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Deriving differential equations for free rotation

  1. May 3, 2013 #1
    I was asked to formulate the equations governing the rotation of a body moving without any external moments acting about its centre of mass in terms of a coupled system of first order, nonlinear differential equations. I decided to go with the Euler equations, and I ended up with this:

    \begin{equation} \label{symdif}
    \begin{array}{l l l}
    \dot{\omega}_x=\frac{I_{yy}-I_{zz}}{I_{xx}}\omega _y\omega _z\\
    \dot{\omega}_y=\frac{I_{zz}-I_{xx}}{I_{yy}}\omega _z\omega _x\\
    \dot{\omega}_z=\frac{I_{xx}-I_{yy}}{I_{zz}}\omega _x\omega _y
    \end{array}
    \end{equation}

    \begin{equation}
    (I_{xx}=I_{yy}<I_{zz})
    \end{equation}

    This indicates that $$\omega_z=constant$$which makes it possible to solve the system of differential equations, but I wonder how one would end up with the differential equations explicitly asked for.
     
  2. jcsd
  3. May 4, 2013 #2

    vanhees71

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    2016 Award

    I guess you look for the solution of the differential equation. Since [itex]\omega_z=\text{const}[/itex] we can indeed solve the remaining system of two coupled linear differential equations. To make it a bit less tedious to write we set
    [tex]\Omega=\frac{I_{zz}-I_{xx}}{I_{yy}} \omega_z.[/tex]
    Since for a symmetric top [itex]I_{xx}=I_{yy}[/itex] the system reads
    [tex]\dot{\omega}_x=-\Omega \omega_y, \quad \dot{\omega}_y=\Omega \omega_x.[/tex]
    To solve this equation, you can either use matrix-exponential functions or, much simpler in this case, the trick to introduce a complex variable
    [tex]u=\omega_x+\mathrm{i} \omega_y.[/tex]
    Then from the equations of motion you get
    [tex]\dot{u}=\Omega (-\omega_y+\mathrm{i} \omega_x)=\mathrm{i} \Omega (\omega_x+\mathrm{i} \omega_y)=\mathrm{i} \Omega u.[/tex]
    The general solution is immediately given by
    [tex]u(x)=u_0 \exp(\mathrm{i} \Omega t)[/tex]
    with [itex]u_0[/itex] integration constants fixed by the initial conditions [itex]\omega_x(0)=\omega_{x0}[/itex], [itex]\omega_y(0)=\omega_{y0}[/itex]. To get the angular velocities you only have to take real and imaginary part of our solution with
    [tex]u_0=\omega_{x0}+\mathrm{i} \omega_{y0}[/tex]
    leading to
    [tex]\omega_x(t)=\omega_{x0} \cos(\Omega t)-\omega_{y0} \sin(\Omega t),[/tex]
    [tex]\omega_y(t)=\omega_{y0} \cos(\Omega t)+\omega_{x0} \sin(\Omega t).[/tex]
    That tells you that the angular velocity rotates around the [itex]z[/itex] axis sweeping out the polhode cone (seen from the reference frame fixed with the spinning body, where the above Euler equations hold).

    Seen from the inertial frame, of course the total angular momentum is conserved, defining a fixed axis, around which both the angular velocity (sweeping out the herpole cone) and the symmetry axis of the body (sweeping out the nutation cone) rotate.
     
  4. May 4, 2013 #3
    Hey

    No, I am unfortunately not looking for the answer to the equations.

    I am asked to come up with a specific set of differential equations to later apply to a dynamic system, where the latter is not of any interest for the moment. However, the equations I derived are not nonlinear, and in that sense better than the equations asked for. My question was about what equations the latter would be and being a common subject in mechanics I was hoping that someone had a quick answer lying around, since deriving them would just be a waste of effort forasmuch as I already have calculated the angular velocity.
     
  5. May 4, 2013 #4

    vanhees71

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    I don't understand what you have to do. What is the precise question asked?

    The Euler equations are, of course non-linear first-order differential equations. Only for the symmetric case they effectively simplify to the linear one as discussed.
     
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