Rotating unbalance:

An electric motor, a turbine blade, and in fact all other rotating machines have an amount of unbalance left in them even after correcting their unbalance on precision balancing machine.

The final unbalance is measured in terms of $m_0$ rotating with its center of gravity at a distance e from the axis of rotation. Let the machine be elastically supported and rotating at w rad/s.

Mass $m_o$ makes an angle ‘wt’ with the reference axis, at any instant. Then the centrifugal force $m_0 ew^2$ acts outwards from the enter of rotation as shown in figure 7(a). The equation of motion in the vertical direction can now be written as follows:

$(m \ – \ m_0) \frac{d^2x}{dt^2} + m_0 \ \frac{d^2}{dt^2} (x \ + \ e \ sin \ wt) = -kx \ – c \ \frac{dx}{dt}$

$\therefore$ $m \frac{d^2x}{dt^2} = m_0 \frac{d^2x}{dt^2} + m_0 \frac{d^2x}{dt^2} – m_0 \ ew^2 \ sin \ wt + cx + kx = 0$

Or $m \frac{d^2x}{dt^2} + c\frac{dx}{dt} + kx = m_0 \ e \ w^2 \ sin \ wt$ - - - -(16)

If we compare this equation with equation for forced damped vibrating system, i.e. $mx + cx + kx = \ F_0 \ sin \ wt$

the only difference is that $F_0$ is replaced by $m_0 \ e \ w^2$. Everything else remains the same. The steady state amplitude is given by

$X = \frac{m_0 ew^2/k}{\sqrt{ ( 1 - \frac{mw^2}{k})^2 + (\frac{cw}{k})^2}} = \frac{(\frac{m_0ew^2}{k}) (\frac{m}{m})}{\sqrt{ (1 - \frac{mw^2}{k})^2 + (\frac{cw}{k})^2}}$

The above equation reduces to the following dimensionless equation.

$\frac{X}{(\frac{m_0e}{m})} = \frac{(\frac{w}{w_n})^2}{\sqrt{ 1 – (\frac{w}{w_n})^2]^2 + [ 2 \zeta \frac{w}{w_n}]^2}}$ - - - -(17)

$\frac{x}{(\frac{m_0e}{x})} = \frac{r^2}{\sqrt{ (1-r^2)^2 + (2 \zeta r)^2}}$

$\frac{x}{(\frac{m_0e}{m})} = \frac{1}{\sqrt{ (\frac{1}{r^2} – 1)^2 + (\frac{2 \zeta }{r})^2}}$

The equation gives the dimensionless steady state amplitude as a function of frequency ratio and damping factor.