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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.

Because the contact is frictionless, the horizontal normal force is ${\mathbf{F}}_h=N\mathbf{n}$, with $N$ a signed outward component. Any vertical contact component balances Newtonian gravity and does not enter the horizontal equations. With $\mathbf{\omega }=\omega \hat{\mathbf{z}}$, the rotating reference frame equation givesSince $\hat{\mathbf{z}}\times \mathbf{\tau }=-\mathbf{n}$, $\mathbf{r}\cdot \mathbf{n}=2a{\cos }^2\theta$, and $\mathbf{r}\cdot \mathbf{\tau }=-a\sin 2\theta$, its normal and tangential components areHence the requested horizontal force vector isIts magnitude is the absolute value of the scalar coefficient. That coefficient need not always be negative: the constraint can exert either sign of normal force, depending on the bead's relative velocity. The tangential equation, which does not contain the reaction, is $\boxed{\ddot{\theta }=-({\omega }^2/2)\sin 2\theta }$.

Let $e_i^{\prime }(t)$ be a Cartesian basis fixed in the rotating frame. A vector $A=A_i^{\prime }{e}_i^{\prime }$ then satisfiesbecause a derivative of a body-fixed basis vector is ${\dot{e}}_i^{\prime }=\omega \times e_i^{\prime }$. Applying this transport formula twice to $r$ gives, for constant $\omega$,The second and third terms correspond to Coriolis acceleration and centrifugal acceleration in the rotating description.

For the bead, use cylindrical unit vectors $(e_r,e_{\varphi },e_z)$ with $e_z$ upward andProjection of $m\ddot{r}=-mg{e}_z+N$ along the wire tangent $\cos \theta e_r+\sin \theta e_z$ eliminates the smooth-wire normal force $N$ and gives

The circular constraint isIts tangent in the $xz$-plane is $(\cos \theta ,0,-\sin \theta )$, while the normal force is radial. Projecting the equations from part (a) onto this tangent eliminates the normal force and givesBecause the ramp is translation-invariant along $y$, $N_y=0$; using $\dot{x}=R\dot{\theta }\cos \theta$ gives

For rest in the rotating frame, $\dot{\theta }=\dot{y}=\ddot{\theta }=\ddot{y}=0$. Assuming $\Omega \ne 0$, the second equation gives $y=0$, and the first givesOn the semicircular ramp the rest points are thereforeand, when ${\Omega }^{2}R\ge g$,

On the surface of revolution, a normal vector is proportional to $-H^{\prime }(r){\mathbf{e}}_r+{\mathbf{e}}_z$. Thus the normal force and gravity have no ${\mathbf{e}}_{\theta }$ component. The ${\mathbf{e}}_{\theta }$ component of Newton's second law, using the given linear drag force, isMultiplication by $r$ therefore givesThe printed target equation omits the factor $1/m$. The matching original PDF, page 5, gives both ${\mathbf{F}}_D=-\alpha \dot{\mathbf{r}}$ and $\dot{h}=-\alpha h$, so this is a source inconsistency. If its $\alpha$ in the target equation is intended to mean the drag coefficient per unit mass, put $\lambda =\alpha /m$ and write $\dot{h}=-\lambda h$. Equivalently, if the intended force was ${\mathbf{F}}_D=-m\alpha \dot{\mathbf{r}}$, the requested convention givesThese formulas determine $\dot{\theta }$ once $r(t)$ is known; they do not assume a constant radius.

For a smooth surface the friction force vanishes. The ${\mathbf{e}}_{\theta }$ component of Newton's second law then gives $r^2\dot{\theta }=h$, constant. Along the constraint $z=H(r)$, the chain rule givesThe acceleration in cylindrical coordinates isTake the dot product of Newton's second law with the tangent vector ${\mathbf{e}}_r+H^{\prime }{\mathbf{e}}_z$. The normal force disappears, while the gravity term is $-mg{H}^{\prime }$. Dividing by $m$ givesUsing $\dot{\theta }=h/r^2$ yields

Continue with $f=p\sin \varphi$ and $q=p\cos \varphi$. The tangent vector to the hoop is ${\mathbf{r}}_{\varphi }=f^{\prime }{\mathbf{e}}_1-q^{\prime }{\mathbf{e}}_3$. Smooth contact means that the hoop's normal force has zero dot product with this tangent vector. The remaining applied force is $-mg{\mathbf{e}}_3$, whose dot product with ${\mathbf{r}}_{\varphi }$ is $mg{q}^{\prime }$.

Since $\mathbf{\omega }=\omega {\mathbf{e}}_3$, the Coriolis force is perpendicular to the hoop's plane and hence to ${\mathbf{r}}_{\varphi }$. AlsoTake the dot product of the rotating-frame Newton's second law with ${\mathbf{r}}_{\varphi }$, substitute a, and divide by $m$:DefineThen $K^{\prime }/2=f^{\prime }{f}^{\prime \prime }+q^{\prime }{q}^{\prime \prime }$ and $J^{\prime }=-{\omega }^{2}f{f}^{\prime }-g{q}^{\prime }$, soExpanding the squares makes the cancellation of the mixed terms explicit:The strict inequality follows from $p>0$. Here $J$ is the rotating-frame effective potential per unit mass. Multiplication of the equation by $\dot{\varphi }$ gives the useful first integral $\frac{1}{2}K(\varphi ){\dot{\varphi }}^2+J(\varphi )=\text{constant}$.

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

The lines of centres from either lower cylinder to the upper cylinder make $60^{\circ }$ with the horizontal. Let $N_L,N_R$ be the upward normal forces on the top cylinder from the left and right cylinders, and let $Q$ be the lower cylinders' mutual normal force. While all three contacts remain closed, the triangular configuration translates horizontally with common acceleration $a$; frictionless normals exert no torques about the cylinder centres.

For the top cylinder, Newton's second law givesThus $N_L=m(g/\sqrt{3}+a)$ and $N_R=m(g/\sqrt{3}-a)$. The right lower cylinder has horizontal equation $N_R/2+Q=ma$, soA contact may push but cannot pull: $N_L,N_R,Q\ge 0$. The limiting inequalities are thereforeAt the lower bound, $Q=0$ and the lower pair are just losing compression; at the upper bound, $N_R=0$ and the upper-right contact is just losing compression. All three pairwise normal forces are strictly positive only inside the open interval. The ground normal forces are positive throughout this range. The applied horizontal force on the left cylinder is $F=3ma$, as follows from the whole system's horizontal momentum balance.

In the frame following the turn, the outward inertial force $m{v}^2/R$ acts at height $h$ and tends to unload the inner rail. The two upward normal forces satisfyThe second equation is moment balance about the midpoint between the rails. ThusFull contact requires a nonnegative inner normal force. Its limiting value is zero, giving the toppling speed on level railsFor strictly positive contact force on both rails, $v<v_0$. This ideal result assumes level rails, negligible suspension motion and enough lateral contact force to follow the curve; lateral sliding could impose a different limit.

Take positive $x$ horizontally toward the downhill end of the plane, with $0\le \theta <\pi /2$. The normal force is $N=mg\cos \theta$, since there is no acceleration perpendicular to the stationary plane. Along the plane, the only tangential force is $mg\sin \theta$, so Newton's second law gives $a_{\parallel }=g\sin \theta$.

Its horizontal projection is thereforeEquivalently, the normal force has horizontal component $N\sin \theta$ while gravity has none, giving $m{a}_x=N\sin \theta$. For a plane whose height increases in the positive $x$ direction, the answer instead carries a minus sign. A horizontal plane gives zero acceleration; the horizontal projection also tends to zero for a vertical plane.

Choose positive horizontal velocity to the right and positive spin angular velocity anticlockwise, as in the source drawing. At the bottom of the sphere, the rotational velocity is then to the right: the initial tangential contact-point velocity isFor the illustrated positive $v_1,{\Omega }_1$, the contact slips to the right, so the tangential contact force acts leftward. The upward normal force acts through the sphere centre and produces no spin torque. The tangential contact force acts at lever arm $a$ and produces a clockwise torque, reducing the positive spin. The weight acts vertically downward through the centre of mass.

During the short impact the contact impulses dominate the weight's impulse. In an ideal elastic rough contact, the tangential deformation can store and return energy; its contact force need not be a dissipative sliding-friction law throughout the collision. For other signed incoming states, reverse the initial tangential force if $w_1<0$; if $w_1=0$, no tangential impulse is required.

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.

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.

The component of magnetic force along the slope has magnitude $IBs\cos \varphi$. The rate at which the sliding bar does work against this magnetic braking isBy Ohm's law, $\mathcal{E}=IR$, soThis exactly equals the Joule heating in the resistor in the negligible-self-inductance approximation. The normal force does no work because it is perpendicular to the velocity.

The gravitational power is instead $mgv\sin \varphi$. By the work-energy theorem,It equals the Joule heating only at constant speed. For a concrete counterexample, during the initial acceleration the kinetic energy increases, so some gravitational power remains mechanical.

Faraday's law of induction states that the oriented electromotive force around a closed material circuit is minus the rate of change of its linked magnetic flux:The surface orientation and circuit direction obey the right-hand rule. The minus sign expresses Lenz's law: the induced electric current, when allowed to flow, produces a magnetic field opposing the change of linked magnetic flux. For a stationary circuit this reduces to $\oint \mathbf{E}\cdot d\mathbf{\ell }=-d{\Phi }_B/dt$.

The source's “Lorenz force” is the Lorentz force, $q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$. A wire element of cross-section $A_c$ contains $n{A}_{c}d\ell$ carriers of charge $q$ with drift velocity ${\mathbf{v}}_d$. Adding their magnetic forces and using $I=nq{A}_c{v}_d$ givesfor a straight wire in a uniform magnetic field. The length vector follows conventional electric current, irrespective of the carrier sign.

For the unlabelled upward-motion request, define $\hat{\mathbf{t}}=(\cos \varphi ,0,-\sin \varphi )$ down the slope, $\hat{\mathbf{b}}=(0,1,0)$ across the rails and $\hat{\mathbf{n}}=(\sin \varphi ,0,\cos \varphi )$ out of their plane. Choose the vertical magnetic field as $\mathbf{B}=B\hat{\mathbf{z}}$. An electric current $I$ across the bar in direction $-\hat{\mathbf{b}}$ gives a horizontal magnetic force $-IBs\hat{\mathbf{x}}$. The other forces are weight $-mg\hat{\mathbf{z}}$ and the rail normal force $N\hat{\mathbf{n}}$. Uniform upward velocity requiresConsequentlyEach symmetric rail supplies half of $N$. The electric current must make $I\mathbf{L}\times \mathbf{B}$ point horizontally toward the uphill side; for the coordinates just defined it flows along $-\hat{\mathbf{b}}$. Reversing the vertical magnetic field reverses this current direction. The external supply must maintain that electric current against the motional electromotive force during ascent. For $\cos \varphi =0$, no finite horizontal magnetic force can support uniform motion along the vertical rails.

The PDF's apex angle is $30^{\circ }$, so each contacted face makes $\alpha =15^{\circ }$ with the vertical. Let N be the inward normal force per face. Its vertical component is downward, $-N\sin \alpha$; limiting upward static friction contributes $\mu N\cos \alpha$. Symmetric contacts cancel horizontal forces and torque. Incipient lift consequently requiresThis is the usual normal gripping force convention, using $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. Friction can lift the prism only if $\mu >\tan \alpha$, which is satisfied here.

If “force applied” means the magnitude of the entire finger-contact resultant, including its tangential component, the normal versus resultant force in a frictional pinch givesThe optimal resultant lies at the limiting friction angle $\arctan \mu$ from the inward normal. Stating which force is being measured avoids confusing these two legitimate quantities.

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.