Planetary gear train in a vertical plane

How to go from a sketch of the mechanism to a nonlinear differential equation of motion: choosing the generalized coordinate, writing the energies, applying the Lagrange equation, building a Simulink model and analysing the response.

Problem statement

A planetary gear train lies in a vertical plane. Gear 2 (radius $r_2$, weight $G_2$) meshes with the internal teeth of the fixed gear 1 (radius $r_1$) and rolls inside it without slipping.

Gear 2 is driven by the crank $OA$ (weight $G$), which is acted upon by a couple of moment $M_0$. The crank is pinned at the fixed point $O$; its other end carries the axle of gear 2 at point $A$.

Assumptions:

gear 2 is a homogeneous circular disk,
crank $OA$ is a thin homogeneous rod,
friction is neglected (conservative system + external moment).

Find: define a generalized coordinate and derive the equation of motion of the system.

Figure 1. Fixed gear 1 (ring), gear 2, crank OA with midpoint C, and angle φ

Choosing the generalized coordinate

The mechanism has one degree of freedom: once the position of the crank is known, so is the position of the centre of gear 2, and, because it rolls without slipping, also how far the gear has turned. The crank angle $\varphi$ is therefore chosen as the generalized coordinate, measured from the horizontal $x$ axis, positive counter-clockwise.

The centre $A$ of gear 2 is always at distance $r_1-r_2$ from $O$, so the crank length is:

$$L_{\text{rod}} = \overline{OA} = r_1 - r_2,\qquad \overline{OC} = \frac{r_1-r_2}{2}$$

The spin angle of gear 2 (denoted $\Theta$ in the paper) is not an independent coordinate; it is expressed through $\varphi$ (see sections 3 and 7). This is the key idea of the Lagrangian approach: every energy is written in terms of a single coordinate $\varphi$ and its derivative $\dot\varphi$.

TerminologyThe paper uses the terminology of the course: attractive energy $\mathcal{A}$ = potential energy of the system, repulsive coenergy $\mathcal{R}^*$ = kinetic (co)energy. The Lagrangian is $\mathcal{L} = \mathcal{R}^* - \mathcal{A}$.

Energies of the system (as derived in the paper)

3.1 Attractive (potential) energy

Gravitational potential energy equals weight times the height of the centre of mass. The rod's centre of mass is at $C$ (height $y_C = \tfrac{r_1-r_2}{2}\sin\varphi$), and the gear's is at $A$ (height $y_A = (r_1-r_2)\sin\varphi$):

(1.2)$$\mathcal{A}_{\text{rod}} = G\,y_C = \tfrac12 G (r_1-r_2)\sin\varphi$$

(1.3)$$\mathcal{A}_{\text{gear}} = G_2\,y_A = G_2 (r_1-r_2)\sin\varphi$$

(1.4)$$\mathcal{A} = \mathcal{A}_{\text{rod}}+\mathcal{A}_{\text{gear}} = (r_1-r_2)\sin\varphi\left(G_2+\tfrac12 G\right)$$

3.2 Repulsive coenergy of the rod

The rod rotates about the fixed point $O$ with angular velocity $\dot\varphi$. The moment of inertia of a thin homogeneous rod about its end follows from integration ($dm = \rho\,dx$, $m=\rho L$):

$$I_O = \int_0^L x^2\rho\,dx = \rho\frac{L^3}{3} = \frac{mL^2}{3}$$

(1.6)$$\mathcal{R}^*_{\text{rod}} = \tfrac12 I_O\dot\varphi^2 = \frac12\cdot\frac{G}{g}\frac{(r_1-r_2)^2}{3}\dot\varphi^2 = \frac16\frac{G}{g}(r_1-r_2)^2\dot\varphi^2$$

3.3 Repulsive coenergy of the gear

Gear 2 undergoes planar motion: its centre $A$ moves along a circle ("revolution", like the Earth around the Sun) while the disk simultaneously spins about its own axis ("rotation"). Its kinetic energy is therefore the sum of a translational and a rotational part:

(1.7)$$\mathcal{R}^*_{\text{gear}} = \mathcal{R}_{\text{gear}}^{(T)} + \mathcal{R}_{\text{gear}}^{(R)}$$

The velocity of the centre is $v_A = \dot\varphi\,(r_1-r_2)$. The paper writes the translational part as:

(1.7.1)$$\mathcal{R}_{\text{gear}}^{(T)} = m_{\text{gear}}\,v_A^2 = \frac{G_2}{g}(r_1-r_2)^2\dot\varphi^2 \quad\color{#c27a00}{\text{(the factor }\tfrac12\text{ is missing, see sec. 7)}}$$

The moment of inertia of a homogeneous disk about its central axis is $I_{\text{gear}} = \tfrac12 m r_2^2 = \tfrac12\frac{G_2}{g}r_2^2$. For the relation between the spin angle $\Theta$ and $\varphi$, the paper uses equal arc lengths $r_2\Theta = r_1\varphi$:

$$\Theta = \frac{r_1}{r_2}\varphi,\qquad \dot\Theta = \frac{r_1}{r_2}\dot\varphi$$

(1.7.2)$$\mathcal{R}_{\text{gear}}^{(R)} = \tfrac12 I_{\text{gear}}\dot\Theta^2 = \frac12\cdot\frac12\frac{G_2}{g}r_2^2\left(\frac{r_1}{r_2}\dot\varphi\right)^2 = \frac14\frac{G_2}{g}r_1^2\dot\varphi^2$$

(1.7′)$$\mathcal{R}^*_{\text{gear}} = \frac{G_2}{g}\left[(r_1-r_2)^2 + \tfrac14 r_1^2\right]\dot\varphi^2$$

3.4 Total coenergy and the Lagrangian

(1.8)$$\mathcal{R}^* = \frac16\frac{G}{g}(r_1-r_2)^2\dot\varphi^2 + \frac{G_2}{g}\left[(r_1-r_2)^2+\tfrac14 r_1^2\right]\dot\varphi^2$$

(1.9)$$\mathcal{L} = \mathcal{R}^*-\mathcal{A} = \frac16\frac{G}{g}(r_1-r_2)^2\dot\varphi^2 + \frac{G_2}{g}\left[(r_1-r_2)^2+\tfrac14 r_1^2\right]\dot\varphi^2 - (r_1-r_2)\sin\varphi\left(G_2+\tfrac12G\right)$$

Lagrange equation and equation of motion

The moment $M_0$ is an external (non-conservative) input and enters as the generalized force associated with the coordinate $\varphi$ (its virtual work is $M_0\,\delta\varphi$):

(2.1)$$\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial\dot\varphi}\right) - \frac{\partial\mathcal{L}}{\partial\varphi} = M_0$$

Required derivatives:

(2.2)$$\frac{\partial\mathcal{L}}{\partial\dot\varphi} = \left\{\frac13\frac{G}{g}(r_1-r_2)^2 + \frac{2G_2}{g}\left[(r_1-r_2)^2+\tfrac14r_1^2\right]\right\}\dot\varphi$$

(2.3)$$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot\varphi} = \left\{\frac13\frac{G}{g}(r_1-r_2)^2 + \frac{2G_2}{g}\left[(r_1-r_2)^2+\tfrac14r_1^2\right]\right\}\ddot\varphi$$

(2.4)$$\frac{\partial\mathcal{L}}{\partial\varphi} = -(r_1-r_2)\cos\varphi\left(G_2+\tfrac12G\right)$$

Substituting into (2.1), the paper obtains the equation of motion:

(2.5)$$\underbrace{\left\{\frac{G}{3g}(r_1-r_2)^2 + \frac{2G_2}{g}\left[(r_1-r_2)^2+\frac{r_1^2}{4}\right]\right\}}_{J_{\text{paper}}}\ddot\varphi \;-\; (r_1-r_2)\cos\varphi\left(G_2+\frac{G}{2}\right) = M_0$$

This is a second-order nonlinear differential equation (the nonlinearity comes from $\cos\varphi$). It has the same form as the equation of a physical pendulum: $J\ddot\varphi \pm c\cos\varphi = M_0$, where $J$ is the equivalent moment of inertia and $c = (r_1-r_2)(G_2+G/2)$ is the amplitude of the gravitational moment.

For Simulink, the equation is solved for the highest derivative:

(2.6)$$\ddot\varphi = \underbrace{\frac{1}{J_{\text{paper}}}}_{\to K_1}M_0 \;+\; \underbrace{\frac{(r_1-r_2)\left(G_2+\frac{G}{2}\right)}{J_{\text{paper}}}}_{\to K_2}\cos\varphi$$

CautionThe sign in front of the $\cos\varphi$ term in (2.5) is wrong: $-\partial\mathcal{L}/\partial\varphi$ gives a plus. The coefficient $J_{\text{paper}}$ also contains two errors from section 3. Everything is explained and corrected in section 7. The simulator offers both models.

Model in MATLAB / Simulink

Equation (2.6) is implemented with the classic two-integrator scheme: the input of the first integrator is $\ddot\varphi$, its output is $\dot\varphi$, and the output of the second is $\varphi$. From $\varphi$ we compute $\cos\varphi$, multiply it by $K_2$ and feed it back to the summing junction, where it is added to $K_1 M_0$.

Figure 2. Simulink diagram from the paper (simplified). The initial condition φ(0) is set in block Integrator1, φ̇(0) in block Integrator.

$$K_1 = \frac{1}{J_{\text{paper}}}\ (\text{multiplied by } M_0),\qquad K_2 = \frac{(r_1-r_2)\left(G_2+\frac{G}{2}\right)}{J_{\text{paper}}}$$

The constants are not written inside the blocks because the expressions are long; they are precomputed in the MATLAB workspace.

Parameter	Value in the paper
$r_1$	15
$r_2$	8
$G$	20
$G_2$	12
$g$	10
$J_{\text{paper}}$, $K_2$	285.27 · 0.540 s⁻²

Simulation results from the paper

The paper presents three analyses. The button under each case loads the same conditions into the simulator (the "paper" model), so the plots can be compared directly with the originals.

Case 1 · equilibrium

$M_0=0$, $\varphi(0)=-\pi/2$, $\dot\varphi(0)=0$. The gear is at its lowest position and the system stays at rest: $\varphi(t)=-\pi/2$ at all times. The paper states that $\varphi=+\pi/2$ is an unstable equilibrium.

Case 2 · free oscillations

$M_0=0$, $\varphi(0)=0$, $\dot\varphi(0)=0$. The crank is released from the horizontal position; the system oscillates like a large-amplitude pendulum, undamped (no friction). In the paper's plot $\varphi$ goes from 0 to $+\pi$, with a period of ≈ 10–12 s.

Case 3 · pulse excitation

$M_0$ is a pulse train (period 2 s, 50 % duty), $\varphi(0)=0$. While a pulse acts, the angle grows faster; between pulses the system keeps moving on its stored energy. The angle reaches ≈ 13.5 rad in 10 s – the crank goes into continuous rotation.

Reconstructing case 3In the paper's plot the moment is drawn with amplitude 1 (normalized). With $M_0=1$ the system could not pass over the top position (the gravitational moment reaches $c=154$). A numerical reconstruction shows that the paper's response (≈1.3 rad at 2 s; 3.8 at 4 s; 5.8 at 6 s; 9.5 at 8 s; 13.5 at 10 s) corresponds to a pulse amplitude of ≈ 112, which is the value used in the simulator.

Critical review: three errors and the corrected model

The procedure in the paper is methodologically sound (choice of coordinate, energies, Lagrange equation, Simulink), but the derivation contains three errors. They are good examples of what to watch out for and can be used as a "spot the mistake" exercise.

Missing factor ½ in the translational energy

Translational kinetic energy is $\tfrac12 m v^2$, not $m v^2$. Correct:

$$\mathcal{R}^{(T)}_{\text{gear}} = \tfrac12\frac{G_2}{g}(r_1-r_2)^2\dot\varphi^2$$

Wrong angular velocity of the gear

The relation $\dot\Theta = \tfrac{r_1}{r_2}\dot\varphi$ gives the rotation of the gear relative to the crank. The kinetic energy requires the absolute angular velocity. The contact point $P$ between gear 2 and the fixed gear 1 is the instantaneous centre of zero velocity ($v_P=0$), so:

$$|\omega_2| = \frac{v_A}{r_2} = \frac{r_1-r_2}{r_2}\dot\varphi,\qquad \omega_2 = -\frac{r_1-r_2}{r_2}\dot\varphi \;\Rightarrow\; \omega_2-\dot\varphi = -\frac{r_1}{r_2}\dot\varphi\ \ (\text{relative, as in the paper})$$

$$\mathcal{R}^{(R)}_{\text{gear}} = \frac12\cdot\frac12\frac{G_2}{g}r_2^2\left(\frac{r_1-r_2}{r_2}\right)^2\dot\varphi^2 = \frac14\frac{G_2}{g}(r_1-r_2)^2\dot\varphi^2$$

Sanity check: as $r_2\to r_1$ the gear can neither move nor turn, so its energy must vanish – the correct expression gives that, while the paper's expression ($\tfrac14\tfrac{G_2}{g}r_1^2\dot\varphi^2$) does not.

Wrong sign of the gravity term

From (2.4), $-\partial\mathcal{L}/\partial\varphi = +(r_1-r_2)\cos\varphi(G_2+G/2)$, so (2.5) should have a plus. With the minus sign gravity would "pull upwards": the stable equilibrium would be $\varphi=+\pi/2$ (crank pointing up) and $-\pi/2$ would be unstable – the opposite of the physics and of the paper's own text.

The error also shows up in the results: for case 2 the text says the system oscillates between 0 and $-\pi$ (below the horizontal), while the Simulink plot shows 0 to $+\pi$ (above the horizontal). The plot faithfully reflects the model with the wrong sign. In case 1 the system stays at $-\pi/2$ only because it is numerically exactly in equilibrium – the slightest perturbation would make it fall away (try $\varphi(0) = -\pi/2+0.001$ in the simulator).

Corrected equation of motion

$$\mathcal{R}^* = \frac12\cdot\frac{(r_1-r_2)^2}{g}\left(\frac{G}{3}+\frac{3G_2}{2}\right)\dot\varphi^2,\qquad \mathcal{A} = (r_1-r_2)\left(G_2+\frac{G}{2}\right)\sin\varphi$$

$$\boxed{\;\frac{(r_1-r_2)^2\,(2G+9G_2)}{6g}\,\ddot\varphi + (r_1-r_2)\,\frac{G+2G_2}{2}\,\cos\varphi = M_0\;}$$

For the paper's parameters: $J = 120.87$ (instead of 285.27), $c = 154$. The period of small oscillations about $\varphi=-\pi/2$ is $T_0 = 2\pi\sqrt{J/c} \approx 5.57$ s. With a constant moment the equilibrium satisfies $\cos\varphi^* = M_0/c$, so for $M_0 > c$ no equilibrium exists and the crank rotates continuously.

In the Simulink diagram this means: $K_1 = 1/J$ and negative feedback $-K_2\cos\varphi$ (a "−" sign on the second input of the summing junction).

Interactive simulator

The equation of motion is solved numerically (4th-order Runge–Kutta, 2 ms step). Change the parameters and look at the response, the phase portrait, the energy balance and the animation of the mechanism. Angle inputs also accept expressions such as -pi/2 or pi/4+0.01.

Examples from the paper

Model

Also show the other model (comparison)

System parameters

r₁ [m]

r₂ [m]

G (crank) [N]

G₂ (gear) [N]

g [m/s²]

Initial conditions

φ(0) [rad]

φ̇(0) [rad/s]

Duration [s]

Input M₀(t)

Shape

Amplitude [N·m]

Period [s]

Pulse width [%]

Delay [s]

Angle φ(t)

Input M₀(t)

Angular velocity φ̇(t)

Phase portrait φ̇(φ)

Energy balance (selected model)

Without friction, $\mathcal{R}^*+\mathcal{A} - (\mathcal{R}^*+\mathcal{A})_0 = W_{M_0}$ must hold – a good check of the numerics.

Mechanism animation

Results table (every 0.5 s)

Review questions

Why does the system have only one degree of freedom?

The position of centre $A$ is determined by the crank angle ($\overline{OA}=r_1-r_2$ is constant), and rolling without slipping ties the gear's rotation to $\varphi$. All coordinates can be expressed through a single parameter.

How does the moment $M_0$ enter the Lagrange equation?

As the generalized force $Q_\varphi$: the virtual work of the moment is $\delta W = M_0\,\delta\varphi$, so $Q_\varphi = M_0$, which appears on the right-hand side of equation (2.1).

What are the equilibrium positions for $M_0=0$, and which one is stable?

From $\cos\varphi=0$: $\varphi=\pm\pi/2$. Physically, $-\pi/2$ is stable (gear at the bottom, minimum of potential energy) and $+\pi/2$ is unstable. The corrected model reproduces this; the paper's model, because of the sign error, gives the opposite.

Why do the oscillations in case 2 not decay?

Friction is neglected, so the sum of kinetic and potential energy is constant. Adding a term $b\dot\varphi$ (viscous friction) would give damped oscillations about $-\pi/2$.

What constant value of $M_0$ is needed for the crank to rotate continuously?

An equilibrium exists as long as $|M_0/c|\le 1$. For $M_0 > c = (r_1-r_2)(G_2+G/2)$ (for the paper's parameters $c=154$ N·m) there is no equilibrium and the crank rotates continuously. Check it in the simulator with a constant moment of 150 and 160.

Why is the period in case 2 longer than $2\pi\sqrt{J/c}$?

That formula holds only for small oscillations. Releasing from the horizontal means an amplitude of $90^\circ$; the period of a nonlinear pendulum grows with amplitude (for $90^\circ$ it is about 18 % longer).

Parameter	Value in the paper
\(r_1\)	15
\(r_2\)	8
\(G\)	20
\(G_2\)	12
\(g\)	10
\(J_{\text{paper}}\), \(K_2\)	285.27 · 0.540 s⁻²

Explanation of Method

Problem statement

Choosing the generalized coordinate

Energies of the system (as derived in the paper)

3.1 Attractive (potential) energy

3.2 Repulsive coenergy of the rod

3.3 Repulsive coenergy of the gear

3.4 Total coenergy and the Lagrangian

Lagrange equation and equation of motion

Model in MATLAB / Simulink

Simulation results from the paper

Case 1 · equilibrium

Case 2 · free oscillations

Case 3 · pulse excitation

Critical review: three errors and the corrected model

Missing factor ½ in the translational energy

Wrong angular velocity of the gear

Wrong sign of the gravity term

Corrected equation of motion

Interactive simulator

Examples from the paper

System parameters

Initial conditions

Input M₀(t)

Review questions