0
$\begingroup$

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 $m_1$.

mass/force diagram

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?

$\endgroup$
2
$\begingroup$

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.

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

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.