# 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 [Broken]
Can someone please explain to me why ir = cosθ . i + sinθ . j and iθ = -sinθ.i + cosθ. j ?

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

(ir, iθ, i and j are vectors.)

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Homework Helper
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.

Homework Helper
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 ir = cosθ . i + sinθ . j and iθ = -sinθ.i + cosθ. j ?

Why isn't for example iθ = -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?

He does, however, seem to me to be wrong about one point: $-cos(\theta)i+ sin( \theta)j$ 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.