4. Moment of Inertia

Moment of Inertia depends on the distribution of mass around the fixed axis of rotation

I = \sum_i m_i r_i^2

where $r$ is the distance to the axis

Kinetic Energy

K = \dfrac{1}{2}mv^2 = \dfrac{1}{2}m\omega^2r^2

we will assume $\omega$ is constant

K_{\text{total}} = \sum_{i} \dfrac{1}{2}m_i r_{i}^2 \omega_{i}^2 = \dfrac{1}{2} I \omega^2

\boxed{K_{\text{total}} = \dfrac{1}{2} I \omega^2}

For a system of point masses:

I = \sum_i m_i r_i^2

For a continuous body, divide it into small elements $(dm)$ :

I = \int r^2 \, \; dm

Express (dm) in terms of density $(\rho)$ and volume element $(dV)$ :

dm = \rho \, dV

then the moment of inertia becomes:

I = \int r^2 \, \; dm = \int r^2 \rho \, \; dV

we choose $(dV)$ such that all points in the element are approximately the same distance $(r)$ from the axis.