1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Apr 1, 2012 #1
    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 [itex]P[/itex] to [itex]Q[/itex] and [itex]Q[/itex] to [itex]R[/itex] in which [itex]P[/itex], [itex]Q[/itex] and [itex]R[/itex]are not the center of the circle(because [itex]P[/itex], [itex]Q[/itex], [itex]R[/itex] are on the circumference of the circle) and we don't know the equation of the circle.

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

    Is it possible to find the 2D coordinate of [itex]R[/itex]?

    The way I tried to solve it:

    I managed to find the distance [itex]QR[/itex] by using the following calculations.

    We know that:

    [itex]s_{pq}= (u_{p} * t_{pq}) + (\frac{1}{2}*a_{pq}*t_{pq}^2)[/itex]

    By rearranging we get:

    [itex]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)}[/itex]

    Now we know [itex]a_{pq}[/itex] Also we know that:

    [itex]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)}[/itex]

    Plugging in [itex]a_{pq}[/itex] and [itex]v_q[/itex] into the equation below we get the distance between [itex]Q[/itex] and [itex]R[/itex]

    [itex]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)}[/itex]

    So i know the distance between [itex]P[/itex] and [itex]Q[/itex] and the distance between [itex]Q[/itex] and [itex]R[/itex].

    Now how do I use this information to get the coordinate of [itex]R[/itex]? Any thoughts on this? Is it possible to find the coordinate of [itex]R[/itex] in this way?
  2. jcsd
  3. Apr 1, 2012 #2


    User Avatar
    Homework Helper

    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.
  4. Apr 1, 2012 #3
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook