# Is a rotor balanced at constant angular velocity?

I have an asymmetrical rotating part. It vibrates its housing and emits audible noise. I need to add weights to ensure smooth rotation. However, I am constrained in which regions I can add material. I can't exploit rotational symmetry to balance this rotor.

I define a rotor as a rigid system of particles, each with mass $m_n$ and position $r_n \mathbb\in {R}^3$, rotated about $k$ (the z-axis).

What formula describes a rotor that is balanced when spun at constant velocity? According to Update International, a vendor of rotor balancing systems, the problem is broken into static and couple unbalance. Here are my interpretations:

# Static balance

When the angular velocity $\omega\neq0$, a net force acts orthogonal to $k$, through the rotor's center of mass.

Static unbalance is resolved by ensuring that the center of mass $C = \dfrac{\sum m_n r_n}{\sum m_n}$ lies along $k$.

$C \times \hat{k} = 0$

Together with the fraction canceled:

$\sum m_n r_n \times \hat{k} = 0$

Rewritten as a scalar system:

$\begin{cases} \sum m_n r_{n,x} = 0 \\ \sum m_n r_{n,y} = 0 \end{cases}$

# Couple unbalance

When the requirements above are met, a pair of equal and opposite net forces act at different points along the axis. The forces are perpendicular to the axis.

I'm stuck. How do point-masses give rise to couple unbalance?

I figured it out. Couple unbalance is torque. We want a system where the torque on all particles cancels, or $\sum T_n = 0$.

Torque is defined as $T = r \times F$, where $r$ is the vector from the center of mass to the point where the force is applied. The center of mass is already constrained to $C \times \hat{k} = 0$; we'll strengthen that constraint to $C = 0$, which leaves this simple equation for the torque on point-mass $n$:

$T_n = r_n \times F_n$

Centripetal force is defined as $F = m r_\perp \omega^2$ where $r_\perp$ excludes the component parallel to $k$.

$r_{n\perp} = r_n \cdot (\hat{i} + \hat{j})$

Combined:

$T_n = r_n \times (m_n r_{n\perp} \omega^2)$

$\sum r_n \times (m_n r_{n\perp} \omega^2) = 0$

Constant $\omega^2$ is divided out:

$\sum r_n \times r_{n\perp} m_n = 0$

Rewritten as a scalar system (note that $r_{n\perp,z}=0$):

$\begin{cases} \sum (r_{n,z} r_{n,y} - 0) m_n = 0 \\ \sum (0 - r_{n,z} r_{n,x}) m_n = 0 \\ \sum (r_{n,x} r_{n,y} − r_{n,y} r_{n,x}) m_n = 0 \end{cases}$

Simplify further and remove the last equation (an identity):

$\begin{cases} \sum m_n r_{n,x} r_{n,z} = 0 \\ \sum m_n r_{n,y} r_{n,z} = 0 \end{cases}$

So together with the condition $C = 0$, we get a system of linear equations that describes any rotor that is in equilibrium when spun at constant velocity about $k$:

$\begin{cases} \sum m_n r_{n,x} = 0 \\ \sum m_n r_{n,y} = 0 \\ \sum m_n r_{n,z} = 0 \\ \sum m_n r_{n,x} r_{n,z} = 0 \\ \sum m_n r_{n,y} r_{n,z} = 0 \end{cases}$