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Formulating lagrangian for a system

  1. Nov 15, 2012 #1
    1. The problem statement, all variables and given/known data
    http://fotozrzut.pl/zdjecia/ad3bbdd9f6.jpg [Broken]

    Mases are as stated on the picture, I is the moment of inertia of the pulley, angle marked is ω and a is a radius.


    2. Relevant equations
    L=T-V


    3. The attempt at a solution
    I try to formulate a lagrangian for this system, but I don't know what coordinates shall I use. How to relate the angle and position of the mass m?
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Nov 15, 2012 #2

    Dick

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    The distance the mass descends is equal to the amount of string that comes off your pulley when it rotates by angle ω, right? There's only one real coordinate in your lagrangian, ω.
     
  4. Nov 16, 2012 #3
    So is it ok to write:

    T=(dω/dt)^2(a^2)(m+M)/2+I(dω/dt)^2/2

    V=-mgaω-Mga(cosω)

    ?
     
  5. Nov 16, 2012 #4
    seems o.k. to me.
     
  6. Nov 16, 2012 #5
    Ok then, so after writing Euler-Lagrange's equations I get:

    (d^2ω/dt^2)(ma^2+Ma^2+I)=agm-Msinωag

    And I need to know the angular velocity. If i integrate wrt. time i get:

    dω/dt=agm/(ma^2+Ma^2+I)∫dt-gam/(ma^2+Ma^2+I)∫sinωdt

    But it seems strange that angular velocity would depend explicitly on time. Did I mess up somewhere? First term in V depends on ω and not on trig. function of ω, so when I differnentiate the lagrangian wrt. ω it disappears, and so after integration there is 't' in angular velocity.
     
    Last edited: Nov 16, 2012
  7. Nov 16, 2012 #6

    Dick

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    Why would that seem strange? If you put M=0 and analyze that then you just have a constant torque on your pulley. Angular velocity should increase linearly in time, right?
     
  8. Nov 16, 2012 #7
    You're right, it was just my autosugestion as question is formulated:

    find angular velocity in terms of a, m, M, I and theta. So how to do that? The answer that I found is:

    \sqrt{\frac{2ga(m\theta - (M + M)\cos\theta)}{I + Ma^2 + ma^2}

    and ther's no t explicitly.
     
  9. Nov 16, 2012 #8

    Dick

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    You should probably figure how tex tags work https://www.physicsforums.com/showthread.php?t=8997 Here's what you wrote:

    [tex] \sqrt{\frac{2ga(m\theta - (M + M)\cos\theta)}{I + Ma^2 + ma^2}} [/tex]

    Aside from typos (why M+M?) it looks like they used conservation of total energy E=T+V setting the energy E to 0. I don't know why they did that. There should really be an arbitrary constant in there.
     
  10. Nov 16, 2012 #9
    Last edited by a moderator: May 6, 2017
  11. Nov 16, 2012 #10

    Dick

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    Lagrangians aren't the way to go. You won't able to solve the equation of motion in any simple way.
     
  12. Nov 16, 2012 #11
    But if this equation is correct:

    [tex] \frac{dω}{dt}=\frac{agm}{(ma^2+Ma^2+I)}\int dt -\frac{gam}{(ma^2+Ma^2+I)}\int sinωdt [/tex]

    Shouldn't it be able to transform it to required form? I mean, if those equations express the same quantity, why do they differ so much?
     
  13. Nov 16, 2012 #12

    Dick

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    No, you can't get a useful solution that way. What are going to do with the integral sinωdt? sinωdω would be easy but that's not what you have. Just use conservation of energy.
     
  14. Nov 16, 2012 #13

    TSny

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    Go back to the equation of motion (Lagrange's equation)

    [tex](ma^2+Ma^2+I) \frac{d^2ω}{dt^2}=mga -Mgasinω[/tex]

    and multiply through by ##dω/dt## before integrating with respect to ##t##. This will lead to the energy equation.
     
  15. Nov 17, 2012 #14
    Damn, it works :). Thank you very much, I didn't think of doing that trick.
     
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