General Equations

Consider the rotational motion of the top shown below. It is assumed to spin without friction such that the point O on the axis of symmetry is fixed. The only external moment about O is that due to the constant gravitational force mg acting through its center of mass at C.

  

Figure 1: A symmetric top top with a fixed point at O

Let us analyze the motion using Lagrange's equations with Eulerian angles as the co-ordinates. If we choose the fixed point O as the reference point, then the total kinetic energy can be written in terms of Euler angle rates. Noting that I_xx=I_a and I_yy=I_zz=I_t for this case of symmetry about the x-axis, we see that:  

$$T=\frac{1}{2} I_a (\dot{\phi}-\dot{\psi} \sin \theta)^2 + \frac{1}{2} I_t (\dot{\theta}^2+\dot{\psi}^2 \cos^2 \theta)$$ (1)

 

$$V = m g l \sin \theta$$ (2)

$$L=T-V=\frac{1}{2} I_a (\dot{\phi}-\dot{\psi} \sin \theta)^2 ... ...(\dot{\theta}^2+\dot{\psi}^2 \cos^2 \theta) - m g l \sin \theta$$ (3)

The generalized momenta are:

where the total spin $\Omega$ is given by

$$\Omega = \dot{\phi} - \dot{\psi} \sin \theta$$

The standard form of Lagrange's equation

$$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q_i}} \right) - \frac{\partial L}{\partial q_i} = 0$$

is applied together with Eq. (4) to obtain

from which we see that both $p_\psi$ and $p_\phi$ are constant. Hence we find that $\Omega$ is constant for this case where there is no applied moment about the symmetry axis. Also, the precession rate $\dot{\psi}$can be obtained from Eq. (4) with the following result:  

$$\dot{\psi}=\frac{p_\psi+I_a \Omega \sin \theta}{I_t \cos^2 \theta}$$ (4)

Now, let us use the principle of conservation of energy to obtain an integral of the $\theta$ equation of motion. From Eqs. (1) and (2), we see that the total energy is

$$E=T+V=\frac{1}{2} I_a \Omega^2 + \frac{1}{2} I_t (\dot{\theta}^2+\dot{\psi}^2 \cos^2 \theta) + m g l \sin \theta$$ (5)

where $\Omega=\dot{\phi}-\dot{\psi}\sin\theta$ is constant. It follows that the total energy minus the kinetic energy associated with the total spin $\Omega$ is also a constant. Calling this quantity E', we can write

$$E'=E-\frac{1}{2} I_a \Omega^2 = \frac{1}{2} I_t (\dot{\theta}^2+\dot{\psi}^2 \cos^2 \theta) + m g l \sin \theta$$ (6)

Substituting for $\dot{\psi}$ from Eq. (6) and solving for $\dot{\theta}^2$, we obtain  

$$\dot{\theta}^2=\frac{2E'}{I_t}- \left( \frac{p_\psi+I_a \Ome... ...heta}{I_t \cos^2 \theta} \right) - \frac{2mgl}{I_t} \sin \theta$$ (7)

Note that $\theta$ is the only variable on the right-hand side of this equation. Thus we can see from Eqs. (6) and (9) that the precession rate $\dot{\psi}$ and the nutation rate $\dot{\theta}$ can be written as functions of $\theta$alone for any given case. Solving this system of equations thus describes the complete motion of a simple spinning top.