A point mass does not have a moment of inertia around its own axis, but using the parallel axis theorem a moment of inertia around a distant axis of rotation is achieved.

This expression assumes that the rod is an infinitely thin (but rigid) wire. This is a special case of the thin rectangular plate with axis of rotation at the center of the plate, with w = L and h = 0.

This expression assumes that the rod is an infinitely thin (but rigid) wire. This is also a special case of the thin rectangular plate with axis of rotation at the end of the plate, with h = L and w = 0.

This is a special case of a torus for a = 0 (see below), as well as of a thick-walled cylindrical tube with open ends, with r₁ = r₂ and h = 0.

This is a special case of the solid cylinder, with h = 0. That ${\displaystyle I_{x}=I_{y}={\frac {I_{z}}{2}}\,}$ is a consequence of the perpendicular axis theorem.

${\displaystyle I_{z}={\frac {1}{2}}m(r_{1}^{2}+r_{2}^{2})}$

${\displaystyle I_{x}=I_{y}={\frac {1}{4}}m(r_{1}^{2}+r_{2}^{2})}$

${\displaystyle I_{z}={\frac {1}{2}}\pi \rho _{A}(r_{2}^{4}-r_{1}^{4})}$

${\displaystyle I_{x}=I_{y}={\frac {1}{4}}\pi \rho _{A}(r_{2}^{4}-r_{1}^{4})}$

This expression assumes that the shell thickness is negligible. It is a special case of the thick-walled cylindrical tube for r₁ = r₂. Also, a point mass m at the end of a rod of length r has this same moment of inertia and the value r is called the radius of gyration.

This is a special case of the thick-walled cylindrical tube, with r₁ = 0.

${\displaystyle I_{z}={\frac {1}{2}}m\left(r_{2}^{2}+r_{1}^{2}\right)=mr_{2}^{2}\left(1-t+{\frac {t^{2}}{2}}\right)}$   ^{[1] [2]}
where t = (r₂ − r₁)/r₂ is a normalized thickness ratio;
${\displaystyle I_{x}=I_{y}={\frac {1}{12}}m\left(3\left(r_{2}^{2}+r_{1}^{2}\right)+h^{2}\right)}$
The above formula is for the xy plane passing through the center of mass, which coincides with the geometric center of the cylinder. If the xy plane is at the base of the cylinder, i.e. offset by ${\displaystyle d={\frac {h}{2}},}$ then by the parallel axis theorem the following formula applies:
${\displaystyle I_{x}=I_{y}={\frac {1}{12}}m\left(3\left(r_{2}^{2}+r_{1}^{2}\right)+4h^{2}\right)}$

${\displaystyle I_{z}={\frac {\pi \rho h}{2}}\left(r_{2}^{4}-r_{1}^{4}\right)}$

${\displaystyle I_{x}=I_{y}={\frac {\pi \rho h}{12}}\left(3(r_{2}^{4}-r_{1}^{4})+h^{2}(r_{2}^{2}-r_{1}^{2})\right)}$

${\displaystyle I_{\mathrm {solid} }={\frac {1}{20}}ms^{2}\,\!}$

${\displaystyle I_{\mathrm {hollow} }={\frac {1}{12}}ms^{2}\,\!}$ ^[3]

${\displaystyle I_{x,\mathrm {hollow} }=I_{y,\mathrm {hollow} }=I_{z,\mathrm {hollow} }={\frac {39\phi +28}{90}}ms^{2}}$

${\displaystyle I_{x,\mathrm {solid} }=I_{y,\mathrm {solid} }=I_{z,\mathrm {solid} }={\frac {39\phi +28}{150}}ms^{2}\,\!}$ (where ${\displaystyle \phi ={\frac {1+{\sqrt {5}}}{2}}}$) ^[3]

${\displaystyle I_{x,\mathrm {hollow} }=I_{y,\mathrm {hollow} }=I_{z,\mathrm {hollow} }={\frac {\phi ^{2}}{6}}ms^{2}}$

${\displaystyle I_{x,\mathrm {solid} }=I_{y,\mathrm {solid} }=I_{z,\mathrm {solid} }={\frac {\phi ^{2}}{10}}ms^{2}\,\!}$ ^[3]

When the cavity radius r₁ = 0, the object is a solid ball (above).

When r₁ = r₂, ${\displaystyle {\frac {r_{2}^{5}-r_{1}^{5}}{r_{2}^{3}-r_{1}^{3}}}={\frac {5}{3}}r_{2}^{2}}$, and the object is a hollow sphere.

${\displaystyle I_{z}={\frac {3}{10}}mr^{2}\,\!}$

About an axis passing through the tip:
${\displaystyle I_{x}=I_{y}=m\left({\frac {3}{20}}r^{2}+{\frac {3}{5}}h^{2}\right)\,\!}$  ^[4]
About an axis passing through the base:
${\displaystyle I_{x}=I_{y}=m\left({\frac {3}{20}}r^{2}+{\frac {1}{10}}h^{2}\right)\,\!}$
About an axis passing through the center of mass:
${\displaystyle I_{x}=I_{y}=m\left({\frac {3}{20}}r^{2}+{\frac {3}{80}}h^{2}\right)\,\!}$

For a similarly oriented cube with sides of length ${\displaystyle s}$, ${\displaystyle I_{\mathrm {CM} }={\frac {1}{6}}ms^{2}\,\!}$

For a cube with sides ${\displaystyle s}$, ${\displaystyle I={\frac {1}{6}}ms^{2}\,\!}$.

For a cube with sides ${\displaystyle s}$, ${\displaystyle I={\frac {1}{6}}ms^{2}\,\!}$.

${\displaystyle {\mathbf {x} }}$

${\displaystyle \rho ({\mathbf {x} })=m{\frac {\exp \left(-{\frac {1}{2}}{\mathbf {x} }^{\mathrm {T} }{\boldsymbol {\Sigma }}^{-1}{\mathbf {x} }\right)}{\sqrt {(2\pi )^{2}|{\boldsymbol {\Sigma }}|}}}}$

${\displaystyle I={\begin{bmatrix}{\frac {2}{3}}mr^{2}&0&0\\0&{\frac {2}{3}}mr^{2}&0\\0&0&{\frac {2}{3}}mr^{2}\end{bmatrix}}}$

${\displaystyle I={\begin{bmatrix}{\frac {1}{3}}ml^{2}&0&0\\0&0&0\\0&0&{\frac {1}{3}}ml^{2}\end{bmatrix}}}$

${\displaystyle I={\begin{bmatrix}{\frac {1}{12}}ml^{2}&0&0\\0&0&0\\0&0&{\frac {1}{12}}ml^{2}\end{bmatrix}}}$

${\displaystyle I={\begin{bmatrix}{\frac {1}{12}}m(3r^{2}+h^{2})&0&0\\0&{\frac {1}{12}}m(3r^{2}+h^{2})&0\\0&0&{\frac {1}{2}}mr^{2}\end{bmatrix}}}$

${\displaystyle I={\begin{bmatrix}{\frac {1}{12}}m(3(r_{2}^{2}+r_{1}^{2})+h^{2})&0&0\\0&{\frac {1}{12}}m(3(r_{2}^{2}+r_{1}^{2})+h^{2})&0\\0&0&{\frac {1}{2}}m(r_{2}^{2}+r_{1}^{2})\end{bmatrix}}}$