---
title: "Motor with rotating unbalance on springs"
subject: "Theory of Mechanism and Machine II"
chapter: 6
key: "semester_7/theory_of_mechanism_and_machine_ii:rotating-unbalance-motor-amplitude"
source: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/rotating-unbalance-motor-amplitude
site: BCT Engineering
---

# Motor with rotating unbalance on springs

Theory of Mechanism and Machine II · Chapter 6: Vibration of Single Degree of Freedom Systems

- Asked in: year not given, 8 marks
- Video: [[Lecture 7-2: Rotating unbalance & Transducer] Mechanical Vibration by Prof. Yang-Hann Kim at KAIST](https://www.youtube.com/watch?v=d6PSz2d7mlw) — yhklab
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/rotating-unbalance-motor-amplitude
- Whole chapter: https://www.bctengineer.com/md/semester_7/theory_of_mechanism_and_machine_ii/chapter-6

## Question

A motor of total mass 90 kg is mounted on springs of total stiffness 800 kN/m. The damping ratio of the mounting is 0.12. The rotor has an unbalance $m_0e = 0.25$ kg·m and runs at 1200 rpm. Find: (i) the amplitude of steady-state vibration, (ii) the force transmitted to the foundation, and (iii) the speed at which resonance occurs, with the amplitude at that speed (neglect the small shift of the resonant peak).

## Short answer

Steady-state amplitude: 5.88 mm. Force transmitted to foundation: 4935 N. Resonance occurs at the natural speed of 900 rpm with amplitude about 11.6 mm.

## Answer

**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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For a rotating unbalance m0 e at speed w, the exciting force amplitude is:
   A. m0 e w
   B. m0 e w^2
   C. M w^2
   D. m0 e / w^2
   Answer: B. The centrifugal force of the unbalanced mass is F0 = m0 e w^2, so it grows with speed squared. (Key point: Exciting force from unbalance)

2. The amplitude of steady-state vibration for rotating unbalance is X = ?
   A. (m0 e/M) / sqrt((1-r^2)^2 + (2 zeta r)^2)
   B. (m0 e/M) r^2 / sqrt((1-r^2)^2 + (2 zeta r)^2)
   C. (m0 e/M) r^2 / (1 - r^2)
   D. (m0 e/k) / sqrt((1-r^2)^2 + (2 zeta r)^2)
   Answer: B. Since F0 = m0 e w^2, the r^2 factor appears in the numerator: X = (m0 e/M) r^2 / sqrt(...), giving 5.876 mm here. (Key point: Steady-state amplitude formula)

3. Why is the transmitted force F_T = X sqrt(k^2 + (c w)^2) rather than (k + c w) X?
   A. Damper force is negligible so only k matters
   B. The damper force acts only at resonance
   C. Spring and damper forces are 90 degrees out of phase so they add in quadrature
   D. The spring force is always larger so it is taken alone
   Answer: C. The spring force is in phase with displacement and the damper force with velocity, i.e. in quadrature, so they combine as a vector sum. (Key point: Force transmitted)

4. At resonance (r = 1) the motor amplitude is:
   A. (m0 e/M)/(2 zeta) at N = wn x 60/2pi = 900 rpm
   B. (m0 e/M)/(2 zeta) at N = 1200 rpm
   C. infinite at 900 rpm
   D. (m0 e/M) x 2 zeta at 900 rpm
   Answer: A. Resonance occurs when w = wn = 94.28 rad/s (900 rpm). With damping the amplitude is finite: 0.25/90/0.24 = 11.57 mm. (Key point: Resonance speed and amplitude)

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