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For rolling without slipping through height $h$, conservation of energy givesA uniform solid sphere has $I/(m{R}^2)=2/5$, a solid cylinder $1/2$, and a pipe $(1+r_{\mathrm{in}}^2/R^2)/2>1/2$. Thus the ball is fastest, then the solid cylinder, then the pipe. A smaller moment of inertia leaves more energy in translation. This assumes release from rest through the same height, adequate static friction and negligible rolling resistance; a thin pipe has $v^2=gh$.

Separate a ball's orbital motion from its spin. For a rolling ball between concentric races, the ball centre lies $2a$ from the axle axis. Let its tangential centre velocity be $v$ in the direction of axle motion and let its signed spin angular velocity be $\Omega$, positive in the same sense as $\omega$. Rolling without slipping at the moving inner contact and stationary outer contact requiresHenceThe spin is opposite to the axle's rotation, while the centre orbits in the same sense. The total angular momentum about the axle is the sum of its orbital angular momentum and its angular momentum about the centre of mass:For a uniform solid spherical ball, $I_{\mathrm{cm}}=2m{a}^2/5$, soThe solid-ball assumption supplies the usual moment of inertia; the preceding expression in terms of $I_{\mathrm{cm}}$ also states the answer for a different internal mass distribution.

Integrating coaxial cylindrical shells givesRolling without slipping has spin speed $u/a$, so $K=m{u}^2/2+I{u}^2/(2a^2)=3m{u}^2/4$.

For the rolling-cylinder loop contact threshold, set $\ell =R-a$, the radius of the centre's circular path, and let $\theta$ run from the loop bottom. The centre starts $h$ above its lowest centre position. Its height gain from the bottom is $\ell (1-\cos \theta )$. Energy conservation givesAt the top, contact requires $N=m(u_t^2/\ell -g)\ge 0$, hence $u_t^2\ge g\ell$. Inserting the energy relation proves the limiting conditionThe same condition keeps $N\ge 0$ elsewhere on the circle, sinceis smallest at the top.

For the centre's acceleration, the inward normal component is $u^2/\ell$ and the tangential component is $-(2g/3)\sin \theta$. If $A=h/\ell -1\ge 7/4$ and $x=\cos \theta$,This convex quadratic has its largest endpoint value at $x=1$. The maximum on the circular loop is at the bottom, directed vertically upward. For the limiting release height,The approach track's curvature is unspecified, so a global bound for an arbitrary approach shape cannot be proved from the sketch alone; the result above concerns the specified circular part and the usual smooth approach interpretation.

The finite-friction qualification of a marginal rolling loop also matters: the contact threshold presumes the stated no-slip idealization. At equality, near the top, required tangential friction vanishes linearly with angular distance from the top but $N$ vanishes quadratically. Thus any finite coefficient of static friction requires a release height above this limiting bound to maintain rolling everywhere; no coefficient is provided. This does not change the requested ideal contact criterion.

Take $q=\ddot{x}$ positive when the hanging mass descends and the other mass moves uphill. For the sliding block, Newton's second law gives $mg-T=mq$ and $T-Mg\sin \theta =Mq$. ThereforeFor the pulley coupled to a rolling disc, tension and weight act through its centre; the contact normal is perpendicular to the slope. Static friction provides the torque required for rolling without slipping. Define signed $F$ down the slope, so $F>0$ for uphill acceleration.

The rotational equation and rolling constraint give $Fa=Iq/a$, hence $F=Mq/2$. The translational equations become $T-Mg\sin \theta -F=Mq$ and $mg-T=mq$, givingThus the threshold remains $m>M\sin \theta$: the moment of inertia increases the effective mass but does not change the sign of the driving force. For downhill acceleration $F$ reverses. The normal force is $Mg\cos \theta$; static friction does no work at the stationary contact point and sufficient roughness is assumed.

Axial moment of inertia. In the original diagram $R_1$ is the outer radius and $R_2$ the inner radius. For uniform mass density $\rho$, a coaxial annulus of thickness $dr$ has $dm=2\pi \rho Lrdr$. Therefore the axial moment of inertia of a hollow cylinder isSince $M=\pi \rho L(R_1^2-R_2^2)$, factorizing the fourth-power difference givesFor a solid cylinder $R_2\to 0$, giving $I=M{R}_1^2/2$. For a thin cylindrical shell with fixed mass and $R_1,R_2\to R$, $I=M{R}^2$. The latter limit agrees with every mass element being at radius $R$.

Undamped rolling oscillations. Let $x$ be axle displacement from the spring equilibrium. Rolling without slipping imposes $|{\omega }_{\mathrm{roll}}|=|\dot{x}|/R$. The sum of translational and rotational kinetic energy isThe potential energy is $U=k{x}^2/2$. Differentiating $T+U$ in the absence of damping, or applying the force and torque equations below, givesFor the thin shell $I=m{R}^2$, the effective inertia is $2m$, and the angular frequency of the simple harmonic motion isStatic contact friction supplies the required torque but does no work when the contact point is instantaneously stationary.

Damped force balance. Take rightward force as positive and counterclockwise rotation as positive. Write $f$ for the signed static friction at the bottom contact. ThenEliminate $f=-I\ddot{x}/R^2$ and use $I=m{R}^2$:Here $b$ has units $\mathrm{kg}{\mathrm{s}}^{-1}$, so $B$ has units ${\mathrm{s}}^{-1}$ and $C$ units ${\mathrm{s}}^{-2}$. The stated drag force is interpreted as acting through the axle, with no additional viscous torque. An oil bath could also exert such a torque physically, but its magnitude is not specified by this force-only model.

Light damping. Substitution of $z=A{e}^{-pt}$ into the differential equation gives $p^2-Bp+C=0$. In the underdamped regime $B^2<4C$,Taking real linear combinations gives the general damped harmonic oscillator displacementThe constants are fixed by the initial displacement and velocity. For example, release from rest at $x_0$ gives $A_1=x_0$ and $A_2=b{x}_0/(4m{\omega }_d)$. Light damping means $b^2\ll 8mk$, in which case ${\omega }_d\simeq {\omega }_0$ and the envelope decays slowly compared with an oscillation.

The contact radius is $r_1$, so rolling without slipping gives $v=r_1\omega$. With $I=M{r}_2^2/2$, the energy at the minimum of the stated parabolic motion model isConsequentlyThe corresponding spin angular frequency is ${\omega }_{\max }=v_{\max }/r_1=14.8\mathrm{rad}{\mathrm{s}}^{-1}$. It is the rod radius, not the disc radius, that relates translation and rotation; the discs determine the larger rotational inertia.

Constraint and example. A holonomic constraint can be written $f_{\alpha }(q_1,\dots ,q_n,t)=0$, with no irreducible dependence on velocities. A velocity relation is holonomic if it integrates to such an equation. For example, a particle constrained to a stationary circle satisfies $x^2+y^2={\ell }^2$, reducing its two Cartesian degrees of freedom to one angle. A nonholonomic constraint cannot be integrated in this way.

Kinematics and kinetic energy. Set $d=R-r>0$. In the plane of the original diagram, choose the orbital angle $\theta$ positive counterclockwise from vertically downward. The marked radius of the inner cylinder then has absolute orientation $\alpha =\theta -\varphi$ measured from the same downward direction. Its signed spin angular velocity is $\Omega =\dot{\theta }-\dot{\varphi }$, whileThe moment of inertia of a uniform solid cylinder is $I=m{r}^2/2$. Applying the kinetic energy decomposition about the center of mass gives

No slip. With tangential unit vector ${\mathbf{e}}_{\theta }=(\cos \theta ,\sin \theta )$, the inner contact point has velocity $(d\dot{\theta }+r\Omega ){\mathbf{e}}_{\theta }$. The fixed wall requires this to vanish. HenceBecause $r,R$ are constant, integration giveswhere the integration constant is zero with the marked-dot reference in the diagram and continuously unwrapped angles. This is a holonomic constraint, even though it was first obtained from a condition on velocities.

Oscillations. The gravitational potential energy above the lowest position is $V=mgd(1-\cos \theta )$. Eliminating $\varphi$ gives the reduced LagrangianThe Euler-Lagrange equation becomesThus the bottom is a stable equilibrium, and for $|\theta |\ll 1$ there is simple harmonic motion, withThe extra rotational kinetic energy reduces the frequency compared with a particle sliding without friction along the same circular path. The small oscillations inside a circular cylinder formula applies with inertia ratio $\gamma =1/2$.

Friction and energy. Let $F_t$ be the signed tangential contact force in the ${\mathbf{e}}_{\theta }$ direction. Applying Newton's second law to translation and the torque equation to rotation givesUsing the rolling without slipping condition yieldsExcept at special instants, static friction is needed to produce the spin acceleration. There is no sliding at the contact point, whose instantaneous velocity is zero; therefore this force does no total work. Its translational and rotational powers cancel: $F_{t}d\dot{\theta }+F_{t}r\Omega =0$. The stationary normal reaction likewise does no work. With a rigid wall, adequate static friction, and no rolling resistance, conservation of energy gives

Angular momentum. The cylinder's spin angular momentum $I\Omega$ is generally not conserved because the contact friction exerts a torque about its center of mass. About the fixed outer-cylinder center $O$, the angular momentum decomposition about the centre of mass givesGravity and friction exert the external torqueThus the rolling cylinder does not generally conserve angular momentum, although it conserves mechanical energy. There is a special geometric cancellation at $R=3r/2$: then $L_O$ and its torque vanish identically for this planar motion. Angular momentum of a larger isolated system including the wall and Earth is a separate conservation statement.

Scope of the holonomic reduction. The integrated constraint is exact at every angle for planar rolling without slipping with sustained contact; it is not restricted to small oscillations. Completing a full turn requires tracking angle winding rather than reducing each angle separately modulo $2\pi$. This global bookkeeping does not make the local velocity relation nonintegrable.

However, arbitrary initial conditions need not sustain this ideal rolling motion. The inward normal reaction and the required friction areContact requires $N\ge 0$, and a finite coefficient ${\mu }_s$ of static friction requires $|F_t|\le {\mu }_{s}N$. If either fails, the cylinder detaches or slips and the one-coordinate rolling Lagrangian no longer applies. Slipping contact allows $\theta$ and $\varphi$ to be independent; detachment removes the fixed center-distance constraint. These changes alone do not establish a nonholonomic constraint. Allowing general three-dimensional reorientation can introduce nonintegrable rolling constraints, but that motion is outside the diagram's two-angle planar model. Thus the stated planar rolling constraint is holonomic throughout its regime of validity; it cannot be imposed on every physically possible motion.

A modern teaching experiment slows gravity-driven motion enough to resolve it with a photogate. PASCO scientific's Inclined Plane Dynamics Set, ME-6966, Instruction Manual 012-10874A describes “Acceleration Down an Incline” on printed pages 15–16. Its apparatus is a GOcar on a PAStrack with an adjustable incline support, a bracket-mounted photogate, a cart-mounted picket fence, and an interface with recording software. For the interface method, the fence's $1\mathrm{cm}$ pattern is aligned to interrupt the beam. Recording starts before release from rest; the cart is caught at the bottom bumper. The procedure specifies three runs per angle, starting at $14^{\circ }$ and reducing by $2^{\circ }$, with the release position fixed. The readout is velocity against time; its slope estimates acceleration. Mean accelerations are plotted against $\sin \theta$, testing the ideal slope $g$. An added-mass comparison is also requested. These are protocol settings, not observed results.^[ref]

The cart undergoes rolling without slipping at its wheels, whereas the worked block undergoes contact slip. Even ideal wheel moment of inertia changes the cart prediction. For total cart mass $M$, wheel radii $r_i$ and axial moments of inertia $I_i$, rigid wheels with fixed axles and no losses giveTo derive it, combine translational kinetic energy $M{v}^2/2$ with wheel rotational kinetic energy ${\sum }_i{I}_i(v/r_i{)}^2/2$, then equate their rate of change to the release of gravitational potential energy $Mg\sin \theta v$. This is the multiwheel counterpart of rolling acceleration with rotational inertia. Neglecting wheel moment of inertia yields $a\simeq g\sin \theta$; treating a cart as a solid rolling cylinder would give a different correction.

Useful additional controls are repeated releases without a push, checking the fence alignment and spacing, and checking that the recorded velocity trace is straight over the fitted interval. A gently moving cart on a level track can expose losses or unintended tilt. These checks help distinguish an angle or timing error from neglected wheel moment of inertia, bearing friction, or drag force. The three-run spread measures repeatability but cannot by itself reveal a shared calibration error. The inspected protocol makes the gravity component measurable; it does not measure the sliding block's coefficient of kinetic friction.

A uniform disk of mass $2M$ and radius $R$ carries an extra point mass $M$ on its rim. For rolling without slipping, let $\theta =0$ place that mass at the contact. Its position is $[R(\theta -\sin \theta ),R(1-\cos \theta )]$. Combining disk translation, disk rotation and point motion gives the displayed Lagrangian. Its small-amplitude angular frequency is $\sqrt{g/(3R)}$, whereas the same object pivoted at its centre has angular frequency $\sqrt{g/(2R)}$.

A round body of radius $R_0$ in a fixed cylinder of radius $R_1$ has center of mass path radius $a=R_1-R_0$. With orbital angle $\theta$ and spin angular velocity $\Omega$ both positive about the same axis, rolling without slipping gives $a\dot{\theta }+R_0\Omega =0$. For moment of inertia $I=\gamma m{R}_0^2$, its kinetic energy is $\frac{1}{2}m{a}^2(1+\gamma ){\dot{\theta }}^2$. Release from rest at angle $\alpha$ gives, by conservation of energy, a descent time $\sqrt{1+\gamma }$ times that of frictionless sliding along the same center of mass path. The spin sign reverses if its positive direction is chosen opposite to the orbital angle.

For a rigid body rolling from rest down a straight incline with fixed rolling radius $R$ and axial moment of inertia $I$, translation and rotation give $(M+I/R^2)\ddot{s}=Mg\sin \theta$. A smooth slide has $\ddot{s}=g\sin \theta$. Travel over the same distance from rest therefore gives the displayed time ratio. Static friction must suffice for rolling without slipping; its contact point is at rest, so this ideal friction does no work. A uniform solid cylinder has $I=M{R}^2/2$ and ratio $\sqrt{3/2}$.

Let the flat face make angle $\theta$ with horizontal while the rounded face contacts a horizontal plane. The centroid has height $a-(3a/8)\cos \theta$ and contact distance squared $b^2=a^2+(3a/8{)}^2-2a(3a/8)\cos \theta$. For rolling without slipping, its speed is $b|\dot{\theta }|$. Combining translation and rotation gives kinetic energy $\frac{1}{2}(I_G+m{b}^2){\dot{\theta }}^2$. Conservation from rest at $\theta =\pi /2$ gives the displayed result until the flat-face/contact geometry changes.

For a uniform solid cylinder, take $s$ positive uphill and $\omega$ positive for uphill rolling without slipping. The velocity of the contact point is $w=\dot{s}-a\omega$. Uphill kinetic friction gives $\ddot{s}=-g(\sin \alpha -\mu \cos \alpha )$ and $a\dot{\omega }=-2\mu g\cos \alpha$. Starting from $\dot{s}(0)=a\omega (0)$, one obtains $w=-gt(\sin \alpha -3\mu \cos \alpha )$. Thus $\tan \alpha >3\mu$ makes the assumed uphill kinetic friction self-consistent for the entire ascent and descent. Without slip the required uphill static friction is $mg\sin \alpha /3$; it is available exactly when $\tan \alpha \le 3\mu$ under the same limiting coefficient.

For rolling without slipping inside a stationary circular cylinder, take the inner body's axial moment of inertia to be $I=\gamma m{r}^2$. Its center of mass follows a circle of radius $d=R-r$, and the reduced Lagrangian is $L=m(1+\gamma )d^2{\dot{\theta }}^2/2-mgd(1-\cos \theta )$. The Euler-Lagrange equation gives $(1+\gamma )d\ddot{\theta }+g\sin \theta =0$. Linearizing near the lowest position yields the displayed small oscillation frequency. A uniform solid cylinder has $\gamma =1/2$. This reduction requires sustained contact and enough static friction to prevent slipping.

A sphere of radius $a$ touching both faces of a symmetric groove of opening angle $\varphi$ rotates instantaneously about the line joining its contacts. The centre is a distance $h=a\sin (\varphi /2)$ from this line, so rolling without slipping means $v=h\omega$. For a homogeneous sphere, $K=m{v}^2/2+I{\omega }^2/2=m{v}^2[1+2/(5{\sin }^2(\varphi /2))]/2$. Differentiating this kinetic energy with respect to distance down an incline gives the displayed acceleration. Both contact forces do no work for fixed faces and no slip.

Attach a spring of spring constant $k$ to the axle of a round body of mass $m$, radius $R$ and axial moment of inertia $I$. Assume rolling without slipping, a horizontal drag force $-b\dot{x}$ applied through the axle, and no additional resisting torque. The translational force equation is $m\ddot{x}=-kx-b\dot{x}+f$, where $f$ is signed static friction. The rotational equation and rolling constraint give $f=-I\ddot{x}/R^2$, yielding the displayed equation. Equivalently, the kinetic energy is $(m+I/R^2){\dot{x}}^2/2$. This added effective inertia lowers the undamped angular frequency to $\sqrt{k/(m+I/R^2)}$. A distributed viscous torque would add a separate damping term and cannot be inferred from the drag force alone.

Static friction is the tangential contact force that maintains zero contact slip. “Static” describes the contact state: both surfaces can move together in the chosen inertial frame. In the ideal dry-contact model, the actual static friction adjusts to the equations of motion, subject to a bound, rather than always taking its maximum value.^[ref]Here $N$ is the magnitude of the normal force and ${\mu }_s$ the coefficient of static friction. Equality means limiting static friction; it is a threshold, not an instruction to replace every unknown contact force by ${\mu }_{s}N$.

For a stipulated example, place a nonrotating $2\mathrm{kg}$ block initially at rest on a stationary horizontal surface. Take gravitational acceleration $g=9.8m/{\mathrm{s}}^2$ and coefficient of static friction ${\mu }_s=0.40$, with no other vertical force. The vertical force balance gives $N=mg=19.6\mathrm{N}$, so the largest allowed static friction magnitude is $7.84\mathrm{N}$. Choose right as positive and apply a horizontal force $F$. Remaining at rest requires Newton's second law to giveAt $F=2\mathrm{N}$, the static friction is $-2\mathrm{N}$; at $F=6\mathrm{N}$, it is $-6\mathrm{N}$. Both states have zero acceleration. At $F=10\mathrm{N}$, balancing would require $f_s=-10\mathrm{N}$, outside the bound, so this initially resting block cannot remain at rest. Its subsequent acceleration requires a kinetic friction law; setting $f_s=-7.84\mathrm{N}$ after contact slip begins would still use the wrong contact state. At exactly $F=7.84\mathrm{N}$, the ideal inequality still permits rest at the threshold. These are model calculations, not measured material properties.

The toy model used in kinetic friction makes the contact-state distinction explicit: $m=1\mathrm{kg}$, $N=10\mathrm{N}$, ${\mu }_s=0.5$ and coefficient of kinetic friction ${\mu }_k=0.3$. On stationary ground, an initially resting box pushed right with $4\mathrm{N}$ has $f_s=-4\mathrm{N}$. At $6\mathrm{N}$ it cannot remain at rest; once it slides right, $f_k=-3\mathrm{N}$ and $a=3m/{\mathrm{s}}^2$. At $5\mathrm{N}$, rest is allowed with $f_s=-5\mathrm{N}$, whereas an already right-sliding box has $f_k=-3\mathrm{N}$ and $a=2m/{\mathrm{s}}^2$. The applied force alone does not specify the friction branch: contact slip and motion history matter.

For a flat contact with two tangential directions, the bound applies to a vector norm, not separately to each component:For the stipulated $7.84\mathrm{N}$ bound, balancing horizontal components $4\mathrm{N}$ and $4\mathrm{N}$ requires a force magnitude $\sqrt{32}\mathrm{N}\simeq 5.657\mathrm{N}$, which is allowed. Components $6\mathrm{N}$ and $6\mathrm{N}$ require $\sqrt{72}\mathrm{N}\simeq 8.485\mathrm{N}$, which is forbidden even though each component separately is below $7.84\mathrm{N}$. This circular bound assumes the same coefficient of static friction in every tangential direction. Near impending contact slip, the limiting static friction opposes the relative contact motion that would otherwise begin. Once actual contact slip exists, kinetic friction instead opposes its tangential relative velocity, which need not have the direction of the body's center of mass motion.

The normal force must also come from the equations of motion. On horizontal ground, an upward pull of magnitude $P$ at angle $\alpha$ above horizontal gives, for a block remaining at rest with no other forces,The sine component unloads the contact and the cosine component demands tangential support. These equations require $N>0$; if the supporting contact opens, this dry-contact model cannot supply a negative normal force. For the stipulated $2\mathrm{kg}$ block and $\alpha =30^{\circ }$, the positive threshold is $P={\mu }_{s}mg/(\cos \alpha +{\mu }_s\sin \alpha )\simeq 7.3544\mathrm{N}$, with $N\simeq 15.9228\mathrm{N}$ and $|f_s|\simeq 6.3691\mathrm{N}$. A downward pull increases $N$ instead. An inclined plane gives another reason why $N$ is not always $mg$.

To solve a new problem, impose zero contact slip, find the required static friction and normal force from the equations of motion, then check the bound. A demand inside it permits that contact state; a demand outside it rules the state out. For an ideal driven tire undergoing rolling without slipping on stationary ground, a forward static friction can accelerate the vehicle. Its role is to prevent contact slip, rather than to oppose all motion of the whole body.

Nor does the word “static” by itself make friction do zero work. The contact power delivered to a body is $P_s={\mathbf{f}}_s\cdot {\mathbf{v}}_c$, a dot product with its material contact-point velocity in the chosen inertial frame. At an ideal contact undergoing rolling without slipping on stationary ground, ${\mathbf{v}}_c=0$, so this power is zero. For a rigid body, adding the translational and rotational contributions gives ${\mathbf{f}}_s\cdot {\mathbf{V}}_{\mathrm{CM}}+{\mathbf{\tau }}_{\mathrm{CM}}\cdot \mathbf{\omega }={\mathbf{f}}_s\cdot {\mathbf{v}}_c$, where ${\mathbf{\tau }}_{\mathrm{CM}}$ is the contact torque about the center of mass and $\mathbf{\omega }$ its angular velocity. Counting only the first term omits the rotational contribution.

At a moving contact with zero contact slip, the two material contact points share a velocity ${\mathbf{v}}_c$ that can be nonzero. For instance, a nonrotating block moving with an accelerating belt receives positive contact power $f_{s}U$ when the belt's rightward velocity $U$ and rightward static friction $f_s$ are positive. The belt receives $-f_{s}U$ by Newton's third law. These contact contributions sum to zero in the ideal no-slip model, while the block's kinetic energy can increase. Zero contact slip and zero work on each body are different conditions.

An inclined plane tilted slowly from horizontal turns a coefficient of static friction into a critical angle. Assume a nonrotating block on a stationary, rigid, dry surface, with gravitational weight and the contact forces as its only forces. Take $0\le \theta <\pi /2$ to be the angle between the uphill surface direction and horizontal. Let ${\mathbf{e}}_t$ be a unit vector pointing uphill and ${\mathbf{e}}_n$ a perpendicular unit vector pointing outward from the surface. Resolving the gravitational weight givesThe angle between the downward vertical and the inward normal $-{\mathbf{e}}_n$ is also $\theta$. Thus the inward component uses cosine and the downhill component uses sine. These are components of one force, not two additional forces.

Before downhill contact slip, zero acceleration and Newton's second law requireThe required static friction points uphill. Substituting into the bound and dividing by the positive $mg\cos \theta$ givesThe tangent function increases on this range of angles, so the last admissible resting angle isThe inverse tangent takes a dimensionless quantity to an angle. At equality the ideal model still permits a limiting resting state; increasing $\theta$ further excludes it. At a smaller angle, $f_s/N=\tan \theta$ is the demand ratio, not the coefficient of static friction.^[ref]

For the stipulated $m=2\mathrm{kg}$, $g=9.8m/{\mathrm{s}}^2$ and ${\mu }_s=0.40$, the critical angle is approximately $21.8014^{\circ }$. At $\theta =15^{\circ }$, $N\simeq 18.932\mathrm{N}$ and $f_s\simeq 5.073\mathrm{N}$, comfortably below the bound ${\mu }_{s}N\simeq 7.573\mathrm{N}$. At $\theta =25^{\circ }$, maintaining rest would require $f_s\simeq 8.283\mathrm{N}$, but the bound is only $7.105\mathrm{N}$: rest is impossible. At ${\theta }_c$, $N\simeq 18.198\mathrm{N}$ and $f_s\simeq 7.279\mathrm{N}$. Notice that the limiting tangential force here is smaller than the horizontal $7.84\mathrm{N}$ limit because tilting reduces the normal force.

Cancellation of mass makes this relation useful: the ideal critical angle tests a contact coefficient without needing a calibrated force gauge. Doubling the block's mass doubles both demanded static friction and its bound, leaving ${\theta }_c$ unchanged if ${\mu }_s$ is unchanged. This is a testable model prediction. It need not survive a change in surface condition, a direction-dependent coefficient, appreciable deformation, or vibration.

The derivation also assumes continued contact and that sliding occurs before tipping or rolling without slipping. The resultant contact force must act within the supporting footprint so that the block can satisfy torque balance; a tall block may tip first. No extra tension, imposed tangential pull, or lift is included. In a slowly raised apparatus, quasistatic process means that motion of the board contributes negligible inertial forces while locating the threshold. Once downhill contact slip begins, a constant-velocity sliding angle would concern kinetic friction, not the onset of static friction.

For a smooth strictly convex rigid body with a smooth unique contact point, a configuration consists of its orientation $R\in SO(3)$ and horizontal position $c_{\parallel }\in {\mathbb{R}}^2$; its height is fixed by contact. Translation makes this a principal bundle over $SO(3)$ with group ${\mathbb{R}}^2$. If $r(R)$ points from its reference point to contact and $\Omega$ is its spatial angular velocity, rolling without slipping requires $\dot{c}+\Omega \times r=0$. The displayed translation-valued principal connection vanishes on precisely these velocities. It is translation-invariant and equals the identity on vertical translation vectors. A prescribed orientation curve consequently has a unique horizontal lift from a given initial position. No-slip alone leaves spin free and does not specify a connection over the plane with orientation as its fiber; an additional no-twist constraint is needed for that familiar spherical model.