Equation of motion of a particle inside a rotating tube [closed]

Question

I'm trying to solve a problem but I don't know even where to start.

The problem is about a smooth hollow cylinder of mass $M$ rotating about an axis in one of the extremes of the pole with an initial angular velocity of $\omega_0$. Inside of that cylinder there's a particle of mass $m_1$ that can move freely along the longitudinal axis with an initial position of $r_0$ and an initial velocity of $0$. The length of the pole is $l$, therefore the moment of inertia is $I_G=\frac{1}{12}Ml^2$ as the mass is uniformly distributed.

What I have to calculate is the differential equation that relates the distance $r$ with the point $O$ and then solve it for $M\gg m_1$. I've never done an exercise of this type with a moving axis, so I don't really know how to do it.

What I understand here is that as the surface of the pole is smooth, there's no friction force and there would be a normal force in the $\theta$ direction. Therefore, the diagram of the system would be:

This would give us the equations:

$r$ axis: $O_x sin\theta+O_y cos\theta=-M\omega^2\frac{l}{2}-m_1 \frac{\dot{r}^2}{r}$

$\theta$ axis: $N+O_xcos\theta-O_ysin\theta=M\dot{\omega}\frac{l}{2}+m_1\dot{\omega}r$

Moment at $O$: $Nr=M\dot{\omega}\frac{l^2}{4}+m_1\dot{\omega}r^2+\frac{1}{12}Ml^2\dot{\omega}$

Which I suppose it has $N, O_x, O_y, \dot{\omega}, r$ as unknowns, so I cannot solve the system.

If I do it with the Lagrange equation, as there is only Kinetical energy: $L=\frac{1}{2}m_1(\dot{r}^2+(r\omega)^2)+\frac{1}{2}M\omega^2\frac{l^2}{4}+\frac{1}{24}Ml^2\omega^2$, and $Q^{NC}_r=0, Q^{NC}_\theta=Nr$. Therefore:

$$m_1\ddot{r}=m_1\omega^2r$$ $$Nr=m_1\dot{\omega}r^2+\frac{1}{3}Ml^2\dot{\omega}$$

But again, I don't know $N$ so I don't know how to continue. Is there anything wrong that I'm doing or something that I'm missing?

I'll paste the whole text so it's more clear

Answer 1

Solving this using polar co-ordinates will be much easier. Write the equations for conservation of angular momentum, and conservation of energy as follows:

$$I_0 \omega_0 + mr_0^2\omega_0=I_0\omega + mr^2\omega$$ $$\frac{I_0\omega_0^2}{2}+\frac{mr_0^2\omega_0^2}{2}=\frac{I_0\omega^2}{2} + \frac{mr^2\omega^2}{2} + \frac{m \dot r ^2}{2}$$

Manipulate these two equations to get the differetial equation.

Note: Here, $I_0$ is the moment of inertia of the rod, $r_0$ is initial position of particle, and $\omega_0$ is initial angular velocity.

Answer 2

The best way to go about this is using angular momentum conservation around $O$. Indeed, there are no external torques on the system around this point and thus angular momentum conservation is valid. This conservation equation tells us that $$(I_O+m_1r_0^2)\omega_0=(I_O+m_1r^2)\dot{\theta}.$$ On the other hand, the acceleration of the particle in the radial direction is $\ddot{r}-r\dot{\theta}^2$. Since there are no forces in this direction, it must vanish, yielding $\dot{\theta}^2=\ddot{r}/r$. We then conclude that $$(I_O+m_1r_0^2)^2\omega_0^2=(I_O+m_1r^2)^2\ddot{r}/r.$$