Two levers with end masses coupled by springs

Two equal levers with mass, each turning about its own pivot, carry point masses at their ends and are connected by three springs. The paper derives the equations with the Lagrangian method and simulates them in Simulink. It makes a chain of errors: a wrong moment of inertia, spring elongations twice too large, a lost sign in the potential energy, and a reversed sign in Lagrange's equation that turns every spring into a "negative" spring. In the paper's model the levers therefore oscillate around the vertical, and its plot shows exactly that once the true initial angle of 1 rad is used. The chosen springs are also far too weak to hold the levers up. The interactive simulator shows the paper's model, the corrected model, the static equilibrium and the normal modes.

Problem statement

Two equal levers $A_iB_i$ of length $l$ and mass $M$ can turn about the joints $O_i$, $i = 1, 2$. Concentrated masses $m = M/6$ are placed at the ends $A_i$. The levers are connected by springs of stiffness $k$. Determine the equations of motion of the system.

The geometry, read from the figure, with each lever divided into thirds:

Upper lever: mass $m$ at the left end $A_1$. The upper spring (to the ceiling) is attached at $l/3$ from $A_1$. The pivot $O_1$ is at $2l/3$, and the right end $B_1$ is at $l$.
Lower lever: mass $m$ at the left end $A_2$. The middle spring (from $B_1$) is attached at $l/3$, and the lower spring (to the floor) at $2l/3$. The pivot $O_2$ is at the right end.
The springs are vertical and unstretched when both levers are horizontal.

The generalized coordinates are the rotation angles $\theta_1$, $\theta_2$ of the levers from the horizontal. Below, both are taken positive counter-clockwise, so a positive angle lowers the left end of a lever.

Figure 1. Two levers of length l and mass M = 6m with end masses m. The springs attach at thirds of the lever length; θ₁ and θ₂ are positive counter-clockwise.

Energies of the system

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

Kinetic coenergy

The end masses move on circles of radius $2l/3$ (upper) and $l$ (lower). For the levers, the paper derives the moment of inertia of a rod about a point at distance $a$ from its end:

$$I = \frac M3\left(L^2 - 3La + 3a^2\right)$$

This formula is correct. For the upper lever ($L = l$, $a = 2l/3$ or equivalently $l/3$) it gives

$$I_1 = \frac M3\left(l^2 - 2l^2 + \tfrac43l^2\right) = \frac{Ml^2}{9}\qquad\Big(\text{check: } \tfrac{Ml^2}{12} + M\big(\tfrac l6\big)^2 = \tfrac{Ml^2}{9}\Big),\qquad I_2 = \frac{Ml^2}{3}$$

The paper writes $I_1 = Ml^2/27$, an arithmetic slip. With $M = 6m$, the correct total is

(1)$$\mathcal{R}^* = \tfrac12\Big(\tfrac49 + \tfrac69\Big)ml^2\dot\theta_1^2 + \tfrac12\big(1 + 2\big)ml^2\dot\theta_2^2 = \tfrac59ml^2\dot\theta_1^2 + \tfrac32ml^2\dot\theta_2^2$$

The paper has $\tfrac13ml^2\dot\theta_1^2$ instead of $\tfrac59ml^2\dot\theta_1^2$. The upper lever's inertia is 40 % too small.

Spring elongations

A point at horizontal distance $r$ from a pivot moves vertically by $r\sin\theta$ when the lever turns by $\theta$. The paper instead uses the "sine theorem" $s = 2r\sin\theta$, which is the length of the chord doubled. The chord of an arc is $2r\sin(\theta/2)$, and it is not the vertical displacement anyway. All elongations are therefore twice too large, and the spring energies four times too large. With the attachment distances $l/3$ (upper), $l/3$ and $2l/3$ (middle), $l/3$ (lower), the correct elongations are:

$$e_\text{upper} = \frac l3\sin\theta_1,\qquad e_\text{middle} = \frac l3\sin\theta_1 + \frac{2l}{3}\sin\theta_2,\qquad e_\text{lower} = -\frac l3\sin\theta_2$$

The middle spring is stretched when $B_1$ rises ($\theta_1>0$) and when its lower attachment, left of $O_2$, falls ($\theta_2>0$).

Gravity

For positive angles all masses move down: $A_1$ by $\tfrac{2l}3\sin\theta_1$, the centre of the upper lever (at $l/6$ left of $O_1$) by $\tfrac l6\sin\theta_1$, $A_2$ by $l\sin\theta_2$, and the centre of the lower lever by $\tfrac l2\sin\theta_2$. With $M = 6m$:

(2)$$\mathcal{A} = kl^2\Big(\tfrac19\sin^2\theta_1 + \tfrac5{18}\sin^2\theta_2 + \tfrac29\sin\theta_1\sin\theta_2\Big) - \tfrac53mgl\sin\theta_1 - 4mgl\sin\theta_2$$

The paper's expression has all spring coefficients four times larger ($\tfrac49$, $\tfrac{10}9$, $\tfrac89$). Its gravity terms are $+\tfrac53mgl\sin\theta_1$ and $+4mgl\sin\theta_2$. The line before still has $-4mgl\sin\theta_2$, so the sign was lost while collecting terms. The two angles then effectively have opposite senses, which does not match the sign of the coupling term.

Equations of motion

The paper's equations

paper$$\ddot\theta_1 = \frac{4k}{3m}\sin\theta_1\cos\theta_1 + \frac{4k}{3m}\cos\theta_1\sin\theta_2 + \frac{5g}{2l}\cos\theta_1,\qquad \ddot\theta_2 = \frac{20k}{27m}\sin\theta_2\cos\theta_2 + \frac{8k}{27m}\sin\theta_1\cos\theta_2 + \frac{4g}{3l}\cos\theta_2$$

Since $\mathcal{R}^*$ does not depend on the angles, Lagrange's equation reads $I_i\ddot\theta_i = -\partial\mathcal{A}/\partial\theta_i$. The paper writes $I_i\ddot\theta_i = +\partial\mathcal{A}/\partial\theta_i$. Every spring term then has the "wrong" sign: the springs push the levers away from the horizontal instead of pulling them back.

Corrected equations

From (1) and (2), with $I_1 = \tfrac{10}9ml^2$ and $I_2 = 3ml^2$:

(3)$$\begin{aligned}\tfrac{10}{9}ml^2\,\ddot\theta_1 &= -\tfrac29kl^2(\sin\theta_1 + \sin\theta_2)\cos\theta_1 + \tfrac53mgl\cos\theta_1\\ 3ml^2\,\ddot\theta_2 &= -kl^2\Big(\tfrac59\sin\theta_2 + \tfrac29\sin\theta_1\Big)\cos\theta_2 + 4mgl\cos\theta_2\end{aligned}$$

The model assumes that the springs stay vertical and that their ends move only vertically. This is reasonable for small angles, which is the only range where this geometry makes sense.

Small oscillations

For small angles the gravity torques become constant ($\cos\theta\approx1$), and they only shift the equilibrium (section 6). Around that equilibrium:

(4)$$\begin{bmatrix}\tfrac{10}9 & 0\\ 0 & 3\end{bmatrix}ml^2\begin{bmatrix}\ddot\theta_1\\ \ddot\theta_2\end{bmatrix} + \begin{bmatrix}\tfrac29 & \tfrac29\\ \tfrac29 & \tfrac59\end{bmatrix}kl^2\begin{bmatrix}\theta_1\\ \theta_2\end{bmatrix} = \mathbf 0$$

The stiffness matrix is symmetric and positive definite (determinant $\tfrac{6}{81}$), so the equilibrium is stable. The levers are coupled only through the middle spring, which gives the off-diagonal $\tfrac29$. The paper's small-angle equations keep the constant gravity terms and inherit the wrong signs, so $\theta = 0$ is unstable in its linear model too.

Simulink model and results

The Simulink model implements the paper's nonlinear equations: two integrator chains with sin and cos blocks, product blocks and gains $4k/3m$, $5g/2l$, $8k/27m$, $20k/27m$ and $4g/3l$. Parameters: $m = 1$ kg, $k = 1$ N/m, $l = 1$ m, $g = 9.81$ m/s². The paper states that "all initial conditions are zero".

The single plot shows $\theta_1$ and $\theta_2$ over 10 s, oscillating between 1.0 and 2.14 rad with periods of about 1.24 s and 1.71 s. The paper concludes that the system oscillates without damping, because there are no dissipative elements.

Simulation of the paper's equations	Range of θ₁, θ₂	Periods
θ₁(0) = θ₂(0) = 0 (as stated)	0 … 3.14 rad	–
θ₁(0) = θ₂(0) = 1 rad	1.00 … 2.14 rad	1.24 s, 1.71 s

The plot is reproduced exactly with initial angles of 1 rad, not zero. The oscillation is centred on $\pi/2 = 1.57$ rad, i.e. both levers pointing straight up or down. With the paper's sign error, $\theta = 0$ is unstable, and the levers settle around the vertical position where $\cos\theta = 0$ cancels all torques. The model has long left the small-angle range in which it was derived.

Critical review

Sign of Lagrange's equation

$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot\theta} - \frac{\partial\mathcal{L}}{\partial\theta} = 0$ with $\mathcal{L} = \mathcal{R}^* - \mathcal{A}$ gives $I\ddot\theta = -\partial\mathcal{A}/\partial\theta$. The paper drops the minus sign. A spring then acts like a negative spring, and the horizontal position becomes unstable. This error alone makes the simulation results meaningless.

Elongations: $2r\sin\theta$ instead of $r\sin\theta$

The sine theorem gives $BC/\sin\theta = 2R$, where $R$ is the radius of the triangle's circumcircle, not the lever arm $r$. The chord of the arc is $2r\sin(\theta/2)$. What stretches a vertical spring is the vertical displacement $r\sin\theta$. All spring terms in the paper are four times too large.

Moment of inertia of the upper lever

$I_1 = Ml^2/9$, not $Ml^2/27$. The general formula in the paper is right; the substitution is not. The upper lever's total inertia comes out 40 % too small.

Lost sign and inconsistent angle conventions

The term $-4mgl\sin\theta_2$ becomes $+4mgl\sin\theta_2$ when the potential energy is collected. In the paper's final expression, a positive $\theta_1$ raises $A_1$ and a positive $\theta_2$ raises $A_2$. The two levers are then measured in opposite rotational senses, but the middle spring's elongation $s_1 + s_2$ assumes the same sense. One sign convention, stated once and drawn in the figure, would avoid this.

Initial conditions and parameters

The plot does not correspond to the stated zero initial conditions. With $k = 1$ N/m the springs cannot hold $6$ kg levers horizontal at all: the correct static equilibrium would be $\theta_2^* \approx 69$ rad (section 6), i.e. the levers simply fall. A meaningful simulation needs $k$ of the order of $10^3$ N/m for these masses. Alternatively, the springs can be assumed pre-loaded so that the horizontal position is the equilibrium.

What is goodThe general approach is sound: the choice of angles as coordinates, the derivation of the inertia formula by integration, and the treatment of each element's energy separately. The kinetic energies of the end masses are correct, and so is the inertia of the lower lever.

Extension: equilibrium and normal modes

Static equilibrium

If the springs are unstretched with the levers horizontal, gravity turns the levers until the spring torques balance it. For small angles, $\mathbf K\boldsymbol\theta^* = mgl\,[\tfrac53,\ 4]^T$, which gives

$$\theta_1^* = \frac{mg}{2kl},\qquad \theta_2^* = \frac{7mg}{kl}$$

For the paper's values this is 4.9 rad and 69 rad. These are meaningless numbers: the springs are far too soft.
With $k = 1000$ N/m: $\theta_1^* = 0.005$ rad and $\theta_2^* = 0.069$ rad (4°), a sensible small tilt.
If instead the springs are pre-loaded so that the horizontal position is the equilibrium, the gravity terms drop out completely. Gravity and spring pre-load both enter through $\sin\theta$ in this geometry, so they cancel at every angle.

Natural frequencies

From (4), $\omega^2$ are the eigenvalues of $\mathbf M^{-1}\mathbf K$:

$$\omega^2 = \frac km\,\lambda,\qquad \lambda^2 - \tfrac{52}{135}\lambda + \tfrac{2}{90} = 0\ \Rightarrow\ \lambda_1 = 0.0707,\ \lambda_2 = 0.3145$$

Slow mode ($\omega_1 = 0.266\sqrt{k/m}$): $\theta_2/\theta_1 = -0.65$. The levers turn in opposite senses, which hardly stretches the middle spring.
Fast mode ($\omega_2 = 0.561\sqrt{k/m}$): $\theta_2/\theta_1 = +0.57$. The levers turn in the same sense and work against the middle spring.

For the paper's $k/m = 1$ the periods would be 23.6 s and 11.2 s. The paper's 1.24 s and 1.71 s come from the wrong model oscillating around the vertical, where gravity rather than the springs provides the restoring torque.

Interactive simulator

The simulator integrates the equations with 4th-order Runge–Kutta (1 ms step). It contains the paper's nonlinear and small-angle equations and the corrected ones. For the corrected models you can choose whether gravity acts on unstretched springs or whether the springs are pre-loaded so that the horizontal position is the equilibrium.

Cases

Model

Compare with

Springs pre-loaded (θ = 0 is the equilibrium; corrected models)

System parameters

m [kg] (M = 6m)

k [N/m]

l [m]

g [m/s²]

θ₁(0) [°]

θ₂(0) [°]

Duration [s]

Angles accept expressions, e.g. 180/pi for 1 rad. They start from rest.

Lever angles θ₁(t), θ₂(t)

Spring forces (selected model)

Force = k · elongation, from the correct geometry; positive = tension.

Energy (correct expression)

Constant only for the corrected nonlinear model.

Animation (selected model)

Results table (every 0.05 s)

Review questions

How can you check the sign of Lagrange's equation without re-deriving it?

Take a single spring–mass system, $\mathcal{L} = \tfrac12m\dot x^2 - \tfrac12kx^2$. The equation must give $m\ddot x = -kx$. A plus sign would make the mass run away, so any derivation that gives "+" has a sign error.

Why is the vertical displacement $r\sin\theta$ and not $2r\sin\theta$?

The point moves on a circle of radius $r$: its height above the pivot is $r\sin\theta$. The straight-line distance travelled is the chord $2r\sin(\theta/2)$, which for small angles is also $\approx r\theta$. Neither is $2r\sin\theta$.

What is the moment of inertia of a rod about a point at one third of its length?

By the parallel-axis theorem: $\tfrac{ML^2}{12} + M\big(\tfrac L6\big)^2 = \tfrac{ML^2}{9}$.

Why does gravity not change the natural frequencies of the linear model?

In this geometry the gravity torques are proportional to $\cos\theta$, which is constant to first order. They shift the equilibrium but add no stiffness. The springs alone set the frequencies.

How stiff must the springs be for the small-angle model to be valid?

The equilibrium tilt $\theta_2^* = 7mg/(kl)$ must be small, e.g. below 0.1 rad. That needs $k > 70\,mg/l$, about 690 N/m for $m = 1$ kg and $l = 1$ m.

Why do the levers turn in opposite senses in the slow mode?

A positive $\theta_1$ raises $B_1$, and a positive $\theta_2$ lowers the middle spring's lower end. Opposite signs move both ends of the middle spring the same way, so it is hardly stretched. Little stiffness means a low frequency.