A state saturating the uncertainty bound for a specified pair of quantum observables. The ground state of the quantum harmonic oscillator satisfies $\Delta x\Delta p=\hslash /2$. Saturation is a property of the state and the chosen observables.

Chain growth by successive olefin metathesis steps that open cyclic alkenes. Relief of ring strain can offset the loss of monomer translational entropy.

The rotational symmetry number corrects a rotational phase-space count for indistinguishable molecular orientations. A linear molecule with identical ends, such as carbon dioxide with identical oxygen isotopes, has $\sigma =2$; making its ends isotopically distinct gives $\sigma =1$. In the high-temperature linear rigid rotor approximation, $q_{\mathrm{rot}}=8{\pi }^{2}IkT/(\sigma h^2)$. Holding the spectrum otherwise fixed, doubling $\sigma$ lowers molar rotational entropy by $R\log 2$.

For a linear rigid rotor with moment of inertia $I$, the energy eigenvalue at angular momentum quantum number $J$ is ${\hslash }^{2}J(J+1)/(2I)$. Each level has statistical weight $2J+1$ before additional symmetry or nuclear-spin restrictions.

A rigid rotor restricted to one angular coordinate $\varphi$. Single-valued spatial states $e^{im\varphi }$ have $m\in \mathbb{Z}$ and energy $E_m={\hslash }^2{m}^2/(2I)$, with moment of inertia $I$. The zero-angular-momentum level has state degeneracy one; every nonzero $|m|$ has two directions. For labelled molecular ends its high-temperature rotational partition function is $\sqrt{\pi T/{\theta }_{\mathrm{rot}}}$, where ${\theta }_{\mathrm{rot}}={\hslash }^2/(2I{k}_{\mathrm{B}})$. Equivalent orientations and nuclear spin statistics require a further molecular counting convention. One quadratic angular momentum gives rotational heat capacity $k_{\mathrm{B}}/2$ per molecule, rather than $k_{\mathrm{B}}$ for an unrestricted linear rigid rotor.

For a rigid heteronuclear diatomic rotor, $E_J={\hslash }^{2}J(J+1)/(2I)$ and electric-dipole absorption has $\Delta J=+1$. Therefore the $J^{\prime }\to J^{\prime }-1$ line frequency is $h{J}^{\prime }/(4{\pi }^{2}I)$ and adjacent absorption lines are spaced by $h/(4{\pi }^{2}I)$. With reduced mass $\mu$, the bond-length estimate is $r=\sqrt{I/\mu }$. The conversion assumes frequency rather than angular frequency; centrifugal distortion makes real high-J spacings imperfectly uniform.

A rigid body with moment of inertia $I$, constant driving power $P$ and linear resisting torque $-\gamma \omega$ has an exponential approach of ${\omega }^2$ to $P/\gamma$. The ideal drive requires torque $P/\omega$, so starting from rest is a limiting energy model with singular initial torque.

The net external torque about an inertial origin equals the time derivative of angular momentum. For rotation about a fixed axis with constant moment of inertia, its axial component reduces to $I\dot{\omega }=\tau$.

For a body constrained to rotate about a fixed pivot, a perpendicular impulse $K$ delivered at distance $b$ from the pivot produces $\omega =Kb/I_O$. If the centre of mass is distance $h$ from the pivot, its momentum is $M\omega h$. The pivot impulse is $J_O=M\omega h-K$, and vanishes when $b=I_O/(Mh)$. This impact location is the centre of percussion. The result assumes the stated planar rigid-body geometry and a short impulse; it is not a claim that all forces at the pivot vanish afterward.

At fixed angular momentum, rotational energy is minimized by rotation about the largest principal moment of inertia. Indeed, $T={\sum }_i{L}_i^2/(2I_i)\ge {\sum }_i{L}_i^2/(2I_{\max })$. Weak internal dissipation can therefore relax a nearly rigid isolated body towards that axis. A perfectly rigid torque-free body conserves energy and does not settle; both extreme-inertia axes are stable in its ideal dynamics. The maximum-inertia axis need not be the geometrically longest axis.

Let a bell have axial moment of inertia $M{k}^2$ and centre of mass h from its axle. A clapper of point mass m at radius l touches the bell at fixed relative angle $\beta$. While they remain in contact, the clapper angle $\theta$ from the upward vertical obeys $(M{k}^2+m{l}^2)\ddot{\theta }=Mgh\sin (\theta -\beta )+mgl\sin \theta$. The contact force opposing increasing angle is $N=mg\sin \theta -ml\ddot{\theta }$. Eliminating the acceleration givesThe first zero on the physical trajectory marks detachment, because a nonadhesive contact cannot support negative normal force. Selecting the correct trigonometric branch is part of the contact problem.