Rotating spring and Euler-Lagrange equations

Question

I'm trying to solve the following exercise

Consider a point P of mass m under the action of a spring of elastic constant $k>0$, whose origin rotates uniformly on a circle of fixed radius $R$ with constant angular velocity $\omega$.Write the Lagrangian and Lagrange equations wrt the cartesian coordinates $(x_1,x_2,x_3)$ and wrt a system of time-dependent coordinate $(q_1,q_2,q_3)$ that moves according to $C$.

The figure will explain the situation

My attempt

With cartesian coordinates

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I have that the kinetic energy is simply given by $T=\frac{1}{2}m (\dot x_1^2+\dot x_2^2+ \dot x_3^2)$, while the potential energy has to be related to the movement of the spring in the $(x_1,x_2)$ plane. I have the relation $OP=O \bar{O} + \bar{O}P$ and hence

$x_1=R \cos(\omega t)+\tilde{x_1}, x_2=R \sin(\omega t)+\tilde{x_2}, x_3=\tilde{x_3}$ $(\star)$

hence the potential energy is

\begin{align} V(x_1,x_2,x_3)=\frac{k}{2}((x_1-R \cos(\omega t)^2+(x_2-R \sin(\omega t))^2+x_3^2) \end{align}

And hence I have just that $L=T-V$.

With coordinates that moves according to $C$

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Here $P$ has coordinates $\tilde{x_1},\tilde{x_2},\tilde{x_3}$ with respect to the moving system.

Hence, $T=\frac{1}{2}m (\dot x_1^2+\dot x_2^2+ \dot x_3^2)$ and here each $x_i$ is given by $\star$, while the potential energy is $V(\tilde{x_1},\tilde{x_2},\tilde{x_3})=\frac{k}{2}(\tilde{x_1}^2+\tilde{x_2}^2+\tilde{x_3}^2)$

Could it be okay?

Answer 1

Hint.

In a rotating frame with $\vec \Omega = (0,0,\omega)$ the kinetic energy is

$$ E_K = \frac 12m ||\dot P + \Omega\times P||^2 $$

and

$$ E_P = m g (0,0,1)\cdot P + \frac 12 k ||P-P_0||^2 $$

with

$$ P = (x(t),y(t),z(t))\\ P_0 = (R,0,0) $$

so

$$ L(P,\dot P) = E_K - E_P $$

and

$$ \frac{\partial L}{\partial \dot P} = m(\dot P +\vec\Omega\times P)\\ \frac{\partial L}{\partial P} = -m g (0,0,1) - k(P-P_0) + m(\dot P+\vec\Omega\times P)\times\vec\Omega $$