I'm trying to find the Lagrangian for a bead on a rotating circular loop (constant angular velocity $\omega$, radius $a$) in two different ways and I'm unsure why these are giving different answers.

Using spherical polar coordinates in a fixed inertial frame the constraint is $\phi = \omega t$ and we can solve that constraint with usual polar coordinates and substituting $\phi = \omega t$ into the free particle lagrangian with $r=a$ constant. Doing this I obtain the Lagrangian $$L = \frac{1}{2}ma^2(\dot{\theta}^2 + \omega^2 \sin^2\theta) - mga\cos\theta$$

However we also have that the Lagrangian for a particle in a non-inertial frame rotating about the z axis of a fixed inertial frame is given by $L = \frac{1}{2}m(\dot{\vec{r}} + \vec{\omega} \wedge \vec{r})^2 + V$ where $\vec{r}$ has components in the rotating frame.

Now in this case it seems that I should be able to take $\vec{\omega} = (0,0,\omega)$ and in the rotating frame write $x = a\cos\theta, y=a\sin\theta$ and then substitute into the Lagrange equation for a free particle in a rotating frame. However doing this I obtain a different Lagrangian: $$L = \frac{1}{2}ma^2\dot{\theta}^2 + \frac{1}{2}\omega^2 a^2 + m\omega a^2 \dot{\theta}$$

Thanks for any help in where I am going wrong here!


In the rotating frame, write $$ \vec{r}=a \sin \theta \space\hat{i} + a \cos \theta \space\hat{k}$$ Then, back in the inertial frame, $$ \vec{v}=\dot{\vec{r}}+\vec{\omega}\wedge \vec{r} =(a \cos \theta \space\hat{i} - a \sin \theta \space\hat{k})\space\dot{\theta} + a\omega \sin{\theta}\space\hat{j}\\ v^2=a^2(\dot{\theta}+ \omega^2\sin^2{\theta})$$ The two Lagrangians will be the same. In much fewer words, change your rotating frame $x$ to $z$ and $y$ to $x$ and include the potential energy function in your second Lagrangian, then you should be good.

  • $\begingroup$ Thanks that's really helpful - I was just getting my y and z directions round the wrong way! $\endgroup$ – Wooster Apr 23 '15 at 8:43

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