Help me sort this out

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In summary: The statement means that z2 = r2 only on the boundary of the cone. In other words, z2 = r2 only if and only if r = z on the boundary.
  • #1
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NOTE: THIS IS NOT A HW PROBLEM
While calculating the moment of inertia of a cone, I made a mistake and got confused. Although I juggled the numbers and got the right answer, I have a doubt.

Say I have to calculate the centroid of a cone z2 = r2.

0 < z < h

It is quite obvious that the x and y co-ordinates of the centroid are zero. As for the z co-ordinate it can be calculated as follows:

[tex]\int[/tex] [tex]\int[/tex] [tex]\int[/tex][tex]_{V}[/tex] r2 dV =

[tex]\int[/tex] [tex]\int[/tex] [tex]\int[/tex][tex]_{V}[/tex] r2 r dr d[tex]\Theta[/tex] dz =

[tex]\int[/tex] [tex]\int[/tex] [tex]\int[/tex][tex]_{V}[/tex] r3 dr d[tex]\Theta[/tex] dz = [tex]\Pi[/tex]h5/10

But instead, in step 2 if I substituted r = z (from the equation of the cone), I get a completely different answer.

[tex]\int[/tex] [tex]\int[/tex] [tex]\int[/tex][tex]_{V}[/tex] z2 z dr d[tex]\Theta[/tex] dz =

[tex]\int[/tex] [tex]\int[/tex] [tex]\int[/tex][tex]_{V}[/tex] z3 dr d[tex]\Theta[/tex] dz = 2[tex]\Pi[/tex]h5/5

I got a similar doubt while calculating the center of mass, wherein I plugged in z = r in the equation and got a different (and wrong) answer.

Shouldn't I get the same answer either way? Why don't I get it?

Anirudh
 
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  • #2
anirudh215 said:
Shouldn't I get the same answer either way? Why don't I get it?

Hi Anirudh! :smile:

(have a theta: θ and a pi: π :wink:)

You've probably got the limits wrong …

what limits did you use?
 
  • #3
tiny-tim said:
have a theta: θ and a pi: π

:rofl:

For the first one I used:

[tex]\int[/tex][tex]^{2\Pi}_{\Theta = 0}[/tex][tex]\int[/tex][tex]^{h}_{z = 0}[/tex][tex]\int[/tex][tex]^{z}_{r = 0}[/tex] r3 dr d[tex]\Theta[/tex] dz

For the second one I used:[tex]\int[/tex][tex]^{2\Pi}_{\Theta = 0}[/tex][tex]\int[/tex][tex]^{h}_{z = 0}[/tex][tex]\int[/tex][tex]^{z}_{z = 0}[/tex] z3 dr d[tex]\Theta[/tex] dz

I used the same limits because r = z. Is that wrong? (Obviously, it MUST be as the answer isn't right)
 
  • #4
hmm … looking again, these are different integrals …
anirudh215 said:
[tex]\int[/tex] [tex]\int[/tex] [tex]\int[/tex][tex]_{V}[/tex] r2 r dr d[tex]\Theta[/tex] dz =

[tex]\int[/tex] [tex]\int[/tex] [tex]\int[/tex][tex]_{V}[/tex] z2 z dr d[tex]\Theta[/tex] dz = …


∫∫∫ r2 is not the same as ∫∫∫ z2

z2 = r2 only on the boundary, not everywhere in the middle …

(and you can't transform away r dr anyway)
 
  • #5
tiny-tim said:
(and you can't transform away r dr anyway)

Okay I understood that r = z only on the boundary. What does the above statement mean though?
 

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