Why is this method not valid-moments?

[gamesguru]

[SOLVED] Why is this method not valid-moments?

I'm trying to find the moment of inertia for a solid sphere, I've seen and understand the correct method, but I can't see anything wrong with this method, except the answer.
I=\int r^2 dm
m=\delta V=\frac{4}{3}\delta\pi r^3
dm=4\delta\pi r^2 dr
Thus,
I=4\delta\pi\int r^4dr
Which after some work, turns out to be, incorrectly,
I=\frac{3}{5}r^2m.
The correct answer is,
I=\frac{2}{5}r^2m.
Thanks in advance.

 

[rock.freak667]

I'm trying to find the moment of inertia for a solid sphere, I've seen and understand the correct method, but I can't see anything wrong with this method, except the answer.
I=\int r^2 dm
m=\delta V=\frac{4}{3}\delta\pi r^3
dm=4\delta\pi r^2 dr
Thus,
I=4\delta\pi\int r^4dr
Which after some work, turns out to be, incorrectly,
I=\frac{3}{5}r^2m.
The correct answer is,
I=\frac{2}{5}r^2m.
Thanks in advance.

m=\delta V=\frac{4}{3}\delta\pi r^3

How'd you arrive at this?

 

[gamesguru]

m=\delta V=\frac{4}{3}\delta\pi r^3

How'd you arrive at this?

I did this by: \delta is the density. Mass=density x volume, m=\delta V. The volume of a sphere is: V=\frac{4}{3}\pi r^3.
So,
m=\frac{4}{3}\delta \pi r^3.
Take the derivative:
\frac{dm}{dr}=4\delta\pi r^2,
rearrange:
dm=4\delta\pi r^2 dr.

 

[jostpuur]

It could be smart to denote the radius of the sphere with R, and use r for the integration variable. Using the same symbol causes confusion.

The true mistake however is that the r in the original formula is not the same r as the integration variable. The r in the original formula is the component of the \bar{r} vector on the normal plane of the rotation axis.

Would these look good? I cannot guarantee they are right

<br /> I = \int (r\sin(\theta))^2 dm<br />

<br /> dm = 2\pi\delta r^2\sin(\theta) d\theta\; dr<br />

<br /> 0&lt;r&lt;R,\quad 0&lt;\theta&lt;\pi<br />

 

[D H]

I'm trying to find the moment of inertia for a solid sphere, I've seen and understand the correct method, but I can't see anything wrong with this method, except the answer.

Here is what is wrong:

I=\int r^2 dm

That r is the distance of some differential quantity of mass dm from the axis of rotation ...

dm=4\delta\pi r^2 dr

... while that r is the distance of some differential quantity of mass dm from the center of the sphere.

 

[tiny-tim]

m=\delta V=\frac{4}{3}\delta\pi r^3
dm=4\delta\pi r^2 dr

Hi gamesguru!

You're integrating over spherical shells of constant radius.

But Moment of Inertia is about an axis, never about a point.

So you need to integrate over cylindrical shells of constant radius.

 

[gamesguru]

Thanks D H, I see the problem now. And I see why using this method with a cylinder, by chance, works.