Spring-suspended stepped pulley with two hanging masses

A three-degree-of-freedom system: a stepped cylinder hangs on a spring, a mass hangs from its outer rim through a second spring, and a heavier mass hangs from its inner rim. The derivation uses Lagrange's equations, a cyclic coordinate explains the rotation, and both models can be compared in the simulator.

Problem statement

A homogeneous circular cylinder of mass $M$ and radius $2a$ hangs from a spring of stiffness $k_1$ and rotates freely about its centre $C$. It has an inner groove (drum) of radius $a$.

From the outer rim (radius $2a$, left side) a rope goes down to a spring of stiffness $k_2$, which carries a mass $m$.
From the inner drum (radius $a$, right side) a rope goes down to a mass $2m$.

Find: the equations of motion.

There is no friction and no external input, so the system is conservative and the right-hand sides of all Lagrange equations are zero.

Figure 1. Stepped cylinder (M, outer radius 2a, inner radius a) on spring k₁; mass m on spring k₂ from the outer rim; mass 2m from the inner drum. Coordinates x₁, x₂ (downward) and φ (counter-clockwise).

Generalized coordinates and kinematics

The paper chooses three independent coordinates:

$x_1$: elongation of spring $k_1$, which is also the downward displacement of the centre $C$,
$x_2$: elongation of spring $k_2$,
$\varphi$: rotation angle of the cylinder, positive counter-clockwise.

The free lengths of the springs and ropes are constants. They drop out of all derivatives, so they can be set to zero without loss of generality. The paper confirms this by trying constant offsets.

Positions of the masses (measured downward)

A counter-clockwise rotation $\varphi$ unwinds the rope on the left side of the outer rim, lowering it by $2a\varphi$. It winds up the rope on the right side of the inner drum, raising it by $a\varphi$:

$$y_C = x_1,\qquad y_m = x_1 + 2a\varphi + x_2,\qquad y_{2m} = x_1 - a\varphi$$

$$v_1 = \dot x_1,\qquad v_2 = \dot x_1 + \dot x_2 + 2a\dot\varphi,\qquad v_3 = \dot x_1 - a\dot\varphi,\qquad \omega=\dot\varphi$$

Where the paper differsThe paper writes $v_3 = \dot x_1 + a\dot\varphi$, with a plus, but uses $x_1 - a\varphi$ for the same mass in the potential energy. The two cannot both be right. The geometry gives the minus sign (see section 7).

Energies (as derived in the paper)

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

3.1 Repulsive (kinetic) coenergy

It consists of the translational and rotational energy of the cylinder and the translational energy of both masses. The moment of inertia of a homogeneous cylinder of radius $2a$ is

$$I_C = \tfrac12 M(2a)^2 = 2Ma^2,\qquad \tfrac12 I_C\dot\varphi^2 = Ma^2\dot\varphi^2$$

(1)$$\mathcal{R}^* = \tfrac12 Mv_1^2 + \tfrac12 I_C\omega^2 + \tfrac12 mv_2^2 + \tfrac12 (2m)v_3^2$$

(7)$$\mathcal{R}^* = \tfrac12 M\dot x_1^2 + Ma^2\dot\varphi^2 + \tfrac12 m(\dot x_1+\dot x_2+2a\dot\varphi)^2 + m(\dot x_1 \color{#c27a00}{+} a\dot\varphi)^2$$

3.2 Attractive (potential) energy

Spring energies plus the gravitational energy of the three bodies:

(8)$$\mathcal{A} = \tfrac12 k_1x_1^2 + \tfrac12 k_2x_2^2 \color{#c27a00}{+} 2mg(x_1 - a\varphi) \color{#c27a00}{+} mg(x_1+x_2+2a\varphi) \color{#c27a00}{+} Mgx_1$$

Sign of gravityThe coordinates point downward, so a body that moves down loses potential energy: $V = -mg\,y$. The paper uses $+mg\,y$, which makes gravity act upward in the model (see section 7).

The $\varphi$ terms cancel: $-2mga\varphi + 2mga\varphi = 0$. The pulley is statically balanced, because $m\cdot 2a = 2m\cdot a$. Gravity exerts no torque on it, and $\varphi$ does not appear in $\mathcal{A}$ at all.

3.3 Lagrangian

(10)$$\mathcal{L} = \tfrac12 M\dot x_1^2 + Ma^2\dot\varphi^2 + \tfrac12 m(\dot x_1+\dot x_2+2a\dot\varphi)^2 + m(\dot x_1+a\dot\varphi)^2 - \Big[\tfrac12k_1x_1^2 + \tfrac12k_2x_2^2 + 2mg(x_1-a\varphi) + mg(x_1+x_2+2a\varphi) + Mgx_1\Big]$$

Lagrange equations (paper)

Coordinate $x_1$

$$\frac{\partial\mathcal{L}}{\partial\dot x_1} = M\dot x_1 + m(\dot x_1+\dot x_2+2a\dot\varphi) + 2m(\dot x_1+a\dot\varphi),\qquad \frac{\partial\mathcal{L}}{\partial x_1} = -(k_1x_1 + 2mg + mg + Mg)$$

(*)$$(M+3m)\ddot x_1 + m\ddot x_2 + 4am\ddot\varphi + k_1x_1 + (M+3m)g = 0$$

Coordinate $x_2$

$$\frac{\partial\mathcal{L}}{\partial\dot x_2} = m(\dot x_1+\dot x_2+2a\dot\varphi),\qquad \frac{\partial\mathcal{L}}{\partial x_2} = -k_2x_2 - mg$$

(**)$$m\ddot x_2 + m\ddot x_1 + 2am\ddot\varphi + k_2x_2 + mg = 0$$

Coordinate $\varphi$

$$\frac{\partial\mathcal{L}}{\partial\dot\varphi} = 2am(\dot x_1+\dot x_2+2a\dot\varphi) + 2am(\dot x_1+a\dot\varphi) + 2a^2M\dot\varphi,\qquad \frac{\partial\mathcal{L}}{\partial\varphi} = 2mag - 2mga = 0$$

(***)$$(6a^2m + 2a^2M)\ddot\varphi + 4am\ddot x_1 + 2am\ddot x_2 = 0$$

Given the paper's $\mathcal{R}^*$ and $\mathcal{A}$, these derivatives are computed correctly. The errors are in the energies themselves (section 7).

Simulink model

The paper solves each equation for its own highest derivative ($\ddot x_1$ from (*), $\ddot x_2$ from (**), $\ddot\varphi$ from (***)) and builds three integrator chains. The gains in its diagram are:

Chain	Gains used in the paper
$\ddot x_1$	$\dfrac{k_1}{M+3m}$, $\dfrac{m}{M+3m}$ (from $\ddot x_2$), $\dfrac{4am}{M+3m}$ (from $\ddot\varphi$), constant $g$
$\ddot x_2$	$\dfrac{k_2}{m}$, $2a$ (from $\ddot\varphi$), $\ddot x_1$ directly, constant $g$
$\ddot\varphi$	$\dfrac{4am}{6a^2m+2a^2M}$ (from $\ddot x_1$), $\dfrac{2am}{6a^2m+2a^2M}$ (from $\ddot x_2$)

Every chain needs the second derivatives of the others, so the diagram contains algebraic loops, which Simulink has to solve iteratively at each step. A cleaner implementation writes the equations in matrix form and inverts the (constant) mass matrix once:

$$\mathbf{M}\begin{bmatrix}\ddot x_1\\ \ddot x_2\\ \ddot\varphi\end{bmatrix} = \begin{bmatrix}-k_1x_1 + f_1\\ -k_2x_2 + f_2\\ 0\end{bmatrix}\quad\Longrightarrow\quad \ddot{\mathbf{q}} = \mathbf{M}^{-1}\mathbf{F}(\mathbf{q})$$

Figure 2. Vectorized Simulink model without algebraic loops: one gain block holds the inverse mass matrix.

Simulation results from the paper

All simulations start from zero initial conditions. The reference parameters are $k_1=20$, $k_2=15$, $M=5$, $m=1$, $a=1$, $g=9.81$ (SI units). The paper then changes one parameter at a time. Each "Load" button reproduces that plot in the simulator with the paper's model; the curves match the paper exactly.

Case	Change	Paper's result (x₁ / x₂ / φ)	Paper's interpretation
1	reference	0…−7.9 / 0…−1.3 / 0…2.1	Motion without excitation, caused by gravity; undamped; negative elongations attributed to the reference level.
2	$a=2$	same / same / 0…1.06	Larger radius → larger moment of inertia → smaller rotation; further increase would stop rotation.
3	$m=5$	0…−19.6 / 0…−6.5 / 0…11.3	Heavier masses → larger rotation and larger elongation of both springs.
4	$M=15$	0…−17.7 / ≈ same / ≈ same	Heavier cylinder → larger elongation of k₁; "too large M makes the system unstable (spring breaks)".
5	$k_2=5$	≈ same / 0…−3.9 / 0…2.45	Softer spring k₂ → larger x₂.
6	$k_1=40$	0…−3.95 / ≈ same / 0…1.1	Stiffer spring k₁ → smaller x₁.

The parameter trends for $x_1$ and $x_2$ are correct: with zero initial conditions each elongation swings between 0 and twice its static value, $2(M+3m)g/k_1$ and roughly $2mg/k_2$. The behaviour of $\varphi$, however, is mainly a consequence of the kinematic sign error, as shown next.

Critical review and corrected model

Wrong sign in the velocity of mass 2m

The rope of mass $2m$ leaves the right side of the inner drum. A counter-clockwise rotation winds it up, so $y_{2m} = x_1 - a\varphi$ and $v_3 = \dot x_1 - a\dot\varphi$. The paper's own potential energy uses $x_1 - a\varphi$, but its kinetic energy uses $\dot x_1 + a\dot\varphi$. This inconsistency creates a spurious coupling term $4am\ddot x_1$ in equations (*) and (***).

Wrong sign of gravity

With downward coordinates the gravitational potential is $V=-mg\,y$. With $+mg\,y$, as in the paper, all bodies are pushed up, and the springs end up compressed. This is the real reason why the elongations in the paper's plots are negative. It is not the choice of reference level. Physically the springs stretch: the static elongations are $x_1^* = (M+3m)g/k_1 = 3.92$ m and $x_2^* = mg/k_2 = 0.65$ m for the reference parameters. In a linear system this error only mirrors the equilibrium. The first error, however, changes the dynamics.

Units and the "instability" remark

The plots label $x$ in cm, but with SI parameters ($k$ in N/m, masses in kg) the elongations come out in metres. The reference springs are very soft (x₁ swings by almost 8 m). The remark that a too-large $M$ makes the system "unstable" is not something this model can show: it is linear and always stable, and any $M$ only shifts the equilibrium. A spring breaking lies outside the model's assumptions.

Corrected equations

$$\mathcal{R}^* = \tfrac12 M\dot x_1^2 + Ma^2\dot\varphi^2 + \tfrac12 m(\dot x_1+\dot x_2+2a\dot\varphi)^2 + m(\dot x_1 - a\dot\varphi)^2,\qquad \mathcal{A} = \tfrac12 k_1x_1^2 + \tfrac12 k_2x_2^2 - (M+3m)g\,x_1 - mg\,x_2$$

$$\begin{aligned}(M+3m)\ddot x_1 + m\ddot x_2 + k_1x_1 &= (M+3m)g\\ m\ddot x_1 + m\ddot x_2 + 2am\ddot\varphi + k_2x_2 &= mg\\ 2am\ddot x_2 + (2M+6m)a^2\ddot\varphi &= 0\end{aligned}$$

The coupling between $x_1$ and $\varphi$ disappears. Lowering the whole pulley does not make it turn, because the two hanging masses balance each other ($m\cdot2a = 2m\cdot a$).

The key insight: φ is a cyclic coordinate

$\varphi$ does not appear in $\mathcal{L}$ (only $\dot\varphi$ does), so $\partial\mathcal{L}/\partial\varphi = 0$. The generalized momentum is conserved: this is the angular momentum of the system about $C$.

$$p_\varphi = \frac{\partial\mathcal{L}}{\partial\dot\varphi} = \text{const}$$

Starting from rest ($p_\varphi=0$) and integrating once gives $\varphi$ algebraically in terms of $x_1, x_2$:

$$\text{paper:}\quad \varphi = -\frac{m\,(2x_1 + x_2)}{a\,(M+3m)}$$

$$\text{corrected:}\quad \varphi = -\frac{m\,x_2}{a\,(M+3m)}$$

This explains all of the paper's observations about $\varphi$. In the paper's model the large $x_1$ (≈ −7.9 m) drives the rotation: $\varphi\approx(2\cdot7.9+1.3)/8\approx2.1$ rad. Rotation is inversely proportional to $a$, so $a=2$ halves it. This is angular momentum conservation, not just a larger moment of inertia. In the corrected model only the stretching of spring $k_2$ turns the pulley, and $\varphi$ stays within about −0.18 rad for the reference parameters.

Natural frequencies

Eliminating $\varphi$ leaves a 2-DOF oscillator (the third "mode" is a free rotation with zero frequency). For the reference parameters the corrected model gives $f_1 = 0.249$ Hz and $f_2 = 0.789$ Hz. The paper's model gives 0.268 and 0.732 Hz. The two frequencies are incommensurate, which is why the curves look complex-periodic.

Interactive simulator

Both models are linear with a constant mass matrix. The simulator inverts it once and integrates with 4th-order Runge–Kutta (2 ms step). The initial velocities are zero.

Cases from the paper

Model

Also show the other model

System parameters

M (cylinder) [kg]

m [kg]

a (inner radius) [m]

k₁ [N/m]

k₂ [N/m]

g [m/s²]

Initial conditions

x₁(0) [m]

x₂(0) [m]

φ(0) [rad]

Duration [s]

Spring elongations x₁(t), x₂(t)

Cylinder rotation φ(t)

Dotted: φ predicted from x₁, x₂ by conservation of angular momentum. It lies exactly on the simulated curve.

Energy balance (selected model)

$\mathcal{R}^*+\mathcal{A}$ must stay constant.

Animation (selected model)

Results table (every 0.1 s)

Review questions

Why can the free lengths of springs and ropes be ignored?

They are constants. In the kinetic energy they vanish on differentiation, and in the potential energy they only add constant terms or shift the equilibrium. Neither changes the equations of motion for the elongations.

How do you check the sign of a kinematic relation like $y_{2m} = x_1 \pm a\varphi$?

Imagine a small positive rotation and look at which way the rope moves. Counter-clockwise rotation lifts the right side of the drum, so the mass on the right rises: $y_{2m} = x_1 - a\varphi$. The kinetic and potential energies must use the same relation.

Why does gravity not appear in the φ equation?

The two hanging masses produce opposite torques: $mg\cdot2a$ and $2mg\cdot a$. These cancel, so the pulley is statically balanced and $\varphi$ does not appear in $\mathcal{A}$.

What is a cyclic coordinate and what does it give us here?

A coordinate that does not appear in the Lagrangian. Its generalized momentum is conserved; here that is the angular momentum about $C$. It lets us express $\varphi$ directly through $x_1$ and $x_2$, reducing the problem to two degrees of freedom.

With zero initial conditions, why does each elongation swing between 0 and twice its static value?

For a linear undamped system released from rest away from equilibrium, the motion is symmetric about the equilibrium. Starting at 0 with equilibrium $x^*$ gives an oscillation between 0 and $2x^*$. Coupling makes this only approximate for $x_2$.

Can the system become unstable if $M$ is very large?

Not within the model. It is linear with positive-definite mass and stiffness matrices, so all motions are bounded oscillations. A large $M$ only increases the static elongation $x_1^*$. A real spring would eventually break or leave its linear range, but that is outside the model's assumptions.

Chain	Gains used in the paper
\(\ddot x_1\)	\(\dfrac{k_1}{M+3m}\), \(\dfrac{m}{M+3m}\) (from \(\ddot x_2\)), \(\dfrac{4am}{M+3m}\) (from \(\ddot\varphi\)), constant \(g\)
\(\ddot x_2\)	\(\dfrac{k_2}{m}\), \(2a\) (from \(\ddot\varphi\)), \(\ddot x_1\) directly, constant \(g\)
\(\ddot\varphi\)	\(\dfrac{4am}{6a^2m+2a^2M}\) (from \(\ddot x_1\)), \(\dfrac{2am}{6a^2m+2a^2M}\) (from \(\ddot x_2\))