Torque on a constantly rotating rod

Question

I recently came across artwork that, in its simplest form, looks something like this. Please ignore the text in the video and the fact that the falling blocks are magnets. For this discussion, let's assume the blocks shown are masses with no magnetism. It cannot be seen in the video that torque is provided at the two ends of the tube-like structure, keeping the angular velocity constant. Also, the rotating structure is attached to a stand, for support, with an axle that goes through the centre of mass of the tube (the pivot). The rotation occurs in the vertical plane. I do not have the lengths, weights, etc. for this configuration but I believe that shouldn't affect the analysis too much and so is not an issue

I was wondering how could one go about calculating torque in this situation. My current understanding/analysis (and questions I am raising) can be seen below

1. In a linear system, one needs to overcome friction to make an object travel with constant velocity. Does that apply here too, meaning does the torque only need to overcome friction in the axle to achieve constant angular velocity. Or are there more forces the torque has to "overcome" to make it work
2. How do I calculate friction. In a linear system, I can just use $\mu N$. How to achieve the same for this system
3. Does gravity play any role here. It can be seen that the combined centre of gravity displaces as the masses inside the tube move
4. It can also be seen that the moment of inertia changes due to moving masses. Does that affect the torque calculation

Please let me know if you need more details. Thanks in advance

Answer 1

To answer your equations first I will write down the equation of motion, assuming that the tube is rotate with constant angular velocity $~\omega$

The EOM: $$m\,\ddot{s}+F_\mu-\omega^2\,m\,s-\sin(\omega\,t -\varphi_0)\,m\,g=0$$

where

• $F_\mu~$ friction force
• $s~$ the motion of the block
• $~m$ mass of the block
• $~\varphi_0$ initial angle of the tube

with $F_\mu=\mu\,N~$ ,where $~N~$ is the contact force between the block and the tube.

$$N=m\,2\,\omega\,\dot s+\cos(\omega\,t -\varphi_0)\,m\,g$$

Case I $~\omega=0~$ inclined plane

$$N=m\,g\,\cos(\varphi_0)$$ the EOM: $$ m{\ddot s}+ \left( \mu\,\cos \left( \varphi _{{0}} \right) m+\sin \left( \varphi _{{0}} \right) m \right) g=0\tag 1 $$

so you obtain the acceleration of the block $~\ddot s=\text{const.}~$ thus the velocity is linear $~\dot s=c\,t$

Case II $~\omega\ne 0~$ rotation of a inclined plane

The EOM

$$m{\ddot s}-{\omega}^{2}ms+2\,\mu\,m{\dot s}\,\omega+ \left( -\sin \left( \omega \,\tau-\varphi _{{0}} \right) +\mu\,\cos \left( \omega\,\tau-\varphi _ {{0}} \right) \right) mg =0 $$

in this case if the acceleration of the block is depending on the position and the velocity of the block $~\ddot s=\ddot s(s\,,~\dot s)$ , thus the velocity of the block not linear.