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

  • #1

Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
14,406
5,988
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
$$|\vec{v}|=r \omega.$$
 
  • #3
QuantumQuest
Science Advisor
Insights Author
Gold Member
911
475
The first relation you write is angular frequency not velocity: ##\omega = \frac{2\pi}{T}##. Then ##\upsilon = \frac{2\pi}{T} R = \omega R##
 
  • #4
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
14,406
5,988
Ok, to be very precise, the angular velocity in my example is ##\vec{\omega}=\omega \vec{e}_z##.
 
  • #5
Thanks all, :)
 

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

Replies
11
Views
8K
Replies
4
Views
564
Replies
7
Views
3K
Replies
2
Views
646
Replies
4
Views
8K
Replies
0
Views
4K
Replies
5
Views
4K
Replies
9
Views
1K
Replies
16
Views
9K
Replies
3
Views
1K
Top