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?
 A: 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 $k$:
$\begin{cases} \sum m_n r_{n,x} = 0
            \\ \sum m_n r_{n,y} = 0
            \\ \sum m_n r_{n,z} = 0
            \\ \sum m_n r_{n,x} r_{n,z} = 0
            \\ \sum m_n r_{n,y} r_{n,z} = 0 \end{cases}$
Further reading:


*

*Torque - Wikipedia

*Rigid body dynamics - Wikipedia

*Example where angular momentum and angular velocity are not parallel - Physics StackExchange

*On the relation between angular momentum and angular velocity - J. P. Silva and J. M. Tavares
