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malawi_glenn
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1 kg bead and 1000 J initial energy, quite tall ramps :)
Yes. When I re-ran to look at elapsed time, separation distance and total bead travel, I scaled that back to a 1 meter launch height. Expected values (over a distribution of mass ratios) for elapsed time and separation distance probably diverge (integral of an approximately harmonic function). Typical elapsed times are modest and are dominated by the final trip. For instance, about 2.5 minutes total for the first 34 one way trips and about 15 minutes additional for the last.malawi_glenn said:1 kg bead and 1000 J initial energy, quite tall ramps :)
Umm... By "internal energies" here, do you refer to the whole system's (slide A,B and mass C)?malawi_glenn said:Will there be any deformation of the objects causing internal energies to be altered?
Pathfinder.malawi_glenn said:Just out of curiosity, which book?
The Individual objectsRikudo said:Umm... By "internal energies" here, do you refer to the whole system's (slide A,B and mass C)?
Pathfinder.
Google suggests: https://www.amazon.com/dp/B072QR5CWH/?tag=pfamazon01-20malawi_glenn said:Never heard of pathfinder, which author and publisher?
The thing is, you cannot assume that the bead will have the same velocity as one of the ramps. These are extreme cases of the possible solutions. The range of solutions include the bead having any speed between these extreme solutions. You cannot know which without performing a series argument with a cutoff condition.haruspex said:Unfortunately it is yet another case of guessing intent from subtle detail in the wording.
Assuming the bead matches the velocity of one slide allows two slide speeds to be deduced, the one the bead last encountered and the one it is vainly chasing.
What we cannot do is say which is A and which is B. They will take it in turns to be the faster.
The question says "Find expressions for the speeds". I note that a) it says speeds, not speed, and b) it is not entirely clear that it requires us to identify which slide has which speed.
I'm fairly confident that such a solution would be more accurate than ##\mathcal O(m/M)##. I tried to start with the exact solution then approximate, but I get a 3x3 transition matrix, so it's a bit messy. There's probably an easier way.
As to which is intended, I toss a coin.
I think one of those should be a left.kuruman said:The orange line is the speed of the right slide and the blue line is the speed of the right slide.
I see that now. In post #22 I gave a reason why I thought it would always be close to a ramp's speed, but, as I acknowledged in post #32, @jbriggs444 proved me wrong.Orodruin said:you cannot assume that the bead will have the same velocity as one of the ramps
You're right. It's fixed, thanks.Orodruin said:I think one of those should be a left.