Lagrangian of rotating springs I'm trying to construct the Lagrangian for the following scenario. A turntable of radius $R$ is rotating at angular velocity $\omega$, maintained by a motor. Two springs with Hooke's constant $k$ are attached to opposing points on the outer edge of the turntable and the other end of each spring is attached to a single point mass $m$. The mass is free to move in any direction along the turntable, but is restricted to remain in contact with it. I've drawn the initial setup below.

In this case I've taken $x$ and $y$ to be inertial coordinates (not rotating with the turntable). The kinetic energy is simply $T = \frac 12 m (\dot x ^2 + \dot y ^2)$. Now, supposing that the springs are initially aligned with the $x$ axis, we can examine the force due to some virtual displacement from the center of the turntable to derive the potential. If the mass is displaced by some amount $x \hat i + y \hat j$ then the force due to the spring on the left is $\vec F_1 = -k(x+R) \hat i - ky \hat j$ and the force due to the spring on the right is $\vec F_2 = -k(x-R) \hat i - ky \hat j$ so the net force due to the two springs is $$\vec F = \vec F_1 + \vec F_2 = -2kx \hat i -2ky \hat j.$$ A potential for this system is then $V = k(x^2+y^2)$ as then $-\nabla V = \vec F$. For some time $t >0$, we can rotate the springs with the turntable to get that the new forces $$\vec F_1 = -k[(x+R\cos \omega t)\hat i +(y+R \sin \omega t) \hat j]$$ and $$\vec F_2 = -k[(x-R\cos \omega t)\hat i +(y-R \sin \omega t) \hat j]$$ resulting in a net force $$\vec F = -2kx \hat i -2ky \hat j$$ and thus a time independent potential $V = k(x^2+y^2)$. The Lagrangian is therefore $$\mathcal L = T-V = \frac 12 m(\dot x^2 + \dot y ^2)-k(x^2+y^2)$$ which is independent of the rotation of the turntable. This doesn't make much sense to me, as even if the turntable is undergoing angular acceleration, the equations of motion for the mass are the same as if the turntable is at rest. Have I missed something here? I believe the Lagrangian should have some dependence on the turntable rotation.
 A: Your intuition is probably correct.  In general, the Lagrangian will depend on the rotation.  There is one thing which you forgot in your derivation, and that is the natural length of the springs.
Let's denote the displacement between the end of the spring attached to the turntable to the mass as $\vec{r}_1$ for spring 1 and $\vec{r}_2$ for spring 2.  The mass will be located at $\vec{r}=x\hat{i}+y\hat{j}$.  Then we have
$$
 \vec{r}_1=(x-L\cos(\omega t))\hat{i}+(y-L\sin(\omega t))\hat{j}.
$$
We get a similar equation for $\vec{r}_2$:
$$
 \vec{r}_2=(x+L\cos(\omega t))\hat{i}+(y+L\sin(\omega t))\hat{j}.
$$
When we calculate the force on the mass, we have
$$
  \vec{F}=\vec{F}_1+\vec{F}_2
$$
where
$$
  \vec{F}_{1,2}=-k(r_{1,2}-L_0)\hat{r}_{1,2}.
$$
The total force on the mass is
$$
  \vec{F}=-k(\vec{r}_1+\vec{r}_2)-L_0(\hat{r}_1+\hat{r}_2)\\
  \vec{F}=-2k\left(\vec{r}+L_0\frac{\hat{r}_1+\hat{r}_2}{2}\right)
$$
In the case where the natural length of the spring is zero, you get your result:
$$
  \vec{F}=-2k\vec{r}.
$$
In this very special case, this force is independent of the orientation of turntable, and thus the rotation of the turntable doesn't matter.
In general, however, we have $L_0 \neq 0$, and the Lagrangian will depend on the orientation of the turntable.  Note that since the orientation is time dependent, the potential term will be generally non-conservative (sometimes you'll have to do some work to keep the turntable rotating at a constant rate).
Does this answer your question?
