A bead of mass $m$ is threaded around a smooth spiral wire and slides downwards without friction due to gravity. The $z$-axis points upwards vertically. Suppose the spiral wire is rotated about the $z$-axis with a fixed angular velocity $\Omega$. Determine the Lagrangian and the equation of motion.

This is related to a previous problem in which the wire shape is given as $$ z = k\psi, \hspace{3mm} x = a\cos\psi, \hspace{3mm} y = a\sin\psi $$ where $a$ and $k$ are both positive.

My attempt at a solution: We still have $z = k\psi$, but now $x = a\cos(\psi + \Omega t)$ and $y = a\sin(\psi + \Omega t)$. This gives $\dot{z} = k\dot{\psi}$, $\dot{x} = -a(\dot{\psi} + \Omega)\sin(\psi + \Omega t)$ and $\dot{y} = a(\dot{\psi} + \Omega)\cos(\psi + \Omega t)$. Then the kinetic energy is $$ T = \frac{1}{2}m\left(\dot{x}^2 + \dot{y}^2 + \dot{z}^2\right) = \frac{1}{2}m\left(a^2\left(\dot{\psi}^2 + 2\dot{\psi}\Omega + \Omega^2\right) + k^2\dot{\psi}^2\right) $$ and the potential energy is $V = mgz = mgk\psi$. Then the Lagrangian becomes: $$ L = T - V = \frac{1}{2}m\left(a^2\left(\dot{\psi}^2 + 2\dot{\psi}\Omega + \Omega^2\right) + k^2\dot{\psi}^2\right) - mgk\psi. $$ This gives $$ \frac{\partial L}{\partial \psi} = -mgk, \hspace{3mm} \frac{\partial L}{\partial \dot{\psi}} = m\left(a^2\dot{\psi} + a^2\Omega + k^2\dot{\psi}\right), \hspace{3mm} \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\psi}}\right) = m\left(a^2 + k^2\right)\ddot{\psi} $$ so plugging into Lagrange's Equation gives $-mgk = m\left(a^2 + k^2\right)\ddot{\psi}$ or $\ddot{\psi} = -\frac{gk}{a^2 + k^2}$, which is the exact same equation of motion as in the case with the coil not rotating. Obviously this isn't correct. I think the problem might be related to my choice of coordinates (in particular, having $\psi$ rotating), but I'm not sure what a better choice of coordinates would be.

  • $\begingroup$ Hi Noah - just asking where you're going wrong isn't the kind of question that's appropriate for this site. Could you edit the question to identify more specifically what concept you're confused about? After you do that, I'll be happy to reopen this. $\endgroup$
    – David Z
    Feb 3, 2014 at 2:41
  • 1
    $\begingroup$ OK, that helps. I changed your title to match. $\endgroup$
    – David Z
    Feb 3, 2014 at 4:17

1 Answer 1


Yeah I got the same homework on my problem set. I wonder if it's a coincidence...

Anyway, I got the same answer as you did. I thought about it a bunch and I think it's correct. Here is how I thought about it: (This might be wrong)

So let's think about it without Lagrangian stuff.

Work in cylindrical coordinates $(r,\theta,z)$. Notice that $r=constant$ for all $\psi$. This means that the direction of motion of the bead is purely in the $\hat{\theta}$ and $\hat{z}$ directions.

When we spin the system, the bead feels a centripetal force in the $\hat{r}$ direction. Therefore the centrepetal force is perpendicular to the bead's direction of motion. Thus, the centrepetal force should not affect the bead's motion along the wire. It is completely canceled by the normal force of the wire on the bead.

This seems counterintuitive, but I think that's because we can't help but to imagine the system with friction, in which case, spinning the wire would certainly affect the motion of the bead


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.