Helicopter rotor assembly on a spring–damper mount

The rotor, gearbox and motor form a mass $m$ mounted on the fuselage $M$ through a spring and a damper, and the rotor is shaken by a harmonic disturbance. We derive the equations of motion with Lagrange's equations. We then correct the generalized forces, show that the paper's Simulink model differentiates instead of integrating, and study vibration isolation in an interactive simulator.

Problem statement

The rotor–gearbox–motor assembly of total mass $m$ is attached to the helicopter fuselage of mass $M$ by a spring of stiffness $k_1$ and a damper with damping coefficient $B_1$. The rotor is subjected to the disturbing force

$$F(t) = F_0\sin\omega t$$

Find: the equations of oscillation of the system.

The motion is vertical (axis $x$ points up). Below, $k\equiv k_1$ and $B\equiv B_1$.

Figure 1. Fuselage M and the rotor assembly m connected by spring k₁ and damper B₁. The disturbance F(t) acts on the rotor. x₁ is the fuselage displacement; x₂ is the displacement of m relative to the fuselage.

Generalized coordinates and energies

The paper chooses two independent coordinates:

$x_1$: displacement of the fuselage $M$,
$x_2$: displacement of the rotor assembly $m$ relative to the fuselage, equal to the spring deformation (as drawn in the figure).

The absolute displacement of $m$ is therefore $x_1+x_2$, and the velocities are $v_1 = \dot x_1$ and $v_2 = \dot x_1 + \dot x_2$.

TerminologyAs in the course: attractive energy $\mathcal{A}$ = potential energy, repulsive coenergy $\mathcal{R}^*$ = kinetic coenergy, $\mathcal{L}=\mathcal{R}^*-\mathcal{A}$.

$$\mathcal{A} = \tfrac12 k x_2^2 + mg(x_1+x_2) + Mgx_1,\qquad \mathcal{R}^* = \tfrac12 Mv_1^2 + \tfrac12 mv_2^2 = \tfrac12 M\dot x_1^2 + \tfrac12 m(\dot x_1+\dot x_2)^2$$

$$\mathcal{L} = \tfrac12 M\dot x_1^2 + \tfrac12 m(\dot x_1+\dot x_2)^2 - \tfrac12 kx_2^2 - mg(x_1+x_2) - Mgx_1$$

The paper writes the spring energy as $\tfrac12 ky_2^2$, a typo for $\tfrac12 kx_2^2$. The gravity terms have the correct sign for an upward axis.

Lagrange equations

$$\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial\dot x_i}\right) - \frac{\partial\mathcal{L}}{\partial x_i} = \Xi_i,\qquad i=1,2$$

Coordinate $x_1$

$$\frac{\partial\mathcal{L}}{\partial\dot x_1} = m(\dot x_1+\dot x_2) + M\dot x_1,\qquad \frac{\partial\mathcal{L}}{\partial x_1} = -mg - Mg$$

paper$$m(\ddot x_1+\ddot x_2) + M\ddot x_1 + (m+M)g = F_0\sin\omega t$$

Coordinate $x_2$

$$\frac{\partial\mathcal{L}}{\partial\dot x_2} = m(\dot x_1+\dot x_2),\qquad \frac{\partial\mathcal{L}}{\partial x_2} = -kx_2 - mg$$

paper$$m(\ddot x_1+\ddot x_2) + kx_2 + mg = -B\dot x_2\quad\color{#c27a00}{(+F_0\sin\omega t\ \text{missing})}$$

Solved for the highest derivatives, as the paper does for Simulink:

$$\ddot x_1 = \frac{F_0\sin\omega t}{M+m} - g - \frac{m\,\ddot x_2}{M+m},\qquad \ddot x_2 = -\frac{B\dot x_2}{m} - \frac{kx_2}{m} - g - \ddot x_1$$

Each equation contains the second derivative of the other coordinate, which creates an algebraic loop in Simulink (see section 4).

Simulink model

The paper's block diagram has two chains, one per coordinate. Each starts with a summing junction ($F_0\sin\omega t\cdot\frac{1}{M+m}$, the constant $g = 9.81$, and the coupling and spring/damper gains $\frac{1}{M+m}$, $B/m$, $k/m$). The chain then continues through two "Derivative" (du/dt) blocks to $x_1$ and $x_2$. The two outputs go to one scope, and their product goes to a second scope labeled "oscillation of the whole system".

This does not simulate the equationsGoing from $\ddot x$ to $x$ requires two integrators (1/s), not two differentiators. With du/dt blocks the diagram computes the second derivative of the accelerations, a completely different (and noise-amplifying) quantity. See section 6 for the evidence in the paper's own plots.

A correct, loop-free implementation writes the equations in matrix form and inverts the constant mass matrix:

$$\begin{bmatrix}M+m & m\\ m & m\end{bmatrix}\begin{bmatrix}\ddot x_1\\ \ddot x_2\end{bmatrix} = \begin{bmatrix}F(t)\\ F(t) - B\dot x_2 - kx_2\end{bmatrix}\quad\Longrightarrow\quad \ddot x_1 = \frac{kx_2 + B\dot x_2}{M},\quad \ddot x_2 = \frac{F(t)}{m} - \frac{M+m}{Mm}\left(kx_2 + B\dot x_2\right)$$

These are written from the static equilibrium (gravity removed, see section 6) and with the corrected force term.

Figure 2. Correct structure with integrators (1/s): the relative motion x₂ is driven by F(t) and held back by the spring/damper force, which in turn accelerates the fuselage.

Simulation results from the paper

Three cases are shown, all with $F = 10\sin5t$:

Case	M	m	k	B	Paper's observation
1	50	2	1	2	M oscillates sinusoidally with amplitude ≈ 5; m much less. The "whole system" (product) shows a transient.
2	50	30	1	2	Comparable masses: a more pronounced transient; M ≈ ±3, m ≈ ±8.
3	50	2	8	10	Stiffer spring and more damping: M unchanged, m's amplitude much smaller.

The paper concludes that the amplitudes are proportional to the force, that a larger mass ratio M/m makes the system "more stable", and that damping mainly affects the relation between the two masses' oscillations.

Critical review

The force is missing from the second equation

The disturbance acts on the rotor, whose absolute displacement is $x_1+x_2$. Its virtual work is $F\,\delta(x_1+x_2)$, so it contributes to both generalized forces. The damper acts on the relative velocity $\dot x_2$, so it appears only in the second:

$$\Xi_1 = F(t),\qquad \Xi_2 = F(t) - B\dot x_2$$

(With the Rayleigh function $\mathcal D = \tfrac12 B\dot x_2^2$: $\Xi_i = Q_i^F - \partial\mathcal D/\partial\dot x_i$.) The paper puts $F$ only in the first equation, as if the force acted on the fuselage.

Differentiators instead of integrators

The paper's diagram turns $\ddot x$ into "$x$" with two du/dt blocks. The paper's own plots confirm it. Feeding $\frac{F_0}{M+m}\sin\omega t$ through two differentiators gives an amplitude of $\frac{F_0\omega^2}{M+m}$:

Case	$F_0\omega^2/(M+m)$	Amplitude of M in the paper's plot
M = 50, m = 2	10·25/52 = 4.81	≈ 4.9
M = 50, m = 30	10·25/80 = 3.13	≈ 3.1

A correct simulation gives a fuselage motion of only a few millimetres to centimetres for these parameters. The differentiators also remove the constant $g$ (its derivative is zero), which hides the next problem. Finally, the "whole system" signal is the product $x_1\cdot x_2$, which has no physical meaning. A meaningful overall quantity is the motion of the centre of mass, $x_{cm} = x_1 + \frac{m}{M+m}x_2$.

Gravity and what holds the fuselage up

In the paper's equations nothing supports the helicopter. Adding them gives $(M+m)\ddot x_{cm} = F - (M+m)g$, so the whole system falls freely. A realistic model needs a support: either the lift of a hovering helicopter balancing $(M+m)g$, or the landing gear holding the fuselage on the ground. In both cases, measuring the coordinates from the static equilibrium removes all gravity terms. The spring's static deflection cancels $mg$.

Corrected equations (hovering helicopter, measured from equilibrium)

$$\begin{aligned}(M+m)\ddot x_1 + m\ddot x_2 &= F_0\sin\omega t\\ m(\ddot x_1+\ddot x_2) + B\dot x_2 + kx_2 &= F_0\sin\omega t\end{aligned}$$

Subtracting the equations gives $M\ddot x_1 = kx_2 + B\dot x_2$: the fuselage is driven only by the spring and damper force. Eliminating $\ddot x_1$ shows that the relative motion is a single damped oscillator with the reduced mass $\mu = \frac{Mm}{M+m}$:

$$\mu\,\ddot x_2 + B\dot x_2 + kx_2 = \frac{M}{M+m}F_0\sin\omega t,\qquad \omega_n = \sqrt{\frac{k}{\mu}},\qquad \zeta = \frac{B}{2\sqrt{k\mu}}$$

For case 1, $\mu = 1.92$ kg, $\omega_n = 0.72$ rad/s and $\zeta = 0.72$. The disturbance at $\omega = 5$ rad/s is far above $\omega_n$, so the mount isolates the fuselage well.

The centre of mass is driven directly: $(M+m)\ddot x_{cm} = F$. Starting from rest, its mean velocity is not zero ($F_0/(\omega(M+m))$), so a hovering helicopter also drifts slowly. This is a rigid-body mode. In the simulator it can be removed by starting with the matching initial velocity.

If the fuselage stands on the ground ($x_1 \equiv 0$), only the second equation remains: $m\ddot x_2 + B\dot x_2 + kx_2 = F_0\sin\omega t$, the classic forced damped oscillator.

Vibration isolation

The quantity that matters for the crew and the structure is the force transmitted to the fuselage, $F_T = kx_2 + B\dot x_2$. Its ratio to the disturbance (the transmissibility) is below 1 only for $\omega > \sqrt2\,\omega_n$. In that range a softer mount isolates better, and more damping actually makes isolation slightly worse. The paper's case 3 ($k=8$, $B=10$) raises $\omega_n$ to 2.04 rad/s and therefore transmits more force to the fuselage than case 1, even though the rotor's relative motion is smaller. The simulator's frequency-response chart shows this.

Interactive simulator

The simulator integrates the corrected equations (4th-order Runge–Kutta, 1 ms step) from equilibrium. It can also run the paper's equations (with gravity and the force only in the first equation, integrated correctly) to show the free fall.

Cases from the paper

Model

Start without centre-of-mass drift

System parameters

M (fuselage) [kg]

m (rotor assembly) [kg]

k [N/m]

B [N·s/m]

F₀ [N]

ω [rad/s]

Duration [s]

Absolute displacements

Spring deflection x₂(t)

Forces: disturbance F vs transmitted F_T

Frequency response (corrected model, steady state)

Transmissibility |F_T|/F₀ (force reaching the fuselage) and relative deflection k|X₂|/F₀. Isolation (|F_T| < F₀) needs ω > √2·ωₙ. Dashed line: current ω.

Animation (displacements exaggerated)

Results table (every 0.1 s)

Review questions

Why does a force on the rotor appear in both Lagrange equations?

Its point of application moves by $\delta(x_1+x_2)$ for a variation of either coordinate. The generalized force $\Xi_i = F\,\partial(x_1+x_2)/\partial x_i$ is therefore $F$ for both.

Why does the damper appear only in the second equation?

It acts between the two bodies, on the relative velocity $\dot x_2$. Its dissipation function $\tfrac12 B\dot x_2^2$ depends only on $\dot x_2$. The damper force does act on the fuselage too, but that enters through the first equation's coupling with the second.

How can a plot reveal that Derivative blocks were used instead of integrators?

The response amplitude scales with $\omega^2$ instead of $1/\omega^2$, constant inputs such as $g$ leave no trace, and the numbers match $F_0\omega^2/(M+m)$. A quick check is to set $\omega$ very high: a correct model responds less, a differentiating one more.

Why is the relative motion governed by the reduced mass?

The spring pushes both bodies in opposite directions. The fuselage recoils, so the effective mass "seen" by the spring is $\mu = Mm/(M+m)$, which is smaller than $m$. For $M\gg m$, $\mu\approx m$ and the fuselage hardly moves.

Does a stiffer mount always reduce vibration?

No. Above $\sqrt2\,\omega_n$ a stiffer mount raises $\omega_n$ and increases the transmitted force. For isolation you want $\omega_n$ well below the disturbance frequency, i.e. a soft mount, with enough damping to limit the response when passing through resonance during start-up.

What changes if the helicopter stands on the ground?

The fuselage is fixed ($x_1=0$) and the system reduces to one degree of freedom: $m\ddot x_2 + B\dot x_2 + kx_2 = F_0\sin\omega t$, with $\omega_n = \sqrt{k/m}$.