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    Not my strongest topic, and I'm stuck on part (b).

    So I know that L and T are conserved, and for rotations we have L=I\omega and T=\frac{1}{2}I\omega^2.

    The answer says that it should be \omega = \dfrac{I+ma^2}{I+ml^2}\omega_0 but shouldn't it be capital M instead? Since we are considering the tube rather than the bead?

    Also, here are my thoughts; we let I 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 I_0 = I+Ma^2.
    Then we do the same thing for when the bead is at a distance l from the centre of the tube and we get that I(l) = I+Ml^2.
    Therefore for angular momentum we got L_0 = (I + Ma^2)\omega_0 and L(l) = (I+Ml^2)\omega.
    Since angular momentum is conserved we get that L_0=L(l) which implies that \omega = \dfrac{I+Ma^2}{I+Ml^2}\omega_0.
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    (Original post by RDKGames)
    Not my strongest topic, and I'm stuck on part (b).

    So I know that L and T are conserved, and for rotations we have L=I\omega and T=\frac{1}{2}I\omega^2.

    The answer says that it should be \omega = \dfrac{I+ma^2}{I+ml^2}\omega_0 but shouldn't it be capital M instead? Since we are considering the tube rather than the bead?

    Also, here are my thoughts; we let I 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 I_0 = I+Ma^2.
    Then we do the same thing for when the bead is at a distance l from the centre of the tube and we get that I(l) = I+Ml^2.
    Therefore for angular momentum we got L_0 = (I + Ma^2)\omega_0 and L(l) = (I+Ml^2)\omega.
    Since angular momentum is conserved we get that L_0=L(l) which implies that \omega = \dfrac{I+Ma^2}{I+Ml^2}\omega_0.
    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 L_z, 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.
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    (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 L_z, 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!
    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? This is why it's m rather than M too it seems.
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    (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?
    That's my thinking.

    This is why it's m rather than M too it seems.
    Yep, and the result follows.
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    (Original post by ghostwalker)
    That's my thinking.



    Yep, and the result follows.
    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 v into it.
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    (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 v into it.
    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.
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    (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.
    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.
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    (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.
    Although I said I'd not studied this - I haven't at uni. level. At A-level we got as far as instantaneous centres of rotation, but that was decades ago. So, what I'm saying is a mix of A-level 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.
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    (Original post by ghostwalker)
    Although I said I'd not studied this - I haven't at uni. level. At A-level we got as far as instantaneous centres of rotation, but that was decades ago. So, what I'm saying is a mix of A-level 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.
    I was wondering if there is an easy way to do part (b), and whether I have overcomplicated with my solution or not.



    From part (a) both L (ang. momentum) and T (K.E.) are conserved, and initially they are:

    L_0 = (I+ml^2)\omega_0

    T_0 = \frac{1}{2}(I+ml^2)\omega_0^2 (this is also the total energy of the system)

    Here I=\frac{1}{2}Ml^2

    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 y>0 above its initial position.
    2. Letting x denote the horizontal displacement from the axis of rotation for this position with height y, such that |x|< l, and \omega_x the angular vel. at this point, the total initial energy at this point must be equal to \frac{1}{2}(I+mx^2)\omega_x^2 + mgy which is precisely \frac{1}{2}(I+ml^2)\omega_0^2.
    3. Use inequalities to get to down to mgy < \omega_0^2-\omega_x^2
    4. Show that the RHS is -ve by considering the angular momentum throughout motion, which yields that \omega_x = \dfrac{I+ml^2}{I+mx^2}\omega_0. Since |x| < l we get that the fraction is > 1 hence \omega_x > \omega_0 thus \omega_0^2 - \omega_x^2 < 0. (WLOG we have \omega_x > \omega_0 > 0 since the direction of the rotation does not change)
    5. Hence mgy < 0 which means that y<0 - contradiction.
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    (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 y>0 above its initial position.

    ]2. Letting x denote the horizontal displacement from the axis of rotation, such that |x|\leq l, and \omega the angular vel. at some point the total initial energy along any point must be equal to \frac{1}{2}(I+mx^2)\omega^2 + mgy (*) which is precisely \frac{1}{2}(I+ml^2)\omega_0^2.
    In (*) you seem to be ignoring any "non-rotational" 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.

    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 x = \sqrt{l^2-y^2} (it's probably clearer with a diagram, of course).

    3. Use inequalities to get to down to mgy < \omega_0^2-\omega^2
    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.

    4. Show that the LHS is -ve by considering the angular momentum throughout motion, which yields that \omega = \dfrac{I+ml^2}{I+mx^2}\omega_0. Since |x| \leq l we get that the fraction is \geq 1 hence \omega \geq \omega_0 thus \omega_0^2 - \omega^2 \leq 0.
    5. Hence mgy < 0 which means that y<0 - contradiction.
    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 a \implies b \implies c, 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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    (Original post by RDKGames)
    ..
    Tagging + quoting you because I messed up the quote initially above and I don't think TSR correctly informs you you were quoted if it's only fixed in an edit.
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    Thanks for the reply!

    (Original post by DFranklin)
    In (*) you seem to be ignoring any "non-rotational" 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.
    Agreed. The tangential velocity of the particle completely escaped my mind and I forgot to consider it. Though I am taking the point (x,y), y>0, 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?

    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 x = \sqrt{l^2-y^2} (it's probably clearer with a diagram, of course).
    Aha, yes, I re-read my argument on that post and noticed that this does not flow as well, so I've changed it and now it stands that x,y define a specific point on the circle with x^2+y^2=l^2 and y>0.

    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.
    I got it from:

    \frac{1}{2}(I+mx^2)\omega_x^2 + mgy = \frac{1}{2}(I+ml^2)\omega_0^2

    \implies mgy = \frac{1}{2}(I+ml^2)\omega_0^2 - \frac{1}{2}(\underbrace{I+mx^2}_  {< I+ml^2})\omega_x^2

    (since |x|<l in my edited argument)

    \implies mgy < \frac{1}{2}(I+ml^2)\omega_0^2 - \frac{1}{2}(I+ml^2)\omega_x^2 = \frac{1}{2}(I+ml^2)(\omega_0 - \omega_x^2)

    \implies mgy < \omega_0 - \omega_x^2

    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 a \implies b \implies c, you write "we're going to show c by showing b by showing a", which can be hard to follow.
    Sorry, yes I've meant to say RHS, which has been changed in my edits.

    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.
    Fair enough!
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    (Original post by RDKGames)

    \implies mgy < \frac{1}{2}(I+ml^2)\omega_0^2 - \frac{1}{2}(I+ml^2)\omega_x^2 = \frac{1}{2}(I+ml^2)(\omega_0 - \omega_x^2)

    \implies mgy < \omega_0 - \omega_x^2
    Bullet point 3 was troubling me.

    My pennyworth: :eek: to the last line - doesn't follow. Missing a postive constant multiplier.
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    (Original post by ghostwalker)
    Bullet point 3 was troubling me.

    My pennyworth: :eek: to the last line - doesn't follow. Missing a postive constant multiplier.
    Ah yeah, forgot to consider the possibility that 0 < \frac{1}{2}(I+ml^2) \leq 1 whereby then the last line is not implied.

    So then if I leave it as y < \underbrace{\frac{I+ml^2}{2mg}}_  {>0}(\omega_0^2-\omega_x^2) then show that \omega_0^2 - \omega_x^2 < 0 then it should be OK?
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    (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 (x,y), y>0, 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?
    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 \dot y is still 0 at the top, \dot x 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.

    I got it from:

    \frac{1}{2}(I+mx^2)\omega_x^2 + mgy = \frac{1}{2}(I+ml^2)\omega_0^2

    \implies mgy = \frac{1}{2}(I+ml^2)\omega_0^2 - \frac{1}{2}(\underbrace{I+mx^2}_  {< I+ml^2})\omega_x^2

    (since |x|<l in my edited argument) (*)

    \implies mgy < \frac{1}{2}(I+ml^2)\omega_0^2 - \frac{1}{2}(I+ml^2)\omega_x^2 = \frac{1}{2}(I+ml^2)(\omega_0 - \omega_x^2)
    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 |x|<l in my edited argument, we have:".
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    (Original post by DFranklin)
    This looks wrong. By replacing (I+mx^2) by (I+ml^2) you're making something bigger.
    :facepalm2: Don't know how i missed that - must be time for a break.
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    (Original post by ghostwalker)
    :facepalm2: Don't know how i missed that - must be time for a break.
    I'm somewhat in the same position as you in terms of "I last did this at A-level... " (although I have a bit of experience of writing simple physics-like stuff for games as well).

    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 \omega^2, 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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    (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 \dot y is still 0 at the top, \dot x 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 |x|<l in my edited argument, we have:".
    OK, that's a big error I didn't even notice

    So then, here's my third go at the whole thing:


    Let us assume that the particle reaches some maximum height on the circle 0<y \leq l above its initial position. Let |x|<l denote its horizontal displacement from the axis of rotation to this point, so that x^2+y^2=l^2. Let \omega_x and v denote the angular and tangential velocities of the particle at this point.
    Let I=\frac{1}{2}Ml^2

    Due to conservation of angular momentum, we get that (I+mx^2)\omega_x = (I+ml^2)\omega_0 \iff \omega_x = \dfrac{I+ml^2}{I+mx^2}\omega_0. Since |x|<l we get that \dfrac{I+ml^2}{I+mx^2} > 1 and also \omega_x > \omega_0

    Also, due to conservation of potential energy, we get that mgy + \frac{1}{2}(I+mx^2)\omega_x^2 + \frac{1}{2}mv^2 = \frac{1}{2}(I+ml^2)\omega_0^2

    \begin{aligned} & \implies 2mgy + (I+mx^2)\omega_x^2 \leq (I+ml^2)\omega_0^2 \\ & \implies 2mgy + (I+mx^2)\left( \dfrac{I+ml^2}{I+mx^2}\omega_0 \right)^2 \leq (I+ml^2)\omega_0^2 \\ & \implies 2mgy \leq (I+ml^2)\omega_0^2 - \frac{(I+ml^2)^2}{I+mx^2}\omega_  0^2 \\ & \implies 2mgy \leq \underbrace{(I+ml^2)\omega_0^2}_  {\geq 0} \underbrace{\left[ 1-\dfrac{I+ml^2}{I+mx^2} \right]}_{<0} \leq 0 \\ & \implies y \leq 0

    Hence we get a contradiction to what we assumed, and so the particle does not reach any height y>0 above it's initial position in the process of motion.


    Does that seem OK now??
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    (Original post by RDKGames)
    Does that seem OK now??
    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.
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    (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.
    Ah yes, got so focused on the math checking out that I didn't pay attention to my energy labels

    Flawless proofs are hard it seems!

    Thanks
 
 
 
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