2024/8/9, Hiizu Nakanishi

Spinning Motion of Symmetric Top

with the Axis of Finite Thickness

Simulation Speed: 1.0
Display:
Axis Thickness: $r_P =$ 0.01
Friction Coefficient: $\; k =$ 0.5
Spin Speed: $\omega_Z$ = 0 rot/time
Axis Tilt Angle: $\theta$ = deg
Possible Steady Precession Mode :
$\omega_X$ = 0 rad/time
$\omega_Y$ = 0 rad/time
Simulator for spinning motion of a top with finite axis radius $r_P$. The tip of the axis is modeled as a hemisphere of radius $r_P$. Friction at the contact point between axis and floor is described with the coefficient $k$, and the tip P receives a resistive force proportional to its velocity $\vec v_P$ \[ \vec F = -k\, \vec v_P. \] If $k = \infty$, the contact point does not slip, and the center of mass velocity $\vec v_G$ satisfies the no-slip condition: \[ \vec v_G = \vec R_C \times\vec\omega \] Here, $\vec\omega$ is the angular velocity vector, and $\vec R_C$ is the vector pointing from the center of mass to the contact point.

When both $r_P$ and $k$ are finite, both nutation and precession decay over time, and the top eventually settles into an apparently upright resting state, known as a "sleeping top". Though it takes time to reach this state in simulation, increasing the simulation speed makes the process clearer.

If the axis has finite thickness but $k = \infty$, precession does not seem to decay.

Equations of Motion for a Symmetric Top with Finite Axis Thickness (Japanese)

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