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Using Lagragian equations to find accelerations

  1. May 9, 2013 #1
    1. The problem statement, all variables and given/known data
    A mass m1 slides on a frictionless horizontal table. it is attached by a massless cord passing over a massless pulley to a mass, m2. A cylinder of mass m3, radius r, and moment of inertia 1/2(m3*r^2) rests on m1.
    (a) Choose and specify generalized coordinates (two are required)
    (b) Write the Lagrangian in terms of the generalized coordinates and their derivatives
    (c) FInd the Lagrange's equations of motion
    (d) Find the acceleration of m1 and the acceleration of the center of mass of the cylinder, m3.


    2. Relevant equations
    T=(1/2)mv^2
    V=mgh
    L=T-V or L=KE-PE
    d/dt(∂L/∂x')=∂L/∂x


    3. The attempt at a solution
    I am not sure how to deal with the cylinder on top of m1.
    I have for the Lagrangian without the cylinder first. I choose my coordinates as x1 to be on the table with m1 and x2 going down the string with m2. So T=(1/2)m1(x2')^2 +(1/2)m2(x1')^2+(1/2)I(x2)'/(a^2) or T=(1/2)(x')^2 (m1+m2+(I/a^2). Since x1=x2
    V=0-m2g(l-∏a-x). l is the length of the string and a is the radius of the pulley.

    Therefore
    L=T-V=(1/2)(x')^2 (m1+m2+(I/a^2)+m2g(l-∏a-m2gx)
    d/dt(∂L/∂x')=∂L/∂x
    m1+m2+(1/a^2)=-m2g
    x''=m2g/(m1+m2+(I/a^2))

    However I know I need to add in for the cylinder so what would T be for the cylinder and does T for m1 change because the cylinder is on top of it? For instance do I just add their masses together for the T? I believe T=Iw^2 for cylinder and w=v/r so I=(1/4)mv^2. Do I just add this do my above T equation and work it the same way?
    Thanks!!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 10, 2013 #2

    BruceW

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    T for the object m1 does not change because this expression only depends on the mass m1, right? and you are keeping the cylinder as a separate object, so m1 remains unchanged.
    Also, I don't think they want you to include the inertia of the pulley. (I am guessing this is what you mean by the term I/a^2 ?) I don't think you need to include this because they say the pulley is massless. (Well, the value of 'I' is going to be zero, since it is the inertia of the massless pulley).

    I think you're missing a factor of a half in the equation for the kinetic energy. Also I don't understand where you got I=(1/4)mv^2 from? I think you mean T, not I. In that case, I agree (and you do seem to have taken account of the half that should be in the equation for kinetic energy). So yes, just add it in to the lagrangian. But also, you need to remember that there will be another term due to the kinetic energy associated with the movement of the centre of mass of the cylinder. (You need to have a careful think about the velocity of the cylinder relative to the mass m1 and the actual velocity of the cylinder, which is relative to the fixed ground).
     
  4. May 10, 2013 #3
    Yes, I did mean T not I. I guess I am just confused about what all I have to add to the KE. I thought that the ((1/4)mv^2) was the term due to the KE associated with the movement of the centre of mass.
     
  5. May 10, 2013 #4
    Also, I do not need to add anything to the potential energy right?
     
  6. May 10, 2013 #5

    BruceW

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    no, this is the term due to just the movement of the cylinder around the centre of mass. (i.e. this is the KE if the centre of mass of the cylinder was not moving). So you need to also add another term for the KE due to the movement of the centre of mass with respect to a fixed reference point (i.e. the ground).
    that's correct. as long as the cylinder doesn't fall off the mass m1, its height is constant, so its PE is constant. Also, I just realised, have you remembered to include the KE of the falling mass?
     
  7. May 12, 2013 #6
    So for the added KE of cylinder due to the ground would be (1/2)mv^2?

    Also, I was wondering if I need to deal with the fact that the cylinder is on m1 so it is a distance above the table which is equivalent to the height of that m1.
     
    Last edited: May 12, 2013
  8. May 12, 2013 #7

    BruceW

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    no, you need to be careful here. if you are using (1/4)mv^2 as the KE due to the rotation of the cylinder, then v is the speed of the cylinder with respect to the object it is resting on, i.e. the block, right? (due to the cylinder not slipping). And then also, the block is moving relative to the ground. So the velocity of the centre of momentum of the cylinder with respect to the ground will not be v.

    well, the velocity is not affected, since it is not moving vertically. And the vertical position is constant. So does this affect the lagrangian? If not, then you don't need to worry about it.
     
  9. May 12, 2013 #8
    Right so I defined the v in 1/4 to be x2' since it is moving with the block and then then v in 1/2(m3)x3'^2 so since x2' and x1' are equivalent I replaced them in the L to be x' and then left x3' so my L=(1/2)(x')^2 (m1+m2+(1/4)m3)+(1/2)m3(x3')+m2g(l-∏a-x2)
     
  10. May 12, 2013 #9

    BruceW

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    If the cylinder moves with the block, it does not rotate, so the v in 1/4 would be zero. But, the cylinder does not move exactly with the block. To begin with, the only thing you can assume is that the cylinder is not slipping on the block. But it can roll. You can use the 'no-slip' condition to find the equation between x3' and x' and the v you need to use in the 1/4 term.
     
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