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

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?

## Answers and Replies

vanhees71
Science Advisor
Gold Member
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
$$|\vec{v}|=r \omega.$$

• Superposed_Cat and QuantumQuest
QuantumQuest
Science Advisor
Gold Member
The first relation you write is angular frequency not velocity: ##\omega = \frac{2\pi}{T}##. Then ##\upsilon = \frac{2\pi}{T} R = \omega R##

• Superposed_Cat
vanhees71
Science Advisor
Gold Member
Ok, to be very precise, the angular velocity in my example is ##\vec{\omega}=\omega \vec{e}_z##.

Thanks all, :)