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For a fixed attachment ratio $\delta >0$, making the disc much heavier than the rod givesIf the disc is also small relative to the attachment distance, $\alpha /\delta =a/d\to 0$, its moment of inertia about its own center is negligible compared with $M{d}^2$. ThusThis is a simple pendulum with a point mass at the disc's center and effective length $d$; it is length $\ell$ only if $d=\ell$. Even before the small-angle approximation, its equation tends to $\ddot{\theta }=-(g/d)\sin \theta$.

If $\delta$ is allowed to tend to zero together with the other parameters, the bare conditions $\alpha \ll 1$ and $\mu \gg 1$ do not alone justify that reduction. The needed dominance conditions are $\mu \delta \gg 1$, $\mu {\delta }^2\gg 1$, and $\alpha /\delta \ll 1$, so that the rod contributes negligible gravitational torque and moment of inertia, and the disc behaves as a point mass. They follow automatically in the combined limit with fixed $\delta >0$.

For $i\ne j$, the inertia tensor component is $I_{ij}=-\rho {\int }_V{x}_i{x}_{j}dV$. Reflect either coordinate appearing once: the cylindrical domain and density are unchanged, while the integrand changes sign. Thus every off-diagonal component vanishes. Rotational symmetry also implies $I_{11}=I_{22}$.

Use cylindrical coordinates $x_1=r\cos \theta$, $x_2=r\sin \theta$, $x_3=z$, with volume element $rdrd\theta dz$. The axial moment of inertia isThe two contributions to a transverse component areThereforeThe axial value $M{R}^2/2$ and transverse value $M(3R^2+h^2)/12$ for $R=1$, $h=4$ give an independent check on both diagonal entries.

Use the mode shapes from part (b), and put $\omega =\sqrt{2k/m}$. The initial velocity decomposes asThere is no antisymmetric component because the initial state treats the equal masses identically. Initial displacements are zero, so the zero mode contributes a linear drift and the nonzero normal mode contributes a sine:In terms of actual angular positions,The conserved angular momentum is initially $2m\Omega$ and the total moment of inertia is $4m$, so the common angular velocity must be $\Omega /2$, agreeing with the drift. The relative displacements remain of order $\Omega /\omega$, even when the common angle becomes large. Hence small $\Omega /\omega$ preserves the cyclic ordering and small spring extensions at arbitrarily late times. In the ideal arc-spring model the displayed energies are exactly quadratic in the relative angles, so this motion remains exact while that ordering persists.

For the stated square side length $2a$, the volume is $4a^{2}h$ and the uniform mass density is ${\rho }_1=m_1/(4a^{2}h)$. Direct integration about its central vertical axis gives the moment of inertia of a uniform square plate:Thus the PDF's coefficient $1/6$ is inconsistent with its stated side length. The coefficient $1/6$ would apply to a square of side $a$.

For the disc, use cylindrical coordinates and $m_2={\rho }_2\pi a^{2}h$:For the sphere, use spherical coordinates and $m_3=4\pi {\rho }_3{a}^3/3$:These agree with the moment of inertia of a uniform disc and the moment of inertia of a uniform solid sphere. The square's thickness does not enter its axial moment of inertia once the mass is fixed.

Measure each generalized coordinate ${\theta }_i$ from an unstressed configuration. The diagram has one torsional spring joining the frame to object 1, one joining objects 1 and 2, and one joining objects 2 and 3. The rotational kinetic energy and elastic potential energy areEach spring contribution follows by integrating its restoring torque: ${\int }_0^{\Delta \theta }{k}_{j}sds=k_j(\Delta \theta {)}^2/2$. ThusThe Euler-Lagrange equations giveThese hold for either interpretation of the square's moment of inertia; the geometry-based value from (a) is used first below.

Let $\mathcal{I}$ be the sphere's mechanical moment of inertia about its rotation axis, and let ${\omega }_i=\omega (0)$. For an undriven rigid sphere the rotational energy is $E_{\mathrm{rot}}=\mathcal{I}{\omega }^2/2$. In the weak-radiation-loss approximation, conservation of energy givesIntegrating $d({\omega }^{-2})/dt=2K/\mathcal{I}$ givesEquivalently, the characteristic spin-down time isThe radiative spin-down of a rotating electric dipole therefore has angular frequency that decays as $t^{-1/2}$ at late times within this model, rather than exponentially. The original paper gives no mass or mass-density profile, so $\mathcal{I}$ cannot be determined from $R$ and the relative permittivity. If a homogeneous solid sphere of mass $M$ is additionally assumed, insert $\mathcal{I}=2M{R}^2/5$; uniform dielectric response alone does not imply this mechanical mass distribution.

The approximation is important. If the dipole angle is $\varphi (t)$ with $\dot{\varphi }=\omega (t)$, direct differentiation gives $|\ddot{\mathbf{p}}{|}^2=p_0^2({\omega }^4+{\dot{\omega }}^2)$. Using the constant-frequency radiation law instantaneously neglects the second term and assumes $|\dot{\omega }|\ll {\omega }^2$. This gives the standard leading slow spin-down solution above. An exact nonadiabatic radiation reaction evolution would require the mechanical and near-field dynamics beyond the information supplied, rather than treating this leading law as exact at arbitrarily rapid damping.

Let $s$ be distance along the rod from its fixed end. The position of that material point is $sn(t)$, where $n=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,-\cos \theta )$ is a unit vector. Its velocity is $s\dot{n}$, and direct differentiation givesThe uniform linear mass density is $M/a$. Integrating the elemental kinetic energy therefore proves, rather than assumes, the rod formula:Equivalently, its moment of inertia about a perpendicular axis through the pivot is $M{a}^2/3$.

Use coordinates with the downward vertical as reference. The first rod's center is at $(a/2)(\sin \theta ,-\cos \theta )$, so its translational plus rotational kinetic energy is $m{a}^2{\dot{\theta }}^2/6$. The second center isIts squared speed is $a^2{\dot{\theta }}^2+a^2{\dot{\varphi }}^2/4+a^2\dot{\theta }\dot{\varphi }\cos (\theta -\varphi )$. Adding its center-of-mass rotation, using moment of inertia $m{a}^2/12$, gives the totalThe two center heights give $V=-mga(3\cos \theta +\cos \varphi )/2$. Hence $L=T-V$ is exactly the stated Lagrangian.

Small stable oscillations are about the downward equilibrium $\theta =\varphi =0$. Retaining quadratic terms gives $T_2=\frac{1}{2}{\dot{q}}^{T}M\dot{q}$ and $V_2=\frac{1}{2}{q}^{T}Kq$, whereA normal mode $q=v{e}^{i\omega t}$ satisfies $(K-{\omega }^{2}M)v=0$. With $\kappa =a{\omega }^2/g$, the determinant equation isThus the angular frequencies areBoth squared frequencies are positive. The amplitude ratio is $\varphi /\theta =(9-8\kappa )/(3\kappa )$: it is positive in the lower-frequency mode and negative in the higher-frequency mode. The rods therefore swing in phase in the first mode and in opposite phase in the second.

A positive-parity $K=0$ nuclear rotational band in an even-even nucleus has $J=0,2,4,\dots$ and excitation energiesThe four nonzero levels give $E_J/[J(J+1)]=15.17,14.85,14.48,14.01\mathrm{keV}$. These approximately constant coefficients, together with the even-spin positive-parity sequence, support a rotational interpretation. The roughly eight-percent decline means a constant-inertia rigid rotor is only an approximation: the effective moment of inertia increases along the sequence.

Using the lowest excited level givesUsing the highest level instead gives $2.48\times 10^{-54}\mathrm{kg}{\mathrm{m}}^2$, showing the scale of the departure from a constant inertia. This is a comparison of nuclear rotational inertia relative to a rigid sphere, not evidence for an exactly spherical rotating intrinsic state.

For a conventional size estimate, explicitly take a uniform sharp radius $R=r_0{A}^{1/3}$ with $r_0\simeq 1.2\mathrm{fm}$ and mass $M\simeq A{m}_u$. These are standard scale approximations, not additional measured properties of this isotope supplied by the paper. They give $R\simeq 6.70\mathrm{fm}$ andThus the fitted rotational inertia is substantially smaller than the rigid-body scale. Nuclear pairing provides a physical reason collective nuclear inertia need not equal the classical rigid-body value. The comparison's numerical radius dependence is explicit; it is not an independently determined radius.

The parallel axis theorem states $I=I_G+M{d}^2$ for parallel axes separated by $d$, with the reference axis through the centre of mass. The perpendicular axis theorem states $I_z=I_x+I_y$ for a planar lamina in the $xy$ plane, about mutually perpendicular axes through one point. It does not apply directly to a three-dimensional solid.

For the transverse moment of inertia of a uniform solid cylinder, divide the cylinder into thin disks at axial positions $z\in [-h/2,h/2]$. Symmetry and the perpendicular axis theorem give the diametral moment of inertia $r^{2}dM/4$ of each disk; the parallel axis theorem adds $z^{2}dM$. HenceThe thin-disk limit is $M{r}^2/4$, while the slender-rod limit is $M{h}^2/12$.

Let $\theta$ measure the rim point mass's displacement from vertically below the fixed axis. The disk has moment of inertia $I_D=\frac{1}{2}(2M)R^2=M{R}^2$, and the rim mass contributes another $M{R}^2$. Its varying gravitational potential energy is $-MgR\cos \theta$. ThusThe disk's centre of mass remains fixed, so its own gravitational contribution is a constant and has been omitted. The total fixed-axis moment of inertia is $2M{R}^2$.

Use the angle $\theta$ from the vertical as the generalized coordinate. For a uniform ladder, its centre of mass is $(l\sin \theta ,l\cos \theta )$, so $v_G^2=l^2{\dot{\theta }}^2$. Its central moment of inertia is $I_G=m(2l{)}^2/12=m{l}^2/3$. With ideal contact constraints that do no work, the Lagrangian, conjugate momentum, and Euler-Lagrange equation areRelease from rest at ${\theta }_0$ also gives ${\dot{\theta }}^2=3g(\cos {\theta }_0-\cos \theta )/(2l)$ by conservation of energy. These results hold while both contacts persist. The ideal wall reaction is normal to the wall; an unspecified tangential wall friction would require an additional generalized force and would prevent determination from the supplied data alone.

Choose parallel axes along $z$, one through the centre of mass and one displaced transversely by $\mathbf{a}$. Write the perpendicular position as ${\mathbf{r}}_{\perp }={\mathbf{r}}_{\perp }^{\prime }+\mathbf{a}$. Direct expansion of the moment of inertia givesThe integral in the cross term is zero by the definition of the centre of mass. Therefore the parallel axis theorem isThe axes must be parallel and the reference axis must pass through the centre of mass; the formula is not generally valid between arbitrary parallel axes using either axis as the reference.

A straight diameter at nonzero angle $\chi$ to the rotation axis moves the body toward that axis during the inward half of its trip. With no external axial torque, the total axial angular momentum of the Earth and body is conserved. The body's contribution to the moment of inertia decreases toward the centre, so the Earth spins slightly faster; the reverse happens on the outward half, restoring the original rotation rate at the far surface.

For a body of mass $m$ constrained to follow a tunnel fixed in the Earth, an illustrative angular-momentum balance isHere $I_E$ is the Earth's axial moment of inertia, treated as constant, and $x$ is the signed distance along the diameter. A frictionless guide can still exert normal forces: these supply the transverse acceleration and exchange axial angular momentum with the rotating Earth. Simply allowing an unconstrained body to fall would not keep it inside a narrow nonpolar straight tunnel. The effect vanishes for $\chi =0$ and is largest for an equatorial diameter.

Choose the smaller disc's initial sense as positive. For either uniform disk, the axial moment of inertia is $I=\frac{1}{2}m{r}^2$, so $I_1=I_2/2$. The signed initial angular velocities are ${\Omega }_1=2{\omega }_2$ and ${\Omega }_2=-{\omega }_2$, giving total axial angular momentum zero. The frictionless axle supplies no axial torque, and the pins' negligible mass supplies no extra stored kinetic energy. ThusThese equations allow the original state or its simultaneous reversal. A nontrivial elastic pin impact selects reversal of the relative contact velocity, rather than continued interpenetration. ThereforeThe requested magnitudes are unchanged: $|{\Omega }_1^{\prime }|={\omega }_1$ and $|{\Omega }_2^{\prime }|={\omega }_2$. The intersecting pin paths make the impact possible; no additional lever arm is needed once axial angular momentum and energy conservation have been imposed.

Let $\theta$ measure the spoke's displacement from downward equilibrium. The total moment of inertia about the axle is $I=M{R}_2^2+m{R}_1^2$: the rim is a thin thin hoop, and the attached body is a point mass. The rim's weight has no axle torque, while the attached mass gives restoring torque $-mg{R}_1\sin \theta$. HenceEquivalently conservation of energy gives $\frac{1}{2}I{\dot{\theta }}^2=mg{R}_1(\cos \theta -\cos {\theta }_0)$. For small angles, $\sin \theta \simeq \theta$, so the motion is approximately $\theta ={\theta }_0\cos (\omega t)$ with ${\omega }^2=mg{R}_1/I$. The small-amplitude period isAt finite amplitude the nonlinear pendulum equation remains valid but this period approximation does not. The $M\to 0$ limit recovers a simple pendulum of length $R_1$.

For the eccentric circular-hole disk inertia, let $D=\pi \rho R^2$ be the mass of the complete disc before removing material, $h=\pi \rho r_o^2$ the missing mass, and $W=D-h$ the wheel's remaining mass. Denote the complete disc's centre by C and the axle/hole centre by H; $CH=r_o$. Assume the hole fits inside the disc, $0<r_o<R/2$. The parallel axis theorem gives the full disc's moment of inertia about H as $\frac{1}{2}D{R}^2+D{r}_o^2$. Subtract the absent little disc about its own centre H:The axle is a support, not part of the rotating wheel's stated mass. Its frictionless bearing does not alter this moment of inertia. For fixed-axis angular velocity $\omega$,The missing material must be subtracted with its actual displacement relative to the axis; treating the wheel as a concentric annulus would give the wrong result.

Using rotational kinetic energy $K=I{\omega }^2/2$ at the two steady endpoint configurations,At fixed angular momentum $J$, $K=J^2/(2I)$ increases when the moment of inertia decreases. The skater supplies work by pulling the arms inward; chemical energy in the muscles is converted into additional rotational kinetic energy, with possible extra dissipation in a real skater. This does not require an external axial torque.

For the ideal rotating point masses, an inward radial force supplies the required centripetal acceleration. During contraction both that inward force and the radial displacement point inward, so their scalar product gives positive work. At the initial and final steady configurations radial motion has ceased; any transient radial kinetic energy must also be included in a full intermediate-time balance. The quoted difference compares the rotational endpoint energies only.

The parallel axis theorem relates the moment of inertia about an axis through the centre of mass to that about a parallel axis displaced by perpendicular distance $d$:The reference axis must pass through the centre of mass. To see why, write the displacement from it as $\mathbf{r}=\mathbf{R}+\mathbf{d}$ in the plane perpendicular to the axes. Expanding $\int |\mathbf{R}+\mathbf{d}{|}^{2}dm$ gives $I_G+M{d}^2$ because $\int \mathbf{R}dm=0$.

For a thin planar mass distribution, let $x,y$ be perpendicular axes in its plane and $z$ the normal axis through their same intersection. The perpendicular axis theorem statesIndeed, $I_z=\int (x^2+y^2)dm$, while $I_x=\int y^{2}dm$ and $I_y=\int x^{2}dm$. Their common intersection need not be the centre of mass, but an arbitrary three-dimensional body does not satisfy the planar equality.

Take $\theta$ from the downward vertical. Under the conventional light-rod idealization, only the endpoint point mass contributes appreciably. Its speed is $l\dot{\theta }$, and its gravitational potential energy is $-mgl\cos \theta$. ThereforeAn additive constant in the potential energy has no effect on the Euler-Lagrange equation.

Rod-mass qualification: “thin” and “uniform” do not imply negligible mass. If the uniform rod has a finite unspecified mass $M_r$, its pivot moment of inertia is $M_r{l}^2/3$ and its centre of mass lies $l/2$ from the pivot. For a pendulum with a uniform rod and an endpoint mass, the general answer isThis formula covers either convention without inventing a rod mass. A result involving only the specified $m,l,g$ uses $M_r=0$ unless an additional relation for $M_r$ is stated.