In the images below I have drawn a double pendulum system with the pivot hub and a stopper A which is imagined to be fixed to some large wall. In both scenarios 1 and 2 the system is initially at rest and there is a wedge and square stopper fixed at the end of the proximal arm to prevent clockwise rotation.

Both frictionless pivots and stoppers are assumed to have negligible mass.

The proximal arm is a rigid rod whose mass is w and not negligible.

The peripheral arm is assumed to be a thin rigid rod with a non negligible mass m attached to its end.

Let's assume that both systems are set in motion by the same torque T applied at the central pivot hub.

I have attempted to draw all the forces when the system is set in motion by torque T for both scenarios:

Tangential forces F1 and F2.

Centripetal forces C1 and C2.

Stopper reaction force R1 and R2 (at a point d/2) whose line of force will be through the stopper COM.

Weight of proximal and peripheral arms wg and mg.

Tension force S1 and S2 in the peripheral arm for scenarios 1 and 2.

Question: For both scenarios 1 and 2 , how would I be able to determine the angle the proximal arm makes with the +x axis at the exact point when the peripheral arm loses contact with the stoppers?

Please note that I'm not insisting on a mathematical proof and would be happy with just an intuitive explanation as to why the peripheral arm might lose contact with the stopper earlier/later (or maybe identical) relative to the position of the proximal arm angle made with the +x axis.

enter image description here

  • $\begingroup$ I added the tension forces S1 and S2 so edited my comment and diagram. I also added a comment in my original question to say that I'd be satisfied with just an intuitive explanation than some rigorous mathematical proof. This is not homework but related to my hobby studying the biomechanics of the golf swing. $\endgroup$
    – Dubious
    Commented Jan 26 at 17:59
  • $\begingroup$ How is the proximal arm connected to the double pendulum? $\endgroup$ Commented Jan 26 at 23:37
  • $\begingroup$ Hi John - its just another frictionless pivot P2 in the 1st diagram. So P1 pivot is fixed to a wall while P2 is just like the linked image .tutelman.com/golf/swing/golfSwingPhysics/swing.gif $\endgroup$
    – Dubious
    Commented Jan 28 at 13:51
  • 1
    $\begingroup$ The triangular and square wedges in scenario 1 and 2 are fixed to the proximal arm to stop the peripheral arm from jackknifing (rotating clockwise). $\endgroup$
    – Dubious
    Commented Jan 28 at 14:10
  • $\begingroup$ None of the scenarios above can be at rest. This is because the combined center of mass needs to be directly under the pivot for the system to be at rest. I feel like there is more to this that is not shared. $\endgroup$ Commented Jan 29 at 16:28

1 Answer 1


My interpretation of your question is as follows:


We do a static analysis to find the stopper forces $R_x$ and $R_y$ when a torque $\tau_1$ is applied.

  • Proximal Arm

    $$ \begin{aligned} Fx_1 - R_x - Fx_2 & = 0 \\ Fy_1 - R_y - Fy_2 - m_1 g & = 0 \\ \tau_1 + (r_1-d_1) R_x - r_1 Fx_2 &= 0 \\ \end{aligned}$$

  • Peripheral Arm

    $$ \begin{aligned} R_x + Fx_2 & = 0 \\ Fy_2 + R_y - m_2 g &= 0 \\ d_1 \tan \beta R_y - \ell_2 m_2 g \tan \beta + d_1 R_x &= 0 \end{aligned}$$

The static solutions to the above force balance equations are

$$\begin{aligned} Fx_1 &= 0 \\ Fy_1 &= (m_1 + m_2) g \\ Fx_2 & = \frac{\tau_1}{d_1-2 r_1}\\ Fy_2 & = m_2 g + \frac{\tau \cot \beta}{d_1-2 r_1} - \frac{\ell m_2 g}{d_1} \\ R_x & = \frac{\tau_1}{d_1 - 2 r_1}\\ R_y & = \frac{\tau_1 \cot \beta}{d_1 - 2 r_1} + \frac{\ell_2 m_2 g}{d_1} \\ \end{aligned}$$

The conditions for the stop to be active are $R_x>0$ and $R_y>0$

But once motion is involved the above equations are no longer valid, because you have to consider the dynamics of the situation (acceleration of the center of mass) and the general solution would be too long to write out here.


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