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I Why is ##2 \pi /T## multiplied by R for v?

  1. Jun 7, 2016 #1
    Hey all, $$v = 2 \pi f =2 \pi \frac{1}{T} =\frac{2 \pi }{T} $$ but why is it multiplied by $$R$$? Any help appreciated.


    upload_2016-6-7_16-31-27.png

    I see now why R is multiplied in now, but why inst L multiplied in in the analogous pendulum equation?
     
  2. jcsd
  3. Jun 7, 2016 #2

    vanhees71

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    2016 Award

    A particle moving with constant angular velocity ##\omega## on a circle is decribed by the position vector
    $$\vec{x}=\begin{pmatrix}
    r \cos(\omega t) \\ r \sin (\omega t) \\ 0
    \end{pmatrix}.$$
    The time derivative gives the velocity
    $$\dot{\vec{x}}=\vec{v}=\begin{pmatrix}
    -r \omega \sin(\omega t) \\ r \omega \cos(\omega t)
    \end{pmatrix}.$$
    The magnitude thus is
    $$|\vec{v}|=r \omega.$$
     
  4. Jun 7, 2016 #3

    QuantumQuest

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    The first relation you write is angular frequency not velocity: ##\omega = \frac{2\pi}{T}##. Then ##\upsilon = \frac{2\pi}{T} R = \omega R##
     
  5. Jun 7, 2016 #4

    vanhees71

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    Ok, to be very precise, the angular velocity in my example is ##\vec{\omega}=\omega \vec{e}_z##.
     
  6. Jun 7, 2016 #5
    Thanks all, :)
     
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