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Angular Velo, Tangential Acceleration

  1. Nov 28, 2008 #1
    I have computed angular speed, centripetal acceleration, and torque for a uniform circular motion.

    What is the tangential acceleration?

    I am leaning towards tangential a = r(alpha) where alpha is d(omega)/dt.
    since I have calculated omega and a = dv/dt - I think its safe to
    say that as dt cancels dv = r d(omega) = a tangential?

    Any tips?
     
  2. jcsd
  3. Nov 29, 2008 #2
    This is a very strange mathematical manipulation, because dt, is not actually an unknown by itself, and not directly "cancel-able". Tangential a, for uniform circular motion is given by r(omega)2 or v2/r depending on what information you have.
     
  4. Nov 29, 2008 #3

    alphysicist

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    This is not correct; for one thing the units are not the same (m/s^2 on the right hand side, m/s on the others).

    If you start with:

    [tex]
    \frac{dv}{dt} = r \frac{d\omega}{dt} = a_t
    [/tex]

    (where v is the speed along the circular path and [tex]a_t[/itex] is the tangential acceleration), then you can rewrite this as:

    [tex]
    dv = r\ d\omega = a_t\ dt
    [/tex]

    so the dt term is now associated with the tangential acceleration.


    It's not clear from your post what other information you might have; but you can relate the torque to the total acceleration, and then relate total acceleration to the tangential acceleration.






    I don't believe those equations are correct.
     
  5. Nov 29, 2008 #4
    Ah, sorry, my mistake... I was using the equations for radial acceleration. Thanks for pointing it out.
     
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