A uniform top of mass $M$ with its lower end fixed to the ground is rotating on its axis of symmetry with angular velocity $\Omega$, initially in a vertical position ($\theta=0, \dot \theta=0$). The main moments of inertia are $I_3,I_1=I_2$. The center of mass is located at a distance $a$ from the bottom point of the top. I need to find the quantities conserved in the system based on the initial conditions and calculate the maximum angle that the top can be tilted.

I tried to solve the problem without initial conditions (Lagrange Top) and obtaining the conserved quantities from the Lagrangian, which I have left in terms of Euler's angles, but now I don't know when to use the initial conditions, if it is from the start in the problem statement or once you have the Lagrangian just substitute the initial condition values?

  • $\begingroup$ I don't know why the top would tilt at all with those initial conditions. It should remain upright, at least without any perturbations. $\endgroup$ – Puk Jun 24 at 5:34

I think you are over-complicating the problem by using the Lagrangian.
I'll give you a few pointers that should allow you to do the maths yourself.

Use conservation of energy: $$ K + U = E = \mathrm{constant}, $$ where $K$ is the kinetic and $U$ the potential energy.

$U = mga(1-\sin\theta)$, so that at $\theta = 0$, with the top upright, the (gravitational) potential energy is given by $mga$, and when the top has fallen to the ground $\theta = \pi/2$ the energy is $0$ which we set as our reference.

Because the end of the top is fixed, there will be no linear motion, only rotational motion about the pivot, so the kinetic energy associated with the "falling down under gravity" is given by: $$ K = \frac{1}{2}I_1 \dot\theta^2,$$ which, using your initial condition $\dot\theta = 0$, starts at $0$.

Lastly, you know angular momentum about the pivot is not conserved, as the pivot gravity is providing an external force. (edit: however, some components of $L$ perpendicular to the external torque are still conserved). This will cause the top to precess with angular velocity $\omega_p$: $$ \omega_p = \frac{mga}{I_3 \Omega} .$$

This rotational motion will have an associated kinetic energy. So the top will fall to an angle $\theta_p$ such that the lost potential energy can support a circular motion with $\omega_p$.

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  • $\begingroup$ The pivot itself applies no torque, because any force it applies is applied at the pivot. The external torque is due to the weight of the top. However the components of the angular momentum along the vertical and along the axis of symmetry are still constant because the torque is orthogonal to these. $\endgroup$ – Puk Jun 24 at 5:37
  • $\begingroup$ You are right. Corrected the answer. Thanks. $\endgroup$ – SuperCiocia Jun 24 at 5:41

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