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Write $r=(x_1,x_2,x_3{)}^T$ for position relative to the centroid. The inertia tensor there isIn particular its diagonal entries integrate the squares of distances from the corresponding axes, and its off-diagonal entries are the negative products of coordinates.

Insert ${\mathbf{v}}_Q={\mathbf{v}}_G+\mathbf{\omega }\times r$ into $T=\frac{1}{2}{\int }_A\rho |{\mathbf{v}}_Q{|}^{2}dV$. The cross term vanishes because ${\int }_A\rho rdV=0$. Using $|\mathbf{\omega }\times r{|}^2=|\mathbf{\omega }{|}^2|r{|}^2-(\mathbf{\omega }\cdot r{)}^2$ givesIf $a$ is the vector from $G$ to $O$, positions relative to $O$ are $r-a$. Expanding the defining tensor integral with this substitution again removes the terms linear in $r$. Thus the tensor parallel axis theorem isSince $O$ is fixed, ${\mathbf{v}}_G=\mathbf{\omega }\times (-a)$, so $m|{\mathbf{v}}_G{|}^2=m{\mathbf{\omega }}^T(|a{|}^{2}I-a{a}^T)\mathbf{\omega }$. Substitution into the energy decomposition yields

Differentiating orthonormality gives ${\dot{e}}_i\cdot e_j=-e_i\cdot {\dot{e}}_j$. With $\omega =\frac{1}{2}{\sum }_i{e}_i\times {\dot{e}}_i$, the vector triple product yieldsThe body coordinates $x_i$ of a fixed material point are constant. Differentiating its position $R_G+{\sum }_i{x}_i{e}_i$ therefore givesThis identifies $\omega$ as the angular velocity of the rigid body.

Because $G$ is the center of mass, $\int \rho rdV=0$, so the translational contribution to the angular momentum about $G$ vanishes. The remaining vector triple product givesThis symmetric matrix is the inertia tensor about $G$. The angular-momentum balance about the center of mass is $dL/dt={\tau }_{\mathrm{ext},G}$ in an inertial frame. With no external forces, that torque is zero, so the inertial vector $L$ is constant. Its components in rotating body axes need not individually be constant.

For the tool, let the disc lie at $z=0$ and the rod extend to $z=l=\sqrt{3}a$. The total mass is $3M$ and its center is at $z_G=l/3$. Axial symmetry makes the inertia tensor diagonal with $I_x=I_y$.

The standard disc moments are $M{R}^2/2$ about its normal and $M{R}^2/4$ about a diameter. A thin rod has moment $m{l}^2/12$ about a transverse axis through its center. The parallel axis theorem adds $m{d}^2$ for a parallel shift of distance $d$. ThereforeandThe rod contributes zero about its own longitudinal axis. Thus $\boxed{J=2M{a}^2{I}_3}$, an isotropic inertia tensor even though the shape is not spherical. This gives a nonspherical rigid body with isotropic inertia. Consequently $L=2M{a}^2\omega$. Conservation of $L$ in torque-free motion proves the angular velocity is constant for every initial orientation and spin.

Centre the square on the $xy$ plane, with uniform surface mass density $m/a^2$. For an in-plane axis at angle $\alpha$ to the $x$ direction, the perpendicular distance of $(x,y)$ from the axis is $-x\sin \alpha +y\cos \alpha$. Therefore its moment of inertia isThe mixed integral vanishes by symmetry, while the $x^2$ and $y^2$ integrals are equal. ThusThis is the in-plane isotropy of the square's inertia tensor: its two central in-plane principal moments coincide. It does not assert isotropy about arbitrary three-dimensional axes, nor about parallel axes that miss the centre.

Relative to an origin, the inertia tensor is $I_{ij}=\int (r^2{\delta }_{ij}-r_i{r}_j)dm$, with $\mathbf{J}=\mathsf{I}\mathbf{\omega }$. Its orthogonal eigenvectors are principal axes, with eigenvalues the principal moments of inertia. Different components of $\mathbf{\omega }$ are multiplied by different eigenvalues, so angular momentum need not point along angular velocity.

For a uniform thin disc, radial integration gives $I_z=\int r^{2}dm=m{a}^2/2$. The perpendicular axis theorem and symmetry then give $I_x=I_y=I_z/2$, henceFor the elastic eccentric impact of a thin disc, neglect gravity's impulse and take the pole's impulse as vertical. Let $z$ point up and let the contact position relative to the centre be $r\hat{x}$. An upward impulse $P$ givesFor an elastic impact with a fixed pole, the contact point reverses its normal velocity, so $-u+P/m+r^{2}P/I_d=+u$. ThereforeHere $v_C>0$ denotes upward motion; the centre can still move down if $r>a/2$. Substitution givesThis also follows by combining conserved angular momentum about the fixed contact with conserved kinetic energy and selecting the separating, nonzero-impulse solution. At $r=0$ the disc simply rebounds without rotation; at $r=a/2$ its centre is instantaneously stationary. The rebound result concerns the contacting material point immediately after this ideal impact, not its subsequent contact history.

For the inertia tensor of a uniform cube, reflection symmetry makes every off-diagonal element of the central inertia tensor vanish. Permutation of the three axes makes its diagonal elements equal. Since $M=8\rho a^3$ and the mean of each squared coordinate in the cube is $a^2/3$,Every central axis is therefore a principal axis, including the body diagonal. The angular momentum is parallel to the angular velocity, and its magnitude is

Choose body-fixed principal axes through a fixed point or the centre of mass, with principal moments $I_1,I_2,I_3$. Angular momentum has components $L_i=I_i{\omega }_i$. The inertial derivative is the body derivative plus $\mathbf{\omega }\times \mathbf{L}$, so Euler's equations readThey avoid differentiating the rotating inertia tensor in laboratory coordinates. Torques must refer to the chosen point, and the displayed fixed moments require a rigid body.

For zero torque, laboratory angular momentum is constant, while its components can change in the body frame. Both $L^2$ and rotational kinetic energy $T=\frac{1}{2}\sum I_i{\omega }_i^2$ are conserved. The intersections of their constant-value surfaces constrain the motion. For distinct $I_1<I_2<I_3$, steady rotation about either extreme principal axis is stable against small perturbations, whereas rotation about the intermediate axis is unstable. For example perturbing rotation $\Omega$ about axis 2 giveswhose coefficient is positive. A symmetric body has additional simplifications and can precess. Conservation of laboratory angular momentum does not imply constant body-frame angular velocity.

For a rigid body about a chosen origin,The inertia tensor is real and symmetric, so orthogonal principal axes diagonalize it. Angular momentum is parallel to angular velocity for rotation about a principal axis, but need not be parallel in a general direction. Its eigenvalues are the principal moments of inertia.

If the origin is translated from the centre of mass by $\mathbf{a}$, the parallel-axis theorem becomes $I_{ij}=I_{ij}^{\mathrm{CM}}+M(a^2{\delta }_{ij}-a_i{a}_j)$. In body axes, Euler's equation is $\mathbf{\tau }=(d\mathbf{L}/dt{)}_{\mathrm{body}}+\mathbf{\omega }\times \mathbf{L}$. Thus the full tensor is needed whenever the rotation is not confined to one fixed principal axis.

For positions measured from the chosen origin, the inertia tensor isIt relates angular momentum and rotational kinetic energy to angular velocity:The real symmetric tensor has orthogonal principal axes, in which $\mathbf{I}=\mathrm{diag}(I_1,I_2,I_3)$. Its eigenvalues are nonnegative moments of inertia. Rotation about a principal axis has $\mathbf{L}\parallel \mathbf{\omega }$; a general rotation does not. The moment about a unit axis $\mathbf{n}$ is $I_{\mathbf{n}}=\mathbf{n}\cdot \mathbf{In}$.

Changing from the centre of mass to a parallel origin displaced by $\mathbf{a}$ gives the tensor parallel-axis theorem,For example, a uniform solid sphere has $\mathbf{I}=(2M{R}^2/5)1$, while a thin rod along $x$ has principal moments $0,M{l}^2/12,M{l}^2/12$ in the ideal zero-radius limit. The tensor is needed to describe off-axis rotation and torque-free precession; a single scalar inertia only suffices after an axis is specified.^[ref]

For a body-fixed mass element at position $\mathbf{r}$ relative to the fixed tip, $\mathbf{v}=\mathbf{\omega }\times \mathbf{r}$. ThereforeThe inertia tensor is real symmetric. Its orthogonal eigenvectors define principal axes and its eigenvalues are principal moments of inertia. angular momentum is parallel to angular velocity when rotation is along a principal axis, but generally not for arbitrary rotation.

Write $\mathbf{n}={\hat{\mathbf{e}}}_3$ and $s=I_3{\omega }_3$. Axial symmetry gives $\mathbf{L}=I_1\mathbf{\omega }+(I_3-I_1){\omega }_3\mathbf{n}$. Since $\dot{\mathbf{n}}=\mathbf{\omega }\times \mathbf{n}$, the vector triple-product identity gives $\mathbf{n}\times \dot{\mathbf{n}}=\mathbf{\omega }-{\omega }_3\mathbf{n}$. HenceDifferentiate in the inertial frame, using constant $s$ and $\dot{\mathbf{n}}\times \dot{\mathbf{n}}=0$. The gravitational torque about the fixed tip is $(\ell \mathbf{n})\times (-mg\hat{\mathbf{z}})$, yielding the symmetry-axis equation of a heavy symmetric top:The tip reaction exerts no moment about that tip.

For steady precession, $\mathbf{n}=(\sin \theta \cos {\Omega }_{P}t,\sin \theta \sin {\Omega }_{P}t,\cos \theta )$. At azimuth zero, put $s_{\theta }=\sin \theta$ and $c_{\theta }=\cos \theta$. Then $\dot{\mathbf{n}}=(0,{\Omega }_P{s}_{\theta },0)$, $\ddot{\mathbf{n}}=(-{\Omega }_P^2{s}_{\theta },0,0)$, and the $y$ component gives $s{\Omega }_P{s}_{\theta }-I_1{\Omega }_P^2{s}_{\theta }{c}_{\theta }=mg\ell s_{\theta }$. For a nonvertical axis this becomesReal steady branches require a nonnegative discriminant. In particular, for a top pointing above the horizontal there is a spin threshold for these fixed-angle solutions. At $\cos \theta =0$ use the quadratic's linear limit ${\Omega }_P=mg\ell /s$, rather than dividing by zero. For a vertical axis, azimuth does not describe a physical conical motion and the division by $\sin \theta$ is inappropriate.

For rapid positive spin $s$ with $s^2\gg 4mg\ell I_1\cos \theta$, the slow and fast steady precession roots areThe approximation assumes a nonsingular fixed inclination. For $g=0$, the quadratic instead gives ${\Omega }_P=0$ or ${\Omega }_P=s/(I_1\cos \theta )$. The zero branch has a stationary symmetry axis and pure axial spin. On the other branch angular momentum is constant and parallel or antiparallel to the chosen $z$ axis; $L_z=I_1{\Omega }_P$ and $s=L_z\cos \theta$. This is torque-free precession in space, not the different rate at which transverse components rotate in body coordinates.

Inertia and principal directions. About the centre of mass, the inertia tensor isIt relates angular momentum and angular velocity by $\mathbf{J}=\mathsf{I}\mathbf{\omega }$ and gives rotational kinetic energy $T=\mathbf{\omega }\cdot \mathsf{I}\mathbf{\omega }/2$. The principal axes are orthogonal eigenvector directions of this real symmetric tensor; in that basis it is diagonal, with entries the principal moments of inertia $I_1,I_2,I_3$.

Derivatives in a rotating frame. Write $\mathbf{J}={\sum }_i{J}_i{\mathbf{e}}_i$ in a body-fixed orthonormal basis. Each basis vector rotates according to $(d{\mathbf{e}}_i/dt{)}_0=\mathbf{\omega }\times {\mathbf{e}}_i$. Differentiating the components and basis separately proves the rotating-frame derivative formula,The body derivative differentiates the scalar components in that basis. A fixed pivot can replace the centre of mass when the corresponding angular-momentum balance is valid.

Euler equations. The inertial angular-momentum equation is $d\mathbf{J}/dt=\mathbf{G}$. Since the principal moments are constant in the body frame, substitute $J_i=I_i{\omega }_i$:These are Euler's equations in the principal body frame.

Linear perturbation of free rotation. After the impulse, $\mathbf{G}=0$. Let ${\omega }_3=\Omega$ to leading order and ${\omega }_1,{\omega }_2$ be small. The third Euler equation changes ${\omega }_3$ only at quadratic order, so the linear transverse equations areFor ${\omega }_1=a{e}^{pt}$ and ${\omega }_2=b{e}^{pt}$, eliminate $a,b$ fromThis givesIts units are inverse square seconds. Positive $p^2$ gives a growing real exponential; negative $p^2$ gives bounded oscillatory perturbations.

Inertia is a tensor. For a rigid body, distances between material points are fixed. Write each position relative to the centre of mass as $\mathbf{r}$, so its rotational velocity is $\mathbf{\omega }\times \mathbf{r}$. Integrating $\mathbf{r}\times (\mathbf{\omega }\times \mathbf{r})$ gives angular momentumThe rotational kinetic energy isThe inertia tensor is real and symmetric, so orthogonal principal axes diagonalize it. Angular momentum is parallel to angular velocity only along a principal axis, or within a subspace of equal principal moments. For translation plus rotation about the centre of mass, $T=M{v}_{\mathrm{CM}}^2/2+T_{\mathrm{rot}}$. Moving the reference to a parallel axis adds $M(d^{2}1-\mathbf{d}{\mathbf{d}}^{\mathsf{T}})$ to the tensor, the tensor form of the parallel-axis theorem.

Euler's equations. The external torque about the centre of mass obeys $\mathbf{\tau }=(d\mathbf{L}/dt{)}_{\mathrm{space}}$. In body axes, the transport rule givesThe principal moments are constant in that frame, henceThese are Euler's equations. With zero torque, spatial angular momentum and kinetic energy are conserved, although the body components generally vary. For distinct moments, small perturbations about rotation around the largest or smallest principal moment oscillate, whereas those about the intermediate moment grow. This is the intermediate-axis instability.

A lamina. Choose the lamina's plane as $z=0$. Then $I_{xz}=I_{yz}=0$, so the normal is a principal axis, andThis perpendicular-axis theorem applies to a planar mass distribution; it is not a general identity for a three-dimensional body. Rotate the in-plane axes to diagonalize the remaining symmetric two-by-two block. For a uniform rectangular lamina with sides $a,b$ and mass $M$, its central principal moments are $I_x=M{b}^2/12$, $I_y=M{a}^2/12$ and $I_z=M(a^2+b^2)/12$. If $a>b$, then $I_x<I_y<I_z$: torque-free rotation about the long in-plane axis or the normal is stable, while that about the short in-plane axis is unstable. Coincident moments in a square require the degenerate, rather than strict intermediate-axis, case. Under a constraint to rotate only about the normal, the motion reduces to the scalar equation ${\tau }_z=I_z{\dot{\omega }}_z$.

A symmetric top without torque. Let $I_1=I_2=I_{\perp }$ and $I_3=I_{\parallel }$. Euler's equations reduce toThus ${\omega }_1+i{\omega }_2$ rotates at signed body-frame frequency $\nu$, while the axial component stays fixed. The symmetry axis has constant angle to fixed spatial angular momentum. Indeed $\mathbf{L}=I_{\perp }\mathbf{\omega }+(I_{\parallel }-I_{\perp }){\omega }_3{\mathbf{e}}_3$, soIt precesses in space about $\mathbf{L}$ with angular speed $L/I_{\perp }$. This spatial precession and the body-frame frequency $\nu$ describe different motions and should not be identified.

A heavy top. If a symmetric top is pivoted a distance $\ell$ from its centre of mass, gravity gives torque ${\mathbf{r}}_{\mathrm{CM}}\times M\mathbf{g}$. Here the relevant moments are about the fixed pivot. The axial angular momentum and the vertical angular-momentum component are conserved, because gravity supplies neither an axial torque nor a torque about the vertical. Using tilt $\theta$, precession angle $\varphi$ and spin angle $\psi$, its Lagrangian isEliminating the two conserved cyclic momenta leaves one-dimensional motion in an effective tilt potential; bounded oscillation in tilt is nutation. Steady precession at $\dot{\varphi }=\Omega$ and fixed tilt satisfiesFor a rapidly spinning top on the slow branch, the first term is small andGravity then changes the direction of angular momentum rather than simply tipping the top straight down. The approximation fails near conditions where nutation or the fast-precession branch matters; the full equations retain those possibilities.

The principal moments of inertia are the eigenvalues of the symmetric inertia tensor. In units of $\mathrm{kg}{\mathrm{m}}^2$, its characteristic determinant isTherefore the requested algebraic answer isAn orthonormal basis of principal axes is $(1,1,0)/\sqrt{2}$, $(1,-1,0)/\sqrt{2}$, and $(0,0,1)$; the last two span the degenerate eigenspace and may be replaced by any orthonormal basis of it.

There is a physical inconsistency in calling the supplied matrix a rigid body's inertia tensor. The moment-of-inertia triangle inequalities require each principal moment to be no greater than the sum of the other two. Here $3>1+1$. In principal coordinates, $I_2+I_3-I_1=2\int x_1^{2}dm$ cannot be negative for a nonnegative mass distribution. Thus the eigenvalue calculation is well defined, but no ordinary rigid mass distribution realizes this tensor. The positive second-moment matrix test for an inertia tensor makes the same obstruction explicit.

Translation and rotation. For particles constituting a rigid body, the total linear momentum isThus the total external force governs the motion of the centre of mass. About that centre of mass, the angular momentum and applied torque arewhere primes denote relative positions and velocities. The derivative of the vector is taken relative to nonrotating axes. The inertia tensor relates angular momentum to angular velocity, ${\mathbf{L}}_{\mathrm{CM}}=\mathsf{I}\mathbf{\omega }$. For rotation about a fixed unit direction $\mathbf{n}$, the moment of inertia is $I_{\mathbf{n}}=\int r_{\perp }^{2}dm=\mathbf{n}\cdot \mathsf{I}\mathbf{n}$; it measures the resistance to changing that rotation and gives rotational kinetic energy $I_{\mathbf{n}}{\omega }^2/2$. The simpler vector relation $\mathbf{L}=I\mathbf{\omega }$ applies when the rotation axis is a principal axis, not for every orientation of an arbitrary rigid body. These translational and rotational balances must be applied together; a nonzero torque does not replace the net-force equation.

Gyroscope precession. A rapidly spinning gyroscope has a large angular momentum approximately along its spin axis. The torque component parallel to this vector changes its magnitude, while the perpendicular component turns its direction. If the vector undergoes precession with angular velocity $\mathbf{\Omega }$, then over $dt$ its change is the infinitesimal rotation $d\mathbf{L}=(\mathbf{\Omega }dt)\times \mathbf{L}$. HenceThis describes direction-changing precession at constant $|\mathbf{L}|$; any spin-up contribution must be added separately. For perpendicular torque and angular momentum, the precessional component perpendicular to $\mathbf{L}$ is ${\mathbf{\Omega }}_{\perp }=\mathbf{L}\times \mathbf{G}/L^2$. The familiar tendency of a supported gyroscope to turn sideways is this vector balance, not the absence of gravitational force.

Wheel-lift threshold without the flywheel. Let $h=0.50\mathrm{m}$ be the centre of mass height, $a=1.0\mathrm{m}$ half the wheel separation, and $R_b=10\mathrm{m}$ the bend radius. The required inward force is $F=M{v}^2/R_b$, provided by static friction. Let $N_o,N_i$ be the summed outer and inner wheel normal forces. Vertical balance gives $N_o+N_i=Mg$. Moment balance about the centre of mass givesThe inner wheels lift when $N_i=0$. Thus the vehicle rollover threshold isThe stated mass cancels. This assumes the rigid geometry and sufficient static friction specified in the paper; the threshold is set by loss of contact, not by sliding.

Engine flywheel. The moment of inertia of a uniform disc givesIts spin angular speed, spin angular momentum, and turn angular speed at $40\mathrm{km}{\mathrm{h}}^{-1}$ areThe transverse flywheel axis is perpendicular to the vertical turn axis, soTo specify its direction, use right-handed axes $+x$ forward, $+y$ to the car's left, and $+z$ upward, and consider a left turn. Then $\mathbf{\Omega }=+\Omega \hat{\mathbf{z}}$. Choose the flywheel angular momentum to the car's right, ${\mathbf{L}}_f=-L_f\hat{\mathbf{y}}$. The car must applyto obtain the required precession. This gyroscopic roll torque in a turning vehicle opposes the outward roll tendency in a left turn. By the right-hand rule, the flywheel must turn counterclockwise viewed from the car's right side toward its centre, opposite to the rotation of wheels rolling forward. Reversing flywheel spin would aggravate the outward roll. Reversing the turn reverses both roll and gyroscopic torques, so the same flywheel spin remains helpful.

For comparison, the outward inertial roll moment at the stated speed is $Mh{v}^2/R_b=6173\mathrm{N}\mathrm{m}$, and the limiting gravitational restoring moment is $Mga=9810\mathrm{N}\mathrm{m}$. The gyroscopic reaction is about $30\%$ of the former and $19\%$ of the latter, so it is appreciable for this large flywheel. If this ideal model is extended to include the favorable gyroscopic reaction in the wheel-lift condition, it gives $Mh{v}^2/R_b-L_{f}v/R_b=Mga$, or about $56.8\mathrm{km}{\mathrm{h}}^{-1}$ rather than $50.4\mathrm{km}{\mathrm{h}}^{-1}$. This quantifies a useful but finite increase in the rollover threshold.

The inertia tensor about the centroid has entries $J_{ij}={\int }_B\rho (r^2{\delta }_{ij}-x_i{x}_j)dV$. Reflection symmetry makes every off-diagonal entry zero. Under $x=a\xi$, $y=a\eta$, $z=c\zeta$, the ellipsoid becomes the unit ball. Its three coordinate second moments are equal, andUsing the stated mass gives ${\int }_B\rho x^{2}dV=M{a}^2/5$, the same for $y$, and ${\int }_B\rho z^{2}dV=M{c}^2/5$. ThereforeFor the parallel coordinate axes based at $P=(0,0,c)$, the parallel axis theorem in tensor form is $J^{(P)}=J^{(G)}+M(|d{|}^{2}I-d{d}^T)$, where $d=P-G$. Thus

For a nonnegative mass distribution relative to an origin, the second-moment matrix is $\mathsf{S}=\int \mathbf{r}{\mathbf{r}}^{\mathsf{T}}dm$ and the inertia tensor is $\mathsf{I}=\mathrm{tr}(\mathsf{S})1-\mathsf{S}$. Taking the trace gives the displayed inverse formula, which must be positive semidefinite. Conversely, any such $\mathsf{S}$ can be realized by pairs of point masses along its orthogonal principal directions. Thus this is also a realizability test when total mass and dimensions are unrestricted. In principal coordinates it is equivalent to the moment-of-inertia triangle inequalities: $I_2+I_3-I_1=2S_{11}\ge 0$, with cyclic counterparts. A symmetric positive definite matrix alone need not be a physical inertia tensor.

A principal axis is an eigenvector direction of the inertia tensor. Choosing three orthogonal principal axes diagonalizes the inertia tensor.

The principal moments of inertia about a specified point $O$ are the eigenvalues $I_1,I_2,I_3$ of the inertia tensor ${\mathbf{I}}_O$. Their associated unit vectors that are eigenvectors specify principal axes through $O$. The spectral theorem for real symmetric matrices guarantees an orthonormal basis of these eigenvectors, even for an irregular rigid body. Choose their signs to make a right-handed basis and collect them as the columns of an orthogonal matrix $Q$:Thus diagonalization removes the off-diagonal coupling in the rotational kinetic energy. The eigenvalues belong to the mass distribution and reference point, independently of which orthonormal basis initially describes them. ^[ref]

In this principal body frame, write angular velocity as $\mathbf{\omega }={\sum }_i{\omega }_i{\mathbf{e}}_i$. ThenUse these as total angular momentum and kinetic energy only under the reference-point conditions stated under inertia tensor. Along a single principal axis with $I_i>0$, angular momentum is parallel to angular velocity. For a mixture of directions with different principal moments of inertia, it generally is not. A principal axis need not be a geometric symmetry axis: it is selected by the complete mass distribution. A principal axis also need not give stable free rotation; for three distinct principal moments of inertia, the intermediate axis theorem supplies the stability distinction.

For any unit vector $\mathbf{n}={\sum }_i{n}_i{\mathbf{e}}_i$,The axial moment of inertia is a weighted average of the principal moments of inertia. At fixed angular speed, the least and greatest rotational kinetic energies occur in the corresponding eigenspaces. At fixed nonzero angular momentum magnitude $L$, instead, $K=\frac{1}{2}{\sum }_i{L}_i^2/I_i$, so the largest principal moment of inertia gives the smallest rotational kinetic energy. These comparisons hold the reference point fixed.

There is an additional physical restriction beyond positive semidefiniteness. In principal axes coordinates,with two analogous cyclic inequalities. Therefore three arbitrary positive numbers do not always represent a possible inertia tensor. For a distribution in the $xy$ plane through $O$, $I_3=I_1+I_2$, as in the planar inertia tensor of paired point masses. A uniform solid ellipsoid of mass $M$ and semiaxes $a,b,c$ supplies a continuous three-dimensional example: using its mass second-moment tensor, its central principal moments of inertia are $M(b^2+c^2)/5$, $M(a^2+c^2)/5$, $M(a^2+b^2)/5$. Setting $a=b=c=R$ recovers the moment of inertia of a uniform solid sphere.

A rigid rotor models rotational motion with fixed internal separations. Its use in the synthetic Stark effect figure omits vibrational motion and additional interactions in real molecules.

Rigidity fixes the molecular geometry, not the orientation, and does not require a spherical mass distribution. Remove motion of the centre of mass, choose principal axes of the inertia tensor, and write the classical rotational kinetic energy asHere $I_a,I_b,I_c$ are the principal moments of inertia, ${\omega }_a,{\omega }_b,{\omega }_c$ components of angular velocity, and $L_a,L_b,L_c$ components of angular momentum. Quantization replaces these last components by angular momentum operators. Different principal moments of inertia produce different spectra: the single-constant linear rigid rotor is a special case, not a formula for every molecule.^[ref]

PASCO's PASPORT Rotary Motion Sensor Experiment Guide, Experiment 1, specifies a PS-2120A sensor, a rod carrying two equal-radius weights, a three-step drive wheel, a thread and hanging mass, a clamp-on guide wheel, a support, a balance and calipers. The thread leaves the drive wheel tangentially. The sensor records angular velocity against time; a linear regression over the accelerating interval supplies angular acceleration $\alpha$. The guide measures the drive radius with calipers and repeats the run with the weights removed to subtract the apparatus's own moment of inertia. Its tables are blank: it prescribes a measurement and does not report an observed numerical agreement. ^[ref]

With hanging mass $m_h$ including its hanger, effective drive radius $b$, no thread slip and negligible guide-wheel inertia, Newton's second law gives thread tension $F=m_h(g-b\alpha )$. The axial torque is $bF$. Under negligible friction, the inference is thereforeRepeat the inference for the unloaded assembly and subtract: $I_{\mathrm{weights}}=I_{\mathrm{loaded}}-I_{\mathrm{empty}}$. Omitting the $m_h{b}^2$ term incorrectly treats thread tension as the hanging weight. Comparing with $(M_1+M_2)R^2$ tests the point mass approximation; finite-sized weights also have their own moments of inertia about their centres of mass. ^[ref]

Useful proposed checks go beyond that ideal inference. Verify the rotary encoder's angle scale by turning the shaft through a marked full revolution; check the recorded time against an independent clock and the balance against a known mass. Measure $b$ at the thread's running surface, including thread thickness; avoid overlapping wraps that change $b$. Release without a push, inspect the angular velocity record for a straight interval, and stop before the hanger lands or the thread runs out. Repeat with several hanging masses, weight radii and rotation directions. A consistent inferred moment of inertia across drive torques supports the model; curvature, direction dependence or drift indicates a limitation rather than a new principal moment of inertia.

For approximately constant resisting torque ${\tau }_f\ge 0$ in one rotation direction, an original calibration model isFit the left-hand quantity against $\alpha$ across several drives: the slope estimates the axial moment of inertia and the intercept estimates ${\tau }_f$. This calibration is conditional on friction remaining constant over the fitted angular speeds; a freely slowing run can expose speed dependence. Sensor differentiation noise, guide-wheel moment of inertia, thread slip, flexing connectors and uncertain weight radii limit the inference. Removing weights can also change bearing loads, so empty-apparatus subtraction does not automatically cancel friction. These are proposed controls and a model extension, not results claimed by PASCO.

A fixed-axis readout measures ${\mathbf{n}}^T{\mathbf{I}}_O\mathbf{n}$, not the full inertia tensor. In a proposed generalization, repeat about $x$, $y$, $z$ and the three diagonal directions $({\mathbf{e}}_x+{\mathbf{e}}_y)/\sqrt{2}$, $({\mathbf{e}}_x+{\mathbf{e}}_z)/\sqrt{2}$, $({\mathbf{e}}_y+{\mathbf{e}}_z)/\sqrt{2}$, all through the same $O$. For example,Six independent scalar readings then determine the six entries of a symmetric matrix. Each mounting needs its own apparatus correction, and different reference points require the tensor parallel-axis theorem before the readings can be combined.

Rotational kinetic energy is the kinetic energy of a rigid body's rotation. For pure rotation about an axis fixed in an inertial frame,Here $r_{\perp }$ is the perpendicular distance of a mass element $dm$ from that axis, $I$ is the moment of inertia about that same axis, and $\omega$ is the common angular speed, measured in radians per second. The rigid body assumption fixes the relative positions of the mass elements; they share angular velocity but have different speeds $v=\omega r_{\perp }$. Summing their kinetic energies $dm{v}^2/2$ gives the formula. This is a nonrelativistic model; deformation or internal motion can contribute additional kinetic energy.^[ref]

For a computed example, take a uniform disc and an ideal thin ring, each of mass $M=2\mathrm{kg}$ and radius $R=0.30\mathrm{m}$, rotating about its central perpendicular axis at angular speed $10rad/s$. The moment of inertia of a uniform disc is $M{R}^2/2=0.090\mathrm{kg}{\mathrm{m}}^2$, while the ring has moment of inertia $M{R}^2=0.180\mathrm{kg}{\mathrm{m}}^2$. Their rotational kinetic energies are therefore $4.5\mathrm{J}$ and $9.0\mathrm{J}$. The ring stores twice the energy at the same angular speed, because more mass lies farther from the axis. At the same nonzero rotational kinetic energy, its angular speed would instead be $1/\sqrt{2}$ times the disc's. This makes radial placement useful for storing energy in a rotating wheel, although a real design must also withstand the stresses of rotation.

For fixed-axis rotation with constant moment of inertia, write the signed rotation rate as $\Omega =\dot{\theta }$, with angle $\theta$ in radians. The axial component of net external torque obeys ${\tau }_{\mathrm{axis}}=I\dot{\Omega }$, so the work-energy theorem becomesAn opposing torque removes rotational kinetic energy; bearing friction must be included in the net torque. The signed $\Omega$ matters when the direction reverses. Rotation about a fixed axis need not be about a principal axis: the scalar energy formula still holds, but transverse bearing torques can be needed to maintain that axis.