# Why is this method not valid-moments?

1. Jun 1, 2008

### 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,
$$I=\frac{3}{5}r^2m$$.
$$I=\frac{2}{5}r^2m$$.

2. Jun 1, 2008

### rock.freak667

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

How'd you arrive at this?

3. Jun 1, 2008

### gamesguru

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:
$$dm=4\delta\pi r^2 dr$$.

4. Jun 1, 2008

### 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

$$I = \int (r\sin(\theta))^2 dm$$

$$dm = 2\pi\delta r^2\sin(\theta) d\theta\; dr$$

$$0<r<R,\quad 0<\theta<\pi$$

5. Jun 1, 2008

### D H

Staff Emeritus
Here is what is wrong:
That $r$ is the distance of some differential quantity of mass $dm$ from the axis of rotation ...
... while that $r$ is the distance of some differential quantity of mass $dm$ from the center of the sphere.

6. Jun 1, 2008

### tiny-tim

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.

7. Jun 1, 2008

### gamesguru

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