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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.

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 $M{h}^2/12$.

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.

For the shorter uniform rod, the parallel axis theorem gives $I=m(a^2/12+x^2)$. Its equivalent simple pendulum length is $I/(mx)=x+a^2/(12x)$. The rod of length $2a$ pivoted at an end has equivalent length $2(2a)/3=4a/3$. Matching the periods therefore requiresA pivot on the shorter rod requires $0<x\le a/2$, leavingEither side of the rod's centre of mass gives this distance. The other algebraic root is beyond the rod and would require a support extension of negligible mass.

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.

For a freely rolling rigid body, decompose the kinetic energy into translation of its centre of mass G and rotation about G. The centre of mass after removing material lies on the side opposite H:Another use of the parallel axis theorem givesThus at the same angular speed,This compares the rotational component, not the total rolling kinetic energy. The total also contains $\frac{1}{2}W{v}_G^2$ and need not be smaller than $T_H$. Describing all rolling motion as instantaneous rotation about its contact point includes that translational contribution; it is not the rotational component intended by this comparison.

Take the square to have uniform surface mass density $\sigma =M/a^2$, and put its centre at $x=y=0$. For an in-plane axis through the centre parallel to a side, direct integration givesThe same holds for $I_y$, and the mixed integral $\int xydm$ vanishes by reflection symmetry. The squared perpendicular distance to an axis making angle $\psi$ with the $x$ axis is $(-x\sin \psi +y\cos \psi {)}^2$. Its integral therefore givesThus a diagonal through opposite corners and its perpendicular through the centre haveThis is the in-plane inertia isotropy of a square plate.

Axis-location qualification: perpendicularity fixes a direction, not a line's position. If the second axis is instead shifted by distance $d$ from the centre, the parallel axis theorem gives $I_2=M{a}^2/12+M{d}^2$. A perpendicular through a corner on that diagonal has $d=a/\sqrt{2}$ and hence $I_2=7M{a}^2/12$, a concrete different value satisfying the stated directional description. If “through the corners” means an edge rather than a diagonal, that edge has $I=M{a}^2/3$; two adjacent edges both have that value. The general displaced-axis formula and these choices make the location convention explicit rather than silently assigning a unique value to an unspecified second axis.

Use two central perpendicular axes in the plate. The perpendicular axis theorem and the preceding moments of inertia giveAlternatively, axes along adjacent edges have moments of inertia $M{a}^2/3$ each. The normal axis through their shared corner therefore has $2M{a}^2/3$. It lies a distance $a/\sqrt{2}$ from the parallel normal axis through the centre, so the parallel axis theorem yields $I_{z,G}=2M{a}^2/3-M(a^2/2)=M{a}^2/6$. This normal central result is independent of the in-plane axis-location ambiguity.

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.

The disc carries mass $M_d=2M/3$ and the thin rim carries $M_r=M/3$. About the wheel's central normal axis, add the moment of inertia of a uniform disc and the thin rim's moment of inertia:Treat the wheel as a thin planar lamina, as implied by the disc-and-thin-rim model. Interpret the requested perpendicular axis as perpendicular to the wheel's original axis and tangent to the rim: if the wheel lies in the $xy$ plane, choose the tangent line $x=R,z=0$, parallel to the $y$ diameter. By rotational symmetry, the two diametral moments of inertia are equal. The perpendicular axis theorem therefore givesThe parallel axis theorem then gives the tangent-axis inertia of an axisymmetric lamina:The axis direction matters. If “perpendicular” were intended to mean perpendicular to the wheel's plane, an axis through a rim point would instead be parallel to the original wheel axis, giving $I_{\mathrm{rim},\mathrm{normal}}=I_z+M{R}^2=5M{R}^2/3$. The $4M{R}^2/3$ answer uses a tangent axis perpendicular to the original axis; the $5M{R}^2/3$ value belongs to the different, explicitly identified normal axis. Finite wheel thickness would require additional information for the in-plane moment of inertia.

The parallel axis theorem gives $I=I_{\mathrm{CM}}+M{d}^2$ for parallel axes separated by $d$, with the reference axis through the centre of mass. For a plane lamina, the perpendicular axis theorem gives $I_z=I_x+I_y$ for mutually perpendicular axes meeting at one point, with $x,y$ in the plane.

Treat the removed material as negative mass. With the hole's centre at $b\hat{\mathbf{x}}$, its moment of inertia about $O$ is $m(a^2/2+b^2)$. The remaining moment of inertia and centre of mass areFor a uniform original disk, $m/M=a^2/R^2$. Write $d_1=mb/(M-m)$ and measure $\varphi$ from the downward equilibrium direction of its centre of mass. The physical pendulum then has LagrangianThe Euler-Lagrange equation is $I_1\ddot{\varphi }+mbg\sin \varphi =0$. For small oscillations, $\sin \varphi \simeq \varphi$, givingNow add the removed small disk on the opposite side, with its centre a distance $c=R-a$ from $O$ so that it is internally tangent to the large disk. The total mass returns to $M$, while its first moment about $O$ is $-m(b+c)\hat{\mathbf{x}}$. ThusReplacing both the inertia and gravitational restoring moment givesThe same first-moment method applies to a general perforated-disc pendulum. The added disk is attached to the lamina and shares its rotation; treating it as a massless patch or updating only the inertia would omit its restoring torque.

The parallel axis theorem relates the moment of inertia $I$ about an axis to $I_{\mathrm{CM}}$ about a parallel axis through the centre of mass, separated by perpendicular distance $d$:In the defining mass integral, expanding the squared distance produces a cross term whose first mass moment about the centre of mass is zero.

For a thin planar rigid body, the perpendicular axis theorem applies to two perpendicular axes $x,y$ in its plane and the perpendicular axis $z$, all meeting at the same point:Indeed, $I_x=\int y^{2}dm$, $I_y=\int x^{2}dm$, and $I_z=\int (x^2+y^2)dm$. The planar condition is essential; this is not an identity for an arbitrary three-dimensional rigid body.

Use the usual dumbbell geometry: the uniform thin rod joins the disc centres and is perpendicular to both disc faces. The required axis through the rod midpoint lies in a direction parallel to a diameter of each disc. For either disc, rotational symmetry and the perpendicular axis theorem giveEach disc centre lies a distance $l/2$ from the requested axis. The parallel axis theorem therefore gives a disc contribution $M{a}^2/4+M(l/2{)}^2$. The rod contributesAdd the two disc contributions and the rod's moment of inertia:Using $M{a}^2/2$ for each disc's central contribution would instead use an axis perpendicular to its face, which is parallel to the rod and is not the requested axis.

For a planar lamina in the $xy$-plane and three mutually perpendicular axes through one common point, the perpendicular axis theorem is $I_z=I_x+I_y$. Indeed $I_z=\int (x^2+y^2)dm$, while $I_x=\int y^{2}dm$ and $I_y=\int x^{2}dm$. Planarity and the common axis intersection are essential.

For two parallel axes, one through the centre of mass $G$ and one a perpendicular distance $d$ away, the parallel axis theorem iswhere $M$ is the total mass. Expanding the squared distance in the defining moment of inertia integral leaves a cross term proportional to $\int {\mathbf{r}}_{G}dm=0$. The second axis need not pass through the body, and uniform density is not required, but the reference axis for $I_G$ must pass through its centre of mass.

Let $O$ be the original disk centre and $C$ the hole centre. The source geometry has $OS=a$, $OC=CS=a/2$, all collinear. The removed disk has mass $m/4$, since its area is one quarter of the original disk's. Use the parallel axis theorem for both the original disk and the removed disk:Subtracting the missing material's moment of inertia givesHere $m$ continues to denote the original unperforated disk mass; the earring itself has mass $3m/4$.

The moment of inertia of a uniform thin ring about its perpendicular central axis is $M_a{a}^2$. Its two in-plane diameters are equivalent by rotational symmetry, so the perpendicular axis theorem gives $M_a{a}^2/2$ about either diameter. The central rod contribution about an axis perpendicular to its length is $M_l{l}^2/12$.

The requested axis is parallel to these central axes and passes through the whole racquet's centre of mass. Applying the parallel axis theorem separately givesAn equivalent reduced expression isBoth refer to the in-plane axis perpendicular to the handle; the ring contribution would be different for an axis normal to the racquet face.

The parallel axis theorem relates the moment of inertia about an axis through the centre of mass to that about a parallel axis a perpendicular distance $d$ away:To see the cross term vanish, write each perpendicular displacement as its displacement from the centre of mass plus the fixed offset. In the defining mass integral, the first moment about the centre of mass is zero.

For a thin planar rigid body in the $xy$ plane, the perpendicular axis theorem sayswhere all three axes pass through the same point, $x,y$ lie in the plane, and $z$ is perpendicular to it. This follows directly from $x^2+y^2$ being the squared distance from the $z$ axis. The thin-planar condition and common intersection point are essential; the formula is not a general identity for three-dimensional bodies.

The parallel axis through the centre is the rod's own long axis, with zero moment of inertia in the thin-rod approximation. By the parallel axis theorem, the displaced parallel axis hasEquivalently, every mass element is the same perpendicular distance $D$ from that axis, so the defining integral immediately gives this result.

Use Cartesian coordinates along the edges from the pivot. With surface mass density $M/L^2$, the moment of inertia isEquivalently, the moment of inertia of a uniform square plate about its central perpendicular axis is $M{L}^2/6$, and the parallel axis theorem adds $M(L/\sqrt{2}{)}^2$.

The source figure measures $\theta$ from the upward vertical to $OC$, with $OC=L/\sqrt{2}$. The centre's height above the pivot is $(L/\sqrt{2})\cos \theta$. The smooth fixed pivot does no work. Conservation of energy, with the initially upright configuration as reference, givesFor rotation in the positive $\theta$ direction on the first descent, take the positive root. Its equivalent form is $\dot{\theta }=\sqrt{3\sqrt{2}g/L}\sin (\theta /2)$ for $0<\theta <\pi$.

Strictly, release from rest at a small angle ${\theta }_0$ replaces $1-\cos \theta$ by $\cos {\theta }_0-\cos \theta$. The displayed result is the ${\theta }_0\to 0$ limiting energy relation. Exactly upright and exactly at rest is an unstable equilibrium; without a perturbation it stays there. This distinction prevents the infinitesimal displacement convention from being mistaken for spontaneous motion.

Inertia is resistance to a change of motion. Translational inertia is measured by mass in Newton's second law; rotational inertia about a specified fixed axis is measured by the moment of inertiawhere $r_{\perp }$ is perpendicular distance from that axis. For a thin uniform rod, take axes perpendicular to the rod. With linear mass density $\lambda =m/l$,The second result uses the parallel axis theorem; direct integration over $0\le x\le l$ gives the same value. The axis specification is essential: an ideal thin rod has zero moment of inertia about its own longitudinal axis.

During the short elastic collision, neglect the impulse from gravity. The pivot can exert an external impulse, so total linear momentum of rod plus particle need not be conserved. Its torque about the pivot is zero, so conservation of angular momentum about that point applies. Let $w$ be the particle's signed outgoing horizontal velocity, and $\omega$ the rod's outgoing angular velocity, positive towards the incident motion. ThenSubstituting $\omega =3(v_0-w)/l$ into the kinetic energy equation givesDiscarding the zero-impulse branch $w=v_0$, the actual collision givesThe rod's lower end initially moves forward at $l\omega =3v_0/2$, so the bodies separate. Afterwards the free particle undergoes projectile motion. With the pivot as origin and $z$ downward, its trajectory is $x=(v_0/2)t$, $z=l+g{t}^2/2$, where $t=0$ at the collision. The rod instead swings about the fixed pivot in a vertical plane as a physical pendulum.

The rod's initial rotational kinetic energy is $I_{\mathrm{end}}{\omega }^2/2=3m{v}_0^2/8$. Reaching the horizontal raises its centre of mass by $l/2$, increasing gravitational potential energy by $mgl/2$. Consequently,For the threshold case, measure the rod's angle $\theta$ from the downward vertical and reset $t=0$ at its first horizontal turning point, so $\theta (0)=\pi /2$ and $\dot{\theta }(0)=0$. The restoring torque about the pivot gives the exact equation of motionAt the horizontal, $\ddot{\theta }=-3g/(2l)$, so the rod immediately swings back. Conservation of mechanical energy yieldsOn the initial descent its explicit time dependence is specified by the definite integralIt continues through the vertical to $-\pi /2$, reverses, and repeats without damping. An alternative expression valid throughout the oscillation uses the Jacobi elliptic sine. Put $k=1/\sqrt{2}$, ${\omega }_r=\sqrt{3g/(2l)}$, and let $K(k)$ denote the complete elliptic integral of the first kind. ThenThis is not simple harmonic motion at an amplitude of $\pi /2$. Its oscillation period exceeds its own small-amplitude value $2\pi \sqrt{2l/(3g)}$ by the factor $2K(1/\sqrt{2})/\pi \simeq 1.18034$.

A simple pendulum of length $l/2$ obeys $\ddot{\theta }+(2g/l)\sin \theta =0$. At the same amplitude $\pi /2$, its oscillation period is $T_p=4\sqrt{l/(2g)}K(1/\sqrt{2})$, soThe rod swings more slowly. Its equivalent simple pendulum length is $I_{\mathrm{end}}/[m(l/2)]=2l/3$, not the distance $l/2$ to its centre of mass. The same oscillation period ratio holds in the common small-amplitude limit, but replacing the actual 90-degree motion by that limit would underestimate both oscillation periods.