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Rigid body -badly worded problem -

  1. Feb 14, 2012 #1


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    Rigid body -badly worded problem :( -

    1. The problem statement, all variables and given/known data
    Consider a symmetric trompo in a uniform gravitational field, with its inferior point fixed.
    a)Write down the Lagrangian. Determine the conserved quantities and from them study the motion of the body.
    b)Determine the stability condition for the rotation of the trompo around a vertical axis.
    c)Determine the motion of the trompo in the case in which the kinetic energy of rotation around its axis of symmetry is big compared to its energy in the gravitational field (fast trompo).

    2. Relevant equations

    L=T-V. Euler's angle, tensor of inertia, etc.

    3. The attempt at a solution
    I don't understand the situation for part a). I think I should model the trompo as a cone whose vertex is the "inferior" point. Now I do not know if it's rotating (I'd assume that no but then they ask me to study its motion... making me think it should rotate) and if it rotates, around a vertical axis? Does this vertical axis of rotation correspon to the axis of symmetry of the trompo? If the axis of symmetry of the trompo does not match a vertical axis, is the trompo spinning on itlsef and rotating around a vertical axis? How is the motion? I really don't get it.
  2. jcsd
  3. Feb 14, 2012 #2


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    Re: Rigid body -badly worded problem :( -

    Ok I've assumed that the trompo is a cone with its vertex fixed. Its symmetry axis does not coincides with a vertical axis, is spinning on itself around its symmetry axis and rotates around a vertical axis. I've been inspired by Goldstein's book page 164 in the first edition.
    Since it has a fixed point (its vertex), the kinetic energy of the rigid body does not involve a translational energy; only a rotational one.
    I've calculated the center of mass of such a cone and it's at a height [itex]\frac{3H}{4}[/itex] from the vertex to the base of the base. In my current problem it means at a height [itex]\frac{3H \cos \theta }{4}[/itex] over the ground, considering [itex]\theta[/itex] as an Euler angle, the angle which is between the vertical z-axis and the axis of symmetry of the trompo, the Z-axis.
    I need the height of the center of mass with respect to the ground because I need the potential energy for the Lagrangian. [itex]V=\frac {H3mg \cos ( \theta )}{4}[/itex].
    The rotational energy is worth [itex]\frac{I_X}{2}(\omega _X ^2 + \omega _Y ^2 )+\frac{I_Z}{2} \omega _Z ^2[/itex], notice that [itex]I_Y=I_X[/itex] because the cone is symmetric. Using some relations found in Golstein's book, I can introduce the Euler angles and I get that [itex]L= \frac{I_X}{\dot \theta ^2 + \dot \phi ^2 \sin ^2 \theta }+\frac{I_Z}{2} (\dot \psi + \dot \phi \cos \theta )^2[/itex].
    Now L=T-V and using the fact that it's a cone, I get that [itex]L=\frac{3m}{10} \left ( \frac{R^2}{4}+H^2 \right ) (\dot \theta ^2 + \dot \phi ^2 \sin ^2 \theta )+\frac{3mR^2}{20} (\dot \psi + \dot \phi ^2 \cos \theta )-\frac{ 3Hmg\cos ( \theta )}{4}[/itex].
    As Golstein, I notice that [itex]\psi[/itex] and [itex]\phi[/itex] are absent from the Lagrangian and hence their generalized momentum are constant. This imply that the angular momentum (I'm not sure that the angular momentum is the sum of the 2 generalized momenta... I don't think so but I don't understand why) is conversed. Energy and linear momentum are conserved, I don't think I need to prove it.
    I realize I still didn't put in gravity. I guess it impact on the Euler angles... going to think about this. If you have any comment, feel free to post.
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