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1.Rigid rotator
A special case of a symmetric top is consisting of a set of particles
in a straight line. The moment of inertia about perpendicular axes are equal.
In this case the moment of inertia of about the line, passing
through the particles, vanishes. Its principal axes are any two mutually perpendicular
axes also perpendicular to the line joining the particles. The line joining the
particles is the third principal axis. In this case we have \(I_1=I_2=I\) and \(I_3=0\)
Such a system is called a rigid rotator.
In this case it is meaningless to talk of rotations about the line joining
the particles. The angular velocity has only two components and the angular
momentum,
\begin{equation}
\vec{L} = (I_1\omega_1,I_2\omega_2,0) = I (\omega_1, \omega_2,0)
\end{equation}
is again parallel to the angular velocity.
2. Symmetric top
We choose the symmetry axis of the top as the third axis. Symmetric top means
two of the principal moments of inertia are equal \(I_1=I_2\).
The torque of gravitational forces about the centre of mass vanishes. Therefore
the Euler's equations take the form
\begin{eqnarray}
I_1\dd[\omega_1]{t} + (I_3-I_1) \omega_2\omega_3 = 0\label{EQ10}\\
I_2\dd[\omega_2]{t} + (I_1-I_3) \omega_3\omega_1 = 0\label{EQ11}\\
I_3\dd[\omega_3]{t} = 0\label{EQ12}
\end{eqnarray}
\EqRef{EQ12} gives \(\omega_3=\) constant. Introducing the notation
\begin{equation}
\Omega =\omega_3(I_3-I_1)/{I_1}
\end{equation}
we rewrite first two Euler equations as
\begin{equation}
\dd[\omega_1]{t}=-\Omega \omega_2, \qquad \dd[\omega_2]{t}= \Omega \omega_1.
\end{equation}
This leads to the equation
\begin{equation}
\DD[\omega_1]{t} + \Omega^2 \omega_1 = 0
\end{equation}
and hence
\begin{equation}\label{EQ16}
\omega_1(t)= A \cos\Omega t + B \sin \Omega t
\end{equation}
where \(A,B\) are real constants. Knowledge of initial values of
\(\omega_1, \omega_2\) is requires to fix the constants \(A,B\).
Knowing \(\omega_1\) as function of time gives
\begin{equation}\label{EQ17}
\omega_2(t) =-\frac{1}{\Omega}\dd[\omega_1]{t}= A \sin \Omega t - B\cos \Omega t.
\end{equation}
This shows that \(\sqrt{\omega_1^2+ \omega_2^2}\), being equal to \(\sqrt{A^2+B^2}\),
is a constant of motion.
The component \(\omega_3\) is also constant, thus the vector \(\omega(t)\) rotates
about the axis of the top. These results, along with \EqRef{EQ16}-\eqRef{EQ17},
imply that the angular velocity vector \(\vec{\omega}\) rotates about the
symmetry axis of top and the its magnitude \(|\vec{\omega}|\) remains constant.
We leave it for the reader to verify that the angular momentum vector also
rotates around the axis of the top.
Question for you
Verify that angular momentum vector also rotates about the axis of the top.
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4727:Diamond Point
