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Lagrangian of a falling rod

  1. Apr 27, 2014 #1


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    1. The problem statement, all variables and given/known data
    A rigid rod of length ##\ell## and mass ##m## has its lower end in contact with a frictionless horizontal floor. Initially, the rod is at an angle ##\alpha_o## to the upward vertical when it is released from rest. The subsequent motion takes place in a vertical plane.

    a)Taking as generalized coordinates ##x##, the horizontal displacement of the centre of the rod and ##\alpha##, the angle between the rod and the upward vertical, show that the Lagrangian is $$L = \frac{1}{2}m\dot{x}^2 + \frac{1}{24}m\ell^2 \dot{\alpha}^2 (1+3\sin^2 \alpha) - \frac{1}{2}mg\ell \cos \alpha$$
    b)Deduce the Euler-Lagrange equations, showing that ##x## remains constant in the motion.

    2. Relevant equations
    Lagrange's Equations, kinetic and potential energies.

    3. The attempt at a solution
    a)I don't see why ##x## and ##\alpha## are independent coordinates. If ##\ell/2## is the distance of the centre of the rod from the contact point, then it seems to me that the x component of the centre of the rod can be adequately described by simply ##\frac{\ell}{2}\sin \alpha##. Since they are related, I don't see why they can both be generalized coordinates in the problem.

    b)I solved for the EoMs and did obtain that ##x## is a constant. But I can't seem to picture this. Initially, ##x_{\text{centre of rod}} = \frac{\ell}{2}\sin \alpha_o## As ##\alpha## increases, ##\sin \alpha## increases since it makes sense for ##0 \leq \alpha \leq \pi/2 ##. Therefore ##x## must increase.

    Perhaps I am misinterpreting something in the question.

    Many thanks
  2. jcsd
  3. Apr 27, 2014 #2


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    The rod could move horizontally, so x can change even without α changing. In addition, the rod can lose contact to the surface, then α can change without x changing.
  4. Apr 27, 2014 #3


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    My interpretation of the question was that one end of a rod is on a horizontal surface and the other end is at an angle π/2 - α from this surface. So when it is released, the point of contact with the floor can move? I think this is what I misinterpreted - I assumed this point acted like a pivot.
  5. Apr 27, 2014 #4
    An unconstrained rod: how many degrees of freedom?

    A rod constrained to be and move in a plane: how many degrees of freedom?

    A rod with one end further constrained to slide along a curve: how many degrees of freedom?
  6. Apr 27, 2014 #5


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    Hi voko,
    3. Can be treated as a rigid body and any point on the body can be described via the Euler angles.

    This is what I envisaged the problem to be initially. In which case, only 1.

    The curve in this case being along the x axis say. That would be 2: one for describing the translation and one for describing the CoM rotation of the rod. So that is x and α in this problem.
  7. Apr 27, 2014 #6
    How so? The Euler angles describe the orientation of the body. The body can also be parallel-transported in any direction, it is not constrained at all.

    Really? How come, then, that by further constraining the body, as in the case below, more degrees of freedom are present? Think about a flat figure allowed to move freely on a plane: how many degrees of freedom?

    Yes. Even though it strange that you are wrong about the less constrained situations above.
  8. Apr 27, 2014 #7


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    The Euler angles would only describe the orientation of a point on the rod (different orientations come about by moving the rod around in one place.) But then we also have translations in the three coordinate directions, so the total is six. (3 rotational, 3 translational)

    I thought when you said 'a rod constrained to be' that that meant one point fixed. Otherwise, there are now only two translational directions and one rotational, so three in total.

    By constraining the body to move only along the horizontal surface, we reduce the # of the translational degrees of freedom by 1, and so the number of degrees of freedom reduces to (3-1 = 2.)
  9. Apr 27, 2014 #8
    Very well. Is there still anything you are not happy with in this problem?
  10. Apr 27, 2014 #9


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    Yes, just getting the right kinetic energy term. Here is my work so far:
    As always with these sorts of problems, the kinetic energy is the translational kinetic energy of the CoM of the rod wrt an external inertial frame + kinetic energy of the CoM about some point. (which I took to be a frame instantaneously coinciding with the contact point on the surface)

    The latter component is probably best dealt with in polar coordinates, so to an inertial frame, the total velocity vector is $$\vec v = \dot{x} \hat{x} + \dot{r} \hat{r} + r \dot{\hat r}$$ The term ##\dot{r}\hat{r}## is zero and in the given set up, ##\dot{\hat r} = \frac{\ell}{2} \dot{\alpha} \left(\cos \alpha \hat{x} - \sin \alpha \hat{y}\right)##. But when I take the scalar product, I end up with cross terms that are not in the solution.
  11. Apr 28, 2014 #10
    I am not sure what you mean by
    both parts seem to be the same?

    Total kinetic energy is the kinetic energy of the CoM (having the entire mass of the system) in an inertial frame and the energy in the CoM frame (which is pure rotation). The CoM is the centre of the rod, and ##x## is its horizontal displacement. What is its vertical displacement? What is its velocity then?
  12. Apr 28, 2014 #11


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    Yes, sorry I did not say it clearly enough.

    So in the CoM frame or frame coinciding with the centre of the rod, we have a rotation about the contact point on the surface. If ##x## is the horizontal displacement then ##y = \frac{\ell}{2}\cos \alpha## is the vertical displacement.

    So ##\vec{v_{\text{rel to inertial frame}}} = \dot{x} \hat{x} - \frac{\ell}{2} \dot{\alpha} \sin \alpha \hat{y}##.

    So the velocity vector of the centre of the rod to an inertial frame is the above.
  13. Apr 28, 2014 #12
    Which I hope leads you to the correct Lagrangian :)
  14. Apr 28, 2014 #13
    No, this is not correct. In the CoM frame, what point of the rod is static? That's your centre of rotation.
  15. Apr 28, 2014 #14


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    It does. But I would like to check something conceptually. The kinetic energy in the CoM frame is purely rotational so it is of the form 1/2 I w2, where I is the moment of inertia about an axis through the point of rotation (which is the centre of the rod) and w is the angular velocity of the CoM. Is this w measured relative to an inertial frame? In which case ##\dot{\alpha}## does the job fine.

    I think I am still getting my head around our discussion in the previous thread about the form of the kinetic energy in a co-rotating frame.
  16. Apr 28, 2014 #15
    As (I think) I said in the previous thread, angular velocity should always be assumed as measured w.r.t. an inertial frame. All the (usual) equations involving angular velocity have this assumption. Angular velocity thus defined is the same in all inertial frames (https://archive.org/stream/Mechanics_541/LandauLifshitz-Mechanics#page/n103/mode/2up). Only when it is explicitly said (and the measurement procedure is fully specified) that angular velocity is measured w.r.t. a non-inertial frame, should you treat it as such and proceed with extreme caution.
  17. Apr 28, 2014 #16


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    Thanks, that looks like a nice text there. I had a question about another part of the problem: Using conservation of energy, it can be shown that $$\dot{\alpha}^2 = \frac{12g}{\ell} \left(\frac{\cos \alpha_o - \cos \alpha}{1+3\sin^2 \alpha}\right)$$ I now want to expand for small ##\alpha_o## and ##\alpha##.

    Doing so, I need to get it into a form like ##\dot{\alpha} = \lambda \sqrt{\alpha^2 - \alpha_o^2}.## What I get is $$\dot{\alpha} = \sqrt{\frac{6g}{(1+3\alpha^2)\ell}} \sqrt{\alpha^2 - \alpha_o^2},$$ which is of the required form.

    But the next part of the questions asks to evaluate the above first order differential equation. I am wondering if I made a small mistake somewhere because the integral does not seem elementary. I was hoping that the parameter λ would be a constant.
  18. Apr 28, 2014 #17
    The equation is not in the required form, unless ##\alpha## in the denominator is supposed to be ##\alpha_0##.

    In the latter case, the integral is elementary.
  19. Apr 28, 2014 #18


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    Do you mean that I should have ##1+3\alpha_o^2##? All I did was expand ##1+3\sin^2\alpha \approx 1+3 \alpha^2##, but that is ##\alpha## and not ##\alpha_o##.
  20. Apr 28, 2014 #19
    I mean that $$\dot{\alpha} = \sqrt{\frac{6g}{(1+3\alpha^2)\ell}} \sqrt{\alpha^2 - \alpha_o^2}$$ is not in the required form, because the first radical has ##\alpha##, while it should not.

    You should expand the entire function, not just parts of it.
  21. Apr 28, 2014 #20


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    Ofcourse, thanks voko!
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