MHB Proofs regarding uniform circular motion in vectors

AI Thread Summary
The discussion focuses on deriving the acceleration vector in uniform circular motion and proving its relationship with the position and velocity vectors. The user successfully calculated the position and velocity vectors but struggled to express the acceleration vector as a sum of an anti-parallel vector to the position vector and a parallel vector to the velocity vector. Clarifications were provided that the coefficients for these vectors must be time-dependent functions rather than constants. The user realized that multiplying by functions of time allows for the necessary components of the acceleration vector to be included. Ultimately, the user gained clarity on the concept of anti-parallel and parallel vectors in this context.
skate_nerd
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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.
 
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skatenerd said:
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:
$$\vec{p}(t)=-aRcos(\theta(t))\hat{i}-aRsin(\theta(t))\hat{j}, a>0$$

skatenerd said:
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. :)
 
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?
 
skatenerd said:
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.
 
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.
 
skatenerd said:
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.
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