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1. Dec 13, 2017

### chrononaut 114

1. The problem statement, all variables and given/known data

3. The attempt at a solution

So I first tried by saying consider a time t in which mass m is directly above the origin O. I.e., mass m at the Cartesian coordinate (0, 4l/3). I wrote a = a(t) as the extension function of the spring, which has 0 natural length. So, I applied the cosine rule and found:

But then how do I find the coordinate of the 4m mass?

Then I'll need some help in going about some other parts of the question, but first need help with finding the coordinate of 4m...
Thanks

2. Dec 13, 2017

### Orodruin

Staff Emeritus
This is not necessarily the case according to the assignment. The mass on the table can be at any point. Take two steps back and think about what your generalised coordinates are going to be. Note that these should be able to describe the configuration of the system regardless of where the masses are. Once you have done that you can start trying to express the position of the masses in terms of these coordinates.

Also, please write out your work rather than posting it in image form.

3. Dec 13, 2017

### chrononaut 114

So I've had another thought about it.. so for the (two) generalised coordinates, what about $\theta = \theta(t)$ being the angle that the spring makes with the horizontal x-axis, and $\phi = \phi(t)$ the angle between the line that joins from the origin to mass m and the vertical y axis? Sorry but I have to add this photo, below, to show what I mean by $\theta$ and $\phi$ in the context of the question:

Would these be the correct generalised coordinates?

4. Dec 13, 2017

### Orodruin

Staff Emeritus
They would make a possible set of generalised coordinates. However, I think this set of coordinates would be quite bulky to work with and they certainly are not the generalised coordinates that I would choose. Just as an example, the possible range of your angle $\theta$ would depend on the value of the angle $\phi$. It is also going to be rather cumbersome to write the positions of the masses in terms of those coordinates, in particular since the spring does not have a fixed length. I suggest that you try to find a set of coordinates that do not depend on the spring at all (i.e., remove the spring when you think about the configuration space). Adding the spring does nothing to the configuration space itself as it stretches between two points that should be fixed by the system's configuration. The angle $\phi$ is also not the most straightforward way to keep track of the position of the mass on the table.

Edit: Also, your coordinate system would generally not uniquely identify a point in configuration space (think about why) so it could only be used locally.

5. Dec 13, 2017

### chrononaut 114

Thanks for your response. So I tried a couple different ways and they all were also convoluted also; basically I don't really know how to properly implement what you're explaining. I'm just still confused as to what would be the most efficient generalised coordinates relative to the fixed point O stated in the question; which I'm assuming they want me to use as the origin.. So could you please show me the best generalised coordinates you have in mind, since my exam is in approx 1.5 days and I really need some urgent help, maybe with some sort of a brief diagram? Even with MS Paint, anything, I don't mind. That would be awesome and greatly, greatly appreciated..Thanks

6. Dec 14, 2017

### Orodruin

Staff Emeritus
This is the thing with generalised coordinates. It makes no difference whatsoever where you place your origin. You just need two numbers that accurately describe the configuration of the system. That you then relate these numbers to a Cartesian coordinate system is a different issue (but again it makes no difference whatsoever where you place the origin of that system).

To be honest, referring to your imminent exam is one of the biggest turn-offs for homework helpers as it gives the impression that you are not interested in learning the subject itself but rather just in passing your exam. Saying that your problem is "urgent" just puts unnecessary pressure on homework helpers and will generally just lead to them not helping at all. Homework helpers are here on a pro bono basis, spending our free time trying to help people understand what they are doing. Furthermore, having such "urgent" issues essentially gives us the impression that you waited to the last moment in starting to study for the exam and did not put enough effort in throughout the course. Finally, posting solutions to homework problems before the OP has solved it is against forum rules so what you are essentially asking is that I break forum rules because you have an exam tomorrow. This is not going to happen. I am happy to give input on what you have done as long as you are actually willing to learn and actually provide us with the work you have done, what you have thought about, and what conclusions you have drawn.

Note that I wrote "gives the impression" above. I have no way of knowing if it is actually true or not, but it is the impression that will affect peoples propensity to provide you with help.

7. Dec 14, 2017

### chrononaut 114

I see, that's fair, my apologies. I didn't know that was a forum rule and will read the rules more closely now. I've actually been studying for it throughout the past few weeks but since this is probably one of the hardest subjects I've encountered I've been panicking, and with some other exams as well, it's been intense. Thanks for the heads up.

Yea, just before reading your latest reply I realised that I could simply change the origin, to then introduce more effective coordinates; it's just that since the question was written in such a way that they have the 'fixed point', O, making it appear as though like they want you to use that as the origin, would've been less confusing if they'd used a different letter haha, but it's a good way of testing our understanding anyway. I made a thing on my computer, using Word, to show my new coordinates, looks a bit nicer better than my notebook. I think they're pretty useful, I should be alright from now on I think..

Cheers

Last edited: Dec 14, 2017
8. Dec 14, 2017

### Orodruin

Staff Emeritus
These are indeed better generalised coordinates, but there is no requirement to move the origin in order to use them. Of course, nothing stops you from doing so either.

I would have used the angle of the rod to the vertical instead of to the horizontal, but this is of course only a matter of taste.

9. Dec 14, 2017

### chrononaut 114

Ok, yeah that makes sense now, thanks

10. Dec 15, 2017

### chrononaut 114

Hi Orodruin, I forgot to ask another question (much simpler than before) regarding a different exam problem, and I thought I'd just write it here to save some time...

So when writing the kinetic energy for the Lagrangian $( L = T - V )$ of this system (diagram below), I got a little confused with whether I should use the centre of mass ($CM$) for expressing the transnational component of the kinetic energy $( T_t )$, or the 'point of suspension' (denote this by $PoS$) instead of the $CM$.
The reason I was confused was that clearly the rotational component of the kinetic energy $( T_r )$ was taken such that its rotational axis should be (into/ out of the page) at the point of suspension where the rod connects with the curved wire.
(i.e. I'd use the moment of inertia of the rod with the axis not at the $CM$ but at one of the ends - which gives
$$I = (1/3)ml^2$$
rather than
$$I = (1/12)ml^2$$
the latter of which would be if the rotational axis was into the page, through the $CM$ of the rod.
(using $( T = T_t + T_r )$ )

To put it in other words, should $T_t$ be:

$$T_t = (1/2)m( \mathbf {\dot r_{CM}} )^2 .......(1)$$
or should it be
$$T_t = (1/2)m( \mathbf {\dot r_{PoS}} )^2 .......(2) ?$$

Whilst at the same time $T_r$ has to be taken at the point of suspension irregardless of what we do for transnational $T_t$:

i.e. $$T_r = (1/2)I(\omega)^2$$
where $\omega = \dot { \theta}$ , with $\theta$ being the angle between the rod and the vertical. And here $I = (1/3)ml^2$. (Rotational axis being into the page through the point of suspension)

Also, the potential energy is purely gravitational, right?

Cheers

Last edited: Dec 15, 2017
11. Dec 15, 2017

### Orodruin

Staff Emeritus
Generally start a new thread for new questions. You will not save time by using the same thread as nobody else will look at it.

You cannot split rotational and translational kinetic energy about any arbitrary point. The only point you can do this for is the centre of mass or a fixed point (for which there is no translational energy).