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Let $x$ be the cylinder centre's small displacement away from the wall. Rolling without slipping gives rotation angle of magnitude $x/r$. The material point initially at the top moves horizontally by another $x$ relative to the centre to first order, so the spring extension is $2x$, not $x$. Its small vertical displacement affects the extension only at higher order.

The kinetic and potential energies areThe linearized equation is $(3m/2)\ddot{x}+4kx=0$, givingStatic friction enforces the rolling constraint and does no work at the instantaneous contact point. Omitting rotational kinetic energy, or attaching the spring to the centre instead of the top material point, would change the result.

A modern teaching experiment slows gravity-driven motion enough to resolve it with a photogate. PASCO scientific's Inclined Plane Dynamics Set, ME-6966, Instruction Manual 012-10874A describes “Acceleration Down an Incline” on printed pages 15–16. Its apparatus is a GOcar on a PAStrack with an adjustable incline support, a bracket-mounted photogate, a cart-mounted picket fence, and an interface with recording software. For the interface method, the fence's $1\mathrm{cm}$ pattern is aligned to interrupt the beam. Recording starts before release from rest; the cart is caught at the bottom bumper. The procedure specifies three runs per angle, starting at $14^{\circ }$ and reducing by $2^{\circ }$, with the release position fixed. The readout is velocity against time; its slope estimates acceleration. Mean accelerations are plotted against $\sin \theta$, testing the ideal slope $g$. An added-mass comparison is also requested. These are protocol settings, not observed results.^[ref]

The cart undergoes rolling without slipping at its wheels, whereas the worked block undergoes contact slip. Even ideal wheel moment of inertia changes the cart prediction. For total cart mass $M$, wheel radii $r_i$ and axial moments of inertia $I_i$, rigid wheels with fixed axles and no losses giveTo derive it, combine translational kinetic energy $M{v}^2/2$ with wheel rotational kinetic energy ${\sum }_i{I}_i(v/r_i{)}^2/2$, then equate their rate of change to the release of gravitational potential energy $Mg\sin \theta v$. This is the multiwheel counterpart of rolling acceleration with rotational inertia. Neglecting wheel moment of inertia yields $a\simeq g\sin \theta$; treating a cart as a solid rolling cylinder would give a different correction.

Useful additional controls are repeated releases without a push, checking the fence alignment and spacing, and checking that the recorded velocity trace is straight over the fitted interval. A gently moving cart on a level track can expose losses or unintended tilt. These checks help distinguish an angle or timing error from neglected wheel moment of inertia, bearing friction, or drag force. The three-run spread measures repeatability but cannot by itself reveal a shared calibration error. The inspected protocol makes the gravity component measurable; it does not measure the sliding block's coefficient of kinetic friction.

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]

For strictly positive principal moments of inertia and a fixed rotational kinetic energy $K_0>0$, the allowed angular velocity components form an ellipsoid:Its semiaxis lengths $s_i$ have units of angular speed. The longest semiaxis corresponds to the smallest principal moment of inertia: that direction permits the greatest angular speed for the same rotational kinetic energy. This is a surface in angular velocity space, not the physical outline of the body. If one principal moment of inertia vanishes in an ideal line-mass model, this constant-energy surface is unbounded in that direction and is no longer an ellipsoid.

The rotational energy ellipsoid gives the geometry behind Poinsot's ellipsoid, but a constant-energy surface alone does not determine a free-rotation trajectory. A torque-free trajectory must also satisfy the conserved angular momentum constraint and the Euler equations for a torque-free rigid body.

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.

For general rigid body motion, use the inertia tensor ${\mathsf{I}}_G$ about the centre of mass $G$:This is the kinetic-energy decomposition about the centre of mass. In a principal body frame, the rotational term is $\frac{1}{2}(I_1{\omega }_1^2+I_2{\omega }_2^2+I_3{\omega }_3^2)$, using the principal moments of inertia. Away from a principal axis, angular momentum and angular velocity need not be parallel. A freely rotating asymmetric rigid body can change the direction of angular velocity relative to the body even while rotational kinetic energy remains constant; Euler equations for a torque-free rigid body govern that motion. One constant axial moment of inertia is then insufficient.^[ref]

A rigid body's axial moment of inertia $I$ and a constant resisting torque ${\tau }_f>0$ give constant angular acceleration $-{\tau }_f/I$. Hence $t_{\mathrm{stop}}=I{\omega }_0/{\tau }_f$ and the angular displacement is the displayed expression. Equivalently, the lost rotational kinetic energy $I{\omega }_0^2/2$ equals the resisting work ${\tau }_f\Delta \varphi$. The total number of revolutions is $\Delta \varphi /(2\pi )$; the number of whole revolutions completed is its floor function.

A useful readout of a tumble is the alignment of the intermediate principal axis ${\hat{e}}_2$ with the conserved angular momentum direction $\hat{L}$. In the principal body frame, write $L_i=I_i{\omega }_i$ and $L=|\mathbf{L}|$. ThenA change from $c_2\simeq +1$ to $c_2\simeq -1$ means that the directed body axis has turned nearly $180^{\circ }$ relative to $\mathbf{L}$. It does not mean that $\mathbf{L}$ reversed in the fixed space frame. The unsigned line through that principal axis hides this distinction, so a marked end matters in an actual demonstration.

For the normalized intermediate-axis model with $\mathbf{\omega }(0)=(0.01,1,0)$, the rotational kinetic energy and squared angular momentum are $E=1.00005$ and $L^2=4.0001$. Hence $L^2-2E{I}_2=-0.0001$: this initial condition lies close to, but off, the intermediate-axis separatrix. An independent calculation evolves ${\dot{\mathbf{L}}}_{\mathrm{body}}={\mathbf{L}}_{\mathrm{body}}\times I^{-1}{\mathbf{L}}_{\mathrm{body}}$ and a rotation matrix taking body components to space components. The intermediate principal axis crosses $c_2=0$ at normalized times approximately $10.3773$ and $31.1319$, alternating long nearly aligned intervals with quicker reversals.

Both calculations use step size $0.005$ without rescaling either invariant. Over $0\le t\le 50$, the tumbling calculation changes rotational kinetic energy and squared angular momentum by less than $4\times 10^{-13}$ relative to their initial values. Reconstructing $\mathbf{L}$ in the fixed space frame gives relative error below $3\times 10^{-12}$. Halving the step changes the body components and rotation matrix entries by less than $3\times 10^{-10}$. These checks distinguish physical tumbling from numerical drift. On the exact intermediate-axis separatrix, approach to an intermediate-axis rotation takes infinite time; the recurrent off-separatrix example here is different.