Derivation of the moment of inertia of a cylinder

  • Thread starter SHawking
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  • #1
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First, this is not a homework assignment even though it may seem like a homework type question.

I have no problem with the derivation of the moment of inertia of a cylinder but am having more trouble with a sphere. I completely understand the often referenced disk method, but, I would like to take a shell method approach. THe following work is wrong conceptually, I think mathematically it is fine, and I hope someone can point me in the right direction. I dont see what one would need to do it using discs, and am assuming that some slight variation on the following should work:
I=[tex]\int[/tex]r2dm
p=dm/dv
dvp=dm
I=[tex]\int[/tex]r2pdv
v=4/3[tex]\pi[/tex]r3
dv=4[tex]\pi[/tex]r2dr
I=[tex]\int[/tex]r2p4[tex]\pi[/tex]r^2dr
I=4[tex]\pi[/tex]p[tex]\int[/tex]r^4dr (Evluated between 0 and R)
I=4[tex]\pi[/tex]pR5/5
p=m/v=m/(4/3 [tex]\pi[/tex] r3)
Plugging that in and simplifying I get
I=3/5mr2, though I need 2/5mr2.
I think the problem lies in my set up, conceptually, and that I need to set up an equation somewhere subtracting two values (R from r?) though I am not sure.

If you could help me out I would really appreciate it.


And again, I understand and have no problem with the disk method for this, I am just trying to figure out what this method is not working.


Thanks!
 

Answers and Replies

  • #2
Doc Al
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The problem is that you are mixing up two meanings of 'r'.

I=[tex]\int[/tex]r2dm
Here 'r' refers to the distance from the axis of rotation.
v=4/3[tex]\pi[/tex]r3
Here 'r' refers to the radius of the sphere. Big difference!

That's why using a 'shell method' is not such a good idea. What's the moment of inertia of a shell? The various parts are all at different distances from the axis, so it's not a trivial integration.
 
  • #3
arildno
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For the record, here is one way how you should proceed to determine the moment of inertia of a ballof radius R:

Volume element:
[tex]dv=r^{2}\sin\phi{d\phi}{d\theta}dr[/tex]
Distance d to axis:
[tex]d=r\sin\phi[/tex]
Thus,
[tex]I=\int{d}^{2}dm=\rho\int_{0}^{R}\int_{0}^{\pi}\int_{0}^{2\pi}r^{4}\sin^{3}\phi{d\theta}{d\phi}dr=2\pi\rho\int_{0}^{R}\int_{0}^{\pi}r^{4}(\sin\theta-\sin\theta\cos^{2}\theta)d\phi{dr}=2\pi\rho\int_{0}^{R}r^{4}(2-\frac{2}{3})dr=\frac{8\pi\rho}{3}\int_{0}^{R}r^{4}dr=\frac{2}{5}Mr^{2}[/tex]
 
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