# Coupling of angular velocities

TubbaBlubba

## Homework Statement

We consider a pendulum of length L hanging along the z-axis with a mass (taken to be unity) at the end, attached to an arm of length R free to rotate about the z-axis but restricted to the xy-plane. The system is completely described by the angle of the pendulum rod with respect to the z-axis (theta), and the angle between the arm with respect to the x-axis (phi), and their velocities.

The objective is to derive the mass' equations of motion when subject to gravity and I'd prefer to use Lagrangian formalism; my problem in doing this is specifically is coupling the angular velocity vectors to reduce degrees of freedom (the angle phi disappears). I have spent a completely absurd amount of time on this that I really need for other things, attacking the problem in any number of ways.

## Homework Equations

The most important quantities appear to be the angular momentum about the z-axis, which is given by:
##A_z = \dot{\phi}(R^2 + L^2\sin^2\theta) + \dot{\theta}RL\cos\theta##, and the velocity squared:
##v^2 = \dot{\phi^2}(R^2 + L^2\sin^2\theta) + \dot{\phi}\dot{\theta}RL\cos\theta + \dot{\theta^2}L^2##
(Both of these I have painstakingly derived using cartesian coordinates and confirmed in various other ways).

## The Attempt at a Solution

The equations are nice and neat and precisely what you would expect, but solving the Euler-Lagrange equations is impossible if we cannot differentiate ##\dot{φ}##. The line of reasoning that has made the most sense to me so far is this: Any acceleration in the "theta direction" should by geometric argument cause an acceleration ##\ddot{\theta}(L/R)\cos\theta## in the phi direction, and the centripetal acceleration perpendicular to the z-axis (i.e. the force conserving angular momentum) an acceleration ##-\dot{\theta^2}(L/R)\sin\theta##, giving us ##\ddot{\phi} = \frac{L}{R}(\ddot{\theta}\cos\theta - \dot{\theta^2}\sin\theta)##, with the merciful solution ##\dot{\phi} = \frac{L}{R}(\dot{\theta}\cos\theta) + C##. While the EoM looked pretty sensible, solving it using these results resulted in one among many utter abominations. My basic test is just to "shut off" gravity and see if conserved quantities are conserved, which none of my attempts have passed so far.

What am I missing? I'm sure it's something relatively elementary and that there is a much simpler way of solving this (Total angular momentum?). I have done every single differentiation of every single quantity I have been able to think of (to exaggerate slightly), but nothing really works. Or is this reasoning correct? (In that case, it could be a mistake in my solution of the EL equations or my implementation.).

Last edited by a moderator:

## Answers and Replies

Related Introductory Physics Homework Help News on Phys.org
haruspex
Homework Helper
Gold Member
a pendulum
with a mass
attached to an arm
What is attached to the arm, the mass or the top of the pendulum, or...?

TubbaBlubba
What is attached to the arm, the mass or the top of the pendulum, or...?
Ah, sorry, not the easiest contraption to describe! There is an arm, length R, free to rotate about the origin in the xy-plane (rotation specified by angle phi). From the end of this arm hangs a simple pendulum of rod length L, so the mass is a distance L from the end of the arm, and the pendulum's axis of rotation is parallel to the arm, ie, the pendulum is always perpendicular to the arm, and its axis of rotation can lie anywhere in the xy-plane. When the angle theta is 0, the arm is parallel to the z-axis, the z-coordinate of the mass is -L. So gravity points in the negative z direction.

Um, not sure that made it clearer... I think the answer to your question would be "the top of the pendulum".

haruspex
Homework Helper
Gold Member
Ok, that's much clearer.
For v2, I get ##\dot\theta^2L^2\cos^2(\theta)+2\dot\theta\dot\phi LR\cos(\theta)+\dot\phi^2(R^2+L^2\sin^2(\theta))##.
I did not understand your reasoning to get ##\ddot\phi##. Doesn't conservation of angular momentum immediately give you an equation expressing ##\dot \phi## in terms of ##\theta## and its derivatives? It looks different from the equation you wrote.

TubbaBlubba
TubbaBlubba
Ok, that's much clearer.
For v2, I get ##\dot\theta^2L^2\cos^2(\theta)+2\dot\theta\dot\phi LR\cos(\theta)+\dot\phi^2(R^2+L^2\sin^2(\theta))##.
I did not understand your reasoning to get ##\ddot\phi##.
How do you get the term ##\dot\theta^2L^2\cos^2(\theta)##? We get ##\dot\theta^2L^2\sin^2(\theta)## from ##\dot{z^2}## plus ##\dot\theta^2L^2\cos^2(\theta)## from #\dot{x^2} + \dot{y^2}##. You are right about the factor 2 in front of the cross term however, no idea how I missed that...
Doesn't conservation of angular momentum immediately give you an equation expressing ##\dot \phi## in terms of ##\theta## and its derivatives? It looks different from the equation you wrote.
I feel like I'm not thinking super clearly here, but if we set angular momentum about z to zero, we get
##\dot{\phi} = -\frac{\dot{\theta}RL\cos\theta}{R^2 + L^2\sin^2\theta}##.
Is that sufficient - can I just use this to get my derivatives without regard to what the initial angular momentum had been and whatever an external torque imposes? Won't I get an awkward term with the initial angular momentum over ##R^2 + L^2\sin^2\theta##? Similarly, if I take the time derivative of angular momentum... I did do an implementation using that relation, but it turned out not to conserve angular momentum. However, the derivatives with regard to theta are pretty headache-inducing, so I could have overlooked something, and I missed the factor 2 in the velocity squared.

I have gone in so many circles around this I think I've lost my direction a bit...

Last edited by a moderator:
haruspex
Homework Helper
Gold Member
How do you get the term ##\dot\theta^2L^2\cos^2(\theta)##?
I took xy coordinates along the arm line ##\phi=0## and perpendicular to that, origin at the z axis.
I found the coordinates to be ##L\sin(\theta)\cos(\phi)+R\sin(\phi)## and ##R\cos(\phi)-L\sin(\theta)\sin(\phi)##.
Differentiating, squaring and adding, the ##\dot\theta^2## terms all have a factor ##\cos^2(\theta)##.

TubbaBlubba
I took xy coordinates along the arm line ##\phi=0## and perpendicular to that, origin at the z axis.
I found the coordinates to be ##L\sin(\theta)\cos(\phi)+R\sin(\phi)## and ##R\cos(\phi)-L\sin(\theta)\sin(\phi)##.
Differentiating, squaring and adding, the ##\dot\theta^2## terms all have a factor ##\cos^2(\theta)##.
Right, but we have the z-velocity squared as well, which has a factor ##sin^2(\theta)##, so we just end up with ##L^2##.

Am I making a mistake in thinking about this as a three-dimensional problem? Of course the z is completely determined by theta and its time derivative...

TubbaBlubba
OK, right, my expression for phi-dot in terms of theta and derivatives needs to have a factor for the initial value of the angular momentum. That's something I've overlooked. And of course gravity exerts no torque about the z-axis, no need to worry about that.

But can I really stick to kinetic energy in the xy-plane?

EDIT: No matter how I go about this, the (R^2 - L^2 sin2(theta)) term in the denominator with the cosine in the numerator gives me absurdly complex derivatives with respect to theta. I simply struggle to believe that the equations of motion could be this complex...

EDIT: I looked back at the original problem formulation. I solved this earlier (I had a lot of time to do it), interpreting the problem formulation as the limiting case of the length of the arm R being 0. This greatly simplifies the equations of motion, but it results in coordinate singularities around theta = 0, which is computationally awkward (phi-dot tries to go to infinity while angular momentum tries to stay constant) which is why I wanted to find the general equations for any R. But with how complicated that gets, it really looks like the intended task takes R = 0 for granted (I picked it from a list of problems - it doesn't have the clearest illustration in the world, but they do specify a "hinge" rather than an arm).

Last edited by a moderator:
TSny
Homework Helper
Gold Member
If the horizontal arm has no inertia, then the behavior of the system might be a little odd. For ##\theta \neq 0##, there will be a torque about the z-axis on the arm which would produce infinite angular acceleration of the arm.

I think your expressions for the angular momentum about the z-axis and for ##v^2## in your first post are correct except for the missing factor of 2 in ##v^2## that was pointed out by haruspex. It doesn't surprise me that the differential equations are impossible to solve analytically, except for the solution where ##\theta = 0## and ##\dot{\phi} = ## constant.

TubbaBlubba
If the horizontal arm has no inertia, then the behavior of the system might be a little odd. For ##\theta \neq 0##, there will be a torque about the z-axis on the arm which would produce infinite angular acceleration of the arm.

I think your expressions for the angular momentum about the z-axis and for ##v^2## in your first post are correct except for the missing factor of 2 in ##v^2## that was pointed out by haruspex. It doesn't surprise me that the differential equations are impossible to solve analytically, except for the solution where ##\theta = 0## and ##\dot{\phi} = ## constant.
They are not just impossible to solve analytically - merely writing down the force equation to solve it numerically is a monumental task for R =/= 0. I cannot imagine that the author of the problems expected me to do that, since that would be far harder than solving one of the "especially difficult" problems I have solved earlier.

So I did the derivation for the artificial R = 0 case and punched that through MATLAB as I had done earlier (You get the angular momentum, and some cotangents and cosecants). Yep, if you try to solve it as two coupled differential equations, it behaves fine until theta = 0, at which point the angular momentum explodes and it all gets absurd. However, in this case, ##\dot{\phi}## only depends on ##\theta##, so at least we only need to nudge our way past one singularity - there's no chaotic behaviour. It actually behaves... kind of reasonably. We get weird spikes in ##\dot{\phi}##, but they are so minor that the motion, overall, is fairly "realistic".

TSny
Homework Helper
Gold Member
"Find the equations of motion" is often interpreted to mean find the differential equations for the motion. So, you would not be expected to solve the differential equations. However, there are people who interpret "equations of motion" as being the solutions of the differential equations.

TubbaBlubba
"Find the equations of motion" is often interpreted to mean find the differential equations for the motion. So, you would not be expected to solve the differential equations. However, there are people who interpret "equations of motion" as being the solutions of the differential equations.
I understand what you are saying - the motive here is to find the differential equations and solve them numerically. The thing is, the Euler-Lagrange equation itself for R =/= 0 (where phi-dot couples to theta AND theta-dot) ended up being absurdly complex to the degree where any analysis of them would be a hopeless endeavour. It would reduce to an exercise in writing down an exceptionally nasty force equation and performing the Herculean task of correctly expressing it in both LaTeX and MATLAB, which isn't the objective of the course (or so I hope).

The R=0 case is contrived and not very useful, but it is an interesting object of study as an equation of motion.

TSny
Homework Helper
Gold Member
How do you define ##\phi## when ##R = 0##?

TubbaBlubba
How do you define ##\phi## when ##R = 0##?
The angle of the axis of rotation for ##\theta## relative to the x-axis.

ETA: Hm, the equations are behaving a bit more explosively than last time I tried. I'm going to have to recheck it all. I really need to learn to save my intermittent work...

ETA 2: Yeah, the infinite impulse at theta = 0 seems to screw it all up no matter how I fudge with the parameters. Ugh, I feel like I'm back at square one. I'm definitely missing something here...

Could I derive some kind of approximate enough EqM by considering a small R or something? Oh dear... It's not like there's a rush, but I have an exam I'd like to study for too, and this stupid pendulum thing is driving me crazy...

Last edited by a moderator:
TSny
Homework Helper
Gold Member
The angle of the axis of rotation for ##\theta## relative to the x-axis.
.
OK, I see.

For ##R = 0## do you get a differential equation for ##\theta## of the form ##\ddot{\theta} = \frac{A_z^2}{L^4} \cot(\theta)\csc^2(\theta) - \frac{g}{L}\sin(\theta)##?

TubbaBlubba
TubbaBlubba
OK, I see.

For ##R = 0## do you get a differential equation for ##\theta## of the form ##\ddot{\theta} = \frac{A_z}{L^2} \cot(\theta)- \frac{g}{L}\sin(\theta)##?
No, I get an additional term from differentiating ##A_z/(L^2\sin^2\theta) = \dot\phi## wrt ##\theta##, so ##\ddot{\theta} = \frac{A_z^2}{L^4}(\cot(\theta)(1 - 2\csc^2\theta) - \frac{g}{L}\sin(\theta)##.

Um... if a velocity is coupled to another coordinate, I should differentiate that velocity with respect to that coordinate, right?

TSny
Homework Helper
Gold Member
I forgot to square a quantity. I'm now getting my corrected equation in post #15. Not quite what you are getting. I'll go back and check over my work.

TubbaBlubba
TubbaBlubba
I forgot to square a quantity. I'm now getting my corrected equation in post #15. Not quite what you are getting. I'll go back and check over my work.
I missed a square as well, corrected now.

No, it's my mistake. I differentiated the same quantity twice instead of cancelling terms.

EDIT: Yep, it works now, and behaves fine as long as L is reasonably long (relative to g), and the initial angular momentum isn't TOO small.

Thank you!

Last edited by a moderator:
TSny
Homework Helper
Gold Member
Yep, it works now, and behaves fine as long as L is reasonably long (relative to g), and the initial angular momentum isn't TOO small.
I don't find any problems with small L or Az when R = 0.

For L = 1 m and Az = 0.2 kg m2/s I get this behavior

For L = 1 m and Az reduced to 0.02 kg m2/s I get

For these graphs I let ##\theta_0 = 0.5## rad, ##\dot{\theta}_0 = 0##.

Last edited:
haruspex
Homework Helper
Gold Member
but we have the z-velocity squared as well
Ahem, yes.

TubbaBlubba
I don't find any problems with small L or Az.

For L = 1 m and Az = 0.2 kg m2/s I get this behavior
View attachment 110957

For L = 1 m and Az reduced to 0.02 kg m2/s I get
View attachment 110958

For these graphs I let ##\theta_0 = 0.5## rad, ##\dot{\theta}_0 = 0##.
Yep, I looked a bit more closely at the parameter space. If you plot it using cartesian coordinates though, you'll see what I mean. The motion goes from smooth to "jerky" rapidly (The sines and cosines of phi go all over the place due to its "stepping" behaviour as in your graph). But indeed, the simulation doesn't break down.

Weird though, those sharp corners in the motion look so unphysical... I might have to see if I can make an animation of it.

EDIT: The animations are very mesmerizing.

Last edited by a moderator: