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Lagrangian mechanics, cone rotating over a plane

  1. Feb 8, 2013 #1
    I wanted to solve the problem of a cone rotating on its side over a table, around an axis that pass through it's apex, like in the figure.

    What I want to find is the angular speed ω, the spin of the solid, such that the cone "stands" over it's apex. I don't know how to set the condition needed. Some how the normal force, at first distributed over the line, will be all applied at the apex when this happens. I can find the kinetic energy and the equations of movement for the cone, but I don't know how to find the angular speed required.

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  3. Feb 8, 2013 #2

    TSny

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  4. Feb 8, 2013 #3
    I think that's a good idea, the cone should get a nutation actually to elevate. I should write the equations of motion and then see what can I do.
     
  5. Feb 9, 2013 #4

    TSny

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    After thinking some about this problem, I'm puzzled :confused:. If I’m not mistaken, when the cone is rolling on the surface about the apex, it rolls around the vertical axis at the apex in a direction opposite to the direction it would want to precess as a top! So, I’m finding it hard to imagine what would cause it to lift off the surface. Maybe friction is the key, as in the flip of a tippe top.

    I’m not too sure what “stands over its apex” means, unless it means that you want the cone to orient itself so the spin axis is vertical and remains vertical (like the flipped tippe top). This is known as a “sleeping top”. There is a fairly simple condition for the angular spin rate of a top to make it sleep. See pages 71 and 72 of this document where the condition for sleeping is at the top of page 72. But the derivation of this condition does not appear to be simple :frown:.
     
  6. Feb 9, 2013 #5
    I want it to stand over the vertex, I don't know if there is any condition on the moment of inertia of the cone for this to happen, and I'm actually not too sure if this would actually happen at all. But I think that it would, because we could think of the cone as a spinning top, we could put it to spin over its vertex, and then, due to friction, it would fall off. Now, there is dissipation in the process for it to fall, due to friction. What I want is go on backwards, starting with the cone as in the figure, and then give it enough angular speed for it to start nutation. Precession is actually happening with the cone rolling I think. I think I need it to nutate. And what I think that happens is that, as the cone is rolling around its vertex, the spinning angular momentum is changing, generating a torque. I think that torque is competing with the weight of the cone (as it happens with the spinning top). Getting the equation of motion by setting Lagrange equations, I think I should be able to state the necessary condition for the cone to stands on its vertex. I'm not sure, but I think it should stand on its vertex, and not on its base. As I said, like starting the final movements of the spinning top with friction, in backwards. I start with zero energy if you want, and then I give it the energy, with a starting velocity, given the constraints, such that it stands. But, actually, now that I think of it, I know that the cone has it's stable position over it's base (the center of mass being closer to it, gives a minimum in potential energy with the cone over the base). And we could put it to spin over the base too, by rolling, like a coin, so I'm not sure on what should actually happens. But the condition I think should be similar.

    Thank you TSny. Perhaps is not a simple problem, I don't know. But its interesting, isn't it? :D
     
    Last edited: Feb 9, 2013
  7. Feb 9, 2013 #6

    TSny

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    It is interesting. I don't see how the cone would get into a sleeping mode from an initially rolling motion. In fact, my guess would be that the rolling motion would tend to flip the cone up on its base rather than apex. But it's only a guess.
     
  8. Feb 9, 2013 #7
    We should determine on which sense the torque is generated to be sure on which side the cone would turn. But I haven't tried to solve this yet. I think I should use Eulerian angles when I get in the mud. I'd let you know, I haven't solved any rigid for a while, but I have to now, so I'll consider this problem. Anyway, I'm pretty much sure, from other problems that I've solved some time ago that the torque goes in the sense I was expecting, but I can't demonstrate it right now, I have to make a review on the subject first.
     
  9. Feb 9, 2013 #8

    TSny

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    OK. Let us know what you find.
     
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