Mass moment of inertia derivations

Definitions

Mass moment of inertia of a body about an axis is $I = \int r^2\,dm$, where $r$ is the perpendicular distance of the element $dm$ from the axis (unit kg·m²). It measures resistance to angular acceleration, as mass does for linear acceleration.

Radius of gyration $k$ is the distance from the axis at which the whole mass could be concentrated to give the same moment of inertia: $I = mk^2$, so $k = \sqrt{I/m}$.

Parallel axis theorem

Statement: the moment of inertia about any axis equals the moment of inertia about a parallel axis through the mass centre plus $md^2$:

$$I = I_G + md^2$$

Proof: take the mass centre G as origin of x-y axes, with the axis G along z and the parallel axis passing through the point $(d, 0)$. For an element $dm$ at $(x,y)$:

$$I = \int [(x - d)^2 + y^2]\,dm = \int (x^2 + y^2)\,dm - 2d\int x\,dm + d^2\int dm$$

The first term is $I_G$. The second is zero because $\int x\,dm = m\bar x = 0$ (G is the origin). The third is $md^2$. Hence $I = I_G + md^2$.

(a) Slender rod, length L, mass m

Mass per unit length $= m/L$. About the transverse axis through the centre:

$$I_G = \int_{-L/2}^{L/2} x^2 \frac{m}{L}\,dx = \frac{m}{L}\cdot\frac{2}{3}\left(\frac{L}{2}\right)^3 = \frac{mL^2}{12}$$

About one end (parallel axis theorem, $d = L/2$):

$$I_{end} = \frac{mL^2}{12} + m\left(\frac{L}{2}\right)^2 = \frac{mL^2}{3}$$

(b) Solid cylinder, radius R, length h, density $\rho$

Take a thin cylindrical shell of radius $r$ and thickness $dr$: $dm = \rho\,(2\pi r\,h)\,dr$.

$$I = \int_0^R r^2\,\rho\,2\pi h\,r\,dr = 2\pi\rho h\,\frac{R^4}{4} = \frac{1}{2}(\rho\pi R^2 h)R^2 = \frac{1}{2}mR^2$$

(independent of the length h).