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Homework Help: Off center circular motion (polar coordinates)

  1. Feb 5, 2013 #1
    1. The problem statement, all variables and given/known data
    A particle moves with constant speed v around a circle of radius b. Find the velocity vector in polar coordinates using an origin lying on the circle.

    Imagine the r starts at (0,0).

    2. Relevant equations

    [tex]\frac{d\vec{r}}{dt}[/tex] = [tex]\dot{r}\hat{r}+r\dot{\theta}\hat{\theta}[/tex]

    3. The attempt at a solution

    We can make a triangle connecting the origin to the center of the circle, to a point where the particle is. the hypotenuse is r

    I assume I need to find the rate of change of r, right? So, could I just do

    [itex]r=b/cos(\theta)[/itex] [tex]\frac{dr}{dt}=\frac{dr}{d \theta}\frac{d \theta}{dt}[/tex]

    [tex]\frac{dr}{dt}=b\frac{tan(\theta)}{cos(\theta)} \frac{d\theta}{dt}[/tex]

    My book doesn't do this, which leads me to believe I've made some horrible mistake.
    Last edited: Feb 5, 2013
  2. jcsd
  3. Feb 5, 2013 #2

    Simon Bridge

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    Science Advisor
    Homework Helper

    Nope. It asks you to find the velocity vector. That's the whole thing, not just the radial component.

    You should already know the velocity vector if the origin in at the center of the circle - so write that down. Then just translate the origin. You don't have to start from scratch.

    Not sure? Then look up the equation of a circle (offset origin) in polar coordinates... work out the time-dependent version of that (constant speed).

    To understand the problem - sketch=ch the circle and put points on the circle for different positions (one at the origin, one opposite etc) and draw a little arrow on each point for the velocity there - write that velocity as a vector in cartesian and polar coordinates.
  4. Feb 5, 2013 #3


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    I think the easiest approach would be to find the motion using rectangular coordinates with the center of the circle at the origin, then translate the coordinates so the point on the circle is at the origin, then convert to polar coordinates.
  5. Feb 5, 2013 #4
    Sorry, I need to head out for a while, I'll work on it some more when I get back!
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