In polar-cylindrical coordinates, what is horizontal velocity?
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OpenStudy (theeric):
Hello! If you're using polar as like...
OpenStudy (theeric):
\(x\) is horizontal and \(y\) is vertical...
OpenStudy (anonymous):
in polar u use, r, theta
OpenStudy (theeric):
And instead you're using \(r\) and \(\theta\)...|dw:1376244337612:dw|
OpenStudy (theeric):
@Mashy :)
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OpenStudy (theeric):
So, @gorica , what do you have to do to find the horizontal part? Any ideas? It'll be good practice to try to think about this, but I'm sure @Mashy and I can help you get there if you're not sure!
OpenStudy (anonymous):
well m sure u alone can handle it :P.. so m out :D
OpenStudy (gorica):
I have position vector \[\vec r=r \vec {e_r}+z \vec {e_z}\]
and I have velocity vector
\[\dot {\vec r}=\dot r \vec {e_r}+r \dot \theta \vec {e_\theta}+\dot z \vec {e_z}\]
OpenStudy (theeric):
Oh, gosh, more complicated than I thought. @Mashy , please check this out when you come back on!
OpenStudy (theeric):
I messaged a person who might be able to understand that.
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