Polar coordinates in mecanics?

[Patrick.Gh]

Ok, here is my problem. I haven't taken anything vector related since at least one year ago. And back then, I wasn't such a good student.. So now my past has come back to haunt me..
I still have some basic notions, but other than that, I pretty much forgot things..

http://img393.imageshack.us/img393/8378/mec1wi0.jpg
Can someone please explain to me why i_r = cosθ . i + sinθ . j and iθ = -sinθ.i + cosθ. j ?

Why isn't for example i_θ = -cosθ.i + sinθ.j?
Our teacher told us that these two are formulas. Or do they vary in each case?

(i_r, i_θ, i and j are vectors.)

Thanks in advance for your help.

 

[CompuChip]

Well, it depends on how you choose your coordinate system. First of all, you want the unit vectors in the r and theta direction to be perpendicular to each other, and the one in the r-direction to point along the line from the origin O to M(t).
Now normally, one would choose the unit vector i along the x-axis, and j along the y-axis. Of course, for theta = 0, you see that the r-unit vector should just be i, and theta- unit vector should be along j. If you plug in theta = 0 in the sine and cosine, you will see that one of the two definitions you gave comes out wrong.

Of course, you are free to choose your i and j vectors differently (for example, i along the y-axis and j along the -x or +x-axis is a possibility, although it goes against the conventions) and you would find another expression for i_theta and i_r. That is why, before solving any mechanics problem, you should always draw a picture with all the relevant basis vectors in it, so no confusion may arise about how you set up the coordinate system.

 

[tiny-tim]

Hi Patrick.Gh!

Ok, here is my problem. I haven't taken anything vector related since at least one year ago. And back then, I wasn't such a good student.. So now my past has come back to haunt me..

ooh … scary!

Can someone please explain to me why i_r = cosθ . i + sinθ . j and iθ = -sinθ.i + cosθ. j ?

Why isn't for example i_θ = -cosθ.i + sinθ.j?

-cosθ.i + sinθ.j is the reflection of cosθ.i + sinθ.j in the x-axis.

you want the perpendicular vector, so you want their dot-product to be zero … in this case, (cosθ . i + sinθ . j).(-sinθ.i + cosθ. j) = cosθsinθ - sinθcosθ = 0.

 

[Patrick.Gh]

Hi Patrick.Gh!


ooh … scary!

I don't know why I have this feeling you are being sarcastic :P Maybe because I'm usually sarcastic, or I have been over dramatic in the post :D

-cosθ.i + sinθ.j is the reflection of cosθ.i + sinθ.j in the x-axis.

you want the perpendicular vector, so you want their dot-product to be zero … in this case, (cosθ . i + sinθ . j).(-sinθ.i + cosθ. j) = cosθsinθ - sinθcosθ = 0.

I see. So if iθ is in the opposite direction, it would still be -sinθ.i + cosθ. j since it's still perpendicular?

 

[HallsofIvy]

No, tiny-tim is never sarcastic (he's much nicer than I am). He really meant it!

He does, however, seem to me to be wrong about one point: $-cos(\theta)i+ sin(<br /> \theta)j$ is the reflection of $cos(\theta)i+ sin(\theta)j$ in the y-axis, not the x-axis. the y coordinate is the same, $sin(\theta)$ in both, only the x coordinate, $cos(\theta)$ is negated. That's a reflection in the y-axis.

 

[CompuChip]

[...] That's a reflection in the y-axis.

Which is consistent with the image you posted.