Motor with rotating unbalance on springs

Data: $M = 90$ kg, $k = 800000$ N/m, $\zeta = 0.12$, $m_0e = 0.25$ kg·m, $N = 1200$ rpm.

Frequencies

$$\omega_n = \sqrt{\frac{k}{M}} = \sqrt{\frac{800000}{90}} = 94.28\ \text{rad/s} \;(N_n = 900\ \text{rpm})$$ $$\omega = \frac{2\pi \times 1200}{60} = 125.66\ \text{rad/s}, \qquad r = \frac{\omega}{\omega_n} = 1.3329$$ $$c = 2\zeta\sqrt{kM} = 2 \times 0.12 \times \sqrt{800000 \times 90} = 2036\ \text{N·s/m}$$

(i) Steady-state amplitude

For the exciting force $F_0 = m_0e\,\omega^2$ rotating with the shaft:

$$X = \frac{(m_0e/M)\,r^2}{\sqrt{(1-r^2)^2 + (2\zeta r)^2}}$$ $$X = \frac{(0.25/90) \times 1.7765}{\sqrt{(1 - 1.7765)^2 + (2 \times 0.12 \times 1.3329)^2}} = 0.005876\ \text{m} = 5.876\ \text{mm}$$

(ii) Force transmitted

The spring and damper forces are in quadrature:

$$F_T = X\sqrt{k^2 + (c\omega)^2} = 0.005876 \times \sqrt{(800000)^2 + (2036 \times 125.66)^2} = 4935\ \text{N}$$

The exciting force amplitude is $F_0 = m_0e\,\omega^2 = 0.25 \times 125.66^2 = 3948$ N, so the transmissibility is $4935/3948 = 1.250$.

(iii) Resonance

Resonance occurs when $\omega = \omega_n$, i.e. $N = 900$ rpm. At $r = 1$:

$$X_{res} = \frac{m_0e/M}{2\zeta} = \frac{0.25/90}{2 \times 0.12} = 0.01157\ \text{m} = 11.57\ \text{mm}$$

Answer: (i) $X = 5.876$ mm; (ii) $F_T = 4935$ N; (iii) resonance at $900$ rpm with amplitude about $11.57$ mm.