Rolling without slipping

Rolling without slipping imposes that the contact point is instantaneously at rest, giving v=R omega.

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 $\sqrt{g/(2R)}$.

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.

Three identical cylinders initially form an equilateral triangle on a horizontal floor and are released from rest. If each has moment of inertia $\beta m{R}^2$, the initial top downward acceleration is $g/(7+6\beta )$ for no-slip rough ground and smooth cylinder contacts, or $g/(7+8\beta )$ for smooth ground and no-slip rough cylinder contacts. In both cases the lower cylinders separate with outward acceleration $u=\sqrt{3}{a}_{\mathrm{top}}$. The corresponding lower-cylinder angular acceleration magnitudes are $u/R$ and $2a_{\mathrm{top}}/R$, with opposite rotation senses between the cases. Sufficient static friction is required; a finite coefficient allowing sliding changes the problem.

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.

A uniform solid sphere released from the top of a fixed spherical surface has $v^2=10g(R+r)(1-\cos \theta )/7$ under the ideal rolling constraint. The normal force is $Mg(17\cos \theta -10)/7$, giving the title's ideal detachment angle. The required static friction is $2Mg\sin \theta /7$. Its ratio to the normal force diverges at detachment, so any finite coefficient of static friction causes prior slipping; a physical subsequent trajectory requires a sliding-friction model.

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 $-(2/3)g\sin \alpha$.

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 $T_{\max }=\mu Mg/(\cos {\theta }_c+\mu \sin {\theta }_c)$.

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.

For a rigid body rolling on a stationary support, the instantaneous contact point $P$ has zero velocity: ${\mathbf{v}}_P={\mathbf{v}}_G+\mathbf{\omega }\times \overrightarrow{GP}=0$. For a wheel of radius $R$ moving horizontally and a counterclockwise angle $\varphi$, this gives $\dot{x}+R\dot{\varphi }=0$, hence $x+R\varphi$ is constant. Maintaining this mechanical constraint generally requires sufficient static friction and a nonnegative normal force.

A solid cylinder rolling inside a circular loop has centre-path radius $\ell =R-a$, kinetic energy $3m{v}^2/4$, and speed $v^2=4g[h-\ell (1-\cos \theta )]/3$ at angle $\theta$ from the bottom. The normal contact force is smallest at the top. Requiring $v_t^2\ge g\ell$ gives the displayed necessary contact threshold under imposed no-slip rolling. The centre's maximum acceleration on the circle is at the bottom, and at the threshold it is $11g/3$ upward. This is a limiting contact condition, not a guarantee for arbitrary finite static friction or an unspecified approach curvature.

At the limiting release height for a solid cylinder, the loop normal force is $N=(7mg/3)(1+\cos \theta )$ and the required static-friction magnitude is $mg|\sin \theta |/3$. Thus $|f|/N=|\sin \theta |/[7(1+\cos \theta )]$ diverges near the top, even though both forces vanish there. A finite friction coefficient therefore requires a height above the ideal contact threshold to maintain no-slip rolling throughout. A question imposing rolling but giving no coefficient determines the ideal limiting bound, not a realistic friction-dependent minimum.

A circular body of radius $r$ rolls inside a circular track of radius $R$. Let $\theta$ be the track's material rotation, $\varphi$ the absolute angle of the line joining the two centres, and $\psi$ the inner body's absolute spin angle, all with one consistent orientation. Equating contact-point velocities gives the displayed constraint. Translation of the track's centre cancels from that equality. A stationary track gives $\dot{\psi }=-(R-r)\dot{\varphi }/r$.

For a uniform sphere struck horizontally with impulse $J$ at height $d$ above its centre, $u_0=J/M$ and clockwise spin ${\Omega }_0=5Jd/(2M{a}^2)$. friction brings the contact slip $u-a\Omega$ to zero. The final rolling speed is $u_f=5J(1+d/a)/(7M)$, independent of a positive constant friction coefficient; friction can accelerate an initially overspinning centre of mass.

For the ideal uniform solid sphere with negligible table impulse during the cue strike, immediate rolling requires $u_0=a{\Omega }_0$, hence the horizontal cue line lies $2a/5$ above the centre. A different inertia distribution changes this height.

A ball bearing transfers loads through rolling spherical elements between two races. The no-slip conditions at both contacts determine the translation and spin of each ball; these are generally not the same angular velocity.

If an inner race of radius $R$ rotates at angular velocity $\omega$ and an outer race is stationary, a ball of radius $r$ rolling at both contacts has centre speed $R\omega /2$ and spin $-R\omega /(2r)$. Its centre follows a circle of radius $R+r$. The angular momentum about the common race axis includes both orbital angular momentum and spin.

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 hanging mass $m$ pulls the centre of a uniform disc of mass $M$ up an incline of angle $\theta$, using a light string and light frictionless pulley. If the disc rolls without slipping and the string is parallel to the slope, its effective translational mass is $M+I/a^2=3M/2$. With downward hanging acceleration $q$, $q=g(m-M\sin \theta )/(m+3M/2)$. Static friction has signed downslope component $Mq/2$ and supplies the rolling torque. The acceleration threshold is unchanged from a sliding block, although rotational inertia reduces the magnitude.

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.

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.

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 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.