Not my strongest topic, and I'm stuck on part (b).
So I know that and are conserved, and for rotations we have and .
The answer says that it should be but shouldn't it be capital instead? Since we are considering the tube rather than the bead?
Also, here are my thoughts; we let be the moment of inertia about the axis of rotation, then the moment of inertia through a parallel axis which goes through the bead is initially .
Then we do the same thing for when the bead is at a distance from the centre of the tube and we get that .
Therefore for angular momentum we got and .
Since angular momentum is conserved we get that which implies that .

RDKGames
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 20042018 21:47
Last edited by RDKGames; 20042018 at 22:50. 
ghostwalker
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 20042018 23:26
(Original post by RDKGames)
Not my strongest topic, and I'm stuck on part (b).
So I know that and are conserved, and for rotations we have and .
The answer says that it should be but shouldn't it be capital instead? Since we are considering the tube rather than the bead?
Also, here are my thoughts; we let be the moment of inertia about the axis of rotation, then the moment of inertia through a parallel axis which goes through the bead is initially .
Then we do the same thing for when the bead is at a distance from the centre of the tube and we get that .
Therefore for angular momentum we got and .
Since angular momentum is conserved we get that which implies that .
I agree with the given answer.
In part a, you had conservation of angular momentum about , i.e. the axis of rotation through the centre of the tube.
In your working you're saying the angular momentum is conserved betweem two different axes, neither of which is the axis of rotation, and that's the flaw, I think!
Edit: Might need one of the physics gurus on this. Don't visit that forum often, so don't know who best to tag.Last edited by ghostwalker; 20042018 at 23:29. 
RDKGames
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 20042018 23:30
(Original post by ghostwalker)
Have to say I struggle to truely pinpoint the flaw in your argument, as my understanding isn't 100%.
I agree with the given answer.
In part a, you had conservation of angular momentum about , i.e. the axis of rotation through the centre of the tube.
In your working you're saying the angular momentum is conserved betweem two different axes, neither of which is the axis of rotation, and that's the flaw, I think! 
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 20042018 23:33
(Original post by RDKGames)
Ah right, yes now that makes more sense, I was looking at the answer and thought that parallel axis theorem was used. So then rather than that, it's just the same working but you stick with the rotational axis, then the MofI for the system is precisely MofI for the tube + MofI for the bead about this axis, right?
Yep, and the result follows. 
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 20042018 23:39
I'm having trouble starting (c) now. I can't seem to wrap my head around it but I believe I have to use conservation of kinetic energy but I'm unsure how to formulate that and involve into it. 
ghostwalker
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 20042018 23:47
(Original post by RDKGames)
Thanks for that!
I'm having trouble starting (c) now. I can't seem to wrap my head around it but I believe I have to use conservation of kinetic energy but I'm unsure how to formulate that and involve into it.
The kinetic energy is conserved. This is the sum of the rotational KE (of tube and bead) and the radial KE (of the bead)
I think that's all you need to lay down an equation.
Last post for today. 
RDKGames
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 21042018 00:03
(Original post by ghostwalker)
Since I've not studied this, I'm making it up as I go. So, that said.
The kinetic energy is conserved. This is the sum of the rotational KE (of tube and bead) and the radial KE (of the bead)
I think that's all you need to lay down an equation.
Last post for today. 
ghostwalker
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 21042018 07:32
(Original post by RDKGames)
That worked out nicely, thanks for that. I wasn't considering the rotational KE and radial KE as two separate things being added so I couldn't make much progress.
With the bead, we're considering its velocity as the sum of the two components, radial and tangential. And, bit of Pythagoras shows that the total KE of the bead is the sum of the KE taking radial velocity, plus the sum of the KE taking the tangential velocity, and the latter is just the rotational KE. 
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 21042018 15:27
(Original post by ghostwalker)
Although I said I'd not studied this  I haven't at uni. level. At Alevel we got as far as instantaneous centres of rotation, but that was decades ago. So, what I'm saying is a mix of Alevel understanding, applying principles, and some intuition. So, here's my take on a bit of background for (c)
With the bead, we're considering its velocity as the sum of the two components, radial and tangential. And, bit of Pythagoras shows that the total KE of the bead is the sum of the KE taking radial velocity, plus the sum of the KE taking the tangential velocity, and the latter is just the rotational KE.
From part (a) both (ang. momentum) and (K.E.) are conserved, and initially they are:
(this is also the total energy of the system)
Here
Then on part (b) my approach was to do the following:
1. Proceed with a proof by contradiction, assuming that the particle reaches a max height above its initial position.
2. Letting denote the horizontal displacement from the axis of rotation for this position with height , such that , and the angular vel. at this point, the total initial energy at this point must be equal to which is precisely .
3. Use inequalities to get to down to
4. Show that the RHS is ve by considering the angular momentum throughout motion, which yields that . Since we get that the fraction is hence thus . (WLOG we have since the direction of the rotation does not change)
5. Hence which means that  contradiction.Last edited by RDKGames; 21042018 at 16:20. Reason: Finer detail, fixes 
DFranklin
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 21042018 15:52
(Original post by RDKGames)
I was wondering if there is an easy way to do part (b), and whether I have overcomplicated with my solution or not. ~snip~
Then on part (b) my approach was to do the following:
1. Proceed with a proof by contradiction, assuming that the particle reaches some height above its initial position.
]2. Letting denote the horizontal displacement from the axis of rotation, such that , and the angular vel. at some point the total initial energy along any point must be equal to (*) which is precisely .
Other than that, this all seems OK, but a little confused  you talk about "some point" and then "any point" and I'm left a little confused as to what you're talking about. I think it would also be clearer if you expressly defined (it's probably clearer with a diagram, of course).
It's slightly unclear from what you're writing (i.e. it may be OK), but note that you have equality in the starting position, so a strict inequality seems a little odd.
Speaking to your actual question: no, I don't think this is particularly "round about". You could probably get something a bit shorter and more direct by simply forming an equation for the total energy and showing that the KE term I described in (*) ends up needing to be ve when y > 0. But I'm not sure it would make much difference.Last edited by DFranklin; 21042018 at 15:53. 
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 21042018 15:54
(Original post by RDKGames)
..Tagged: 
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 21042018 16:12
Thanks for the reply!
(Original post by DFranklin)
In (*) you seem to be ignoring any "nonrotational" KE of the particle  that is, energy due to x or y varying. Of course, any such KE can only make a "there's not enough energy..." argument stronger, but it should at least be mentioned I think.
It's slightly unclear from what you're writing (i.e. it may be OK), but note that you have equality in the starting position, so a strict inequality seems a little odd.
(since in my edited argument)
Does that seem OK?
Did you mean "show that the RHS is ve"? Obviously this also shows the LHS is ve, but it seems weird to pick the side that you're *indirectly* showing is ve. As a general comment, I find you have a bit of a tendency to "argue backwards". That is, given a logical train , you write "we're going to show c by showing b by showing a", which can be hard to follow.
Speaking to your actual question: no, I don't think this is particularly "round about". You could probably get something a bit shorter and more direct by simply forming an equation for the total energy and showing that the KE term I described in (*) ends up needing to be ve when y > 0. But I'm not sure it would make much difference. 
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 21042018 16:29
My pennyworth: to the last line  doesn't follow. Missing a postive constant multiplier.Last edited by ghostwalker; 21042018 at 16:31. 
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 21042018 16:36
(Original post by ghostwalker)
Bullet point 3 was troubling me.
My pennyworth: to the last line  doesn't follow. Missing a postive constant multiplier.
So then if I leave it as then show that then it should be OK?Last edited by RDKGames; 21042018 at 16:37. 
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 21042018 16:41
(Original post by RDKGames)
Agreed. The tangential velocity of the particle completely escaped my mind and I forgot to consider it. Though I am taking the point , , on the circle such that it is the maximum that the particle reaches. At this max point, the velocity of the particle must be zero along w.r.t the circle therefore the tangential KE must be 0 either way, right?
This looks wrong. By replacing (I+mx^2) by (I+ml^2) you're making something bigger that you're subtracting, which means you're making the RHS snaller, which means your inequality is the wrong way around. (You basically need to calculate a fair bit further, using conservation of angular momentum, before drawing conclusions).
Again on a minor stylistic point: I would expect (*) as written to refer to the previous line of calculation. It would be clearer to write something like :"Since in my edited argument, we have:". 
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 21042018 17:12
(Original post by DFranklin)
This looks wrong. By replacing (I+mx^2) by (I+ml^2) you're making something bigger. 
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 21042018 17:25
(Original post by ghostwalker)
Don't know how i missed that  must be time for a break.
But from memory, the way these questions usually go is that you get a change in M of I which means two things (for an increase of MI, say)
(1) The KE for a given angular velocity goes up.
but
(2) Conservation of angular momentum means the angular velocity goes down.
Since KE depends on , fact (2) usually wins, but I think it's fairly unusual to be able to decide if energy goes up or down without considering both points. (Well, on a heuristic basis, you can probably decide (2) wins without considering (1), but still...)
So when I saw RDK *hadn't* considered both, I was pretty suspicious there was something wrong  which obviously helps in finding it. 
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 22042018 11:47
(Original post by DFranklin)
There's a slight issue here. Conceivably, the particle could end up going fast enough to "loop round" at the highest point (the top of the circle), in which case although is still 0 at the top, would not be. So it's probably best to acknowledge the KE due to dot x and dot y, since they don't hurt the argument.
This looks wrong. By replacing (I+mx^2) by (I+ml^2) you're making something bigger that you're subtracting, which means you're making the RHS snaller, which means your inequality is the wrong way around. (You basically need to calculate a fair bit further, using conservation of angular momentum, before drawing conclusions).
Again on a minor stylistic point: I would expect (*) as written to refer to the previous line of calculation. It would be clearer to write something like :"Since in my edited argument, we have:".
So then, here's my third go at the whole thing:
Let us assume that the particle reaches some maximum height on the circle above its initial position. Let denote its horizontal displacement from the axis of rotation to this point, so that . Let and denote the angular and tangential velocities of the particle at this point.
Let
Due to conservation of angular momentum, we get that . Since we get that and also
Also, due to conservation of potential energy, we get that
Hence we get a contradiction to what we assumed, and so the particle does not reach any height above it's initial position in the process of motion.
Does that seem OK now??Tagged:Last edited by RDKGames; 22042018 at 12:11. Reason: Typo 
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 22042018 13:45
(Original post by RDKGames)
Does that seem OK now??
Second conservation is "Conservation of energy", which includes potential and kinetic. 
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 22042018 13:54
(Original post by ghostwalker)
Can't see any issue with the maths  hopefully my brain is working today!
Second conservation is "Conservation of energy", which includes potential and kinetic.
Flawless proofs are hard it seems!
Thanks
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