 1, 2014
  8. May 1, 2014 #7

    CAF123

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    Hi BruceW,
    Ok, so the distance of a piece of string from the axle is ##a + \ell##. But since ##a## is constant I could ignore the constant terms and perhaps the cross terms will cancel later on.

    Right, there is a minus.

    I got it, thanks!

    Indeed not, since we are to assume that there is a sufficient amount of string over the pulley, there will always be a constant length ##a\pi## over the pulley and so we can ignore the potential energy contribution. A small piece of string wrapped around the pulley can be written as ##d\ell = a d\theta## and its potential energy ##\rho g d\ell##. I think this would be the right integral to check: $$U = \rho g a\int_{\pi}^0 \text{d}\theta = \text{const}$$
    Edit: No, wait, that cannot be correct by inspection of dimensions. I seem to be a missing a dimension of length somewhere which I cannot find.

    Not at all!
     
    Last edited: May 1, 2014
  9. May 1, 2014 #8

    BruceW

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    That's not quite right still. But maybe it's best not to dwell on this stuff... It is a 2d problem, and the vertical section of string essentially has a mass per area ##\delta_{(x-a)} \rho## So the 2d equation is ##dU=\delta_{(x-a)} \rho y dx dy## when we integrate over x, we simply get to your equation ##dU=\rho y dy##. So I think you intuitively skipped this first step. Which is not surprising, since teachers skip these kinds of steps all the time. But I think sometimes it's nice to recognise these kinds of steps.


    yeah, exactly. The contribution to the potential energy does not depend on time, or on our variable ##\ell## so we can ignore its contribution.

    umm, hehe you don't really need to find this value, since you've already stated that as long as there is enough string over the pulley, there will always be the same amount of mass here, and in the same position, so the contribution to the potential energy is constant.

    But I guess you can calculate it anyway. uh... it looks like you're calculating the weight, rather than the potential energy. Weight is ##\rho g d \ell## so calculating this won't give you potential energy.
     
  10. May 1, 2014 #9

    CAF123

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    Let me just check the notation - that is a delta function there?
    It doesn't matter about the actual constant, I just wanted to get the integral right. if ##d\ell = ad\theta## is a small segment of string around the pulley and at a height ##\ell## above the axle. Then ##dU = \rho d\ell g \ell = \rho a d\theta g \ell## is its potential energy contribution. Then I would integrate from 0 to pi, to give $$U = \rho a g \int_{0}^{\pi} l(\theta) \text{d}\theta = \rho g a( λ(\pi) - λ(0)) = \text{const}$$##λ## is the antiderivative of ##\ell##. Is that right?
     
  11. May 1, 2014 #10

    BruceW

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    yep, yeah, it looks a little bit weird because I put the argument in subscript for some reason.

    yeah, that looks good to me. Of course, ##d \ell## is not the differential of ##\ell## so the notation is a little confusing. So could use ##dS## instead (just to change notation), then it all looks good.
     
  12. May 2, 2014 #11

    BvU

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    Dumb says as dumb does:
    can't V be simply (l-l') M/L times g times (l+l')/2 ?
    First factor is the excess mass on one side
    Last factor is its (fixed) center-of-mass position

    Re-expressing in the (one single) generalized coordinate l (which was already part of the problem statement !!!) gives something proportional to l plus something constant.
     
  13. May 2, 2014 #12

    CAF123

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    Could you explain why you used a delta function there and the need for integrating over x?
     
  14. May 3, 2014 #13

    BruceW

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    it's a 2d problem, so what's the correct general equation for 2d? and our mass distribution in 2d is a very thin piece of string hanging vertically, so what is the equation for the mass distribution in this case? I said maybe it's not so useful to dwell on this kind of stuff, but I think it's important to recognise this is a 2d problem. So the starting point for this problem is generally to use 2d equations, and then later you reduce to an effectively 1d problem.
     
  15. May 3, 2014 #14

    BvU

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    Dumb speaking up again: Why nor 3D instead of 2D ? From the exercise formulation I like to think they want a 1D approach...
    Am I just lazy, or is there any clue that it should be 2D instead of 3 or 1 ?
     
  16. May 3, 2014 #15

    BruceW

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    you have to think in at least 2d at the start. the problem reduces to an effectively 1d problem. But that will not always happen generally, so you must start by thinking in 2d. For example, if we allowed the end of the string to move over the pulley, then you would need to use 2d.
     
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