Submitted by jetay1 on Thu, 03/23/2017 - 12:14
Question 4 is about non-uniform circular motion. I worked out the $x$ and $y$ coordinates on point $P$ to be \[\mathbf{P}=\begin{bmatrix} a(\theta-\sin\theta) \\ a(1-\cos\theta) \end{bmatrix}\] But the question asks for position of $P$ in cartesian coordinates. Can I have the parameter $\theta$ in 'cartesian coordinates'? I can eliminate $\theta$ and obtain \[x=a\cos^{-1}\left(\frac{a-y}{y}\right)-\sqrt{2ay-y^2}\] but that's not really what I think of as 'coordinates'.
I can go on and differentiate $\mathbf{P}$ and obtain the kinetic energy expression given that $Q$ has the opposite sign cosine, but I'm not sure about the last bit - show that the hoop rolls with constant speed. The preparation section mentions that in the absence of dissipative forces the total energy of the system is conserved, so when I differentiate the total energy I get \[\frac{dE}{dt}=4ma^2\dot{\theta}\ddot{\theta}\] which will be zero if $\ddot{\theta}=0$. $\ddot{\theta}$ is the angular acceleration, which is related to the tangential acceleration. The circle is rotating about the point that is in contact with the ground, so the normal contact force has no effect on the moment and the two weights are equidistant from the centre of rotation and so generate equal moments about the contact point. Hence there is no tangential acceleration and hence $\ddot{\theta}=0$. Is that a plausible argument?
What you initially gave $
What you initially gave $\mathbf{P}$ as *is* Cartesian coordinates. Look, you've even specified "the $x$ and $y$ coordinates..." which is precisely what Cartesian coordinates are, you specify a $x$ coordinate and a $y$ coordinate. It doesn't matter that they depend on $\theta$. [If you're confusing this with polar coordinates just because $\theta$ is there, then this certainly isn't polar coordinates, you haven't specified an angle nor a length, simply two coordinates, which is precisely what the Cartesian system is].
----------------------------------------------------------------------------------------------------------------------------------------------
What you're doing in the last part is a little weird, I'm not sure I really understand. You're differentiating the total energy and then trying to show it's $0$ by showing $\ddot{\theta}$ is $0$. But the question wants you to show that there is constant speed, aka $\dot{\theta}$ is constant. But this is equivalent to $\ddot{\theta}$ being 0.
So if you show that $\ddot{\theta}$ is $0$ through your consideration of 'generate equal moments' then there was no point in differentiating the energy, since you've finished the question at that point.
That said, I don't really follow your argument, it could potentially be correct - but anyway, it seems much more 'physicsy' than actual maths. Instead, consider this:
The total energy of the system is KE + GPE = KE since GPE = 0 ($mga(-\cos \theta) + mga(\cos \theta) = 0$. Hence the total energy is just the kinetic energy. Differentiating this gives $4ma^2 \dot{\theta}\ddot{\theta} = 0$ since total energy is conserved (so it has derivative = 0). But then since none of the other factors in the expression of the derivative are 0 apart from possibly $\ddot{\theta}$ we then have $\ddot{\theta} = 0$ and we are done.
To recapitulate: your approach seemed to be "show that derivative of energy is 0 using $\ddot{\theta} = 0$" when the actual approach you need is "show that $\ddot{\theta} = 0$ using derivative of energy is 0". Also I'm not sure if you proved that GPE = 0 or at least GPE=constant (depending on where you measured it from) before taking the derivative of "total energy", since the expression you derived in the previous paragraph was only kinetic energy, not total energy.
Let me know if any bit here was unclear.
I get it now
Thanks very much. If you'd asked me before, I would have said energy is definitely conserved in an interaction, but for some reason I didn't then make the tiny leap to that meaning $\frac{dE}{dt}=0$.
I did have the GPE term in my total energy equation, I just didn't type it into my original question. I like your idea of using $y=a$ as the baseline - I used $y=0$ as the baseline, which still means the GPE term disappears when you differentiate, but having the GPE terms cancel each other out is more elegant.
I was getting confused between cartesian and polar coordinates. But I was also getting confused with a cartesian equation, which would not contain the parameter.
Thanks again for all your assistance so far.
Ah, yeah. Fair enough. Just
Ah, yeah. Fair enough. Just to spell it out (although I'm sure you already know), something being conserved means that it doesn't change with respect to time. So if you look at it's "rate of change with respect to time" a.k.a dE/dt that necessarily has to be 0, since it doesn't change with respect to time.
Awesome, as long as you did include your GPE and see that it goes away after differentiating, that's awesome.
I see what you mean, yeah. A Cartesian equation will relate $x$ and $y$ together, but Cartesian coordinates can certainly be parametrized as $x = f(\theta), y = g(\theta)$ (this sort of stuff crops up in C4 papers, iirc).
No worries, keep on posting here -- I can certainly see you improving.