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The parallel axis theorem says that for an axis through a body's centre of mass and a parallel axis at perpendicular distance $d$,where $M$ is the total mass. The displacement is the distance between the two parallel axes; neither axis may be arbitrary in place of the center-of-mass axis. Expanding the squared distances of mass elements gives this formula, since the cross term integrates to zero at the centre of mass.

The moment of inertia of the uniform disc about its central perpendicular axis isThe parallel axis theorem gives $I_A=I_O+m{a}^2=3m{a}^2/2$. For angle $\beta$ from the downward resting position, the gravitational torque is $-mga\sin \beta$, so $I_A\ddot{\beta }=-mga\sin \beta$. Linearizing at $\beta =0$ gives $\ddot{\beta }+(2g/(3a))\beta =0$. Thus the uniform disc pivoted at its rim is a physical pendulum with period

Let $\mathbf{r}$ be position relative to the center of mass, and let the new origin have vector $\mathbf{c}$ relative to it. The inertia tensor about this origin isExpanding and using $\int \mathbf{r}dm=0$ proves the tensor form of the parallel axis theorem:For a uniform cube of side $L$ and mass $M$, centered coordinate integrals give $I_0=(M{L}^2/6)I$. The displacement to a vertex is $(L/2,L/2,L/2)$ up to signs. Using axes along its incident edges givesThe principal moments of inertia are $\boxed{M{L}^2/6,11M{L}^2/12,11M{L}^2/12}$. The first principal axis is the body diagonal through the vertex; every direction perpendicular to it has the repeated moment.

Let $M$ be the mass of a complete uniform sphere of radius $a$. The intact marble has $I_1=2M{a}^2/5$. Each removed bubble has mass $M/8$, radius $a/2$, and centre a distance $a/2$ from the rolling axis. By the parallel axis theorem, each removesThus $M_2=3M/4$, $I_2=2M{a}^2/5-7M{a}^2/80=5M{a}^2/16$, and $I_2/(M_2{a}^2)=5/12$. The two accelerations are therefore $(5/7)g\sin \alpha$ and $(12/17)g\sin \alpha$. From rest, distance is proportional to acceleration, so

Let a change of orthonormal coordinates be represented by a rotation matrix $R$. Since both angular momentum and angular velocity are vectors,Using $L_i=I_{ij}{\omega }_j$ in both frames givesfor every vector $\mathbf{\omega }$. Multiplication by $R^{-1}=R^T$ yieldswhich is exactly the transformation law for a rank-two tensor. Thus $I_{ij}$ is a rank-two tensor.

Put $\mathbf{r}=\mathbf{x}-\mathbf{a}$. The vector triple product givesThereforeAt the centre of mass,Expanding the first formula, the terms linear in $x_i$ vanish because the centre of mass is the origin, while $\rho {\int }_{B}dV=M$. Hence the parallel axis theorem in tensor form is

The axis $y=0,z=h$ is parallel to the $x$-axis and lies a distance $|h|$ from it. The parallel axis theorem therefore givesto leading order in the thickness.

The parallel axis theorem says that for an axis through $x$, parallel to the centre-of-mass axis in direction $\hat{d}$,LetMoments about the axis through $x_C$ add, soUsing $M_A=M_C-M_B$ and solving giveswhen the denominator is nonzero. In the degenerate equal-distance case, this inertia equation alone does not determine $M_B$; the centre-of-mass equation $M_C{x}_C=M_A{x}_A+M_B{x}_B$ supplies the remaining information.

A uniform circular disc of radius $a$ and mass $m$, pivoted on a perpendicular axis through its rim, has moment of inertia $I_A=3m{a}^2/2$ by the parallel axis theorem. Its hanging small oscillations have period $2\pi \sqrt{3a/(2g)}$. Measured from the upright position, the falling equation is $\ddot{\theta }=2g\sin \theta /(3a)$. Exact upright rest remains an unstable equilibrium; the energy level of that state describes a limiting separatrix after perturbation, with ${\dot{\theta }}^2=4g(1-\cos \theta )/(3a)$. The center acceleration and pivot reaction follow by resolving radial and tangential components.