Pulley with two springs and a hanging mass

A two-degree-of-freedom system: a pulley on a fixed axle, a rope that does not slip, a spring anchoring one rope end to the floor, and a mass hanging from the other end through a second spring. We derive the equations of motion with Lagrange's equations, correct the sign of gravity, check the rope tensions (a rope can pull but never push), and explore the two normal modes in an interactive simulator.

Problem statement

Determine the equations of motion of the system shown in the figure. The rope does not slip on the cylinder.

The pulley is a homogeneous cylinder of mass $m_2$ and radius $R$ on a fixed, frictionless axle.
The left end of the rope is anchored to the floor through a spring of stiffness $k_2$.
The right end of the rope carries a spring of stiffness $k_1$, from which hangs a mass $m_1$ with weight $G = m_1g$.

The springs are linear and massless. The paper also notes that there is no dissipation, so the system oscillates without damping.

Figure 1. Pulley (m₂, R) on a fixed axle; rope anchored to the floor via k₂ on the left; mass m₁ hanging via k₁ on the right. θ: pulley rotation (clockwise); x: downward displacement of m₁.

Generalized coordinates

Because the rope does not slip, a rotation $\theta$ of the pulley moves the right rope end down by $R\theta$ and the left rope end up by $R\theta$. The elongation of spring $k_2$ is therefore fixed by the pulley angle:

$$x_2 = R\theta$$

The paper decomposes the system (its Figure 2) and chooses two independent coordinates:

$\theta$: rotation angle of the pulley (clockwise),
$x$: downward displacement of mass $m_1$, measured from the position where both springs are unstretched.

The elongation of spring $k_1$ is the difference between the mass's displacement and the rope end's displacement: $x - R\theta$.

Energies of the system

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

3.1 Repulsive (kinetic) coenergy

The mass translates with $v = \dot x$, and the pulley rotates with $\omega = \dot\theta$. The moment of inertia of a homogeneous disk is $I = \tfrac12 m_2R^2$:

$$\mathcal{R}^* = \tfrac12 m_1\dot x^2 + \tfrac12 I\dot\theta^2,\qquad I = \tfrac12 m_2R^2$$

The paper describes $\omega$ as "the angular velocity of mass $m_1$". It is the angular velocity of the pulley.

3.2 Attractive (potential) energy

$$\mathcal{A} = \tfrac12 k_1(x - R\theta)^2 + \tfrac12 k_2(R\theta)^2 \;\color{#c27a00}{+\;Gx}\quad\text{(paper)}$$

$$\mathcal{A} = \tfrac12 R^2\theta^2(k_1+k_2) + \tfrac12 k_1x^2 - k_1R\theta x \;\color{#c27a00}{+\;Gx}\quad\text{(paper, expanded)}$$

Sign of gravity$x$ points down. When the mass moves down it loses potential energy, so the gravity term must be $-Gx$, not $+Gx$. See section 6.

3.3 Lagrangian (paper)

$$\mathcal{L} = \tfrac12 m_1\dot x^2 + \tfrac12 I\dot\theta^2 - \tfrac12R^2\theta^2(k_1+k_2) - \tfrac12k_1x^2 + k_1R\theta x - Gx$$

Lagrange equations

There is no dissipation and no external force, so

$$\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial\dot q}\right) - \frac{\partial\mathcal{L}}{\partial q} = 0,\qquad q\in\{x,\theta\}$$

Coordinate $x$

$$m_1\ddot x - \left(-k_1x + k_1R\theta - G\right) = 0\quad\Longrightarrow\quad m_1\ddot x + k_1x + G = k_1R\theta\quad\text{(paper)}$$

Coordinate $\theta$

$$I\ddot\theta - \left[-R^2\theta(k_1+k_2) + k_1Rx\right] = 0\quad\Longrightarrow\quad I\ddot\theta + R^2\theta(k_1+k_2) = k_1Rx$$

With $I = \tfrac12 m_2R^2$ and dividing by $R$:

$$\tfrac12 m_2R\ddot\theta + R\theta(k_1+k_2) = k_1x$$

The $\theta$ equation is correct. The $x$ equation inherits the wrong sign of $G$ from the potential energy.

Simulink model

Each coordinate gets a chain of two integrators. The gains in the paper's diagram are:

$$\ddot x = \frac{G}{m_1}\cdot(\pm1) - \frac{k_1}{m_1}x + \frac{k_1R}{m_1}\theta$$

$$\ddot\theta = -\frac{2(k_1+k_2)}{m_2}\theta + \frac{2k_1}{m_2R}x$$

Unlike several other assignments, this model has no algebraic loop. The mass matrix is diagonal, so each acceleration depends only on positions. The diagram is a direct translation of the equations.

Figure 2. Simulink structure (as in the paper): two integrator chains with cross-coupling gains k₁R/m₁ and 2k₁/(m₂R), and the constant input G/m₁.

The paper's parameters are $m_1 = m_2 = 1$ kg, $R = 1$ m, $k_1 = k_2 = 5$ (N/m) and $G = 9.81\,m_1$ N. The paper states that "the system is oscillatory" and that, because the components are idealized (no damping, massless springs, no bearing friction), the oscillations never decay. The plots themselves are not included in the document.

Critical review and corrected model

Gravity has the wrong sign

With $x$ measured downward, the gravitational potential energy of $m_1$ is $-Gx$. The paper's $+Gx$ makes gravity push the mass up. In this linear system that does not change the frequencies, but it moves the equilibrium to the wrong side. It also leads to a configuration that a rope cannot sustain (see below).

A rope can only pull

The model treats the rope as an inextensible link that never goes slack. That holds only while both rope segments are in tension:

$$T_1 = k_1(x - R\theta) \ge 0\quad(\text{right side}),\qquad T_2 = k_2R\theta \ge 0\quad(\text{left side})$$

With the paper's sign, the static equilibrium has both springs compressed ($T_1 = T_2 = -G$). The rope would have to push, which is physically impossible. With the correct sign both springs carry the weight in tension.

Corrected equations

$$\begin{aligned} m_1\ddot x + k_1x - k_1R\theta &= G\\ \tfrac12 m_2R^2\ddot\theta + R^2(k_1+k_2)\theta - k_1Rx &= 0\end{aligned}$$

Static equilibrium

Setting the accelerations to zero gives $T_1 = T_2 = G$. The weight hangs from the two springs in series, through the pulley:

$$R\theta^* = \frac{G}{k_2},\qquad x^* = \frac{G}{k_1} + \frac{G}{k_2}$$

For the paper's parameters: $x^* = 3.92$ m and $\theta^* = 1.96$ rad. These are soft springs.

Normal modes

$$\mathbf M = \begin{bmatrix} m_1 & 0\\ 0 & \tfrac12 m_2R^2\end{bmatrix},\qquad \mathbf K = \begin{bmatrix} k_1 & -k_1R\\ -k_1R & R^2(k_1+k_2)\end{bmatrix},\qquad \det(\mathbf K - \omega^2\mathbf M) = 0$$

For the paper's values, $\omega_1 = 1.48$ rad/s (mass and pulley moving together) and $\omega_2 = 4.78$ rad/s (mass and pulley in opposition). The ratio is not an integer, so a general response is a beat-like mixture of the two frequencies: the "oscillatory" behaviour the paper reports.

A subtle point: releasing from the unstretched position

If the mass is released with both springs unstretched ($x=\theta=0$), the response swings around the equilibrium. In this case the left rope tension $T_2$ briefly drops to about −2 N (θ becomes slightly negative), so the left rope would actually go slack for a moment and the linear model is not valid then. Starting from equilibrium with a small extra pull avoids this. The simulator shows both tensions and warns when either becomes negative.

Interactive simulator

The simulator solves the equations numerically (4th-order Runge–Kutta, 1 ms step). Initial positions are measured from the unstretched state. An optional viscous friction $c$ in the pulley bearing (mentioned in the paper's conclusion) adds a torque $-c\dot\theta$.

Cases

Model

Also show the other model

System parameters

m₁ [kg]

m₂ (pulley) [kg]

R [m]

k₁ [N/m]

k₂ [N/m]

g [m/s²]

c (bearing) [N·m·s]

Initial conditions (from unstretched)

x(0) [m]

θ(0) [rad]

ẋ(0) [m/s]

θ̇(0) [rad/s]

Duration [s]

Displacements: mass x(t) and rope end Rθ(t)

Rope tensions T₁ (right, spring k₁) and T₂ (left, spring k₂)

The model is valid only while both stay ≥ 0 (a rope cannot push).

Energy balance (selected model)

Animation (selected model)

Results table (every 0.1 s)

Review questions

Why does the no-slip condition remove one coordinate?

Without slip, every point of the rope in contact with the pulley moves with the rim, so the rope ends move by exactly $\pm R\theta$. The elongation of $k_2$ is then not independent: $x_2 = R\theta$.

How do you check the sign of the gravity term?

Ask what happens when the mass moves a little in the positive direction of its coordinate. Here $x$ points down, the mass descends and its potential energy decreases, so the term must be $-m_1gx$. In the equation of motion gravity then appears as $+G$ on the right-hand side, a force in the direction of increasing $x$.

Why are the two rope tensions different during motion?

The pulley has inertia. Its equation of motion is $I\ddot\theta = R(T_1 - T_2)$, so the tensions are equal only when the pulley does not accelerate, for example in static equilibrium, where $T_1 = T_2 = G$.

Does gravity change the natural frequencies?

No. In this linear system gravity is a constant force; it only shifts the equilibrium (by $G/k_1 + G/k_2$ for the mass). The frequencies depend only on $\mathbf M$ and $\mathbf K$.

When is the linear model invalid?

Whenever a rope segment would have to carry compression ($T<0$). The rope then goes slack, the no-slip relation no longer holds, and the dynamics change. A real analysis would need a piecewise model.

What do the two mode shapes look like?

In the slow mode the mass and the rope end move in the same direction (the pulley follows the mass). In the fast mode they move in opposite directions and spring $k_1$ is strongly stretched and compressed. Try the "Mode 1" and "Mode 2" presets.