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Find my Mistake? (Moment of inertia)

  1. Apr 18, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the moment of inertia of a spherical shell

    2. Relevant equations


    3. The attempt at a solution
    So I've been mulling over this for about two hours now and haven't figured out where I've made a mistake. Here's what I've done:

    I know I can do the integral by doing ##\int ^{r} _{-r} \frac{8}{3}\pi \rho da##, with ##da## a surface element, but I wanted to do it sort of "from scratch" sort of.

    I start with ##\int r^2 dm = \int r^2 A(r) dz## and then sub ##R \cos \theta = r## (the distance from ##dm## to ##z##) and ##A(r) = 2 \pi r \rho = 2 \rho\pi R \cos \theta## (circumference of the disk as a function of theta). We then have ##\int ^{R} _{-R} \rho\pi R^3 \cos ^3 \theta dz##. Next we sub in ##dz = \frac{r d \theta}{\cos \theta}## to get ##\int ^{\pi / 2} _{-\pi/2}2 \pi\rho R^4 \cos ^2 \theta d\theta ##. This doesn't come out to the correct expression, clearly.

    I know there must be some dumb mistake in there, but I just can't seem to find it. Any help would be appreciated.
     
    Last edited by a moderator: Apr 18, 2016
  2. jcsd
  3. Apr 18, 2016 #2

    Charles Link

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    I think your ## R \cos(\theta) ## should be ##R \sin(\theta) ##. Would suggest you use spherical coordinates instead of integrating along z.
     
    Last edited: Apr 18, 2016
  4. Apr 18, 2016 #3

    SammyS

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    How are you accounting for the thickness of the shell ?
     
  5. Apr 18, 2016 #4
    I didn't but I shouldn't have to, right? It's a spherical shell of zero thickness.
     
  6. Apr 18, 2016 #5

    SammyS

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    In that case you will have a hard time if you use a finite density.
     
  7. Apr 18, 2016 #6

    Charles Link

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    Suggestion: You can let the density ## \delta ## of mass per unit area be ## \delta=M/(4\pi R^2) ## .
     
  8. Apr 18, 2016 #7
    Sure we can get it to work with just ##dm = \frac{M}{A} dA = \frac{M}{A}R d \theta 2 \pi r##, but this contradicts what I had found for it, which was (say ##\rho = 1##)##dm = 2 \pi r dz = 2 \pi (R \cos \theta )(\frac{R d \theta}{\cos\theta})## which leads to the integral after ##r^2## is introduced
    $$\int ^{\pi / 2} _{-\pi/2}2 \pi R^4 \cos ^2 \theta d\theta$$. I guess my problem is seeing why ##dz = R d\theta##.
     
  9. Apr 18, 2016 #8

    Charles Link

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    The ## dA ## in spherical coordinates is ## dA=R^2 \sin(\theta) d\theta d\phi ##. The ## \phi ## integral is of course ## 2 \pi ##. The distance from the z-axis ## \rho=R \sin(\theta) ##. The integral is straightforward. (Your mistake is ## \cos(\theta) ## should be ## \sin(\theta) ##). Don't forget to include the density ## \delta=M/(4 \pi R^2) ##. I see you basically derived the ## dA ## for spherical coordinates yourself, but that needs to be ## \sin(\theta) ##. Incidentally, ## 2 \pi \rho dz ## is not equal to ## dA ##. (I'm using ## \rho ## in place of your ## r ##.)
     
    Last edited: Apr 18, 2016
  10. Apr 18, 2016 #9

    SammyS

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    Don't give too much help . Let OP do some of this.
     
  11. Apr 18, 2016 #10

    Charles Link

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    One more input to the OP: (@SammyS I think the OP is trying, but his methodology is clumsy). Would recommend having the ## \theta ## polar coordinate integration go from ## 0 ## to ## \pi ##. That is normally how the spherical coordinates work.
     
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