I was solving the exercise where the massless ring with radius $R$ is rotating around axis (shown in the picture) with angular velocity $\omega$. On the ring is a point-object with mass $m$ which moves freely without friction. I want to write equation for the system with Lagrangian formalism and I have a couple of questions about it.

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I will use $J_{0} = mR^2$ for moment of inertia of the point-object.

If I write the kinetic energy for system with intuition I get $$T = \frac{(J_0 + mR^2)}{2}\omega^2 = \frac{1}{2}m[r^2 + r^2]\omega^2,$$

but if I try to do it in non-inertial frame of reference I get $$T = \frac{1}{2}m[\vec{v_{rel}} + \vec{\omega}\times(\vec{r_{rel}} + \vec{r_0})]$$ where $r_{rel}$ is position of point-object in non-inercial reference frame (with origin in center of ring) and $r_0$ is vector from origin of inertial reference frame to the non-inertial reference frame. $$T = \frac{1}{2}m[\dot r \vec{e_r}+ r\dot\phi \vec{e_\phi} + \omega\vec{e_z}\times ((r-r\cos\phi)\vec{e_r} + r\sin\phi \;\vec{e_\phi})]^2$$ $$T = \frac{1}{2}m[r^2 + r^2 -2rr\cos\phi]\omega^2$$ this expression has one more term that the initial one. What did I do wrong? Did I make mistake when going from inertial to non-inertial reference frame? How could I solve this exercise with use of the moved circle equation $R^2 = (x+R)^2 + y^2$?


1 Answer 1


How could I solve this exercise with use of the moved circle equation

just rotate the position vector to the mass.

$$\vec \rho= \left[ \begin {array}{ccc} \cos \left( \omega\,t \right) &-\sin \left( \omega\,t \right) &0\\ \sin \left( \omega\,t \right) &\cos \left( \omega\,t \right) &0\\ 0&0&1 \end {array} \right] \,\begin{bmatrix} x -R\\ y \\ 0 \\ \end{bmatrix}$$

I assumed that the rotation is about the z axes .

the kinetic energy is now

$$T=\frac m2 \vec{\dot{\rho}}\cdot\vec{\dot\rho}\\\\ \begin{align*} &\vec{\dot{\rho}}= \left[ \begin {array}{c} -\cos \left( \omega\,t \right) {\it \dot{x}}+ \left( \left( R+x \right) \omega-{\it \dot{y}} \right) \sin \left( \omega \,t \right) -\cos \left( \omega\,t \right) \omega\,y \\ -\sin \left( \omega\,t \right) {\it \dot{x}}-\sin \left( \omega\,t \right) \omega\,y+ \left( \left( -R-x \right) \omega+{\it \dot{y}} \right) \cos \left( \omega\,t \right) \\ 0\end {array} \right] \end{align*} $$

and you have one holonomic constraint equation



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