Question

What is position vector equation of a simple pendulum using normal and tangent vector? String has length s and the angle that it makes with vertical axes is θ.

Answer 1

i think we are to work with polar coordinates with origin at the point of suspension.

Answer 2

I would write what I have got, but I don't know how to use LaTex here. I can attach a picture to see what I wrote.

Answer 3

i am having problem using latex too. I want to use unit vectors along length and along increasing value to \(\theta\). -these are unit vectors in polar coordinates.But i don't know how to put "cap" on them!

Answer 4

Do you agree to write ' instead of cap? :)

Answer 5

all right let me try.

Answer 6

may I write first and then you correct my mistakes?

Answer 7

fine

Answer 8

\[e_n=-\sin \theta i+\cos \theta j\]
\[e_t=\cos \theta i +\sin \theta j\]
\[e_n '=-\theta '\cos \theta i-\theta '\sin \theta j=-\theta'e_t\]
\[r=-lcos \theta e_n\]
\[r'=l \theta'\sin \theta e_n-lcos \theta e_n '=l \theta'\sin \theta e_n+l \theta 'e_t\]

If this is good, that, by what my professor wrote, \[l \theta '\sin \theta e_n\] is supposed to be zero.

Answer 9

you guys want this?\(\huge\vec{f}\)

Answer 10

i mean the vector sign over a letter...!

Answer 11

yes, and dot too.

Answer 12

i know only the vector sign..\(\vec{f}\)
slash-first bracket open-slash-"vec"-second bracket open-"f"-second bracket close-slash-first bracket close
slash means "\"
hope u understand....!

Answer 13

for more details in latex....
http://openstudy.com/users/souvik#/updates/51d58ebbe4b059f463fe7608

Answer 14

It is all too confusing! What is 'l' you are using above?

Answer 15

sorry :D I wrote l as I did on paper... it's length of a string. I was supposed to write s.

Answer 16

u can do this...left click on the f vector i posted then "show math as" and then "tex commands "...
here u can see what i wrote..

Answer 17

do not write those what i told..
type this...
\(\vec{f}\]
just write ')' in place of ']'

Answer 18

@gorica

Answer 19

@souvik thanks, I got it :)

Answer 20

:)
too difficult to remember those code words...

Answer 21

What do you mean by "position vector equation" ?
Could you give an example using a simpler case?

Answer 22

This is all i know.

Position vector ,r is written as
r=(distance of point from origin)*unit vector pointing from origin to the point.
here distance is 's' and unit vector u are using is \(\epsilon_{t}\)

So, \(r=s\epsilon_{t}\)
The normal vector, \(\epsilon_{n}\) is perpendicular to \(\epsilon_{t}\) and related as follows (u proved somewhat similar above)
\(\epsilon_{t}'\)=\(\theta '\)\(\epsilon_{n}\) and \(\epsilon_{n}'\)=-\(\theta '\)\(\epsilon_{t}\)
So \(r'=s\theta' \epsilon_{n}\)
Similarly
\(r''=s\theta''\epsilon_{n} -s\theta'^{2}\epsilon_{t}\)

Sorry, i cannot do more!

Answer 23

Is it its equation wrt time?
Is it the (differential) equation of motion?

Answer 24

@Diwakar thank you, that's it. Now I see my mistake, I used a projection on y-axes... Thanks!