Unit tangent vector vs principal normal vector

  • Thread starter chetzread
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  • #76
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If you recognize ##\vec r = (cos\omega t, \sin\omega t) ## as describing the unit circle, you know which of the answers for ##\kappa = {1\over \rho}## is wrong ...
Check your own post #66...

And cry out that I made two huge mistakes in post #72 o:) o:) o:) -- sorry about that !
i still couldnt figure out which part is wrong, can you point out?
 
  • #77
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i still couldnt figure out which part is wrong, can you point out?
A unit circle has a radius 1, so my guess is that the answer 1 is correct and the answer ##|\vec \omega|## is wrong. And I've found why and it's somewhat my fault because in post #72 I made two huge mistakes.



Note that we are doing two things in parallel: working out the simple example ##\vec r = (cos\omega t, \sin\omega t)## and make an inroad for your long-standing question as in post #1, #13 and #63.

- - - - - - - -

I humbly fix #72:

So far we have the unit vectors (:smile:) ##\ \ \vec T = (-\sin \omega t,\cos\omega t)\ \ ## and ##\ \ \vec N = -\vec r\ \ ## Curvature is 1 and radius is 1.

- - - - - - -

In the more general case the acceleration was decomposed into a tangential component by means of projection . You did not have a problem with that step ? Because we are going to use that to understand the magnitude of the normal component "The second equalities will be left as exercises"
 
  • #78
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A unit circle has a radius 1, so my guess is that the answer 1 is correct and the answer ##|\vec \omega|## is wrong. And I've found why and it's somewhat my fault because in post #72 I made two huge mistakes.



Note that we are doing two things in parallel: working out the simple example ##\vec r = (cos\omega t, \sin\omega t)## and make an inroad for your long-standing question as in post #1, #13 and #63.

- - - - - - - -

I humbly fix #72:

So far we have the unit vectors (:smile:) ##\ \
\vec T = (-\sin \omega t,\cos\omega t)\ \ ## and ##\ \ \vec N = -\vec r\ \ ## Curvature is 1 and radius is 1.

- - - - - -

In the more general case the acceleration was decomposed into a tangential
component
by means of projection . You did
not have a problem with that step ? Because
we are going to use that to understand the
magnitude of the normal component "The second equalities will be left as exercises"
You mean k=(T't) /(r't) is wrong?
 
  • #79
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You mean k=(T't) /(r't) is wrong?
I don't know how to divide a vector by a vector. (or perhaps I intentionally misunderstand your notation :smile: )
Anyway, yes: that ##\kappa\equiv \displaystyle {|\vec T'|\over |\vec r'|} ## evaluates to ##|\omega|/ |\omega| = 1 ##.
 
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  • #80
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I don't know how to divide a vector by a vector. (or perhaps I intentionally misunderstand your notation :smile: )
Anyway, yes: that ##\kappa\equiv \displaystyle {|\vec T'|\over |\vec r'|} ## evaluates to ##|\omega|/ |omega| = 1 ##.
T'(t) = (omega ^2) , right?
 
  • #81
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T'(t) = (omega ^2) , right?

No, not if ##\ \ \vec T = (-\sin \omega t,\cos\omega t)\ \ ##...

And:
Please please please become adept at using a sensible and consistent notation. ##\vec T'## is a vector. ##\omega^2## is a number. They can never never never be equal.
 
  • #82
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No, not if ##\ \ \vec T = (-\sin \omega t,\cos\omega t)\ \ ##...

And:
Please please please become adept at using a sensible and consistent notation. ##\vec T'## is a vector. ##\omega^2## is a number. They can never never never be equal.
##\ \ \vec T = (-w\sin \omega t, w\cos\omega t)\ \ ##...
##\ \ \vec T' = (-(w^2)\cos \omega t, (w^2)\sin\omega t)\ \ ##...
why you left out w for ##\ vec T\ ##...
Here's my previous working
 

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  • #83
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In your working (I had seen it and checked it already) in the expression for ##\vec T## you can divide out the ##\omega## in the numerator against the ##\omega## in the denominator. (That way you also get a vector that clearly has length 1). Differentiate wrt time to get ##\vec T' = -\omega \;(\cos\omega t, \sin\omega t)## ...

The expression for ##\vec T## in your working may be right (apart from the incorrect notation), but
you should still correct your expression for ##\vec N##: it is not correct.
 
  • #84
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In your working (I had seen it and checked it already) in the expression for ##\vec T## you can divide out the ##\omega## in the numerator against the ##\omega## in the denominator. (That way you also get a vector that clearly has length 1). Differentiate wrt time to get ##\vec T' = -\omega \;(\cos\omega t, \sin\omega t)## ...

The expression for ##\vec T## in your working may be right (apart from the incorrect notation), but
you should still correct your expression for ##\vec N##: it is not correct.
do you mean by cancelling off the w in the denominator and numerator ##\vec T##
and cancelling off (w^2) in the denominator and numerator ##\vec T'##
eventually, we get magnitude of ##\vec T## and ##\vec T'## equal to 1? k =1 ?
 

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  • #85
801
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In your working (I had seen it and checked it already) in the expression for ##\vec T## you can divide out the ##\omega## in the numerator against the ##\omega## in the denominator. (That way you also get a vector that clearly has length 1). Differentiate wrt time to get ##\vec T' = -\omega \;(\cos\omega t, \sin\omega t)## ...

The expression for ##\vec T## in your working may be right (apart from the incorrect notation), but
you should still correct your expression for ##\vec N##: it is not correct.
tat's weird, how can we do that?
 
  • #86
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do you mean by cancelling off the w in the denominator and numerator ##\vec T##
yes
and cancelling off (w^2) in the denominator and numerator ##\vec T'##
No. You differentiate wrt time . See post #83.
eventually, we get magnitude of ##\vec T## and ##\vec T'## equal to 1? k =1 ?
Yes for ##|\vec T|##, beccause ##\vec T## is a unit vector.
No for ##|\vec T'|## because ##\vec T'## is NOT a unit vector. (but of course ##\vec N \equiv \vec T'/|\vec T'|## IS a unit vector !)

see the example. Very useful example !​


tat's weird, how can we do that?
Because ##\omega## is a scalar. Not to be confused with possible vectors ##\vec \omega##.

In case you insist on a vector ##\vec \omega## in the example: there you have ##\vec \omega = (0,0, \omega)## pointing in the z direction. In short, with the scalar ##\omega##, the norm of ##\vec \omega ## is meant.) I agree that one can become confused, especially when sloppy or inconsistent with notation in general :rolleyes: .​
 

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