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proofs regarding uniform circular motion in vectors

skatenerd

Active member
Oct 3, 2012
114
Just started this Analytical Mechanics class, so I figured this question should go here...
I've been pretty stuck with a problem. I felt like I totally knew what I was doing but I've become very stumped.
We're given the vector for general circular motion,
$$\vec{r}(t)=Rcos(\theta(t))\hat{i}+Rsin(\theta(t))\hat{j}$$
And told to find the velocity vector. Easy enough...
$$\vec{v}(t)=\frac{d\vec{r}(t)}{dt}=-\dot{\theta}(t)Rsin(\theta(t))\hat{i}+\dot{\theta}(t)Rcos(\theta(t))\hat{j}$$
Next we were told to prove that the position and velocity vectors are perpendicular, which was a simple enough dot product of the two that indeed ended up equaling zero.
Then we were told to find the acceleration vector...fine...
$$\vec{a}(t)=\frac{d\vec{v}(t)}{dt}=\frac{d^2\vec{r}(t)}{dt^2}=-R[\ddot{\theta}(t)sin(\theta(t))+\dot{\theta}(t)^2cos(\theta(t))]\hat{i}+R[\ddot{\theta}(t)cos(\theta(t))-\dot{\theta}(t)^2sin(\theta(t))]\hat{j}$$
Phew.
Now here's where I ran into trouble. We have to show that the acceleration vector can be written as a sum of a vector that is anti-parallel to the position vector and another vector parallel to the velocity vector. I've never heard of anti-parallel before, but I'm assuming you just need to multiply each of the vectors components by -1.
Anti-parallel to position vector:
$$\vec{p}(t)=-Rcos(\theta(t))\hat{i}-Rsin(\theta(t))\hat{j}$$
and parallel to velocity vector I feel like would just make sense to be any constant multiple of the velocity vector. For simplicity, I think I can just use the velocity vector...
So \(\vec{p}(t)+\vec{v}(t)\) should = \(\vec{a}(t)\).
But nope, it doesn't. I got
$$-R[cos(\theta(t))-\dot{\theta}(t)sin(\theta(t))]\hat{i}-R[sin(\theta(t))-\dot{\theta}(t)cos(\theta(t))]\hat{j}$$
Any idea of what I am doing wrong? I have a feeling I am missing something conceptually, because the two added vectors are completely missing the whole second time derivative of the theta function. So I feel I am pretty far from the correct answer.
Any help is appreciated.
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Now here's where I ran into trouble. We have to show that the acceleration vector can be written as a sum of a vector that is anti-parallel to the position vector and another vector parallel to the velocity vector. I've never heard of anti-parallel before, but I'm assuming you just need to multiply each of the vectors components by -1.
It's not just -1 you can multiply with. Any negative multiple gives you an anti-paraller vector.

So anti-parallel to position vector:
$$\vec{p}(t)=-aRcos(\theta(t))\hat{i}-aRsin(\theta(t))\hat{j}, a>0$$

And parallel to velocity vector I feel like would just make sense to be any constant multiple of the velocity vector. For simplicity, I think I can just use the velocity vector...
Umm.. you don't have choice here. It's true that a vector parallel to the velocity vector would be a constant multiple of velocity vector but what this constant multiple will actually be is not upto us to choose.

You should work out the equation $\vec a(t)= \lambda \vec p(t) +\mu \vec v(t)$ and get $\lambda<0$ and $\mu>0$. Note that there's is just one pair of $(\lambda,\mu)$ which satisfies the above equation because $\vec p(t)$ and $\vec v(t)$ are linearly independent (You showed this by showing that $\vec p(t)$ is perpendicular to $\vec v(t)$). Whatever that pair comes out to be we have to live with it. :)
 

skatenerd

Active member
Oct 3, 2012
114
Thanks for the reply. I follow what you are implying, but I still don't see how I can end up with the conclusion I need. Any idea where the \(\ddot{\theta}(t)\)'s (that are prevalent in the acceleration vector) will come from?
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Thanks for the reply. I follow what you are implying, but I still don't see how I can end up with the conclusion I need. Any idea where the \(\ddot{\theta}(t)\)'s (that are prevalent in the acceleration vector) will come from?
I see where you are getting confused. The thing is that $\lambda$ and $\mu$ themselves are functions of time. By a 'constant' multiple, here, we don't mean constant with time. By a constant multiple we just mean a scalar. Give it a try. If you get stuck I'll give you the solution in full. Right now I've got to run to class.
 

skatenerd

Active member
Oct 3, 2012
114
Ahhh okay no I have it now. I didn't really figure that multiplying by a function of time and a negative function of time would still give a parallel and anti-parallel vector respectively. That makes sense to me now. Thanks.
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Ahhh okay no I have it now. I didn't really figure that multiplying by a function of time and a negative function of time would still give a parallel and anti-parallel vector respectively. That makes sense to me now. Thanks.
(Yes)