Find my Mistake? (Moment of inertia)

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Homework Statement


Find the moment of inertia of a spherical shell

Homework Equations




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.
 
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  • #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.
 
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  • #3
SammyS
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Homework Statement


Find the moment of inertia of a spherical shell

Homework Equations




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.

Homework Statement


Find the moment of inertia of a spherical shell

Homework Equations




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.
How are you accounting for the thickness of the shell ?
 
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How are you accounting for the thickness of the shell ?
I didn't but I shouldn't have to, right? It's a spherical shell of zero thickness.
 
  • #5
SammyS
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I didn't but I shouldn't have to, right? It's a spherical shell of zero thickness.
In that case you will have a hard time if you use a finite density.
 
  • #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) ## .
 
  • #7
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In that case you will have a hard time if you use a finite density.
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##.
 
  • #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 ##.)
 
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  • #9
SammyS
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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) ##.
Don't give too much help . Let OP do some of this.
 
  • #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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