Is Center of Mass Relevant in Calculating Solid Angle of a Disk?

  • Thread starter jlmac2001
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In summary: For a cylindrical plate, the surface area would be \pi a^2. Therefore, in summary, to find the solid angle subtended at the origin by a thin circular disk of radius a, whose center is a distance b from the origin and where the normal to the disk is parallel, you can use the formula d\Omega = {\frac{dA_s}{r^2}} and approximate the solid angle with \pi {a^2}\over{b^2}. It is not necessary to find the center of mass to solve this problem.
  • #1
jlmac2001
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Question: Find the solid angle subtended at the origin by a thin circular disk of radius a, whose cente is a distance b from the origin and where the normal to the disk is parallel.

Do I have to find the center of mass to solve this question?
 
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  • #2
Originally posted by jlmac2001
Question: Find the solid angle subtended at the origin by a thin circular disk of radius a, whose cente is a distance b from the origin and where the normal to the disk is parallel.

Do I have to find the center of mass to solve this question?

"the normal to the disk is parallel"?? Parallel to what?

The problem would be relatively easy if the normal to the disk were directed at the origin. It's quite a bit harder if the normal is parallel to the xy-plane or parallel to the z-axis.
 
  • #3
If the normal of the disk is parallel to the direction from the orgin, the incremental solid angle is defined by

[tex] d\Omega = {dA_s}\over{r^2) [/tex]

where the numerator is the surface area of the shape and the demoninator is the distance from the orgin. For a cylindrical plate the surface area would be [tex] \pi a^2 [/tex]. For your problem, the solid angle could be approximated by

[tex] \pi {a^2}\over{b^2} [/tex].
 
  • #4
should be

[tex] d\Omega = {\frac{dA_s}{r^2}} [/tex]
 

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