I'm trying to compute the total angular momentum of a 3D multi-body, pendulum-like (as in, each body is connected to another one), mechanical system. Let us consider, for a simpler case, a 2D double pendulum. With some tips, I hope to transfer this to the more difficult 3D case myself.

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Now, I am aware that angular momentum can be computed as $$L = I\omega$$ If we generalize and don't use the standard inertia equation for a beam rotating around its edge, for beam one, we can say $$ L_1 = (I_{1,com} + m_1(\bar{x}_{1,com} - \bar{x}_{origin})^2 )\cdot \dot{\phi}_1 $$ Now, issues arise when considering the second beam. I could simply add $$ L_2 = (I_{2,com} + m_2(\bar{x}_{2,com} - \bar{x}_{origin})^2) \cdot \dot{\phi}_2 $$ but I am not sure if this correct as I need to consider the beam rotating around the origin, as well as the beam rotating around 'itself'. So would I also need to add a term that is $$ L_{2,self} = (I_{2,com} + m_2(\bar{x}_{2,com} - \bar{x}_{1,pinjoint})^2) \cdot \dot{\phi}_2 $$ ? It seems like I'm missing some fundamental understanding so I would really wish to not only obtain the answer but also some reasoning/explaining behind it.

In short, is it $$ L_{total} = L_1 + L_2 $$ is it $$ L_{total} = L_1 + L_2 + L_{2,self} $$ or is it 'something inbetween'....

I appreciate all effort!

  • $\begingroup$ Isn't m2 is rotate about its connection to m1. Once m2 reaches the vertical position, m1 & m2 will rotate together about the origin? $\endgroup$
    – r13
    Aug 31, 2021 at 23:54
  • $\begingroup$ Could you elaborate how you derived and the quantities on $L_1 = I_{1,com}\cdot m_1(\bar{x}_{1,com} - \bar{x}_{origin})^2 \cdot \dot{\phi}_1$? It's not obvious to me. $\endgroup$
    – NMech
    Sep 1, 2021 at 2:07
  • $\begingroup$ @NMech Parallel axis theorem. The subscript com is probably centre of mass. $\endgroup$
    – AJN
    Sep 1, 2021 at 13:23
  • 1
    $\begingroup$ @AJN I still don't understand what is $I$ and $I_{com}$ and how that represents the parallel axis theorem. If $I_{com} $ is the center of mass mass moment of inertia, then -in my mind- it should have been $I= I_{com} + m_1\cdot (x_{1,com}-x_{origin})^2$. I.e. Addition not multiplication. As it is I can't get my head around it. $\endgroup$
    – NMech
    Sep 2, 2021 at 3:35
  • $\begingroup$ @NMech You are right. I didn't notice that it was multiplication. $\endgroup$
    – AJN
    Sep 2, 2021 at 11:58

1 Answer 1


I'm going to give an intuitive explanation.

The idea behind the parallel axis theorem, is that if you have a rigid body, that is rotating about a point on it, other than the center of mass, the equivalent moment of inertia is how an observer on the point about which the body is rotating, will see.

For our case, let's assume that we have an observer at the pin point. in his belief, the rod number 2 is rotating around the pin point. What the observer can't see is his own movement. Now if another observer at point O looks at the hole scenario, will see a different story. each point on the second rod has a 2 speed components, one due to the 2nd rod rotating around the pin point, and the second is the speed at which the entire rod 2 is also moving because of rod 1's motion.

The calculations must be from an observer on the origin. So $$L_1 = (I_{1,com} + m_1(\bar{x}_{1,com} - \bar{x}_{origin})^2 )\dot{\phi}_1$$

Is what the observer at the pin point sees, not the origin. This makes sense since the observer sees a different speed for those points, and assuming that $$L = r.P.sin(angle)$$

You can see that differing the value of the momentum (through changing the speed) changes the finalized value for angular momentum.

Now, why is the total angular momentum not equal to the sum of the two individual ones? think of this case, where the 2 rods are in one line, meaning that

$$\phi1 = \phi2$$

That way the rod looks like one rod with length l1 + l2, and mass m1 + m2. I'm not gonna even go into the maths, but you must know that the angular momentum has a meter squared in it's units, so if for l1 it has l1 squared, and for l2 it has l2 squared, for l1 + l2 it must have that, squared! so $$(l_1+l_2)^2 = l_1^2 + l_2^2 + 2l_1l_2$$

So you can see it does not only rely on the sum of the squared values, rather the squared of the sum... a bit hard to get a hang of!

So now that we are certain that the following does not apply: $$L_t = L_1 + L_2$$

one might ask what is the generalized expression for this setup of pendulums? we can be certain that the initial condition has a major role, since as shown, if the 2 angles are equal, the scenario is totally different in a another scenario (in fact so different it's used as a classic example for chaos).

For that, you need to write down expressions (in their vector form ofc) for the distance of points on the second rod from the origin, and their velocity, and from their you will arrive at your final expression. If there are issues there, probably edit and add here or post a new question.

  • $\begingroup$ I'm not sure if I understand. What is $r$ and $P$ in your second equation? Going from your explanation, it would seem that if $\phi_1 = \phi_2$, that at that instance $L_t = L_1 + L_2$ (as we can add the angular momentum for beam two as if beam 1 is not there, rotating with $\dot{\phi}_2$ around the origin. Perhaps if you could write out what you mean in the equation for the two-pendulum setup, I'll be able to understand better $\endgroup$
    – J.V.
    Sep 24, 2021 at 3:02

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