Moment of inertia of a solid of rotation

[Grand]

Homework Statement

The first quadrant area bounded by the curve
x^3+y^3=8
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:

I_y=M\frac{\int_0^2x^2ydx}{\int_0^2x^2dx}

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

 

[tiny-tim]

Hi Grand!

(write "itex" rather than "tex", and it won't keep starting a new line )

I don't understand what you've done.

Slice it into cylindrical shells of radius x to x+dx, find the https://www.physicsforums.com/library.php?do=view_item&itemid=31" of each shell, and integrate over x …

what do you get? ​

 

[SammyS]

The Moment of Inertial of a cylindrical shell of mass, m, and radius, R, is: m(R²).

 

[Grand]

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
x=\sqrt[3]{(8-y^3)}
which moment of inertia is:
dI=\rho x^2 dy=\rho (8-y^3)^{2/3}dy
For the whole body this is:
I=\int_0^2\rho (8-y^3)^{2/3} dy
I=\frac{m}{V}\int_0^2 (8-y^3)^{2/3} dy

I think this is wrong.

 

[SammyS]

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, \displaystyle \int_0^2 (8-y^3)^{2/3} dy , gives the volume of your solid of revolution.

 

[tiny-tim]

Hi Grand!

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
x=\sqrt[3]{(8-y^3)}
which moment of inertia is:
dI=\rho x^2 dy=\rho (8-y^3)^{2/3}dy

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 says, cylindrical shells would be much easier.

Try again. ​