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

[Superposed_Cat]

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.

I see now why R is multiplied in now, but why inst L multiplied in in the analogous pendulum equation?

 

[vanhees71]

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
$$|\vec{v}|=r \omega.$$

 

[QuantumQuest]

The first relation you write is angular frequency not velocity: $\omega = \frac{2\pi}{T}$. Then $\upsilon = \frac{2\pi}{T} R = \omega R$

 

[vanhees71]

Ok, to be very precise, the angular velocity in my example is $\vec{\omega}=\omega \vec{e}_z$.

 

[Superposed_Cat]

Thanks all, :)