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Centre of mass of a solid hemisphere. What am I doing wrong?

  1. May 28, 2015 #1
    2sayzr6.jpg

    I have to calculate the centre of mass of the drawn hemisphere. I don't understand where I make mistakes. The process I followed on the above image is the following:

    (I) Here I simply translated position vector components from cartesian to polar coordinates.
    (II) Formula of x-component of position vector of centre of gravity (this position vector clearly lies on the x-axis, so I'm interested only in this component)
    (III) This is the volume element for polar coordinates system.

    After that, I just put (III) into (II).

    The problem is that, the result of the last (triple) integral (computable easily with wolfram alpha) seems to result zero! I thought that the mistake could be in the integration limits, but they look to be correct. Where am I doing wrong?

    Thank you very much in advance.
     
  2. jcsd
  3. May 28, 2015 #2

    BvU

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    Hi Marv,

    Check your integration limits anyway. Looks to me that you aren't integrating over the righthand half of the sphere....
     
  4. May 28, 2015 #3
    I tried to understand why what I wrote should be wrong, but I don't see the mistake. Which limits do you think are correct?
     
  5. May 28, 2015 #4

    BvU

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    Well, you havent shown what ##\theta## ande ##\phi## are, so I can't tell, but you can !
    [edit] but it's implicit in your coordinate transformation. Check if Rx > 0 always during the integration...
     
  6. May 28, 2015 #5
    θ = angle between position vector and x-axis
    ϕ = angle between position vector and z-axis
     
  7. May 28, 2015 #6
    I used an ambiguous notation: I should precise, that the initial vector r (see (I) ) is a general position vector. The vector rx I use after that refers to the position vector of the centre of mass (not the general one!).
     
  8. May 28, 2015 #7

    BvU

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    Check if your rx is always positive.
    Check if your ry really runs from -1 to +1
    Check if your rz really runs from -1 to +1

    As an exercise, do the integration yourself, instead of with the Wolfram crutches.
     
  9. May 28, 2015 #8

    BvU

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    Nope. Not if ##z = r\cos \theta##
     
  10. May 28, 2015 #9
    Check terms in $$ r_{cm} = \frac{1}{V}\int\int\int \,r\,dV $$
     
  11. May 28, 2015 #10
    Thanks you all a lot :)
     
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