# How does axial offset effect rotor balance?

I have a thin rod rotating about its primary axis, rigidly attached to mass $m_1$ whose center sits at distance $r_1$ from the axis and position $y_1$ along the axis.

$m_1$ exerts a centripetal force $F_1 = m_1 r_1 \omega^2$ where $\omega$ is angular velocity. This force leaves the rotor unbalanced and will cause the housing to shake, so I must add masses to balance it. I am constrained here because $m_1$ projects light up and out, so I can't put balancing masses near or above it; I can only add balancing masses to the axis below $m_1$.

So I add mass $m_2$ centered at $(r_2, y_2)$. If $F_1+F_2=0$ then the rotor is statically balanced, meaning that it will stand on its end without toppling over. However, because $y_2<y_1$, the rotor has a couple unbalance and will still cause the housing to shake.

So I add mass $m_3$ centered at $(r_3, y_3)$. I can attain static balance by ensuring $F_1+F_2+F_3=0$. I know anecdotally that this 3-mass configuration can remove the couple unbalance, but I'm struggling to express it mathematically.

What is the relation between $m_{1..3}$, $r_{1..3}$, and $y_{1..3}$ that guarantees zero couple unbalance?

To balance the system with respect to the lowest point of the rotating thin rod, ensure the balance of moments with respect to the $y$ axis is zero. That simply yields:

$m_1r_1 + m_3r_3=m_2r_2$.

To ensure no net centripetal force acts on the rod:

$m_1 r_1 \omega^2y_1+m_3 r_3 \omega^2y_3=m_2 r_2 \omega^2y_2$.

$\omega^2$ drops out, so we get:

$m_1 r_1y_1+m_3 r_3y_3=m_2 r_2 y_2$.

Now we have too many degrees of freedom: 3 variables and only 2 equations (I assume $m_1$, $r_1$ and $y_1$ are known!)

To solve this, I would set $m_1=m_3$, $m_2=m_1+m_3$ and $r_1=r_3$.

It's now possible to calculate the remaining unknowns from the knowns and these equations.

• I think these are the same equations that explain forces at different distances along a lever. Makes sense. Thanks! – Neal Ehardt Sep 4 '15 at 22:46