Lagrange top

The simulation displays the evolution of the heavy symmetric top with the following Lagrangian:
L = ½ I_x (φ'² sin² θ + θ'²) + ½ I_z (φ' cos θ + ψ')² - m g R cos θ,
where we use ' for the time derivative, R is the mass-of-center position from the fixed point and R = |R|.

To see the definition of the Euler angles (θ, φ,ψ), click on Variables in the simulation below.
There you can also select the initial values (since φ and ψ are cyclic, there is no restriction in taking φ₀ = ψ₀ = 0) as well as the parameter Izx ≡ I_z/I_x.
The cylinder height is always h = R
The unit length is R.
We also have chosen the unit time so that mgR/I_x = 9.81.

Check Velocities to see the evolution of the precession φ', the nutation θ' and the spin ψ'.
Set N = 0 to remove the trajectory of the axis.
The point of view of the three-dimensional projection can be changed with the cursors or the mouse.
The whole image can be moved with the mouse while pressing Ctrl.
To change the zoom in the projection, press Shift when moving up or down the mouse pointer.
Put the mouse pointer over an element to get information about it.

Activities

Discuss the different evolutions with the default settings and φ'₀= 0, 1, 1.5, 3.
How are precession, nutation and spin in each case?
Use the program to check numerically the condition for motion without nutation, which is given by θ = 0, π or by
φ' = Ω, ψ' = ω, (I_x-I_z) Ω² cos θ = I_z Ω ω - m g R.
Select a small value for θ₀, say 1°, and check the condition for the sleeping top:
I_z² (φ' cos θ + ψ')²> 4 m g R I_x.

This is an English translation of the Basque original for a course on mechanics, oscillations and waves.
It requires Java 1.5 or newer and was created by Juan M. Aguirregabiria with Easy Java Simulations (Ejs) by Francisco Esquembre. I thank Wolfgang Christian and Francisco Esquembre for their help.