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Falling disk with string

  1. Dec 1, 2016 #1
    1. The problem statement, all variables and given/known data
    Let's have a disk of mass ##m## and radius ##a## and massless rope tangled in it. One end of rope is tied to the ceiling and the disk is falling freely down. System has one degree of freedom. As a coordinate we can choose angle ## \phi## which says an angle of rotation from the start position. Find from the Lagrange equations of the first kind equation of motion and binding force.

    2. Relevant equations
    $$m\ddot{x}=F_x+\lambda\frac{\partial G}{\partial x}$$
    $$m\ddot{y}=F_y+\lambda\frac{\partial G}{\partial y}$$
    where G is binding function.

    3. The attempt at a solution
    I see that kinetic translation and rotation energy is connected. Probably by ## y=-a \phi## (##\dot{y}=-a\dot{\phi}## ) but generally I don't know what is my G function. I have seen a problem with pendulum or problem falling from ball where this functions are easy to guess. Please give me a kick.

    Thank you.
     
  2. jcsd
  3. Dec 1, 2016 #2

    TSny

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    For this problem, you are working with the two variables ##\phi## and ##y## rather than ##x## and ##y##.

    You can rearrange your constraint condition as ##y + a \phi = 0##. Thus, you can choose ##G( \phi, y) = y + a \phi ##.
     
  4. Dec 1, 2016 #3

    haruspex

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    Do you mean wrapped around it?
     
  5. Dec 2, 2016 #4
    Yes.
     
  6. Dec 2, 2016 #5
    Ok I have some attemp of solution but I mean the result is wrong:
    ##G=y+a\phi=0## ,+ because of it goes against y direction
    ##m\ddot{y}=F_g+\lambda\frac{\partial G}{\partial y}=-mg +\lambda##
    ##m\ddot{\phi}=0+\lambda\frac{\partial G}{\partial \phi}=\lambda a##
    Now I want find out ##\lambda ## =>
    ##\ddot{y}+a\ddot{\phi}=0 ##
    ##\ddot{y}=-g+\frac{\lambda}{m} ##
    ##a\ddot{\phi}=\frac{\lambda a^2}{m} ##
    ________________________________
    ##\ddot{y}+a\ddot{\phi}=0=-g+\frac{\lambda (a^2+1)}{m}## => ##\lambda=\frac{gm}{a^2+1} ##
    ##m\ddot{y}=-mg +\frac{gm}{a^2+1}##
    ##m\ddot{\phi}=\frac{gm}{a^2+1} a##
     
  7. Dec 2, 2016 #6

    TSny

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    The last equation (for ##\ddot {\phi})## is not correct. ##\phi## is an angular coordinate rather than a Cartesian coordinate, like ##x## or ##y##.
    Do you know how to get the equation of motion for ##\phi## starting from the Lagrangian?
     
  8. Dec 5, 2016 #7
    It seems very reasonable. I believe I can manipulate with lagrangian :) For me is more familiar deal with Lagrange equation of second kind. So in this point of view I take a Lagrangian ##L=T+V=T_t+T_{rot} - V=\frac{1}{2}m\dot{y}^2 + \frac{1}{2}J\dot{\phi}^2 - mgy=\frac{1}{2}m\dot{y}^2 + \frac{1}{4}m a ^2\dot{\phi}^2 - mgy## =>
    ##\frac{d}{dt}\frac{\partial L}{\partial \dot{\phi}} - \frac{\partial L}{\partial \phi} =\frac{1}{2}m a ^2\ddot{\phi}=0 ##. ##\phi## isn't there but I feel it should be there probably due the constrain...or I don't know.
    On the other hand in problem is written they want Lagrange equation of 1. kind...
     
  9. Dec 5, 2016 #8

    TSny

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    Lagrange's equation for ##\phi## with constraint ##G(\phi, y) = 0## is $$\frac{d}{dt} \frac{\partial L}{\partial \dot \phi} - \frac{\partial L}{\partial \phi} = \lambda \frac{\partial G}{\partial \phi}$$
     
  10. Dec 5, 2016 #9
    I see... so my equations of 1. kind are
    ##m\ddot{y}+mg=\lambda ##
    ##\frac{ma^2\ddot{\phi}}{2}=\lambda a##
    ## y+a\phi=0## do you agree?
     
  11. Dec 5, 2016 #10
    ##\ddot{y}=-g+\lambda/m##
    ##a\ddot{\phi}=2\lambda/m##
    ##\ddot{y}+a\ddot{\phi}=-g+3\lambda/m=0##=>##\lambda=gm/3 ##, ##\ddot{\phi}=\frac{2g}{3a} ## and ## \ddot{y}=-\frac{2g}{3}##.
    A binding force is ##\vec{F}_B=\lambda(\frac{\partial G}{\partial y},\frac{\partial G}{\partial \phi})## ?
     
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