# Finding 2d coordinate of a point in a complicated circular motion with acceleration

1. Apr 1, 2012

### jdred23

Hello everyone! I've found a physics problem that i don't know the solution of(maybe because of my limited knowledge). The problem is something like this:

Let's say an object travels in a circular path from $P$ to $Q$ and $Q$ to $R$ in which $P$, $Q$ and $R$are not the center of the circle(because $P$, $Q$, $R$ are on the circumference of the circle) and we don't know the equation of the circle.

We are given the 2D coordinate of $P$ and $Q$, the initial velocity at $P$ is $0$, the object has the same acceleration all the way from $P$ upto $R$ and the object goes from $P$ to $Q$ in $1 \text{ millisecond}$ and $Q$ to $R$ in $2 \text{ millisecond}$

Is it possible to find the 2D coordinate of $R$?

The way I tried to solve it:

I managed to find the distance $QR$ by using the following calculations.

We know that:

$s_{pq}= (u_{p} * t_{pq}) + (\frac{1}{2}*a_{pq}*t_{pq}^2)$

By rearranging we get:

$a_{pq} =\frac{\Large{2(s_{pq} - u_{p}*t_{pq})}}{\Large{t_{pq}^2}}\text{..........(1)}\\ \text{where } a_{pq} = \text{ acceleration between }P\text{ and }Q, s_{pq} = \text{ the distance between } P \text{ and } Q\text{(found using simple vector math) }, u_{p} = 0\text{( initial velocity is 0 given)} \text{ and }t_{pq} = 1\text{ millisecond(given)}$

Now we know $a_{pq}$ Also we know that:

$v_{q} = u_{p} + a_{pq} * t_{pq}\text{................(2)} \\ \text{where } v_{q} = \text{ velocity at } Q, u_{p} = 0\text{ (initial velocity)}, a_{pq} = \text{ found above in eq(1) and } t_{pq} = 1\text{ millisecond(given)}$

Plugging in $a_{pq}$ and $v_q$ into the equation below we get the distance between $Q$ and $R$

$s_{qr} = v_{q} * t_{qr} + \frac{1}{2} * a_{pq} * t_{qr}^2 \\ \text{where }s_{qr} =\text{ distance between } Q \text{ and } R, v_{q} = \text{ is found above in eq(2) }, a_{pq} = \text{ found above in eq(1) and } t_{qr} = 2\text{ millisecond(given)}$

So i know the distance between $P$ and $Q$ and the distance between $Q$ and $R$.

Now how do I use this information to get the coordinate of $R$? Any thoughts on this? Is it possible to find the coordinate of $R$ in this way?

2. Apr 1, 2012

### rcgldr

Re: Finding 2d coordinate of a point in a complicated circular motion with accelerati

You could use polar coordinates where distance = r θ, (in this case r is constant) then
convert to cartesian afterwards. I assume when you stated acceleration is the same,
you meant the tangental component of acceleration, since the radial component of
acceleration increases over time in this case and at any moment = v2 / r.

3. Apr 1, 2012

### jdred23

Re: Finding 2d coordinate of a point in a complicated circular motion with accelerati

Thank you rcgldr so much for your help. That solves my problem. I really appreciate your help.