1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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:

$\endgroup$
  • 1
    $\begingroup$ Yes. It is possible to balance a rotor so that it will not have a preferred direction under gravity, but that is not sufficient to make sure it is balanced when rotating: that requires there to be a mass balance along axes perpendicular to the axis of rotation. It's nicely described in the wiki article on wheel balancing $\endgroup$ – Floris Dec 28 '15 at 19:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.