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For the complete uniform disc, the surface density is $\sigma =M/(\pi a^2)$. An annular strip of radius $r$ and thickness $dr$ has mass $dm=2\pi \sigma rdr$, so its contribution to the axial moment of inertia is $r^{2}dm$. HenceThis derives the moment of inertia of a uniform disc rather than assuming it.

Removing a concentric disc of radius $a/2$ at the same density removes mass $M/4$ and inertia $\frac{1}{2}(M/4)(a/2{)}^2=M{a}^2/32$. The remaining annulus therefore has mass $m_A=3M/4$ andThe moment of inertia of a uniform annulus gives the same value directly.

Let $A_c$ be the downhill centre-of-mass acceleration and let $F$ denote the magnitude of the uphill static friction. Translation, rotation about the centre, and rolling without slipping giveThus $F=I_A{A}_c/a^2=(5/8)m_A{A}_c$. Eliminating $F$ gives the rolling acceleration with rotational inertiaSubstitution gives the contact forcesThe normal force balances the perpendicular component of gravity because the centre remains at a constant distance from the plane. Although the annulus accelerates downhill, friction must act uphill to supply the torque that increases its rolling angular speed. For a plane with $0\le \alpha <\pi /2$, the assumed no-slip motion requires a coefficient of static friction satisfying ${\mu }_s\ge F/N=(5/13)\tan \alpha$. With that condition, contact friction does no work because the instantaneous contact point is stationary; it redistributes the gravitational energy between translational and rotational kinetic energy.

For the moment of inertia of a uniform solid sphere, write the mass density as $\rho =3m/(4\pi a^3)$. In spherical coordinates about the chosen axis, the squared perpendicular distance is $R^2{\sin }^2\vartheta$. The volume element is $R^2\sin \vartheta dRd\vartheta d\phi$, so

Take $0<\alpha <\pi /2$ and $\ell >0$, with a fixed plane. The normal force has magnitude $N=mg\cos \alpha$ and is perpendicular to the plane. On the perfectly smooth plane there is no friction. Both gravity and the normal force have zero torque about the marble's centre, so a marble released from rest does not rotate. The component of Newton's second law down the plane is $m{A}_1=mg\sin \alpha$, henceThe normal force does no work because the velocity is tangent to the plane. Gravity is conservative, so mechanical energy, the sum of translational kinetic energy and gravitational potential energy, is conserved.

For rolling without slipping, let the static friction force have magnitude $f$, pointing up the plane. It supplies the torque that spins the marble while reducing its translational acceleration. If $\Omega$ is the angular velocity and $A_2$ the acceleration down the plane, the rolling constraint is $a\Omega =\dot{s}$, hence $a\dot{\Omega }=A_2$. The translation and rotation equations areThus $f=I{A}_2/a^2$ andThese are the equations for rolling acceleration with rotational inertia; the plane must be rough enough to supply this static friction.

The point of contact is instantaneously at rest relative to the fixed plane, so static friction does no total work on the rolling marble. Equivalently, its translational power is $-f\dot{s}$ and its rotational power is $fa\Omega =f\dot{s}$, which cancel. The normal force also does no work. Consequently mechanical energy is conserved here too, now with rotational kinetic energy included:where $H$ is the centre's height. No energy loss is implied by the presence of ideal static friction. Constant acceleration from rest gives

For the hollow marble, put $I=\lambda m{a}^2$ in the same equations. The smooth-plane time is unchanged, while the rolling acceleration becomes $g\sin \alpha /(1+\lambda )$. ThereforeThis again assumes rolling without slipping and enough static friction; the ratio depends on the mass distribution through its moment of inertia.

Let the uniform mass density be $\rho =3M/(4\pi a^3)$. In spherical polar coordinates about the rotation axis, the perpendicular distance is $r\sin \theta$ and the volume element is $r^2\sin \theta drd\theta d\varphi$. The moment of inertia is thereforeIn the given rigid body kinetic energy decomposition, $\frac{1}{2}M|\dot{\mathbf{R}}{|}^2$ is the translational kinetic energy of the whole mass moving at its center of mass velocity. The term $\frac{1}{2}I{\omega }^2$ is the rotational kinetic energy about the center of mass. The mixed term vanishes because the mass-weighted relative positions sum to zero.

For rolling without slipping, the point of contact has zero instantaneous velocity relative to the stationary surface. Its rotational velocity is opposite the center of mass velocity and has magnitude $a|\omega |$. Hence $V=a|\omega |$. Substituting this and the moment of inertia of a uniform solid sphere givesThe rotational axis is through the center of mass and parallel to the surface, perpendicular to the direction of rolling.

Choose positive $x$ downhill and positive $\theta$ in the corresponding rolling direction, with $x=\theta =0$ initially. For rolling without slipping, the contact point has zero velocity, so $\dot{x}-b\dot{\theta }=0$. Integrating the rolling constraint givesLet $F$ be the uphill static friction. Along the inclined plane, Newton's second law for the centre of mass givesThe weight passes through the centre of mass, and the normal reaction has its line of action through it. Thus only static friction supplies torque about $G$. The angular velocity is along the cylinder's symmetry axis, so the inertia tensor gives angular momentum $C\dot{\theta }$ in that direction. The torque equation is thereforeEliminating static friction and inserting $C$ givesThe extra term represents the rotational inertia that must be accelerated along with the translational mass.

For a uniform solid cylinder, set $a=0$ in the moment of inertia formula, givingFor the hollow cylinder, $a>0$ makes the denominator larger:on a downward incline with $\sin \phi >0$. Thus the solid cylinder accelerates faster, assuming rolling without slipping for both. More of the hollow cylinder's mass lies far from the rotation axis, increasing its moment of inertia relative to $M{b}^2$.

Apply the general linear momentum equation $d\mathbf{p}/dt=\sum \mathbf{F}$ to the whole rigid body. Its center of mass has acceleration $\ddot{x}{\mathbf{e}}_1$, and each rail contributes its own static friction and normal force. HencesoHere $N$ is the normal force at each rail, not the combined support.

Take torques about the center of mass, with positive angular velocity along ${\mathbf{e}}_3$. The thread leaves the lower side shown in the source diagram: its contact position vector relative to the center isThus $\mathbf{q}\times \mathbf{T}=aT{\mathbf{e}}_3$. Each contact point has relative position vector $-R{\mathbf{e}}_2$, so each friction torque is $-RF{\mathbf{e}}_3$. The gravitational weight acts through the center of mass and the normal forces have zero torque about it. The angular momentum equation becomesThe condition of rolling without slipping makes the contact point instantaneously stationary:Integrating and using $x(0)=\varphi (0)=0$ gives $x=-R\varphi$. Thus rightward translation corresponds to clockwise rotation. Eliminating $F$ from the two dynamical equations with $\ddot{x}=-R\ddot{\varphi }$ yieldsand therefore

Put $D=m{R}^2+{\mathcal{I}}_{33}=m(R^2+a^2/2)$. The angular acceleration is constant. The initial rest conditions and $x(0)=\varphi (0)=0$ giveFor the upward rightward pull in the diagram, $0\le \alpha \le \pi /2$, the sign of $x(t)$ is the sign of $R\cos \alpha -a$. Consequently it rolls towards the horizontal pull when $0\le \alpha <\arccos (a/R)$. At $\alpha =\arccos (a/R)$ it remains at rest, with balanced forces and torques. For $\arccos (a/R)<\alpha \le \pi /2$, it rolls to the left. In particular a horizontal pull, $\alpha =0$, makes it roll to the right, since $R>a$. If other pulling angles are allowed, the general condition for motion in the same horizontal direction as the pull is $\cos \alpha (R\cos \alpha -a)>0$.

The required static friction at each rail follows from the translational equation:For $0\le \alpha \le \pi /2$ it is positive, agreeing with the leftward force convention. The assumed rolling without slipping is physically possible only if the rails can supply this force while maintaining contact. With coefficient of static friction ${\mu }_s$, the necessary contact inequalities areFor the usual pulling range, these becomeIf either fails, the proposed motion requires lift-off or more static friction than the rails provide. Equality $T\sin \alpha =mg$ is also incompatible here: it gives $N=0$ but a nonzero required friction. The question supplies no coefficient of static friction, so it determines the required friction exactly but gives no numerical slip threshold. If $F$ is regarded as a prescribed force rather than an available contact response, it must equal $F_{\mathrm{req}}$ for the rolling constraint to hold.

For the rolling disk with a rim point mass, impose rolling without slipping and take the clockwise angular displacement $\theta$ from the configuration with the extra mass at the bottom. The disk centre has $x_C=R\theta$ and $y_C=R$. The extra mass hasThe disk's translational and rotational kinetic energies sum to $3M{R}^2{\dot{\theta }}^2/2$. Including the rim mass givesThe exact Euler-Lagrange equation isTo first order, $3M{R}^2\ddot{\theta }+MgR\theta =0$. ThereforeAt the bottom the extra mass is instantaneously at the contact point and has no first-order velocity. The increased effective inertia comes from the moving disk. Static friction enforces the rolling constraint without doing work at the stationary contact.

A holonomic constraint can be written as an equation $f(q_1,\dots ,q_n,t)=0$ involving coordinates and possibly time, without an irreducible dependence on velocities. For example a particle confined to a fixed sphere obeys $x^2+y^2+z^2-a^2=0$. A general rolling without slipping constraint in three dimensions need not be integrable into such a coordinate equation.

The supplied PDF labels this equation clarified from original for tutorial purposes. Use the model of an eccentric wheel with missing-mass kinetic energy neglected: neglect the missing disc's contribution to kinetic energy, while retaining its subtraction from gravitational potential energy. Do not silently replace the gravitational mass $W$ by the full-disc mass $D$ as well.

Choose $\theta$ as the clockwise angle of $\overrightarrow{CH}$ from the upward vertical, and set $s=R\theta$. The rolling without slipping constraint makes changes in $s$ equal changes in the distance C travels down the slope. Choose its additive origin so $s=0$ when the hole is vertically above C. Then, up to an irrelevant height constant,The approximate kinetic energy is that of the complete uniform disc rolling at $\dot{s}=R\dot{\theta }$:Subtracting the missing disc's gravitational potential energy givesApply the Euler-Lagrange equation to $\mathcal{L}=T-U$:ThereforeHere $K_1,K_2$ have dimensions of acceleration and $\gamma$ has dimensions of inverse length. The initial conditions are $s(0)=\dot{s}(0)=0$ under the vertical-above interpretation. If an initial orientation is instead specified relative to the slope normal, the same coordinate convention produces a different initial $s$, not different constants. With $r_o\to 0$ the equation becomes $\ddot{s}=(2/3)g\sin \alpha$, the familiar solid-disc result. Retaining the hole's exact kinetic energy would introduce coordinate-dependent inertia and extra ${\dot{s}}^2$ terms; the printed neglect is what permits the requested constant-coefficient form.

Assume the cylinders are released from rest. In each case, “there is friction” is interpreted as sufficient static friction for rolling without slipping at the rough contacts. Without a coefficient or a no-slip assumption, sliding cases would not have a unique specified acceleration. On release the two lower cylinders separate; their mutual contact must not be retained as a rigid constraint.

Let the top cylinder's initial downward acceleration be $w$, and each lower cylinder's outward horizontal acceleration be $u$. The top cylinder initially has no horizontal acceleration. Differentiating the fixed upper-to-lower centre separation twice at the initial zero velocity givesBy reflection symmetry the top cylinder has zero angular acceleration, while the lower cylinders rotate in opposite senses.

The familiar right-angle result requires an elastic collision with conserved horizontal momentum and the appropriate treatment of rotational kinetic energy. In the ideal smooth-sphere model the spins do not change during the collision, so the translational kinetic energy is separately conserved. For equal masses, with initial velocity $\mathbf{u}$ and outgoing velocities ${\mathbf{v}}_1,{\mathbf{v}}_2$,Squaring the first equality and using the second givesThus the two nonzero outgoing directions are perpendicular. In the head-on limiting case one sphere stops, so an angle for its zero velocity is undefined. An ideal elastic model that enforces rolling without slipping at both ends gives the same energy equation: each solid sphere has total kinetic energy $7m{v}^2/10$, and the common factor cancels.

Elasticity is not stated explicitly in the question. It cannot be omitted from the argument: for a smooth equal-mass collision with normal incoming component $u_n$ and coefficient of restitution $e<1$, the outgoing scalar product is $(1-e^2)u_n^2/4>0$. Moreover, after a smooth impact the spheres need not still be rolling; subsequent frictional relaxation can change their directions. The right-angle statement therefore concerns the specified ideal elastic collision, not arbitrary later rolling trajectories.

For the frictional conversion of spin to rolling, consider the horizontal slipping stage on the rough surface, whose contact radius is now $a$, and neglect extra vertical impact or height-change energies at the transition. A friction impulse $J_f$ changes $Mv$ by $J_f$ and clockwise spin $I\omega$ by $-a{J}_f$. Thus $I\omega +Mav$ is conserved. At the end $v_f=a{\omega }_f$, givingBecause $I=M{a}^2$, the final total kinetic energy is $K_f=M{v}_f^2$. Subtraction yieldsThe result is independent of the friction coefficient, provided sufficient contact time allows rolling without slipping to be regained. Initially $v_0=b{\omega }_0<a{\omega }_0$, so the contact point slips backwards. Friction acts forwards and can increase translational kinetic energy, while decreasing rotational kinetic energy by more. A finite vertical drop or inelastic landing would require additional geometric information; the boxed expression describes the horizontal rolling-conversion loss.

Let $\theta$ be the displacement from the upward vertical and $v$ the speed of the ball's centre of mass. Its centre follows a circle of radius $R+r$. The rolling without slipping constraint is $v=r\omega$, with $\omega$ the ball's spin angular velocity. The static contact friction does no work on the ball because the instantaneous contact point is stationary. Thus conservation of energy givesProject Newton's second law toward the hemisphere's centre. The inward component of gravity is $Mg\cos \theta$ and the normal force $N$ points outward, soAt contact loss $N=0$. Eliminating $v$ gives the ideal sphere rolling off a hemisphere resultThe radius cancels, so replacing $R+r$ by $R$ under the stated small-ball approximation leaves the same answer.

Finite-friction qualification. This is an ideal rolling-constraint result. It does not describe uninterrupted pure rolling to detachment with any finite coefficient of static friction. The tangential force and torque equations giveso $f=(2/7)Mg\sin \theta$ acts uphill, whereasConsequently the required ratio isFor a concrete example, a coefficient of static friction of one allows rolling only up to about $47.54^{\circ }$, where $17\cos \theta -10=2\sin \theta$, before sliding begins. A subsequent physical contact-loss angle needs a sliding-friction model. The boxed $53.97^{\circ }$ answers the source's expressly imposed no-slip idealization.

Keep the signs of part i. The tension has horizontal component $T\cos \theta$ but still has counterclockwise torque $Tr$, since it is tangent to the axle. ThusSolving these equations gives the critical pulling angle of a rolling spool lawThe zero-acceleration angle is thereforeAt this angle the line of action of the pull passes through the ground contact, so its torque about that point vanishes. A spool initially at rest stays at rest while the contact constraints hold; one already rolling has zero acceleration, rather than being brought to rest.

For equilibrium the vertical normal force is $N=Mg-T\sin {\theta }_c$ and $|f|=T\cos {\theta }_c$. The static friction inequality $|f|\le \mu N$ becomesHence, for $0<r<R$,At this limit the normal force is still nonnegative; for finite $\mu$ and positive $r$ it is positive. For ${\theta }_c<\theta \le \pi /2$, the numerator of $a$ is negative. Released from rest, the spool rolls leftward, away from the pulling hand, with counterclockwise rotation, provided the assumed rolling without slipping and ground contact remain possible. The nonzero acceleration is not obtained by assuming that friction simply opposes the centre's motion.

Take rightward translation as positive and counterclockwise angular acceleration $\alpha$ as positive. Let $f$ denote signed ground friction, positive to the right. The source's lower-tangent winding makes the tension exert counterclockwise torque $Tr$. For a horizontal pull, translation and rotation obeyThe final relation is the rolling without slipping constraint. Eliminating $f$ and $\alpha$ givesThus the requested magnitude of static friction isIt acts leftward. Released from rest, the spool accelerates rightward, toward the pulling hand, and rotates clockwise. There is no contradiction between leftward friction and rightward motion: static friction prevents slipping of the contact point, and the horizontal tension is larger than $|f|$ because $r<R$.

Interpret $h$ as the vertical drop of the cylinder's centre of mass, and initially take the suspended rod to be vertical. The moment of inertia of a uniform solid cylinder is $I=m{r}^2/2$. Before impact, rolling without slipping gives $\omega =u/r$, so conservation of mechanical energy givesUse the usual instantaneous smooth elastic collision model: the horizontal impact impulse passes through the cylinder's centre of mass, and the floor supplies negligible tangential impulse during impact. The cylinder's angular velocity is then unchanged during impact; its rotational kinetic energy does not transfer to the bob. The rod supplies no horizontal impulse when vertical. If $v$ and $V$ are the signed cylinder and bob velocities immediately afterwards, conservation of momentum and the unit coefficient of restitution giveDuring the subsequent simple pendulum swing, conservation of mechanical energy givesThe cylinder generally slips immediately after this impact, which is consistent with imposing rolling without slipping only on the descent.

Impact convention matters. The floor-impulse dependence of a rolling-cylinder collision gives a different answer if a different constraint is imposed. If instead the floor enforces rolling without slipping throughout the impulsive collision, it supplies a nonzero horizontal impulse. The cylinder's effective translational mass is then $m+I/r^2=3m/2$. Writing $J$ for the impulse delivered to the bob gives $(3m/2)(v-u)=-J$, $2mV=J$ and $V-v=u$. Hence $V=6u/7$ and $\cos {\theta }_{\max }=25/49$, giving $59.32^{\circ }$. The original wording specifies the descent constraint but does not specify an impulsive floor-friction law; the first boxed value uses the smooth-impact convention.

Let $a$ be the center of mass acceleration downhill and $F$ the uphill static friction. Translation and rotation about the center of mass obeyThe rolling without slipping constraint is $a=r\alpha$. For the coin's moment of inertia $I=m{r}^2/2$, this gives $F=ma/2$, soThe normal force is $N=mg\cos \theta$. Static friction can enforce rolling without slipping only if $F\le \mu N$, givingThe static friction acts uphill to provide the torque; it is not automatically equal to $\mu N$, attaining that limiting value only at the threshold. For finite $\mu \ge 0$ the result lies between $0$ and $\pi /2$.

For a Lagrangian $\mathcal{L}=T-U$, the Euler-Lagrange equation is $d(\partial \mathcal{L}/\partial \dot{q})/dt=\partial \mathcal{L}/\partial q$ for each unconstrained generalized coordinate. For example, $\mathcal{L}=m{\dot{x}}^2/2-k{x}^2/2$ gives $m\ddot{x}+kx=0$ for a spring oscillator.

For a uniform solid sphere, slicing into disks normal to the rotation axis givesUse total dumb-bell mass $M$ and $\ell =L-r$. Since both centres lie on its spin axis, its moment of inertia about that axis is $2M{R}^2/5$. If $\theta$ measures displacement from the bottom of the track, rolling without slipping imposes spin speed $|\dot{\varphi }|=\ell |\dot{\theta }|/r$. The rolling inside a fixed circular cylinder therefore hasThe equation of motion and its small-amplitude limit areWith $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and the given lengths, $\boxed{T=2\pi /\omega \simeq 80.5\mathrm{s}}$. If the centre rises by height $h$, conservation of energy at the bottom givesThe speed expression uses exact height-energy conversion within the ideal rolling constraint; only the period formula uses the small-angle approximation.

Rolling without slipping gives ${\omega }^2=g/[(L-r)(1+2R^2/(5r^2))]$. With the supplied dimensions, $T\simeq 80.5\mathrm{s}$ and the maximum centre-of-mass speed for a $10\mathrm{mm}$ rise is $0.0349\mathrm{m}{\mathrm{s}}^{-1}$.

Full derivation: solution.