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Classical mechanics describes the motion of bodies by assigning positions and velocities and specifying how their interactions change those quantities. A point mass is the simplest model: its size and internal structure are ignored, leaving a mass and a moving position. A rigid body retains orientation and rotational kinetic energy. Choosing a model means deciding which features affect the question being asked; a sliding block may be treated as a point mass, whereas predicting its rolling without slipping requires its moment of inertia.

Three complementary questions organize the subject. Newton's second law asks how the net force determines the instantaneous acceleration. The work-energy theorem asks how that force, acting through a displacement, changes kinetic energy. Conservation of mechanical energy asks which combinations of position and velocity are accessible when the interactions admit a time-independent potential energy and no other force supplies net power. These descriptions are compatible, but answer different questions: an energy calculation may find a maximum speed without first solving for position at every time.

The harmonic oscillator below makes this unity visible. A restoring force produces a repeating position trace; the corresponding phase plane has a closed ellipse; kinetic energy and potential energy exchange while their sum stays fixed. Lagrangian mechanics and Hamiltonian mechanics provide further ways to organize the same dynamics, especially when coordinates or constraints make individual forces awkward to track.

The scope of classical mechanics also matters. Newtonian mechanics approximates motion at speeds small compared with the speed of light; special relativity changes that description when this approximation fails. Quantum mechanics is needed for phenomena for which a classical assignment of simultaneous position and momentum is inadequate. These limitations do not prevent classical mechanics from being a useful predictive model within its regime. ^[ref]

The instantaneous velocity of the material point of a rigid body touching another body. In the displayed formula, ${\mathbf{r}}_P$ is its position relative to the centre of mass, and $\mathbf{\Omega }$ is the body's angular velocity. The rotational term is a cross product. This material-point velocity differs from the motion of the geometrical contact location: in rolling without slipping on a stationary plane it vanishes even while the contact location advances. For a sphere touching a horizontal plane, with rightward centre velocity $v$ and anticlockwise spin $\Omega$, its tangential component is $v+a\Omega$.

Contact slip is the tangential relative velocity between the two material points instantaneously in contact. For a nonrotating box on a belt it is $w=v-U$; for a rigid body, use the velocity of its contact point, including the contribution from angular velocity. Zero contact slip defines rolling without slipping at a rolling contact, even when the center of mass moves.

For a spool of outer radius $R$, axle radius $r$, mass $M$ and axial moment of inertia $I$, a lower-tangent pull of tension $T$ at upward angle $\theta$ produces horizontal acceleration $a=TR(R\cos \theta -r)/(I+M{R}^2)$ while rolling without slipping. This follows by combining $Ma=T\cos \theta +f$, $I\alpha =Tr+Rf$ and $a=-R\alpha$, with rightward translation and counterclockwise rotation positive. The acceleration reverses at ${\theta }_c$; zero acceleration means equilibrium only if initially at rest. At that angle the static friction limit gives $T_{\max }=\mu Mg/(\cos {\theta }_c+\mu \sin {\theta }_c)$.

For a uniform solid cylinder of mass $M_c$ and a frictionlessly sliding wedge of mass $M_b$, take the horizontal wedge position $X$ and uphill contact coordinate $s$. Contact and rolling without slipping leave two degrees of freedom. Up to a constant, the Lagrangian is $L=\frac{1}{2}(M_b+M_c){\dot{X}}^2+M_c\cos \alpha \dot{X}\dot{s}+\frac{3}{4}{M}_c{\dot{s}}^2-M_{c}gs\sin \alpha$.

The cyclic coordinate $X$ gives $(M_b+M_c)\ddot{X}+M_c\cos \alpha \ddot{s}=0$. The other Euler-Lagrange equation is $\cos \alpha \ddot{X}+(3/2)\ddot{s}=-g\sin \alpha$. Elimination proves the displayed acceleration, with $\ddot{X}=-M_c\cos \alpha \ddot{s}/(M_b+M_c)$. For $M_b\gg M_c$, the cylinder approaches the fixed-incline acceleration $-(2/3)g\sin \alpha$.

Model an eccentric-hole wheel as a complete uniform disk for kinetic energy only, subtracting the absent disc's potential energy. If $s=R\theta$ and $\theta$ is its hole-centre direction measured clockwise from upward vertical, rolling without slipping gives $T\simeq 3D{\dot{s}}^2/4$. On a slope of angle $\alpha$, its potential energy is $U=-(D-h)gs\sin \alpha -hg{r}_o\cos (s/R)$ up to a constant. Substitution into the Euler-Lagrange equation proves the displayed equation. Treating the hole's kinetic energy exactly introduces angle-dependent inertia, so this simplified equation must retain its explicit modeling assumption.

For rolling without slipping with contact radius $r$, substitute $\omega =v/r$ into translational plus rotational kinetic energy:The coefficient acts as an effective inertia for a translation coordinate. It is not additional gravitational mass. A quadratic potential energy $K{x}^2/2$ then gives small-oscillation frequency ${\omega }_0=\sqrt{K/M_{\mathrm{eff}}}$ when this coefficient is constant.

A smooth central collision leaves a cylinder's angular velocity unchanged if the floor supplies negligible tangential impulse. Imposing rolling without slipping during the impact instead requires a floor impulse and changes the cylinder's effective translational mass to $m+I/r^2$. Consequently a pre-impact rolling condition alone does not determine the outgoing velocities: the impulsive contact law must also be specified. For a uniform solid cylinder striking a stationary point mass $2m$ in an elastic collision, the target acquires $2u/3$ with negligible floor impulse, and $6u/7$ with an enforced rolling constraint, where $u$ is the incoming centre speed.

Friction is a contact force tangent to the surfaces. Kinetic friction opposes contact slip; static friction maintains contact without contact slip, including rolling without slipping. The perpendicular contact component is the normal force. These two components answer different questions: how strongly the surfaces press together, and how strongly they resist sliding past one another.

The ideal Coulomb dry-contact model combines a bound on static friction with a sliding law for kinetic friction. Its parameters are the coefficient of static friction and the coefficient of kinetic friction; constant coefficients are an empirical approximation, rather than a consequence of Newton's laws of motion. Surface condition matters, and the static coefficient need not universally exceed the kinetic coefficient. ^[ref]

The model's historical setting was practical machinery: Coulomb's Théorie des machines simples investigated friction and the stiffness of ropes. The Académie's bibliography records its 1781 prize and its appearance in volume 10, pages 161–332, of the Recueil des savants étrangers. ^[ref]

For a circular rigid body of radius a, mass M and axial moment of inertia I on a fixed horizontal surface, choose positive spin so that rolling means $v=a\omega$. During purely horizontal frictional relaxation, an impulse J changes $Mv$ by J and $I\omega$ by $-aJ$. Hence $I\omega +Mav$ is constant. At final rolling without slipping, $v_f=(Ia{\omega }_0+M{a}^2{v}_0)/(I+M{a}^2)$. Subtracting initial from final translational-plus-rotational kinetic energy gives the title formula. The coefficient of friction controls the relaxation time, but not this loss once the no-slip state is reached. Vertical landing losses and rolling resistance are excluded.

Kinetic friction is friction during nonzero contact slip. For one-dimensional motion along a surface, choose right as positive and write $v$ for the body's contact-point velocity and $U$ for the surface's contact-point velocity, both measured in the same inertial frame. The signed relative velocity is $w=v-U$. The ideal Coulomb law for the tangential force on the body isHere $N\ge 0$ is the magnitude of the normal force, ${\mu }_k\ge 0$ is the coefficient of kinetic friction, and $\mathrm{sgn}$ is the sign function. Thus kinetic friction opposes contact slip, which can differ from the body's direction of motion. In several tangential directions the same law reads ${\mathbf{f}}_k=-{\mu }_{k}N\mathbf{w}/|\mathbf{w}|$ for a nonzero tangential relative velocity $\mathbf{w}$. The magnitude law is documented in Ling, Sanny and Moebs, OpenStax University Physics 1, §6.2: static and kinetic friction.

For a stipulated toy model, take a nonrotating box with mass $m=1\mathrm{kg}$, normal force $N=10\mathrm{N}$, coefficient of static friction ${\mu }_s=0.5$, and coefficient of kinetic friction ${\mu }_k=0.3$. These are assumed model parameters, rather than measured properties of any particular pair of materials; the prescribed normal force is consistent with horizontal support under $g=10m/{\mathrm{s}}^2$ and no other vertical forces. On stationary ground, a box already sliding right has $f_k=-3\mathrm{N}$. With an applied horizontal force $F=6\mathrm{N}$ to the right, Newton's second law gives $a=(F+f_k)/m=3m/{\mathrm{s}}^2$, for as long as rightward contact slip persists.

Direction is clearer by comparing two instantaneous states of this same model. If the box has $v=1m/s$ on stationary ground, $U=0$ and $w=+1m/s$, so kinetic friction acts left with magnitude $3\mathrm{N}$. If the box still has $v=1m/s$ but a belt beneath it has $U=2m/s$, then $w=-1m/s$: kinetic friction acts right with magnitude $3\mathrm{N}$, even while the box moves right.

In the belt case, the power delivered to the box is $P_{\mathrm{box}}=f_{k}v=+3\mathrm{W}$. With no other horizontal force, its kinetic energy therefore increases by the work-energy theorem. By Newton's third law, the box exerts $-f_k$ on the belt, so the contact power delivered to the belt is $P_{\mathrm{belt}}=-f_{k}U=-6\mathrm{W}$. Their sum is $-3\mathrm{W}$: frictional heating produces thermal internal energy at $3\mathrm{W}$. An ideal motor maintaining the prescribed belt velocity supplies $+6\mathrm{W}$ against this contact load, neglecting other losses, so its input divides into $3\mathrm{W}$ of box kinetic energy and $3\mathrm{W}$ of frictional heating. These are instantaneous rates, since the box's velocity subsequently changes. This balance follows directly from power as force times velocity and conservation of energy; the power law is given in Ling, Sanny and Moebs, OpenStax University Physics 1, §7.4: power.

When contact slip vanishes, including a transition to rolling without slipping, the nonzero-slip law no longer applies. Determine the required static friction from the no-slip condition and the equations of motion, then test its bound. Substituting $\mathrm{sgn}(0)=0$ into the sliding law would incorrectly exclude a nonzero static friction.

A uniform solid sphere of mass $M$ and radius $a$ has the indicated moment of inertia about any diameter. Integrating the squared perpendicular distance against uniform mass density gives $I=8\pi \rho a^5/15=2M{a}^2/5$. It is different from the value for a thin spherical shell and determines the rotational contribution to kinetic energy during rolling without slipping.

A nonholonomic constraint is a nonintegrable restriction on allowed velocities: it cannot locally be replaced by relations among generalized coordinates and time. In contrast, a velocity equation $r\dot{\varphi }-R\dot{\theta }=0$ with constant radii integrates to a holonomic constraint $r\varphi -R\theta =C$. Rolling without slipping is therefore not automatically nonholonomic: planar rolling on a prescribed curve can be holonomic, whereas rolling with additional directional freedoms can impose nonintegrable velocity constraints. Unwrapped angles may be needed to record the winding of a rolling trajectory.

Isaac Newton made the greater contribution if the criterion is the scope and explanatory integration of mathematical natural philosophy; Galileo Galilei has the stronger claim if the criterion is the formative transformation of how terrestrial motion could be investigated. The first criterion gives a defensible overall verdict for Isaac Newton, but a serious comparison must show what each changed rather than award credit according to who worked later.

Galileo Galilei's contribution was not simply to attach numbers to familiar events. In the Third Day of Two New Sciences by Galileo Galilei, he defines uniformly accelerated motion by equal additions of speed in equal intervals of time, then derives consequences. Starting from rest, the modern notation is $v=at$; the mean speed over the interval is $at/2$, soHence distances in successive equal intervals have the ratios $1:3:5:7:\dots$. The novelty lies in making a precisely specified kind of motion the subject of mathematical demonstration, with observed falls assessed against its consequences. This is not the same as deriving the underlying physical cause of gravity.^[ref]

The reported apparatus shows how such an idealization could be investigated. A smooth bronze ball descended in a straight grooved wooden channel, raised at one end. Water discharged through a small opening into a vessel during the descent; its collected mass served as a measure of elapsed time. Repeating runs over different portions of the channel compared the predicted squared dependence of time on distance. The inclined plane slowed the event, while repetition and weighing made a temporal comparison possible without a modern stopwatch. The published report is evidence of Galileo Galilei's experimental presentation, not a surviving raw-data archive from which every reported precision can now be reconstructed.^[ref]

A necessary physical qualification is often missed. A ball in rolling without slipping is not a point particle sliding without friction. For a solid sphere on an ideal inclined plane, modern rigid body dynamics gives $a=5g\sin \alpha /7$, rather than $g\sin \alpha$. Both are constant accelerations, so both give $s\propto t^2$. The trial can therefore test the temporal form without directly establishing the free-fall coefficient. This distinction makes the experiment more intelligible: the evidence supports a specified relation under approximations, rather than every detail of a modern idealization.

The Fourth Day extends this approach to projectile motion. Combine uniform horizontal velocity $u$ with constant downward acceleration $g$, taking downward displacement positive. Then $x=ut$ and $y=g{t}^2/2$. Eliminating time givesa parabola. Actual projectiles experience air resistance, but the ideal curve organizes the effects that need separate treatment. Galileo Galilei thus gave a powerful example of how mathematical abstraction and physical observation could cooperate.^[ref]

Isaac Newton went further by connecting kinds of motion through a general account of force. Newton's second law links the change of momentum to the impressed force; for constant mass, its modern differential expression is $\mathbf{F}=m\mathbf{a}$. Newton's third law makes interaction reciprocal, enabling a consistent analysis of interacting bodies. The Philosophiae Naturalis Principia Mathematica by Isaac Newton develops consequences of central forces and then asks which laws fit astronomical phenomena. Mathematical deduction now connects measurable orbits to an underlying force law, rather than just describing each trajectory separately.^[ref]

A simple modern reconstruction shows the increase in explanatory reach. For a circular orbit of radius $r$ and orbital period $T$ about a much more massive central body, centripetal acceleration is $4{\pi }^{2}r/T^2$. Kepler's third law, $T^2\propto r^3$, therefore implies an acceleration proportional to $r^{-2}$. This circular argument is illustrative; Isaac Newton's treatment of elliptical orbits is more general. With Newton's law of universal gravitation,the two-body orbital period is tied to the semimajor axis $a$ and both masses. The same theory connects the Moon's orbital acceleration with terrestrial gravity, and supports further investigations of tides, comets and departures from a simple two-body orbit.^[ref]

This achievement depended on more than two individuals. Johannes Kepler's relations supplied crucial astronomical constraints, and both authors inherited geometry, observational records and technical practices. Nor should the Philosophiae Naturalis Principia Mathematica by Isaac Newton be described simply as the first application of symbolic calculus: its published arguments are largely geometrical. What distinguishes Isaac Newton is the demonstrative network uniting terrestrial and celestial problems, together with explicit idealizations whose failures generated further calculable questions.

On breadth, quantitative unification and the inference from motion to force, Isaac Newton's contribution was greater. Galileo Galilei's work supplied an exemplary way to mathematize terrestrial motion and test its idealized relations. Giving the former the overall verdict should therefore acknowledge the latter's distinct foundational achievement, not portray the Philosophiae Naturalis Principia Mathematica by Isaac Newton as a creation without predecessors.

Choose the cylinder centre's initial position as the coordinate origin, with downhill motion to the right and $0\le \alpha <\pi /2$. For rolling without slipping, the clockwise angular speed is $\omega =v/r$. A point initially vertically above the axis therefore hasA different origin adds constant coordinates without changing the following velocity test. Setting $\theta =vt/r$ gives $\dot{x}/v=\cos \alpha +(a/r)\cos \theta$ and $\dot{y}/v=-\sin \alpha -(a/r)\sin \theta$.

For strictly upward motion, $\dot{x}=0$ and $\dot{y}>0$. The first condition fixes $\cos \theta =-(r/a)\cos \alpha$; choose the negative sine branch to maximize $\dot{y}$. Positivity then requiresAt $a=r$ the relevant point is momentarily stationary, not moving upward. Thus a material point inside or on an ordinary cylinder cannot move exactly vertically upward in this motion. The mathematical extension $a>r$ describes the looping trochoid of a point beyond the rolling radius, illustrated in the figure.

A coaxial shell of radius $r$ and thickness $dr$ has mass $dm=2\pi \rho lrdr$. Integrating squared radius against mass, the moment of inertia of a uniform solid cylinder about its axis isOn the smooth incline, no tangential contact force produces spin, so the initially stationary cylinder simply slides with acceleration $g\sin \theta$. Therefore $t_{\mathrm{slide}}=\sqrt{2L/(g\sin \theta )}$.

For rolling without slipping, let $F$ be uphill static friction and $A$ the downward translational acceleration. Newton's second law, the rotational torque equation, and the contact constraint giveWith $I=M{a}^2/2$, these give $F=MA/2$ and $A=2g\sin \theta /3$. The rolling-time factor on a straight incline is consequentlyThe required friction is $Mg\sin \theta /3$; the stated no-slip motion presupposes sufficient static friction. At an instantaneously stationary contact point it does no work. Gravitational potential energy is shared between translational and rotational kinetic energy, explaining why rolling accelerates more slowly than the smooth slide.

The parallel axis theorem states $I_O=I_G+m{d}^2$ for two parallel axes, one through the centre of mass and the other a perpendicular distance $d$ away. To see this, expand the squared distance to the translated axis in the mass integral; its mixed term vanishes because the centroidal first moment is zero.

For a uniform solid hemisphere, the moment about a diameter in its flat face is half the corresponding full-sphere moment. With full-sphere mass $2m$, it is $\frac{1}{2}[\frac{2}{5}(2m)a^2]=2m{a}^2/5$. The original PDF gives the centroidal displacement $d=3a/8$; the TeX's $3a/5$ is incorrect. Moving the diameter axis to its parallel centroidal axis therefore yieldsThis is the moment of inertia of a uniform solid hemisphere, not that of a thin hemispherical shell.

Take a cross-section perpendicular to the rotation axis. Let $O$ be the underlying sphere centre, $G$ the centre of mass and $P$ the contact point. During rounded-surface contact, $OP=a$ vertically and $OG=d$ points toward the curved face, with downward component $d\cos \theta$. ThereforeFor rolling without slipping, $P$ is instantaneously at rest and the body rotates about it with angular speed $|\dot{\theta }|$. Thus the centroid speed is $b|\dot{\theta }|$. Translation and rotation contributeThe initial vertical base has $\theta =\pi /2$ and $K=0$, so conserved energy is $mga$. Equating the later $K+V$ to this value givesOn the initial roll toward the lowest centroid position, $\dot{\theta }<0$. Static contact friction does no work because the contacting material point is at rest; its coefficient must be large enough for the assumed no-slip motion. The formula describes the rounded-contact phase of the rolling uniform solid hemisphere and should not be continued through a change of contact geometry without another model.

The moment of inertia of a uniform solid cylinder is taken about its symmetry axis. Write its mass density as ${\rho }_c=m/(\pi a^{2}L)$. A coaxial shell of radius $r$ and thickness $dr$ has mass $dm=2\pi {\rho }_{c}Lrdr$. ThusA hollow cylinder would have a different moment of inertia; uniform solid density is the intended assumption.

Let $s$ measure the centre's displacement uphill from its initial position, and choose signed angular velocity $\omega$ so that uphill rolling without slipping has $\dot{s}=a\omega$. Let $F$ be the signed contact force uphill. The translation and rotation equations areThe minus sign in the torque equation follows from the chosen rolling sense: uphill contact force opposes the positive spin. Under rolling without slipping, $\ddot{s}=a\dot{\omega }$. Eliminating $F$ givesUsing $s(0)=0$ and $\dot{s}(0)=V$, the no-slip motion and maximum vertical rise areIndeed the centre turns at $t=3V/(2g\sin \alpha )$, reaching $s_{\max }=3V^2/(4g\sin \alpha )$, whose vertical component is the boxed height. Conservation of energy independently gives the same answer from initial kinetic energy $m{V}^2/2+I(V/a{)}^2/2=3m{V}^2/4$.

The available static friction is at most $\mu mg\cos \alpha$. It can supply the required uphill force exactly when $mg\sin \alpha /3\le \mu mg\cos \alpha$. ThereforeInterpret the stated friction magnitude as limiting static friction and as kinetic friction during slip, with the same coefficient. Take $0<\alpha <\pi /2$, $\mu \ge 0$. In the slipping regime the contact point moves downhill relative to the slope, so kinetic friction acts uphill. To verify this direction rather than assume it unchecked, solve with $F=\mu mg\cos \alpha$. Define $b=g(\sin \alpha -\mu \cos \alpha )>0$. The result isThe centre's speed is $|V-bt|$, with the sign of $v$ specifying ascent or descent. The contact point's relative velocity isThus the assumed uphill kinetic friction is consistent throughout both ascent and descent, including when the centre's velocity changes sign. There is no later return to rolling without slipping before reaching the bottom. This is the characteristic slipping solid cylinder on an incline regime.

The centre turns at $t_{\ast }=V/b$, givingThe formulas apply up to the positive return time $t_b=2V/b$. Substitution into the signed velocity and angular velocity givesThe return spin can have either sign and vanishes at $\tan \alpha =5\mu$. It reverses when $3\mu <\tan \alpha <5\mu$, although the contact point still slips downhill. The unchanged centre speed does not mean that no energy was dissipated: the rotational kinetic energy has decreased. Indeed integrating the frictional power $Fw<0$ giveswhich agrees with the difference of the initial and final rotational kinetic energies.

Let $m$ be the car's mass and let $v(t)$ be its forward speed. Initially the driven contact surfaces slip backward relative to the ground because $v<r\Omega$. Thus kinetic friction on the car acts forward. On the horizontal surface the total normal force is $mg$, so Newton's second law gives $m\dot{v}=\mu mg$. While slipping persists,The contact slip speed is $r\Omega -v(t)$. Its first zero is the onset of rolling without slipping, henceThe engine maintains the prescribed wheel speed, so wheel spin-down is not part of this calculation.

On a slope of angle $\alpha >0$, directed uphill, the normal force is $mg\cos \alpha$ and gravity contributes $-mg\sin \alpha$ along the slope. Therefore the uphill acceleration while slipping is $g(\mu \cos \alpha -\sin \alpha )$. If it is positive, the onset of rolling without slipping occurs atThe inequality follows because $\mu \cos \alpha -\sin \alpha <\mu$. If $\mu \cos \alpha \le \sin \alpha$, the car does not accelerate uphill to $r\Omega$, so there is no finite catch-up time in this sliding-friction model. The finite-time conclusion presumes positive traction sufficient to climb the slope.

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.