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Tangential and radial acceleration

  1. Apr 17, 2012 #1
    Hey

    I have an accelerated circular motion problem.

    I have only the position equation, from which I derived the velocity and acceleration.

    how can I tell what is the tangential acceleration and what is the radial acceleration?

    If you could point me towards a source to read about the subject, it would be great.
     
  2. jcsd
  3. Apr 17, 2012 #2
    The tangential is the usual position, velocity or acceleration. The radial is called the angular postition,velocity or acceleration
    So they are connected by
    [itex] v = \omega * r , a = r*\alpha[/itex]
     
  4. Apr 18, 2012 #3
    Let's see if I understand, here's an example:

    I have:
    [itex]a= -40\pi [ \sin (2\pi t^2 - \frac{\pi}{3}) + 4\pi t^2 \cos (2\pi t^2 - \frac{\pi}{3})] \hat{i} + 40\pi [\cos (2\pi t^2 - \frac{\pi}{3})- t^2 \sin (2\pi t^2 - \frac{\pi}{3})] \hat{j}
    [/itex]

    (I derived it from [itex] r= 10 \cos (2\pi t^2 - \frac{\pi}{3}) \hat{i} + 10\sin (2\pi t^2 - \frac{\pi}{3}) \hat{j}[/itex] )

    I was asked to determine what is the acceleration without direction at [itex]t=\sqrt{2}[/itex]

    and also the tangential and radial acceleration.

    so I just plugged it in, and got about 483.5 m/s^2

    so by what you say, [itex]\sqrt{a^2 + (\frac{a}{R})^2} = 483.5 [/itex] ?





    edit: I looked in wikipedia under Non-uniform circular motion, I understand now - the value I got was for the tangential acceleration, the radial is simply [itex]\frac{v^2}{R}[/itex] I don't know why I thought it's only true for Uniform circular motion
     
    Last edited: Apr 18, 2012
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