enter image description here

So I'm really stumped with this. I have a particle in a cone, like pictured. The particle orbits the z axis on the dotted line for $r$.

So knowing that $\alpha$ and $r$ remain constant in this motion, because as long as the particle is in the cone $\alpha$ won't change and as long as the particle stays in the same $z$ coordinate $r$ won't change.

So this motion would similar to a 3D Pendulum that has a constant $z$ (the height doesn't change)

Would the kinetic energy for the particle be: $$ T=\frac12mr^2\dot\theta^2 $$ or would it be $$T=\frac12mr^2\dot\theta^2\sin^2(\alpha)?$$

I know the potential would be: $$U=-mgr\cos(\alpha)$$

So the Lagrangian would be: $$L=T-U$$

Can anyone check me? I get really confused on polar coordinates.

  • $\begingroup$ Welcome to Physics! Please note that Physics.StackExchange is not a homework help site. Please see this Meta post on asking homework-like questions and this Meta post for "check my work" problems $\endgroup$
    – Kyle Kanos
    Mar 4, 2015 at 4:02
  • $\begingroup$ Is the constant height part of the problem or are you assuming it? If the particle is placed in circular motion it can stay there, of course, but you can also imagine putting it on a sort of wobbly orbit where it oscillates in $z$ $\endgroup$
    – zeldredge
    Mar 4, 2015 at 4:09
  • $\begingroup$ Sorry I didn't know about the Meta, I'll go there next time for sure! @SirElderberry, the particle is on a fixed height so there isn't any wobbling, it's just revolving around z $\endgroup$ Mar 4, 2015 at 4:12

2 Answers 2


In general, when working out the Lagrangian, start in coordinates that you know and then rewrite in generalized coordinates. Kinetic energy in this case is proportional to $v^2 = \dot x^2 + \dot y^2 + \dot v_z^2$. In your spherical coordinates $$x = r \sin \alpha \cos \theta;~ y = r \sin \alpha \sin \theta;~ z = r \cos \alpha.$$ Take full time derivatives of each of these, write them with $\dot r$ and $\dot \theta$, plug them into the above equation, and simplify, simplify, simplify.

For Lagrangian mechanics it is so important to start from a correct Lagrangian that when you are working it out you do not want to be hand-waving about the expressions involved. Do this extra work. It is not that much and you will often be surprised by interesting terms. Do not settle for a guess as guessing a Lagrangian is exactly like guessing an answer, except with no principles to fall back on to verify that answer if it is wrong.

In your case there will be terms with $\dot r^2$ and terms with $\cos(2 \alpha)$; don't stop until you find them.

  • 1
    $\begingroup$ Solid advice! In particular the "no principles to fall back on" - it is very hard to "see" whether the Lagrangian is correct. $\endgroup$
    – Floris
    Mar 4, 2015 at 13:29

The kinetic energy of the particle depends just on the path it is following - if you imagine the cone is suddenly invisible, the particle continues to go around in a circle. That means that there is no reason to add $\sin\theta$ in your expression for the kinetic energy if you used $r$ to mean (as drawn) the distance from the axis of rotation.

Note - your drawing implies you are using a cylindrical coordinate system. If you were using a polar system (as your question implies, but not what your drew), then the distance from the origin (of the cone) to the particle would be $r$ and the distance to the axis would indeed be $r\sin\theta$.

Is it possible that this is the source of your confusion?


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