Moment of Inertia of Hollow Cylinder Derivation

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  • #1
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Main Question or Discussion Point

For a uniform, hollow cylinder, why is this derivation wrong?

M = mass of whole solid cylinder
m = mass of missing cylindrical piece
R = radius of whole cylinder
r = radius of missing cylindrical piece

moment of inertia = moment of inertia of whole cylinder - moment of inertia of missing cylindrical piece

I = MR2/2 - mr2/2

m/M = pi*r2*h/pi*R2*h = r2/R2

m = M*r2/R2

I = MR2/2 - M*r4/2R2

I = MR4/2R2 - M*r4/2R2

I = M/2R2*(R4 - r4)
 

Answers and Replies

  • #2
Delta2
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It seems correct, it is only that the result contains in its expression the mass M of the whole cylinder and not the mass of the corresponding hollow cylinder. If you substitute [itex]M=\frac{M_{h}R^2}{R^2-r^2}[/itex] you ll get the usual expression for the inertia of a hollow cylinder [itex]I=\frac{1}{2}M_h(R^2+r^2)[/itex].
 
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  • #3
olivermsun
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It's fine as written. If you want to express ##I## using the mass of the outer shell only, call it ##M^\prime,## then you have to use ##M^\prime = M - m = M (1 - r^2/R^2)## to get the usual form for ##I##.
 
  • #4
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Thanks guys. Just a matter of confusion due to the specific application of the formula.
 

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