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Prism Mechanics

  1. Apr 17, 2005 #1
    If you have two prisms on top of each other, one smaller than the other, and the one on top glids down to the ground the lower prisms moves to the left.
    The problem is to find how far the lower prism moves.

    How do you set up the Lagrangian and the constraints for this problem?
    Any ideas?
     
  2. jcsd
  3. Apr 17, 2005 #2
    There is no friction between the prisms or between the lower prism and the ground.
     
  4. Apr 17, 2005 #3

    ZapperZ

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    You posted 2 questions in here. You do know that we have a Homework help section here in PF, don't you?

    Zz.
     
  5. Apr 17, 2005 #4
    Johan,

    What exactly do you expect from us ? Haven't you read the PF-guidelines ?

    What have you done up till now in order to solve your problem? What are your suggestions ? let us know and we will help you out as much as is needed. Just don't post such generalized questions. What you are really asking for is a plain complete solution...We cannot give you that according to the guidelines of this Forum

    regards
    marlon
     
  6. Apr 17, 2005 #5
    I have posted in the homework help section before. I post in classical physics when its not homework. If you want i can post in here whenever i have questions about a problem.

    I just wanted some hints on how to attack the problem. Not a complete solution.

    I attached an image of the problem

    If you choose an orthogonal coordinate system with the x-axis parallell to the ground
    i can set up the Lagrangian. If Tpi is the kinetic energy for the i:th prism you get

    L = Tp1 + Tp2 - mgh

    where h is the height to the center of mass of prism 2, the prism on top of the other.
    Then i guess you need a constraint because the upper prism glids down the lower prism. Here im not sure about how to set up the equation for the constraint. The prisms are in contact all the time but i cant see an obvious relation between the center of masses or something like that. If you choose the coordinate system with the x-axis parallell to the "inclined plane" (the lower prism) the upper prism glids down on maybe its easier to set up the constraint equation but on the other hand this inclined plane moves because the lower prism moves to the left.
     

    Attached Files:

  7. Apr 19, 2005 #6
    No idea if this is right but:
    Call M the larger bottom mass, m the small top mass, L the distance the mass has to slide down, and theta the rightmost angle of the bottom prism.
    So....
    m exhibits a force downards on M of which a component of this force is perpendicular to the surface.
    [tex]F_p = mgsin(\theta)[/tex]
    Of this force acting perpendicular to the surface there is a component of it that acts in the negative x direction.
    [tex]F_x = mgsin(\theta)cos(\theta)[/tex]
    This causes an acceleration on M
    [tex]F_x = Ma_x = mgsin(\theta)cos(\theta)[/tex]
    [tex]a_x = \frac{mgsin(\theta)cos(\theta)}{M}[/tex]
    There is also a component of the gravitational force acting parallel to the surface which directs m down M.
    [tex]a_y = \frac{F_{||}}{m} = gcos(\theta)[/tex]
    The time it takes for m to get to the bottom can be found by the equation:
    [tex]L=1/2gcos(\theta)t^2[/tex]
    The distance moved left by the larger prism can now be found.
    [tex]d = 1/2a_x t^2 = 1/2 \frac{mgsin(\theta)cos(\theta)}{M} \frac{2sec(\theta)L}{g}[/tex]
    [tex] = \frac{MLsin(\theta)}{m}[/tex]
     
  8. Apr 20, 2005 #7
    I have tried your approch.
    But it doesnt work...i think.
    I have attached a more correct image of the problem.

    I know the answer. If b1 is the horizontal length of the upper prism and b2 is the horizontal length of the lower prism the distance to the left the lower prism moves when the upper prism hits the ground is: (m1 mass of the upper prism, m2 mass of the lower prism)

    m1/(m1+m2) * (b2 - b1)
     
  9. Apr 20, 2005 #8
    here is the more correct image of the prisms
     

    Attached Files:

  10. Apr 20, 2005 #9

    Doc Al

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  11. Apr 20, 2005 #10

    OlderDan

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    Are you required to use a Langrangian approach to solving this problem? If so, we need to focus on this aspect of the problem.
     
  12. Apr 20, 2005 #11
    yes Doc Al, that is of course the easiest way to solve this problem! Thank you. But i still dont get the right answer tho.

    with i for initial and f for final and 1 for upper prism and 2 for lower i get:

    [m1x(1i) + m2x(2i)]/(m1 + m2) = [m1x(1f) + m2x(2f)]/(m1 + m2)

    d = distance moved left by lower prism = x(2f) - x(2i)

    x(1f) - x(1i) = b2 -b1

    d = m1/m2 * (b2 - b1)

    and the right answer is

    m1/(m1+m2) * (b2 - b1)
     
  13. Apr 20, 2005 #12

    Doc Al

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    No problem so far.

    This is an error. b2 -b1 is how much the top prism moves with respect to the bottom prism. Don't forget that the bottom prism moves also.
     
  14. Apr 20, 2005 #13
    yes of course.

    x(1f) - x(1i) = b2 - b1 - a

    which gives the right answer! thx


    But i still need a lot of training on using lagrange equations. Here is another problem where this approach is needed.

    a "particle" with mass m is attached to a spring on a horizontal rod and another rod with mass m and length 2a, which can move as a pendulum.
    See the image below.

    Set up Lagrange equations for the systems movement. No friction
    For small oscillations find the normal modes and x and phi as functions of time t, if the start is from rest.

    solution:

    If x is the particle's distance from the equillibrium position and phi is the angle for the pendulum you get.

    [tex]
    T = \frac{1}{2}I\dot{\phi}^2 + \frac{m\dot{x}^2}{2}
    [/tex]

    [tex]
    V = -mgacos\phi + \frac{kx^2}{2}
    [/tex]

    [tex]
    L = T- V
    [/tex]

    But this isnt enough. If you solve lagrange equations for this you get a solution for a system where the spring is isolated from the pendulum. You also need this...i think

    [tex]
    akx=I\ddot{\phi}
    [/tex]

    But how should you use this. You cant use it to eliminate one of the coordinates...that would be wrong i think.

    The answer says that lagrange equations gives

    [tex]
    4a\ddot{\phi} + 3\ddot{x}cos\phi + 3gsin\phi = 0
    [/tex]

    [tex]
    a\ddot{\phi}cos\phi + 2\ddot{x}- a\dot{\phi}^2sin\phi + 3gx/a = 0
    [/tex]
     

    Attached Files:

  15. Apr 20, 2005 #14

    OlderDan

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    I'm thinking you are missing something in [itex]T = \frac{1}{2}I\dot{\phi}^2 + \frac{m\dot{x}^2}{2}[/itex]. Since x is time varying, the angular velocity of the rod relative to the mass is not a complete description of its motion. It is going to have a translation component as well as a rotation component. I don't think you have that included.
     
  16. Apr 21, 2005 #15
    yes i missed that.

    [tex]
    T = \frac{1}{2}I\dot{\phi}^2 + \frac{m\dot{x}^2}{2} + \frac{m}{2}(\dot{x}+a\dot{\phi}cos\phi)^2
    [/tex]

    is the correct kinetic energy then...i think.

    But i still dont get the right equations. With k=3mg/a and I=1/3*ma^2 the equations of motion are

    [tex]
    4a\ddot{\phi} + 3\ddot{x}cos\phi + 3gsin\phi = 0
    [/tex]

    [tex]
    a\ddot{\phi}cos\phi + 2\ddot{x}- a\dot{\phi}^2sin\phi + 3gx/a = 0[/tex]

    Am I missing something else?
     
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