Center of mass of solid hemisphere.

  • Thread starter AlchemistK
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  • #1
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Homework Statement


I've just started with center of mass, and instead of the method in the book, I tried solving the center of mass for a solid hemisphere using angle as a variable, but the answer didn't match.
A small disk of mass "dm" is taken, which subtends an angle "dθ" at the center.The total mass of the hemisphere is "M". Please refer to the attachment for the figure and my full attempt.
The solutions elsewhere use distance from the center of the hemisphere as a variable.
Where have i gone wrong?


Homework Equations


y(cm) = ∫y dm / ∫ dm = 1/M (∫y dm)


The Attempt at a Solution


Please refer to the attachment.
 

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Last edited:

Answers and Replies

  • #2
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That is not the right volume element. The radius of the chosen disk is not the "same" at the two sides of the infinitesimal angle as the sphere has a definite curvature. What you calculated would be the volume element of a cylinder.
 
  • #3
vela
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That is not the right volume element. The radius of the chosen disk is not the "same" at the two sides of the infinitesimal angle as the sphere has a definite curvature. What you calculated would be the volume element of a cylinder.
That's not the problem. The effect on the volume element due to the difference in radii is a second-order effect, so it is negligible. The problem is the thickness of the slab dy varies with θ, and that isn't accounted for in the OP's attempt.
 
  • #4
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Yes, Vela that's right :)
 
  • #5
SammyS
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Homework Statement


I've just started with center of mass, and instead of the method in the book, I tried solving the center of mass for a solid hemisphere using angle as a variable, but the answer didn't match.
A small disk of mass "dm" is taken, which subtends an angle "dθ" at the center.The total mass of the hemisphere is "M". Please refer to the attachment for the figure and my full attempt.
The solutions elsewhere use distance from the center of the hemisphere as a variable.
Where have i gone wrong?

Homework Equations


y(cm) = ∫y dm / ∫ dm = 1/M (∫y dm)

The Attempt at a Solution


Please refer to the attachment.
In your work, you have that [itex]y=R\sin(\theta)\,.[/itex]

So, how is the thickness, dy, related to dθ ?
 
  • #6
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Lets zoom in at the curved part of the disk, since dθ is very small, it can be considered as a straight line, there in the triangle formed, dy = R sinθ dθ. Using this as the thickness and R cosθ as the radius we can find the volume of the disk, but there will be the part that is left in the 3D triangular ring that is formed, how do I calculate that part? (please refer to the new attachment for the diagram)
 

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