Find the Lagrangian function for a symmetric top whose pivot is free to slide on a smooth horizontal table, in terms of the generalized coordinates \(X, Y, \varphi, \theta, \psi\), and the principal moments \(I_{1}^{*}, I_{1}^{*}, I_{3}^{*}\) about the centre of mass. (Note that \(Z\) is related to \(\theta .\) ) Show that the horizontal motion of the centre of mass may be completely separated from the rotational motion. What difference is there in the equation (10.15) for steady precession? Are the precessional angular velocities greater or less than in the case of a fixed pivot? Show that steady precession at a given value of \(\theta\) can occur for a smaller value of \(\omega_{3}\) than in the case of a fixed pivot.

Since the pivot is free to slide on a smooth horizontal table, we can ignore the potential energy term and only focus on the kinetic energy terms. The kinetic energy of the object can be divided into two parts: the translational kinetic energy of the center of mass (CM) and the rotational kinetic energy about the CM. For a symmetric top, the translational kinetic energy can be written as (using \(X, Y\) for the CM positions): \(T_{trans} = \frac{1}{2}M (\dot{X}^2+\dot{Y}^2)\) The rotational kinetic energy is given by (using \(\varphi, \theta, \psi\) as the Euler angles): \(T_{rot} = \frac{1}{2} (I_1^* \omega_1^2 + I_1^* \omega_2^2 + I_3^* \omega_3^2)\) To proceed further, we will need to introduce the coordinate transformations that relate the angular velocities \(\omega_1, \omega_2, \omega_3\) to the time derivatives of Euler angles \((\dot{\varphi}, \dot{\theta}, \dot{\psi})\).

We can express \(\omega_1, \omega_2, \omega_3\) in terms of Euler angles: \(\omega_1 = \dot{\varphi}\sin{\theta}\sin{\psi} + \dot{\theta}\cos{\psi}\) \(\omega_2 = \dot{\varphi}\sin{\theta}\cos{\psi} - \dot{\theta}\sin{\psi}\) \(\omega_3 = \dot{\varphi}\cos{\theta}+\dot{\psi}\) Now, we can substitute these expressions into the rotational kinetic energy expression \(T_{rot}\) to get: \(T_{rot}=\frac{1}{2}(I_1^*(\dot{\varphi}\sin{\theta}\sin{\psi}+\dot{\theta}\cos{\psi})^2 + I_1^*(\dot{\varphi}\sin{\theta}\cos{\psi}-\dot{\theta}\sin{\psi})^2 + I_3^*(\dot{\varphi}\cos{\theta}+\dot{\psi})^2)\)

To find the Lagrangian function, \(L\), we simply add the translational and rotational kinetic energy terms: \(L = T_{trans} + T_{rot} = \frac{1}{2}M (\dot{X}^2+\dot{Y}^2) + \frac{1}{2}(I_1^*(\dot{\varphi}\sin{\theta}\sin{\psi}+\dot{\theta}\cos{\psi})^2 + I_1^*(\dot{\varphi}\sin{\theta}\cos{\psi}-\dot{\theta}\sin{\psi})^2 + I_3^*(\dot{\varphi}\cos{\theta}+\dot{\psi})^2)\) This Lagrangian represents the energy of the system and is a function of the generalized coordinates \((X, Y, \varphi, \theta, \psi)\) and their time derivatives.

The Lagrangian function clearly shows that the horizontal motion terms \(\dot{X}^2\) and \(\dot{Y}^2\) are completely separate from the rotational motion terms involving \(\varphi, \theta,\) and \(\psi\). This indicates that the horizontal motion of the center of mass can be decoupled from the rotational motion of the symmetric top.

In the steady precession situation, we have \(\dot{\theta}=0\) and \(\dot{\psi}=\Omega\). The equation for steady precession differs in that we now have a free pivot while sliding on the horizontal table. This additional freedom of movement enables the top to adjust its axis, allowing the precessional angular velocities to be smaller than in the fixed pivot case. In the case of a fixed pivot, a given value of \(\theta\) requires a precise value of \(\omega_3\) for steady precession. With a free pivot, a smaller value of \(\omega_3\) is possible for the same steady precession at a given \(\theta\), which implies that the overall energy needed for precession is lower compared to the fixed pivot scenario.