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?
Best Answer
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.