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Idealize the pole and stick as having negligible transverse thickness. Put the pole at the origin, the stick initially along the $y$ axis, and the struck lower end at $y=-l/2$, as in the source diagram. A short horizontal impulse $J$ along $+x$ gives the centre of mass a constant speed $V=J/m$. The angular impulse about the centre is $Jl/2$, giving constant counterclockwise angular velocityThe pole exerts no further force while the stick moves clear of it. Let $\psi =\Omega t$. The trailing upper end has coordinatesThe pole can lie on the translating straight stick again only when the stick becomes horizontal: for other orientations, its intersection with $y=0$ is the moving midpoint $x=Vt>0$. At the first horizontal orientation, $\psi =\pi /2$, the trailing end is atConsequently the grazing clearance of an impulsively struck rod occurs atFor $\beta >1/(2\pi )$ the whole stick has passed the pole by its first horizontal orientation, and the continually advancing centre makes all later passes safe. Equality is the limiting grazing contact of the endpoint; strict avoidance requires the strict inequality. If $\beta <1/(2\pi )$, the pole lies within the stick at that first horizontal pass, so contact occurs before the freely moving trailing end can pass the pole. A uniform thin stick has $\beta =1/12$, below the threshold: sufficient mass near the ends is needed to increase its moment of inertia and reduce the rotation relative to translation.

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.

Centre the square on the $xy$ plane, with uniform surface mass density $m/a^2$. For an in-plane axis at angle $\alpha$ to the $x$ direction, the perpendicular distance of $(x,y)$ from the axis is $-x\sin \alpha +y\cos \alpha$. Therefore its moment of inertia isThe mixed integral vanishes by symmetry, while the $x^2$ and $y^2$ integrals are equal. ThusThis is the in-plane isotropy of the square's inertia tensor: its two central in-plane principal moments coincide. It does not assert isotropy about arbitrary three-dimensional axes, nor about parallel axes that miss the centre.

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 uniform thin disc, the surface mass density is $\sigma =m/(\pi a^2)$. An annulus of radius $r$ and thickness $dr$ has mass $dm=2\pi \sigma rdr$, all at distance $r$ from the normal central axis. Its contribution to the moment of inertia is $r^{2}dm$, soThis establishes the moment of inertia of a uniform disc; the normal axis and uniform surface mass density are part of the result.

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.

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.

For a uniform disk, the surface mass density is $\sigma =m/(\pi a^2)$. Integrate concentric annuli of mass $dm=2\pi \sigma rdr$ about the central perpendicular axis:The result is specific to a uniform thin disk and that axis; it is not the diameter-axis moment of inertia, which is $m{a}^2/4$ by planar symmetry and the perpendicular axis theorem.

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 $3m/4$.

Take the positive coordinate from $O$ toward $S$. The centre of mass of the remaining material has coordinateThus $SX=a+a/6=7a/6$. If $\varphi$ is the small angular displacement from the stable hanging orientation, gravity gives torque $-(3m/4)g(7a/6)\sin \varphi$ about the suspension point. The physical pendulum equation isUsing the small-angle approximation $\sin \varphi \simeq \varphi$ gives simple harmonic motion withBoth the reduced mass and displaced centre of mass matter; changing only the moment of inertia would give an incorrect frequency.

Treat the ladder as a uniform slender rigid body. While both contacts remain imposed, its centre of mass is $(l\sin \theta ,l\cos \theta )$, with speed $l|\dot{\theta }|$. Its central moment of inertia is $m(2l{)}^2/12=m{l}^2/3$. Adding translational and rotational kinetic energy givesTaking the floor as zero gravitational potential energy, the Lagrangian, conjugate momentum and Euler-Lagrange equation areThe positive sign is appropriate: gravity increases the angle from the vertical. These are the constrained equations of a frictionless sliding ladder. Real wall contact is unilateral: a wall cannot pull the ladder. Once the computed normal force would become negative, the two-contact model ceases to apply; the question explicitly assumes that contact persists.

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.

Measure position $x$ along the ideal thin rod from its centre of mass. Its perpendicular distance from the inclined axis is $|x|\sin \alpha$, and its linear mass density is $M/L$. The moment of inertia is thereforeThis tends to zero for an axis along an infinitely thin rod and to $M{L}^2/12$ for a perpendicular central axis. A finite rod thickness would add a contribution absent from the stated thin-rod model.

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.

Write the incoming particle speed as $u>0$, its final signed speed as $v$, and the rod's final centre-of-mass velocity along the impact direction as $V$. Let $b$ be the signed impact distance from the centre, and choose the sign of the angular velocity $\Omega$ so that the contact point's velocity component is $V+b\Omega$. The rod's central moment of inertia is $I=M{L}^2/12$.

The impulse delivered to the rod is $MV$. Conservation of linear and angular momentum givesUsing these in conservation of kinetic energy,This is the equal-mass instance of the effective contact mass calculation. For $u>0$, both final speeds are nonnegative, so equal speeds require $v=V=u/2$. Therefore $M{b}^2/I=2$ andBoth symmetric impact points lie on the rod, since $L/\sqrt{6}<L/2$.

More generally, the relative contact velocity immediately afterwards isThus its magnitude equals the incoming relative speed $u$ and its direction reverses. At the specified impact point, $V+b\Omega =3u/2$ and $v=u/2$, which makes the reversal particularly explicit.

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.