Analytical Dynamics - Conservation laws

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!

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.

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.

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.

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. ~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 $\frac{1}{2}(I+ml^2)\omega_0^2$.

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

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.

..

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).

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.

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.

Bullet point 3 was troubling me.

My pennyworth: to the last line - doesn't follow. Missing a postive constant multiplier.

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.

Don't know how i missed that - must be time for a break.

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:".

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.