Challenging problem about an impact with a smooth frictionless surface

[pepos04]

It would have to be a consequence of the conservation of mechanical energy.

Excuse me, how is mechanical energy conserved? What equations should be set up to verify what you say?

 

[haruspex]

Excuse me, how is mechanical energy conserved? What equations should be set up to verify what you say?

You stated at the beginning that it is, and we used that in the analysis.

 

[pepos04]

You stated at the beginning that it is, and we used that in the analysis.

Of course, you are absolutely right. But I meant to say, actually: how is the fact that $v_{1y} = |u$| a direct consequence of the conservation of mechanical energy? From what equations should it be deduced? I do not grasp the obvious connection between $v_{1y} = |u|$ and conservation of mechanical energy: could you explain it to me? Imagine two masses and a connecting rod are suspended from a ceiling by a string a bit away from CG and string burnt at t=0: can you intuitively explain $v_{1y} = |u|$ by observing this small experiment?

 

[haruspex]

Of course, you are absolutely right. But I meant to say, actually: how is the fact that $v_{1y} = |u$| a direct consequence of the conservation of mechanical energy? From what equations should it be deduced? I do not grasp the obvious connection between $v_{1y} = |u|$ and conservation of mechanical energy: could you explain it to me? Imagine two masses and a connecting rod are suspended from a ceiling by a string a bit away from CG and string burnt at t=0: can you intuitively explain $v_{1y} = |u|$ by observing this small experiment?

I did not say there was an obvious connection. If ME were not conserved the velocity would have been different, so there is not going to be a connection that does not involve it.

 

[pepos04]

Conservation of energy, conservation of horizontal linear momentum, conservation of angular momentum, constancy of separation. One way in which it will be sure to go through a finite repeating sequence is if the descent after the bounce is the mirror image of the ascent. For that to be the case, at the highest point between first and second bounce the rod must be either horizontal or vertical. If it takes time t to reach the highest point we have $v=gt$ and $\omega t=\theta+n\pi/4$. Plugging in the values found for $v, \omega$ leads to a relationship between $h$ and $\theta$. But the expression I quoted before wasn't quite right: I only considered the rod being horizontal at the top, so I had $n \pi/2$ instead of $n\pi/4$. And it is a four bounce sequence, not a three bounce sequence. There will be many other finite sequences, but they're more complex to analyse.

I appreciated the elucidation. But another observation grips me. I went looking for these other finished sequences, but I couldn't figure out how to find them? Why are they more complex to analyze? What would they be?

 

[haruspex]

I appreciated the elucidation. But another observation grips me. I went looking for these other finished sequences, but I couldn't figure out how to find them? Why are they more complex to analyze? What would they be?

If you want to study that further, the first step would be to get more general equations. We only considered the case in which there is no initial rotation.
From there, you may be able to find two bounce sequences using the same approach of setting the orientation to be either horizontal or vertical at the highest point.
Beyond that, it may be possible to prove the existence of more complicated finite sequences without necessarily finding any.