Precession Rate of a Rolling Disk

Last edited: 2024-11-01 14:20:22

At how small radius can a rolling disk rotate around its precession center? What is the disk's angular velocity and its precession rate? Let us look at the physics that determines these values.

Given Variables

We assume to know the values of the following variables:

• $R$: The distance from the center of precession to the contact point of the disk
• $r$: The radius of the disk
• $g$: The acceleration due to gravity
• $\theta$: The angle between the disk and the vertical to the ground

Variables, Units, and Assumptions for Searched Values

We define the following assumptions and variables which we want to calculate:

• $\omega$: The angular velocity (unit rad/s) around the center of mass of the disk
• $\Omega$: The precession rate (unit rad/s), also known as the precession angular velocity, of the disk
• The minimum value of $R$ for the phenomenon to occur
• Lastly, we assume that the disk is homogeneous.

Relation Between Disk's Angular Velocity and Precession Rate

The speed around the center of mass of the disk, $\omega r$, is the same as that around the precession point, $\Omega R$. This leads to the following:

$$\omega=\frac{\Omega R}{r}.$$

Equation For Precession Rate

Projecting the force of gravity that acts upon the disk on the local coordinate system shown in the figure above the moment around the contact point becomes the following:

$$\vec{M}= r \hat{z} \times mg (-\sin{\theta} \hat{x} - \cos{\theta} \hat{z}) = -mgr \sin{\theta} \hat{y}.$$

The center of mass is closer to the center of precession than the contact point so the the velocity of the center of mass is

$$\vec{v} = \Omega (R - r \sin{\theta}) \hat{y}.$$

The rotational velocity around the center of mass in the vector in the local coordinate system is

$$\vec{\omega} = \begin{bmatrix} \frac{\Omega R}{r} \\ 0 \\ 0 \end{bmatrix} .$$

The precession rate translated to the local coordinate system:

$$\vec{\Omega} = \begin{bmatrix} \Omega \sin{\theta} \\ 0 \\ \Omega \cos{\theta} \end{bmatrix} .$$

We also need the moment of inertia around the center of mass of the disk, which is

$$I = \begin{bmatrix} 2mr^2k & 0 & 0 \\ 0 & mr^2k & 0 \\ 0 & 0 & mr^2k \end{bmatrix},$$

where $k=\frac{1}{4}$. We can now use expressions in vector form so the angular momentum around the contact point becomes

$$\vec{H} = I (\vec{\omega} + \vec{\Omega}) + r \hat{z} \times \vec{v}$$ $$= mr \Omega ((r \sin{\theta} - R)(2k +2) \hat{x} + r(k+1) \cos{\theta} \hat{z}).$$

Then we take the time derivative of the angular momentum at the contact point:

$$\frac{d \vec{H}}{dt} = \frac{\partial \vec{H}}{\partial t} + \vec{\Omega} \times \vec{H}$$

Since the components are not time-dependent, the time-related derivative $\frac{\partial \vec{H}}{\partial t}$ is zero. Therefore the derivative of the angular momentum around the contact point becomes

$$\vec{\Omega} \times \vec{H}$$ $$= -\Omega^2 r m \cos{\theta} (rk \sin{\theta} + (R - r \sin{\theta})(2k + 1)) \hat{y}.$$

The derivative of the angular momentum at the contact point is equal to the moment at the contact point so

$$-mgr \sin{\theta} \hat{y}$$ $$= -\Omega^2 r m \cos{\theta} (rk \sin{\theta} + (R - r \sin{\theta})(2k + 1)) \hat{y}.$$

With that equation, $\Omega$ can be solved and we get the equation for the precession rate:

$$\Omega = \sqrt{\frac{g \tan{\theta}}{rk \sin{\theta} + (R - r \sin{\theta})(2k + 1)}}.$$

Minimum value of $R$

When the equation above becomes undefined, that is, when the denominator becomes zero, the minimum value of $R$ for precession to occur. So we have have:

$$0 = rk \sin{\theta} + (R - r \sin{\theta})(2k + 1).$$

So if we solve for $R$ we get:

$$R = \frac{5}{6} r \sin{\theta}.$$

Analysis of Solution

During these calculations, an assumption regarding friction has been made. Some friction must exist to keep the disk in a circular path. Furthermore, these calculations also assume that the disk has no height, which means that $I_{22}$ and $I_{33}$ in $I$ do not have a height term. On top of that, the edge radius in these calculations cannot exist for the length vectors to be correct and at the same time, a larger edge radius would provide a larger contact area, which would lead to greater maximum friction.

If $\theta = 0$, no moment would arise since the gravitational and normal forces would align and precession could not occur. If the velocity were also very low, only a slight instability would cause a moment to arise and since the time derivative of the angular momentum must equal the moment, this means that a large change occurs in the angular momentum when it is small. This would likely lead to the disk falling.