stylistic:tone
Examples

The moment of inertia of a compound pendulum constructed from a thin disc mounted at the end of a thin rod that oscillates around a pivot at the other end of the rod, begins with the calculation of the moment of inertia of the thin rod and thin disc about their respective centers of mass.

A list of moments of inertia formulas for standard body shapes provides a way to obtain the moment of inertia of a complex body as an assembly of simpler shaped bodies. The parallel axis theorem is used to shift the reference point of the individual bodies to the reference point of the assembly.

As one more example, consider the moment of inertia of a solid sphere of constant density about an axis through its center of mass. This is determined by summing the moments of inertia of the thin discs that can form the sphere whose centers are along the axis chosen for consideration. If the surface of the sphere is defined by the equation x 2 + y 2 + z 2 = R 2, {\displaystyle x^{2}+y^{2}+z^{2}=R^{2},}

then the square of the radius r {\displaystyle r} of the disc at the cross-section z {\displaystyle z} along the z {\displaystyle z} -axis is r (z) 2 = x 2 + y 2 = R 2 − z 2. {\displaystyle r(z)^{2}=x^{2}+y^{2}=R^{2}-z^{2}.}

Therefore, the moment of inertia of the sphere is the sum of the moments of inertia of the discs along the z {\displaystyle z} -axis, I C, sphere = ∫ − R R 1 2 π ρ r (z) 4 d z = ∫ − R R 1 2 π ρ (R 2 − z 2) 2 d z = 1 2 π ρ − R R = π ρ (1 − 2 3 + 1 5) R 5 = 2 5 m R 2, {\displaystyle {\begin{aligned}I_{C,{\text{sphere}}}&=\int _{-R}^{R}{\tfrac {1}{2}}\pi \rho r(z)^{4}\,dz=\int _{-R}^{R}{\tfrac {1}{2}}\pi \rho \left(R^{2}-z^{2}\right)^{2}\,dz\\&={\tfrac {1}{2}}\pi \rho \left_{-R}^{R}\\&=\pi \rho \left(1-{\tfrac {2}{3}}+{\tfrac {1}{5}}\right)R^{5}\\&={\tfrac {2}{5}}mR^{2},\end{aligned}}} where m = 4 3 π R 3 ρ {\textstyle m={\frac {4}{3}}\pi R^{3}\rho } is the mass of the sphere.