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#### skatenerd

##### Active member

- Oct 3, 2012

- 114

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.