# Moment of inertia - formula derivation: Missing $\frac{1}{2}$

I'm trying to deduce the formula of the moment of inertia of an object of rotation. The general formula for the moment of inertia is declared as:

$$J=m*r^2 =\sum{m_i * r_i^2}$$

If I replace $m_i$ of the $\sum{m_i * r_i^2}$ with $\int dm$ (where dm are the masses) and $r_i^2$ with the $\int(y)^2dx$, I get $J=\int y^2 dm = \int y^2*(\rho\space dV)$ (remember: $(\rho\space dV)$ since $\rho=\frac{m}{V}$)

Furthermore $V=\int{\pi*y^2}dx$ leads me to $J = \int y^2*(\rho\space dV) = \int y^2 * \rho*\pi*y^2 dx =$

$$J=\pi*\rho*\int y^4 dx$$

Now my question: If I compare my formula with the formula it should be, I perceive that there is $\frac{1}{2}$ missing.

$$J=\frac{\pi*\sigma}{{\color{red}2}}*\int y^4 dx$$

What mistake did I make?

-
What is the object you are trying to calculate the moment of inertia of? – Rahul Jun 5 '12 at 20:01
By the way, in mathematical notation, it is often preferable to omit the multiplication signs ($\times$), and in any case asterisks ($*$) usually denote something other than multiplication. – Rahul Jun 5 '12 at 20:03
It should be a general formula for every object rotated. – libjup Jun 5 '12 at 20:03
Is this a solid of revolution or a hollow surface of revolution? You use $\rho$ for the mass per unit volume, but the formula you want uses $\sigma$, which usually denotes mass per unit area. – Rahul Jun 5 '12 at 20:43
It's a solid of revolution about an axis. What's the difference between $\rho$ and $\sigma$? Aren't both densities? – libjup Jun 5 '12 at 20:53

The problem is that you didn't trace the meaning of $y$ through the formulas. The $y$ in

$$J=\int y^2\,\mathrm dm$$

is the $y$ coordinate of a mass element in the body, whereas the $y$ in

$$V=\int\pi y^2\,\mathrm dx$$

and in

$$J=\frac{\pi\rho}2\int y^4\,\mathrm dx$$

is the $y$ coordinate of the curve forming the boundary of the body. In replacing $\mathrm dm$ by $\rho\,\mathrm d V$ with $\mathrm dV$ representing an entire infinitesimal disk of the body of rotation, you'd have to average $y^2$ over the disk. However, I get a factor of $1/4$ from that, not $1/2$, so there might be another problem, too.

-
Thank you very much for your prompt answer! Why do the y-coordinates of the mass element and the curve forming the body differ? Don't I need to calculate the mass of the infinitesimal disk? – libjup Jun 5 '12 at 20:45