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Circluar motion, omega

  1. Oct 31, 2009 #1
    circluar motion, "omega"

    hi, I erased the default format by accident, but it's just a quick question:
    i and j are unit vectors, w= omega dtheta/dt ,
    theta= angle dependent on time

    when we have i(-rwsin(theta)) and j(rwcos(theta)) as our velocity vector components,
    and we wish to find the acceleration

    i(-rwsin(theta)) and j(rwcos(theta)) is differentiated to i(-rw^2cos(theta)) and
    j(-rw^2sin(theta))

    the factor of omega remains constant because dtheta/dt will be the same everywhere since it is in uniform motion right? consistent speed?

    thanks

    edit: and if you did not have uniform circular motion, you would get

    i(-r(a/r)wcos(theta)) and j(-rw(a/r)sin(theta)) as acceleration components:

    using
    w = dtheta/dt = v/r
    d(v/r) / dt = 1/r(dv/dt) = a/r
    when differentiating
    -rwsin(theta), getting -r(a/r)wcos(theta) , remembering that theta is a function of time

    thanks
     
    Last edited: Oct 31, 2009
  2. jcsd
  3. Nov 1, 2009 #2

    kuruman

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    Re: circluar motion, "omega"

    You are correct for the uniform circular motion part. For non-uniform (but still circular) motion, ω is not constant so you will need dω/dt terms.
     
  4. Nov 1, 2009 #3
    Re: circluar motion, "omega"


    thanks, dω/dt is d/dt(v/r) right? which is 1/r dv/dt, isn't that a/r?
     
  5. Nov 2, 2009 #4

    kuruman

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    Re: circluar motion, "omega"

    You can say that, but I prefer to think of dω/dt as the angular acceleration alpha and leave it at that.
     
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