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Let $\rho =3m/(4\pi R_0^3)$ be the uniform mass density. In spherical coordinates $(r,\vartheta ,\varphi )$ with the chosen diameter as polar axis, the squared perpendicular distance to that axis is $r^2{\sin }^2\vartheta$. The moment of inertia is thereforeThis derives the moment of inertia of a uniform solid sphere about any central axis by spherical symmetry.

Put $a=R_1-R_0$. The center of mass follows the circle of radius $a$, with position ${\mathbf{r}}_C=a(\sin \theta ,-\cos \theta )$. Let ${\mathbf{e}}_r=(\sin \theta ,-\cos \theta )$ and ${\mathbf{e}}_{\theta }=(\cos \theta ,\sin \theta )$. Its velocity is $a\dot{\theta }{\mathbf{e}}_{\theta }$, and the contact point is displaced from the center of mass by $R_0{\mathbf{e}}_r$. With positive spin angular velocity ${\Omega }_s$ counterclockwise, its contact-point velocity isThe cylinder is fixed, so rolling without slipping requires this velocity to vanish. Consequently ${\Omega }_s=-a\dot{\theta }/R_0$. If instead the spin angle $\psi$ is positive clockwise, opposite to increasing $\theta$, then the printed positive-coefficient convention isThe spin and orbital rotation have opposite directions when both are measured about the same oriented axis; the angular speed is $a|\dot{\theta }|/R_0$. This specifies the sign convention in the printed expression.

The instantaneous contact point is at rest, so the normal reaction and static friction do no work. Conservation of energy equates the loss of gravitational potential energy from release to the total kinetic energy:On the descent from $0<\alpha <\pi /2$, $\theta$ decreases, soIntegrating $dt=-d\theta /|\dot{\theta }|$ from release to the bottom givesThe integral is finite: at the release point its denominator is asymptotic to $\sqrt{\sin \alpha (\alpha -\theta )}$. Contact is maintained, since the inward radial equation of motion gives the positive normal reaction $N=m(g\cos \theta +a{\dot{\theta }}^2)$ throughout this range.

For frictionless sliding, the normal reaction passes through the sphere's centre and produces no torque; Newtonian gravity also exerts no central torque. With initial zero spin, the sphere remains nonrotating. All the lost gravitational potential energy becomes translational kinetic energy, soA hollow spherical shell of the same mass and radius places its material farther from the centre, giving a larger moment of inertia. Under rolling without slipping, a given centre speed therefore requires more rotational kinetic energy; the same gravitational drop yields a smaller speed. Replacing the solid sphere by a hollow shell increases $T_R$ and leaves $T_S$ unchanged. The sliding case remains nonrotating and has the same center of mass path and gravitational drop. This is the general comparison for rolling inside a fixed circular cylinder.

For the moment of inertia of a uniform solid sphere, write the mass density as $\rho =3m/(4\pi a^3)$. In spherical coordinates about the chosen axis, the squared perpendicular distance is $R^2{\sin }^2\vartheta$. The volume element is $R^2\sin \vartheta dRd\vartheta d\phi$, so

Take $0<\alpha <\pi /2$ and $\ell >0$, with a fixed plane. The normal force has magnitude $N=mg\cos \alpha$ and is perpendicular to the plane. On the perfectly smooth plane there is no friction. Both gravity and the normal force have zero torque about the marble's centre, so a marble released from rest does not rotate. The component of Newton's second law down the plane is $m{A}_1=mg\sin \alpha$, henceThe normal force does no work because the velocity is tangent to the plane. Gravity is conservative, so mechanical energy, the sum of translational kinetic energy and gravitational potential energy, is conserved.

For rolling without slipping, let the static friction force have magnitude $f$, pointing up the plane. It supplies the torque that spins the marble while reducing its translational acceleration. If $\Omega$ is the angular velocity and $A_2$ the acceleration down the plane, the rolling constraint is $a\Omega =\dot{s}$, hence $a\dot{\Omega }=A_2$. The translation and rotation equations areThus $f=I{A}_2/a^2$ andThese are the equations for rolling acceleration with rotational inertia; the plane must be rough enough to supply this static friction.

The point of contact is instantaneously at rest relative to the fixed plane, so static friction does no total work on the rolling marble. Equivalently, its translational power is $-f\dot{s}$ and its rotational power is $fa\Omega =f\dot{s}$, which cancel. The normal force also does no work. Consequently mechanical energy is conserved here too, now with rotational kinetic energy included:where $H$ is the centre's height. No energy loss is implied by the presence of ideal static friction. Constant acceleration from rest gives

For the hollow marble, put $I=\lambda m{a}^2$ in the same equations. The smooth-plane time is unchanged, while the rolling acceleration becomes $g\sin \alpha /(1+\lambda )$. ThereforeThis again assumes rolling without slipping and enough static friction; the ratio depends on the mass distribution through its moment of inertia.

Let $M=M_1+M_2$, and let $\mu =M_1{M}_2/M$ be the reduced mass. The stellar distances from the center of mass are $a_1=a{M}_2/M$ and $a_2=a{M}_1/M$. For a circular Kepler orbit, ${\Omega }^2=GM/a^3$, so summing the two orbital contributions gives the circular-binary orbital angular momentumThe separation $a$ is inside the square root. This also has the required dimensions of angular momentum.

For the homologous rotating collapse, label a fluid element by $s=r/R$ and its polar angle $\theta$. Its enclosed mass $m(s)$ is constant during stellar homology, as are the dimensionless density profile and the inertia coefficient $\alpha$. Conservation of angular momentum and solid-body rotation giveThe magnitude of the centrifugal acceleration is ${\Omega }^{2}r\sin \theta$, while the Newtonian gravitational field has magnitude $Gm(s)/r^2$. ThereforeUsing only the radial component of centrifugal force replaces $\sin \theta$ by ${\sin }^2\theta$ and leaves the scaling unchanged. The centrifugal force vanishes on the rotation axis; the central point is understood through the limiting field rather than a ratio of two zero forces.

At the equator of the cloud, critical rotation of a spherical cloud means ${\Omega }_c^2{R}_c=GM/R_c^2$, where $R_c=R_{\mathrm{crit}}$. ConsequentlyImmediately after the first fission, each daughter has mass $M/2$ and radius $a/2$ because the spheres touch. Their orbital moment of inertia is $M{a}^2/4$, while their combined spin moment of inertia is $\alpha M{a}^2/4$. With a common spin and orbital frequency ${\Omega }_f=\sqrt{GM/a^3}$, the synchronous fission model for binary formation givesThe common frequency can change during fission; corotation requires equal instantaneous frequencies, not an unchanged frequency from before the rearrangement. Equating the two values of $J$ yieldsFor a nonzero positive inertia coefficient, the necessary and sufficient condition in this algebraic model for $a<R_c$ isFor example, a moment of inertia of a uniform solid sphere has $\alpha =2/5$ and gives $a/R_c=64/49>1$; it fails the required compact-fission condition. A sufficiently centrally concentrated cloud can have smaller $\alpha$.

For the next collapse, neglect exchange of spin with the unchanged outer orbit, so each daughter's spin angular momentum is separately conserved. Immediately after the first fission that spin isFor a daughter of mass $m=M/2$ to reach critical rotation of a spherical cloud at radius $R_{c,d}$, the same relation gives $S_d=\alpha m\sqrt{Gm{R}_{c,d}}$. HenceEquivalently, its centrifugal-to-gravity ratio starts at $1/4$ when its radius is $a/2$ and reaches one after contraction by a factor four. Applying the same fission rule to this daughter produces an inner pair with separationCorotation at this second fission is local to each newly formed inner pair; it is not a single common frequency for the entire four-star hierarchy.

This spin assumption matters. If tidal locking instead kept each shrinking daughter synchronized to the fixed outer orbital period throughout the intervening collapse, its angular velocity would remain fixed. Its centrifugal-to-gravity ratio would then scale as $R_d^3$ and decrease, so it would never reach the proposed second fission. This is a counterexample to extending the corotation assumption through the whole contraction. The printed ratio describes the separately spin-conserving interpretation of the repeated fission hierarchy.

The model can generate a compact hierarchy algebraically, but is not a general physical account of multiple-star formation. In the allowed range $\alpha <1/3$, one has $a^{\prime }/a<1/8$, suggesting well-separated inner and outer scales. However, rapidly rotating gas deforms, touching daughters are tidally distorted, and pressure, gas flows, dissipation and spin-orbit torques cannot generally be ignored. The assumed identical profiles, equal mass splits, instantaneous corotation and later torque-free contractions are restrictive. Star formation can involve gravitational fragmentation and redistribution of angular momentum; the toy budget neither proves that fission occurs nor predicts the population of real multiple systems.

Let the uniform mass density be $\rho =3M/(4\pi a^3)$. In spherical polar coordinates about the rotation axis, the perpendicular distance is $r\sin \theta$ and the volume element is $r^2\sin \theta drd\theta d\varphi$. The moment of inertia is thereforeIn the given rigid body kinetic energy decomposition, $\frac{1}{2}M|\dot{\mathbf{R}}{|}^2$ is the translational kinetic energy of the whole mass moving at its center of mass velocity. The term $\frac{1}{2}I{\omega }^2$ is the rotational kinetic energy about the center of mass. The mixed term vanishes because the mass-weighted relative positions sum to zero.

For rolling without slipping, the point of contact has zero instantaneous velocity relative to the stationary surface. Its rotational velocity is opposite the center of mass velocity and has magnitude $a|\omega |$. Hence $V=a|\omega |$. Substituting this and the moment of inertia of a uniform solid sphere givesThe rotational axis is through the center of mass and parallel to the surface, perpendicular to the direction of rolling.

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.

For a sphere with uniform mass density, $M=(4\pi /3)\rho a^3$. Its moment of inertia of a uniform solid sphere is $I=2M{a}^2/5$, and rotation about a diameter gives angular momentum magnitude $L=I\omega$. The spheres have the same angular frequency $\omega$, soThe mass grows by a factor of two, and the extra squared lever arm supplies a further factor of four. Equal angular frequencies alone do not mean equal angular momenta.

Take the rotation axis as $z$. A slice of the uniform sphere at coordinate $z$ is a disc of radius $R(z)=\sqrt{a^2-z^2}$ and mass $dm=\rho \pi (a^2-z^2)dz$, where $\rho =3M/(4\pi a^3)$ is its mass density. Applying a to the slice givesThe slices' axes coincide, so no displaced-axis correction is needed. Integrating,Thus the moment of inertia of a uniform solid sphere isThe sphere's rotational symmetry makes this valid about every central axis. A thin spherical shell would have a different mass distribution and a different moment of inertia.

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.