Kinetic Energy of a Rigid Body

Kinetic Energy of a Rigid Body

For a rigid body whose body frame $\{B\}$ is placed with origin at the center of mass (CoM), the kinetic energy decomposes cleanly into translational and rotational pieces:
$T=\frac{1}{2}M\Vert \dot{p}{\Vert }^2+\frac{1}{2}{\omega }^{\top }\mathcal{I}\omega ,$
where $M={\sum }_i{m}_i$ is the total mass, $\dot{p}$ is the spatial velocity of the CoM, and $\omega$ is the body’s angular velocity expressed in body coordinates. The first term is the translational KE of the CoM; the second is the rotational KE about the CoM. The CoM origin makes the cross term vanish; the body frame keeps everything time-invariant.

The inertia tensor $\mathcal{I}\in {\mathbb{R}}^{3\times 3}$ is the body-frame matrix
$\mathcal{I}={\sum }_i{m}_i(\Vert q_i{\Vert }^2{I}_3-q_i{q}_i^{\top }),$
with $q_i$ the body-frame position of mass element $i$. It is symmetric and positive semi-definite. For a continuous body, replace the sum with ${\int }_B\rho (q)(\Vert q{\Vert }^2{I}_3-q{q}^{\top })dV$. Diagonal entries ${\mathcal{I}}_{kk}$ are moments of inertia about each body axis; off-diagonal entries are products of inertia. Aligning body axes with the body’s symmetry axes (the principal axes) makes $\mathcal{I}$ diagonal.

Stacking translational and angular velocity into the body twist $V^b=(v^b,{\omega }^b)\in {\mathbb{R}}^6$ gives the compact quadratic form
$T=\frac{1}{2}(V^b{)}^{\top }\mathcal{M}{V}^b,\mathcal{M}=\left[\begin{array}{cc}M{I}_3& 0\\ 0& \mathcal{I}\end{array}\right]\in {\mathbb{R}}^{6\times 6},$
where $\mathcal{M}$ is the generalized inertia matrix of the body. $\mathcal{M}$ is block-diagonal because of the CoM origin and constant in time because of the body frame. Both blocks are positive semi-definite, so $T\ge 0$ always.

For a system of $n$ rigid bodies (the links of a manipulator), kinetic energies add:
$T_{\text{sys}}={\sum }_{i=1}^n\frac{1}{2}(V_i^b{)}^{\top }{\mathcal{M}}_i{V}_i^b,$
each link evaluated in its own body frame. Substituting the body Jacobian $V_i^b=J_i^b(q)\dot{q}$ for a serial chain produces $T=\frac{1}{2}{\dot{q}}^{\top }M(q)\dot{q}$ with $M(q)={\sum }_i{J}_i^b(q{)}^{\top }{\mathcal{M}}_i{J}_i^b(q)$ — the system inertia matrix.

Don't confuse $\mathcal{I}$, $\mathcal{M}$, and $M(q)$ Three distinct objects with similar names. $\mathcal{I}\in {\mathbb{R}}^{3\times 3}$ is the inertia tensor of one rigid body. $\mathcal{M}\in {\mathbb{R}}^{6\times 6}$ is the generalized inertia of one rigid body, combining mass and $\mathcal{I}$. $M(q)\in {\mathbb{R}}^{n\times n}$ is the system inertia matrix of an $n$-DoF manipulator, aggregating all links via Jacobians.

Intuition

The kinetic energy of a swinging hammer has two pieces: the bulk motion of its center of mass, and the spinning about that center. König’s theorem says these never mix, provided the description is anchored to the CoM. Any other reference point introduces a translation–rotation cross term that does not vanish. The CoM is the unique point that makes the cross term zero, and the body frame is the unique frame in which the spinning component looks the same at every instant. Together they convert what looks like a coupled, time-varying mess into a tidy block-diagonal quadratic form.

The inertia tensor $\mathcal{I}$ is “mass for rotation”: the larger it is, the more torque is needed to produce a given angular acceleration. Its directional dependence (encoded in the matrix structure rather than a scalar) is what makes a figure skater spin faster when they pull their arms in — the same angular momentum corresponds to a higher $\omega$ once $\mathcal{I}$ shrinks.

Properties

Decomposition

(König Decomposition) $T=\frac{1}{2}M\Vert \dot{p}{\Vert }^2+\frac{1}{2}{\omega }^{\top }\mathcal{I}\omega$ requires the body frame to be at the CoM.

The cross term $\dot{p}\cdot (\omega \times {\sum }_i{m}_i{q}_i)$ vanishes because ${\sum }_i{m}_i{q}_i=0$ is the defining property of the CoM in body coordinates. Move the origin elsewhere and the coupling reappears, complicating every algebraic step.

(Frame Independence of $\Vert V\Vert$) Kinetic energy depends only on the magnitude of velocity, not the frame in which it is expressed.

The norm $\Vert {\dot{q}}_i{\Vert }^2$ is invariant under orthogonal change of basis, so $T$ takes the same value in spatial or body coordinates. Pick the frame that makes $\mathcal{I}$ simplest — usually the body frame aligned with principal axes.

Inertia Tensor

(Defining Formula) $\mathcal{I}={\sum }_i{m}_i(\Vert q_i{\Vert }^2{I}_3-q_i{q}_i^{\top })$, summed over all mass elements in body coordinates.

Diagonal entries ${\mathcal{I}}_{kk}={\sum }_i{m}_i(\Vert q_i{\Vert }^2-q_{i,k}^2)$ are moments of inertia; off-diagonals are products of inertia. For a continuous body replace the sum with an integral over volume against density $\rho (q)$.

(Body Frame Constancy) Expressed in body coordinates, $\mathcal{I}$ is time-invariant; in spatial coordinates it is not.

Mass elements $q_i$ are fixed in the body frame, so $\mathcal{I}$ is a constant matrix. In a spatial frame, the mass elements orbit the CoM as the body rotates and $\mathcal{I}$ becomes time-dependent — making $T$ much harder to differentiate when applying Euler–Lagrange. This is the central reason robotics dynamics uses the body frame for kinetic energy.

(Tensor Transformation Law) Under a rotation $R$, the inertia tensor transforms as ${\mathcal{I}}^{\prime }=R\mathcal{I}{R}^{\top }$.

The number of distinct invariants (eigenvalues, determinant, trace) does not change with frame choice. Calling $\mathcal{I}$ a tensor instead of a matrix highlights this frame-independence — the matrix entries depend on the frame, but the underlying object does not.

(Positive Semi-Definiteness) $\mathcal{I}⪰0$, so the rotational KE $\frac{1}{2}{\omega }^{\top }\mathcal{I}\omega \ge 0$ for any $\omega$.

For a body with positive mass and non-collinear mass distribution, $\mathcal{I}$ is in fact positive definite. A negative trace is impossible: $\mathrm{tr}(\mathcal{I})=2{\sum }_i{m}_i\Vert q_i{\Vert }^2$ is twice the polar moment.

(Parallel Axis Theorem) To shift $\mathcal{I}$ from the CoM to another point $P$ at offset $d$:

${\mathcal{I}}_P={\mathcal{I}}_{\text{CoM}}+M(\Vert d{\Vert }^2{I}_3-d{d}^{\top }).$
Useful when an inertia about a non-CoM point is given (or needed) — never re-integrate from scratch.

Generalized Inertia Matrix

(Block Structure) $\mathcal{M}=\mathrm{diag}(M{I}_3,\mathcal{I})\in {\mathbb{R}}^{6\times 6}$, block-diagonal precisely because the body frame is at the CoM.

Off-CoM origins introduce off-diagonal $3\times 3$ blocks involving the cross product with the CoM offset (a “first moment of mass”). For derivation work the CoM origin is overwhelmingly preferred.

(Aggregation Across Bodies) Energies add: $T_{\text{sys}}={\sum }_i\frac{1}{2}(V_i^b{)}^{\top }{\mathcal{M}}_i{V}_i^b$, each link in its own frame.

The coupling between links shows up later, when each $V_i^b$ is expressed via the body Jacobian as $V_i^b=J_i^b(q)\dot{q}$. Substituting yields $T=\frac{1}{2}{\dot{q}}^{\top }M(q)\dot{q}$ with system inertia matrix $M(q)={\sum }_i{J}_i^b(q{)}^{\top }{\mathcal{M}}_i{J}_i^b(q)$.

Standard Inertias (Memorize)

(Solid sphere) $\mathcal{I}=\frac{2}{5}m{R}^2{I}_3$.

Isotropic — the same about every axis through the CoM.

(Thin rod) Length $\ell$, perpendicular axis through center: ${\mathcal{I}}_{\perp }=\frac{1}{12}m{\ell }^2$. About its own axis: zero.

(Thin disk) Radius $R$, symmetry axis: ${\mathcal{I}}_{\text{axis}}=\frac{1}{2}m{R}^2$. Diameter axis: $\frac{1}{4}m{R}^2$.

(Solid cylinder) Radius $R$, length $\ell$, symmetry axis: ${\mathcal{I}}_{\text{axis}}=\frac{1}{2}m{R}^2$. Perpendicular axis through center: $\frac{1}{12}m(3R^2+{\ell }^2)$.

(Rectangular box) Sides $a\times b\times c$ about CoM, principal axes: $\mathcal{I}=\frac{m}{12}\mathrm{diag}(b^2+c^2,a^2+c^2,a^2+b^2)$.

Derivation

From point-mass sum to $T=\frac{1}{2}M\Vert \dot{p}{\Vert }^2+\frac{1}{2}{\omega }^{\top }\mathcal{I}\omega$

The kinetic energy of a collection of point masses is $T={\sum }_i\frac{1}{2}{m}_i\Vert {\dot{q}}_i{\Vert }^2$. For a rigid body, every mass element shares the same angular velocity $\omega$, and the velocity of $q_i$ relative to a body frame attached at the CoM is purely rotational. The full velocity is
${\dot{q}}_i=\dot{p}+\omega \times q_i.$
Expanding the squared norm,
$\Vert {\dot{q}}_i{\Vert }^2=\Vert \dot{p}{\Vert }^2+2\dot{p}\cdot (\omega \times q_i)+\Vert \omega \times q_i{\Vert }^2.$
Summing with the masses and pulling constants out of the cross term,
$T=\frac{1}{2}M\Vert \dot{p}{\Vert }^2+\dot{p}\cdot (\omega \times {\sum }_i{m}_i{q}_i)+\frac{1}{2}{\sum }_i{m}_i\Vert \omega \times q_i{\Vert }^2.$
The CoM origin makes ${\sum }_i{m}_i{q}_i=0$, eliminating the cross term. For the rotational term, the identity $\Vert \omega \times q_i{\Vert }^2={\omega }^{\top }(\Vert q_i{\Vert }^2{I}_3-q_i{q}_i^{\top })\omega$ (verified by component-wise expansion) gives
$\frac{1}{2}{\sum }_i{m}_i\Vert \omega \times q_i{\Vert }^2=\frac{1}{2}{\omega }^{\top }\mathcal{I}\omega ,$
which is the headline result.

Examples

Inertia tensor of a thin rod about its CoM

A uniform rod of mass $m$ and length $\ell$ lies along the body $x$-axis with CoM at the origin, linear density $\lambda =m/\ell$.

Compute via integral $\mathcal{I}={\int }_{-\ell /2}^{\ell /2}\lambda \left[\begin{array}{ccc}0& 0& 0\\ 0& x^2& 0\\ 0& 0& x^2\end{array}\right]dx=\frac{m{\ell }^2}{12}\left[\begin{array}{ccc}0& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right].$ The $(1,1)$ entry is zero because the rod has no extent along $x$ — no resistance to spinning about its own axis. The $(2,2)$ and $(3,3)$ entries match the textbook value $\frac{1}{12}m{\ell }^2$.

Kinetic energy of a uniform disk rolling without slipping

A disk of mass $m$, radius $R$, and inertia ${\mathcal{I}}_{\text{axis}}=\frac{1}{2}m{R}^2$ rolls along a flat surface with the CoM moving at speed $v$. The rolling constraint gives $\omega =v/R$.

Plug into rigid-body KE formula $T=\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{1}{2}m{R}^2\right)(v/R{)}^2=\frac{1}{2}m{v}^2+\frac{1}{4}m{v}^2=\frac{3}{4}m{v}^2.$ The rotational contribution adds an effective $50\%$ to the translational KE — which is why a rolling disk decelerates more slowly than a sliding block.

Generalized inertia matrix of a uniform sphere

For a solid sphere of mass $m$ and radius $R$ centered at the body-frame origin, $\mathcal{I}=\frac{2}{5}m{R}^2{I}_3$ (isotropic), so
$\mathcal{M}=\left[\begin{array}{cc}m{I}_3& 0\\ 0& \frac{2}{5}m{R}^2{I}_3\end{array}\right].$
The rotational KE $\frac{1}{5}m{R}^2\Vert \omega {\Vert }^2$ depends only on the magnitude of $\omega$, not direction — the sphere spins identically about every axis. This is the simplest possible $\mathcal{M}$ and a useful sanity-check reference.