Kinematics and Dynamics II: Gyroscope, Free and Forced Vibration, Damping, Isolation, Resonance and Critical Speeds of Shafts
1. The gyroscope and the gyroscopic couple
A rotor of moment of inertia I spinning at ω has angular momentum H = Iω along its axis. Turning that axis at an angular velocity ω_p (precession) about a perpendicular axis needs a torque that changes the direction of H, so C = I ω ω_p, acting about the axis perpendicular to both the spin and the precession axes. A disc of I = 0.5 kg·m² spinning at 3000 rpm (314.16 rad/s) and precessing at 0.2 rad/s therefore needs C = 0.5 × 314.16 × 0.2 = 31.4 N·m. The reaction of the rotor on its supports is equal and opposite, and it is this reaction couple that loads the bearings of a turbine on a ship or an aircraft.
- The rotor’s spin axis tends to turn towards the axis of the applied torque: the spin vector moves towards the torque vector, which is how the direction of the reaction couple is found.
- A vehicle rounding a bend, a ship pitching, rolling or steering, and an aircraft turning with a spinning engine all impose a precession on rotating parts, and the gyroscopic couple then presses one side of the vehicle down and lightens the other.
2. Free vibration of an undamped single-degree-of-freedom system
A mass m on a spring of stiffness k obeys m ẍ + kx = 0, so it oscillates harmonically at the natural frequency ω_n = √(k/m) rad/s, f = ω_n/2π hertz and period T = 1/f, whatever the amplitude. With m = 10 kg and k = 4000 N/m, ω_n = 20 rad/s and f = 3.18 Hz. In terms of the static deflection δ_st = mg/k the frequency is ω_n = √(g/δ_st): a spring compressed 25 mm under its load has ω_n = √(9.81/0.025) = 19.8 rad/s, and f_n ≈ 0.5/√δ_st hertz with δ in metres is a handy rule.
Springs in parallel add their stiffnesses, k = k₁ + k₂; springs in series add their compliances, 1/k = 1/k₁ + 1/k₂. For torsional vibration J θ̈ + k_t θ = 0 gives ω_n = √(k_t/J), and a simple pendulum has ω_n = √(g/L). The energy method equates the maximum kinetic energy to the maximum potential energy, ½ m ẋ²_max = ½ k x²_max, and avoids writing the equation of motion; Rayleigh’s method uses it with an assumed shape. The natural frequency depends on the system, never on how it was started.
3. Damping and free damped vibration
With a viscous damper c, m ẍ + c ẋ + kx = 0. The critical damping is c_c = 2√(km) = 2mω_n and the damping ratio ζ = c/c_c. For ζ < 1 the motion is an underdamped oscillation x = Xe−ζω_n t sin(ω_d t + φ) at the damped frequency ω_d = ω_n√(1 − ζ²), slightly below ω_n, so the damped period is longer; for ζ = 1 the mass returns fastest without oscillating; for ζ > 1 it creeps back, without oscillating, more slowly. Damping does not change the natural frequency appreciably until ζ is large: at ζ = 0.1, ω_d is 0.995 ω_n.
Damping is measured from the decay of a free oscillation. The logarithmic decrement is δ = ln(x₁/x₂) = 2πζ/√(1 − ζ²) ≈ 2πζ, the natural log of the ratio of successive peaks, and (1/n) ln(x₀/xₙ) over n cycles gives the same value. If each peak is half the previous one, δ = ln 2 = 0.693, and ζ = δ/√(4π² + δ²) = 0.693/6.321 = 0.11. The approximation 2πζ would give 0.110, close here because ζ is small. The quality factor Q = 1/(2ζ) is the same information as the resonant amplification.
4. Forced vibration, resonance and vibration isolation
Under a harmonic force F₀ sin ωt the steady-state amplitude is X = (F₀/k) · M, with the magnification factor M = 1/√[(1 − r²)² + (2ζr)²] and the frequency ratio r = ω/ω_n. The response lags the force by φ = tan⁻¹[2ζr/(1 − r²)]: 0° at low frequency, 90° at r = 1, tending to 180° above resonance. At resonance, r ≈ 1, the amplification is 1/(2ζ), limited only by damping: for ζ = 0.05, M = 10. The amplitude peaks at r = √(1 − 2ζ²), just below 1. Well below resonance M ≈ 1 (stiffness-controlled), well above it M falls as 1/r² (mass-controlled), and near it the damper alone decides (damping-controlled).
Vibration isolation reduces the force a machine transmits to its foundation, or the motion a foundation transmits to an instrument. The transmissibility is TR = √[1 + (2ζr)²]/√[(1 − r²)² + (2ζr)²]. It exceeds 1 up to r = √2, equals 1 at r = √2, and only beyond r = √2 does an isolator isolate; there, less damping gives less transmission, so a soft mounting is chosen and just enough damping to pass through resonance. For an undamped mount TR = 1/(r² − 1). A machine at 1800 rpm (30 Hz) to be isolated to TR = 0.1 needs r² = 11, so the mount’s natural frequency is 30/√11 = 9.05 Hz (a static deflection of about 3 mm).
5. Critical speeds of shafts
A rotor on a flexible shaft, however well balanced, has a small eccentricity, and at one speed its centrifugal force excites the shaft’s own transverse vibration: the shaft whirls and deflects violently. That speed is the critical speed, equal to the natural frequency of transverse vibration, ω_c = √(k/m) = √(g/δ_st) with δ_st the static deflection of the shaft under the rotor’s weight. For δ_st = 1 mm, ω_c = √(9.81/0.001) = 99.05 rad/s, that is N_c = 60 × 99.05/2π = 946 rpm. A stiffer or lighter-loaded shaft has a higher critical speed; a shaft is run well away from it, below (a rigid-shaft design) or, with a quick passage, above (a flexible-shaft design).
Above ω_c the rotor’s centre of mass moves to the inside of the whirl, and the rotor self-centres, rotating about its mass centre with a small load on the bearings. For a shaft with several rotors, Dunkerley’s formula estimates the lowest critical speed from those of each part acting alone: 1/N_c² = 1/N₁² + 1/N₂² + … For rotors alone giving 1000 and 1500 rpm, 1/N² = 10⁻⁶ + 4.44 × 10⁻⁷, so N = 832 rpm, below both. Dunkerley’s value is a lower bound to the true fundamental.
Key takeaways
- Gyroscopic couple C = I ω ω_p, about the axis perpendicular to spin and precession.
- ω_n = √(k/m) = √(g/δ_st); parallel springs add stiffness, series springs add compliance.
- ζ = c/2√(km); ω_d = ω_n√(1 − ζ²); δ = ln(x₁/x₂) = 2πζ/√(1 − ζ²).
- Resonance amplification is 1/(2ζ) with a 90° lag; isolation begins at r > √2, where TR = 1, and damping then hurts.
- Critical speed N_c = (60/2π)√(g/δ_st) rpm; Dunkerley: 1/N² = Σ 1/N_i²; above ω_c the rotor self-centres.
Practice questions (16)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
A rotor of moment of inertia I spins at angular velocity ω and its axis precesses at ω_p, perpendicular to the spin axis. The magnitude of the gyroscopic couple is:
Show answer
Answer: A — I ω ω_p
C = dH/dt = ω_p × H, whose magnitude is Iωω_p when the axes are perpendicular. It grows with both the spin and the precession, and vanishes when either is zero or when the two axes are parallel.A disc of moment of inertia 0.5 kg·m² spins at 3000 rpm and its spin axis is made to precess at 0.2 rad/s about a perpendicular axis. The gyroscopic couple, correct to one decimal place, is ____ N·m.
Numerical answer — type the value.
Show answer
Answer: 31.4
ω = 2π × 3000/60 = 314.16 rad/s, so C = Iωω_p = 0.5 × 314.16 × 0.2 = 31.4 N·m. Using 3000 directly as ω would give 300 N·m.The axis of the gyroscopic couple acting on a spinning rotor that is being precessed is:
Show answer
Answer: A — perpendicular to both the spin axis and the precession axis
The couple is the vector product ω_p × H, which is perpendicular to both factors. This is why a spinning top does not simply fall: gravity’s torque is horizontal and the axis moves sideways.A mass of 10 kg is attached to a spring of stiffness 4000 N/m and set into free vibration without damping. Its natural frequency, correct to two decimal places, is ____ Hz.
Numerical answer — type the value.
Show answer
Answer: 3.18
ω_n = √(k/m) = √(4000/10) = 20 rad/s, and f = ω_n/2π = 3.183 Hz. Giving 20 reports the angular frequency in rad/s, not hertz.A mass supported on a spring deflects the spring statically by 25 mm. Taking g = 9.81 m/s², the natural frequency of the system is ____ rad/s, correct to one decimal place.
Numerical answer — type the value.
Show answer
Answer: 19.8
ω_n = √(g/δ_st) = √(9.81/0.025) = √392.4 = 19.81 rad/s; the mass itself is not needed. Using δ in millimetres would give 0.63.The critical damping coefficient of a system of mass m and stiffness k is:
Show answer
Answer: A — 2√(km)
c_c = 2√(km) = 2mω_n is the damping at which the characteristic roots become equal and real, ζ = 1. Below it the motion oscillates; above it the response is overdamped.In a free damped oscillation each peak is half the amplitude of the preceding peak. The damping ratio of the system, correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.11
The logarithmic decrement is δ = ln 2 = 0.693, and ζ = δ/√(4π² + δ²) = 0.693/√(39.478 + 0.480) = 0.693/6.321 = 0.1097 ≈ 0.11. Taking ζ = δ/π = 0.22 forgets the 2 in 2πζ.A damped single-degree-of-freedom system is driven harmonically at exactly its natural frequency. Compared with the driving force, the steady-state displacement:
Show answer
Answer: A — lags by 90° and has an amplitude limited only by the damping
At r = 1 the spring and inertia forces cancel, so the damping force alone balances the excitation: X = F₀/(cω_n) = (F₀/k)/(2ζ), with φ = 90°. In phase is the low-frequency behaviour and 180° the high-frequency one.A lightly damped system with damping ratio 0.05 is driven at resonance. The magnification factor, the ratio of the dynamic amplitude to the static deflection F₀/k, is ____.
Numerical answer — type the value.
Show answer
Answer: 10
At r = 1 the magnification factor is 1/(2ζ) = 1/0.1 = 10. Undamped it would be infinite; the peak of the curve sits at r = √(1 − 2ζ²) = 0.9975, with a value only slightly above 10.A vibration isolator reduces the force transmitted to the foundation only when the frequency ratio ω/ω_n exceeds:
Show answer
Answer: A — √2
The transmissibility equals 1 at r = √2 for every damping ratio and is below 1 only beyond it. At r = 1, resonance, the force transmitted is amplified.A machine runs at 1800 rpm and its isolation mount, assumed undamped, is to transmit only 10% of the disturbing force. The natural frequency of the mounted system must be ____ Hz, correct to one decimal place.
Numerical answer — type the value.
Show answer
Answer: 9
TR = 1/(r² − 1) = 0.1 gives r² = 11, r = 3.317. The forcing frequency is 1800/60 = 30 Hz, so f_n = 30/3.317 = 9.05 Hz, which is 9.0 Hz to one decimal place. Using TR = 1/r² would give r² = 10 and 9.5 Hz.A rotor on a shaft deflects the shaft statically by 1 mm under its own weight. Taking g = 9.81 m/s², the critical speed of the shaft is ____ rpm, to the nearest whole number.
Numerical answer — type the value.
Show answer
Answer: 946
ω_c = √(g/δ_st) = √(9.81/0.001) = 99.05 rad/s, so N_c = 60ω_c/2π = 945.8, that is 946 rpm. Reporting 99 gives the critical speed in rad/s, not rpm.A shaft carries two rotors. Acting alone, the first would give a critical speed of 1000 rpm and the second of 1500 rpm, the shaft’s own mass being neglected. By Dunkerley’s formula the lowest critical speed of the shaft with both rotors is ____ rpm, to the nearest whole number.
Numerical answer — type the value.
Show answer
Answer: 832
1/N² = 1/1000² + 1/1500² = 1 × 10⁻⁶ + 4.444 × 10⁻⁷ = 1.4444 × 10⁻⁶, so N = 832.05, that is 832 rpm. It lies below both single-rotor values, as it must, since adding a mass lowers a natural frequency.Which of the following statements about vibrating systems are true?
Show answer
Answer: A — The natural frequency of a linear undamped system does not depend on the amplitude of vibration; B — The damped period of an underdamped system is longer than its undamped period; C — A system with damping ratio above 1 does not oscillate when displaced and released
(A) ω_n = √(k/m) has no amplitude in it. (B) ω_d = ω_n√(1 − ζ²) < ω_n. (C) The roots are real and negative. (D) False: TR = 1 at r = √2 exactly.Which of the following statements about the critical speed of a shaft are true?
Show answer
Answer: A — At the critical speed the shaft whirls with large deflections; B — Above the critical speed the rotor tends to rotate about its centre of mass (self-centring); C — The critical speed equals the natural frequency of transverse vibration of the shaft–rotor system
(A) Resonance between the rotation speed and the transverse natural frequency. (B) Beyond it the whirl radius tends to −e. (C) The definition. (D) False: ω_c = √(k/m) rises with k.A viscous damper is added to a well-designed vibration isolator that operates at a frequency ratio of about 4. The transmissibility at the operating speed will:
Show answer
Answer: A — increase, although the resonant peak on run-up falls
For r > √2 the numerator √(1 + (2ζr)²) grows faster than the denominator, so TR rises with ζ; at r = 4 with ζ = 0 it is 1/15 = 0.067, and with ζ = 0.2 about 0.125. Damping earns its place only for the resonance the machine passes through.