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# Polar co-ordinates to Cartesian co-ordinates

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1. Hi,

I'm revising for a mechanics exam by going through past papers.

One of the questions is about a wheel of radius R. It gives me the density P and asks for the mass M.

So P(r,θ) = ae^(-k(r^2)) which is in polar co-ordinates but I think I need to change it to cartesian so then I can use the equation M = ∫∫∫ P(x,y,z) dx dy dz.

It also gives that 0 ≤ θ ≤ 2π. This prob comes into it somehow.

Can anyone help?

Thanks
Amber
2. (Original post by xAmb)
Hi,

I'm revising for a mechanics exam by going through past papers.

One of the questions is about a wheel of radius R. It gives me the density P and asks for the mass M.

So P(r,θ) = ae^(-k(r^2)) which is in polar co-ordinates but I think I need to change it to cartesian so then I can use the equation M = ∫∫∫ P(x,y,z) dx dy dz.

It also gives that 0 ≤ θ ≤ 2π. This prob comes into it somehow.

Can anyone help?

Thanks
Amber
I presume your wheel is a cylinder of fixed width, say h. I think the intention is that you use an alternative version of your integral for M but one that is in cylindrical coordinates rather than 3-dimensional Cartesian coordinates. Does this ring any bells?
3. (Original post by davros)
I presume your wheel is a cylinder of fixed width, say h. I think the intention is that you use an alternative version of your integral for M but one that is in cylindrical coordinates rather than 3-dimensional Cartesian coordinates. Does this ring any bells?
Thanks, I'm not too sure how to go about it but I'll try searching cylindrical coordinates.

I know the answer is M = πa/k [1-e^(-k(r^2))]
4. (Original post by xAmb)
Thanks, I'm not too sure how to go about it but I'll try searching cylindrical coordinates.

I know the answer is M = πa/k [1-e^(-k(r^2))]
Is it actually a wheel or a 2-dimensional disc? Your answer looks right (assuming you mean R not r) for a 2-d integral over r and theta, using the density you've given, but for a 3-d wheel (like a cylinder) I'd expect the thickness to be in there somewhere (unless it has thickness 1 unit).
5. (Original post by davros)
Is it actually a wheel or a 2-dimensional disc? Your answer looks right (assuming you mean R not r) for a 2-d integral over r and theta, using the density you've given, but for a 3-d wheel (like a cylinder) I'd expect the thickness to be in there somewhere (unless it has thickness 1 unit).
Yes it says a wheel of radius R. IT doesn't mention anything about thickness tho.

Question:

A wheel of radius R is made from non-uniform composite material with a density given by P(r,θ) = ae^(-k(r^2)).
Show that the Mass of the wheel is M = πa/k [1-e^(-k(r^2))].
6. (Original post by xAmb)
Yes it says a wheel of radius R. IT doesn't mention anything about thickness tho.

Question:

A wheel of radius R is made from non-uniform composite material with a density given by P(r,θ) = ae^(-k(r^2)).
Show that the Mass of the wheel is M = πa/k [1-e^(-k(r^2))].
OK, well unless there's something missing it looks like your "wheel" is actually a 2 dimensional disk, so you just need to look up how to do an area integral in polar coordinates

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