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Moment of inertia of a solid of rotation

  • Thread starter Grand
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Homework Statement


The first quadrant area bounded by the curve
[tex]x^3+y^3=8[/tex]
is rotated around y-axis to give a solid of rotation. The question asks for an integral which represents the solid's moment of inertia around the axis.

My answer is:

[tex]I_y=M\frac{\int_0^2x^2ydx}{\int_0^2x^2dx}[/tex]

but seems like I'm wrong. Could someone help me out, please?
 

Answers and Replies

  • #2
tiny-tim
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  • #3
SammyS
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The Moment of Inertial of a cylindrical shell of mass, m, and radius, R, is: m(R2).
 
  • #4
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We need to integrate over the height, that is from 0 to 2, and for each y we need to account for a disk of radius
[tex]x=\sqrt[3]{(8-y^3)}[/tex]
which moment of inertia is:
[tex]dI=\rho x^2 dy=\rho (8-y^3)^{2/3}dy[/tex]
For the whole body this is:
[tex]I=\int_0^2\rho (8-y^3)^{2/3} dy[/tex]
[tex]I=\frac{m}{V}\int_0^2 (8-y^3)^{2/3} dy[/tex]

I think this is wrong.
 
  • #5
SammyS
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The Moment of Inertial of a disk of mass, m, and radius, R, is: (½)m(R2).

However, I think it would be better to use cylindrical shells as tiny-tim recommended.

The last integral you give, [itex]\displaystyle \int_0^2 (8-y^3)^{2/3} dy[/itex] , gives the volume of your solid of revolution.
 
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  • #6
tiny-tim
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Hi Grand! :smile:
We need to integrate over the height, that is from 0 to 2, and for each y we need to account for a disk of radius
[tex]x=\sqrt[3]{(8-y^3)}[/tex]
which moment of inertia is:
[tex]dI=\rho x^2 dy=\rho (8-y^3)^{2/3}dy[/tex]
This is wrong for two reasons:

i] the moment of inertia of a disc is 1/2 mr2, not mr2

ii] you can't just use ρ instead of m … you need m for the whole disc, which is ρr2dy, so the whole formula for moment of inertia of the disc is 1/2 ρr4dy

But why are you making things so difficult for yourself? As SammyS :smile: says, cylindrical shells would be much easier.

Try again. :smile:
 

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