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yungman
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This is calculus question, but I don't think calculus really cover this topic in either multi-variables or even vector calculus classes. This is really more common problem in electrodynamics.
Let R be position vector that trace out a circle or radius a with constant velocity. In rectangular coordinates:
[tex] \vec R = \hat x acos \omega t +\hat y a sin\omega t [/tex]
[tex]\vec v = \frac {d\vec R}{dt}= -\hat x a\omega sin \omega t +\hat y a\omega cos\omega t [/tex]This is very straight forward. So I am going to try to do the derivative in cylindrical coordinates. For cylindrical coordinates;
[tex] \vec R =\hat r a [/tex]
How do you take the derivative? There is only a constant. I know [tex] \vec v=\hat {\phi} a\omega [/tex]
The only way I know how to get v in cylindrical coordinates is doing a translation of v from rectangular coordinates to cylindrical coordinate after performing the derivative in rectangular coordinates.
Anyone can tell me a direct method to take the derivative in cylindrical coordinates?
Also for constant angular velocity [itex]\omega[/itex], angular acceleration is supposed to be zero. But if you look at in rectangular coordinates:
[tex] \vec a = \frac {d \vec v}{dt} = -\hat x a \omega^2 cos \omega t \;-\; \hat y a\omega^2sin\omega t [/tex]
Obviously it is not zero, does this mean you don't look at this as angular velocity or acceleration, but instead, look at it as in straight line components x,y?
Let R be position vector that trace out a circle or radius a with constant velocity. In rectangular coordinates:
[tex] \vec R = \hat x acos \omega t +\hat y a sin\omega t [/tex]
[tex]\vec v = \frac {d\vec R}{dt}= -\hat x a\omega sin \omega t +\hat y a\omega cos\omega t [/tex]This is very straight forward. So I am going to try to do the derivative in cylindrical coordinates. For cylindrical coordinates;
[tex] \vec R =\hat r a [/tex]
How do you take the derivative? There is only a constant. I know [tex] \vec v=\hat {\phi} a\omega [/tex]
The only way I know how to get v in cylindrical coordinates is doing a translation of v from rectangular coordinates to cylindrical coordinate after performing the derivative in rectangular coordinates.
Anyone can tell me a direct method to take the derivative in cylindrical coordinates?
Also for constant angular velocity [itex]\omega[/itex], angular acceleration is supposed to be zero. But if you look at in rectangular coordinates:
[tex] \vec a = \frac {d \vec v}{dt} = -\hat x a \omega^2 cos \omega t \;-\; \hat y a\omega^2sin\omega t [/tex]
Obviously it is not zero, does this mean you don't look at this as angular velocity or acceleration, but instead, look at it as in straight line components x,y?
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