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Uniform Circular Motion puzzle

  1. Nov 23, 2014 #1
    The position function for a particle moving on a circle (with constant speed) is:
    ## \vec{r}(t)=\left \langle r\,sin(t), \, r\,cos(t) \right \rangle ##
    Taking the first and second derivatives,
    ## \vec{v}(t)=\left \langle -r\,cos(t), \, r\,sin(t) \right \rangle ##
    ## \vec{a}(t)=\left \langle -r\,sin(t), \, -r\,cos(t) \right \rangle ##
    suggesting that the magnitudes of each are equal (i.e., r = a = v).

    However, I frequently see this equation:
    ## \sum F = ma_{c}=m \frac{v^2}{r} ##

    ...but wouldn't ## a_{c}=\frac{v^2}{r} ## simply be equal to r?

    If so, why express it in the above form?
     
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  3. Nov 23, 2014 #2

    Orodruin

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    Writing down your expressions, you have implicitly assumed that the angular frequency is 1. This is not always the case and is dimensionally inconsistent as the angular frequency should have units s-1. Instead, use a general angular frequency ##\omega## such that the argument of the sines and cosines becomes ##\omega t##.
     
  4. Nov 23, 2014 #3

    Philip Wood

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    sin (t) and cos (t) don't make sense. You can't take the sin or cos of something with units. Try looking up sin (10 second) !
    t must be multiplied by a constant, [itex]\omega[/itex], with units of s-1, or rad s-1.

    [itex]\omega[/itex] has an easy physical interpretation. Agreed?
     
  5. Nov 23, 2014 #4
    Even with the correct arguments, won't the magnitudes of the vectors still be equal?
    I.e., ##\sqrt{(\pm rCos(\omega t))^{2}+(\pm rSin(\omega t))^{2}}=\sqrt{(- rSin(\omega t))^{2}+(\pm rCos(\omega t))^{2}} = r##, right?

    I'm sure that there is something I'm missing, but I don't know what.
     
  6. Nov 23, 2014 #5

    jbriggs444

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    what is the first derivative of ##cos({\omega}t)## with respect to t?
     
  7. Nov 23, 2014 #6
    Yes.
     
  8. Nov 24, 2014 #7

    Philip Wood

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    Jbriggs444 has identified another reason why you're confused. You need to think about the answer to this poster's question. It is not [itex]-sin(\omega t)[/itex].
     
    Last edited: Nov 24, 2014
  9. Nov 24, 2014 #8
    Chain rule. Right. Well that clarifies a lot!
    So the magnitudes end up being:
    ##v=\omega r##
    ##a=\omega^{2}r,## right?

    Excellent. This is a useful result! Thanks for indicating the way out of my confusion.
     
    Last edited: Nov 24, 2014
  10. Nov 24, 2014 #9

    Philip Wood

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    Good!

    It's worth remembering that any function (such as sin, cos, ln, exp) that can be expanded as a power series can only have a pure number as argument, otherwise we'd be adding together terms with different units.
     
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