226Week 4: Systems of Particles, Momentum and Collisionse) In this particular problem one could in principle solve Newton’s second law because the elasticcollision force isknown. In general, of course, it is not known, although for a verystiffspringthis model is an excellent one to model collisions between hard objects. Assuming that thespring is sufficiently stiff that the two masses are in contact for a very short time ∆ , write atsimple expression for theimpulseimparted tom 2andqualitativelysketchF avover this timeinterval compared toF tx( ).
Week 4: Systems of Particles, Momentum and Collisions227Problem 8.mMvvoof vθThis problem will help you learn required concepts such as:•Newton’s Third Law•Momentum Conservation•Fully Inelastic Collisionsso please review them before you begin.In the figure above, a large, heavy truck with massMspeeds through a red light to collide witha small, light compact car of massm. Both cars fail to brake and are travelling at the speed limit( ) at the time of the collision, and their metal frames tangle together in the collision so that afterv 0the collision they move as one big mass.a) Which exerts a larger force on the other, the car or the truck?b) Which transfers a larger momentum to the other, the car or the truck?c) What is the final velocity of the wreck immediately after the collision (please give (v , θf))?d) How much kinetic energy was lost in the collision?e) If the tires blow and the wreckage has a coefficient of kinetic frictionµ kwith the ground afterthe collision,set up an expressionin terms ofv fthat will let you solve for how far the wreckslides before coming to a halt. You do not need to substitute your expression from part c) intothis and get a final answer, but you should definitely beableto do this on a quiz or exam.
228Week 4: Systems of Particles, Momentum and CollisionsProblem 9.xxxmmm rbjbjrThis problem will help you learn required concepts such as:•Center of Mass•Momentum Conservation•Newton’s Third Lawso please review them before you begin.Romeo and Juliet are sitting in a boat at rest next to a dock, looking deeply into each other’seyes. Juliet, overcome with emotion, walks at a constant speedvrelative to the water from her endof the boat to sit beside him and give Romeo a kiss. Assume that the masses and initial positionsof Romeo, Juliet and the boat are (m , xrr), (m , xjj), (m , xbb), wherexis measured from the dockas shown, andD = x j− x ris their original distance of separation.a) While she is moving, the boat and Romeo are moving at speedv ′in the opposite direction.What is the ratiov /v ′?b) What isvrel=dD/dt, their relative speed of approach in terms of .vc) How far has the boat moved away from the dock when she reaches him?d) Make sure that your answer makes sense by thinking about the following “BB” limits:m b≫m r= m j ;m b≪ m r= m j .
Week 4: Systems of Particles, Momentum and Collisions229Problem 10.MmHThis problem will help you learn required concepts such as:•Newton’s Second Law•Gravitation•Newton’s Third Law•Impulse and Average Force•Fully Elastic Collisionsso please review them before you begin.In the figure above, a feeder device provides a steady stream of beads of massmthat fall adistanceHand bounce elastically off of one of the hard metal pans of a beam balance scale (andthen fall somewhere else into a hopper and disappear from our problem).Nbeads per second comeout of the feeder to bounce off of the pan. Our goal is to derive an expression forM, the mass weshould put on the other pan to balance theaverage forceexerted by this stream of beads106a) First, the easy part. The beads come off of the feeder with an initial velocity of~v= v 0 xˆxinthe -direction only. Find the -component of the velocityxyv ywhen a single bead hits the panafter falling a heightH .b) Since the beads bounce elastically, the -component of their velocity is unchanged and thexy-componentreverses. Find the change of themomentumof this bead ∆ during its collision.~ pc) Compute theaverage forcebeing exerted on the stream of beads by the pan over a second(assuming thatN≫1, so that many beads strike the pan per second).d) Use Newton’s Third Law to deduce the average force exerted by the beads on the pan, andfrom this determine the massMthat would produce the same force on the other pan to keepthe scale in balance.106This is very similar (conceptually) to the way a gas microscopically exerts a force on a surface that confines it;we will later use this idea to understand the pressure exerted by a fluid and to derive the kinetic theory of gases andthe ideal gas lawPV=NkT, which is why I assign it in particular now.
230Week 4: Systems of Particles, Momentum and CollisionsProblem 11.RvmMThis problem will help you learn required concepts such as:•Conservation of Momentum•Conservation of Energy•Disposition of energy in fully inelastic collisions•Circular motion•The different kinds of constraint forces exerted by rigid rods versus strings.so please review them before you begin.A block of massMis attached to a rigid massless rod of lengthR(pivot to center-of-mass ofthe block/bullet distance at collision) and is suspended from a frictionless pivot. A bullet of massmtravelling at velocityvstrikes it as shown and is quickly stopped by friction in the hole so thatthe two masses move together as one thereafter. Find:a) The minimum speedv rthat the bullet must have in order to swing through a complete circleafter the collision. Note well that the pendulum is attached to arod!b) The energy lost in the collision when the bullet is incident at this speed.
Week 4: Systems of Particles, Momentum and Collisions231Advanced Problem 12.0L+x2M2Lλ(x) = xThis problem will help you learn required concepts such as:•Center of Mass of Continuous Mass Distributions•Integrating Over Distributionsso please review them before you begin.In the figure above a rod of total massMand lengthLis portrayed that has been machined sothat it has a mass per unit length that increaseslinearlyalong the length of the rod:λ x( ) =2ML 2xThis might be viewed as a very crude model for the way mass is distributed in something like abaseball bat or tennis racket, with most of the mass near one end of a long object and very littlenear the other (and a continuum in between).Treat the rod as if it is really one dimensional (we know that the center of mass will be in thecenter of the rod as far asyorzare concerned, but the rod is so thin that we can imagine thaty≈ z≈0) and:a) verify that the total mass of the rod is indeedMfor this mass distribution;b) findxcm, the -coordinate of the center of mass of the rod.x
232Week 4: Systems of Particles, Momentum and CollisionsOptional ProblemsThe following problems arenot required or to be handed in, but are provided to give yousome extra things to work on or test yourself withaftermastering the required problems and conceptsabove and to prepare for quizzes and exams.Optional Problem 13.MmDµfvbvvik(block at rest)This problem will help you learn required concepts such as:•Momentum Conservation•The Impact Approximation•Elastic versus Inelastic Collisions•The Non-conservative Work-Mechanical Energy Theoremso please review them before you begin.In the figure above a bullet of massmis travelling at initial speedv ito the right when it strikesa larger block of massMthat is resting on a rough horizontal table (with coefficient of frictionbetween block and table ofµ k). Instead of “sticking” in the block, the bullet blasts its waythroughthe block (without changing the mass of the block significantly in the process). It emerges with thesmaller speedv f, still to the right.a) Find the speed of the blockv bimmediately afterthe collision (but before the block has hadtime to slide any significant distance on the rough surface).b) Find the (kinetic) energy lost during this collision. Where did this energy go?c) How far down the rough surfaceDdoes the block slide before coming to rest?
Week 5: Torque and Rotation in One Dimension233Optional Problem 14.vmmvvotbθ uθ bThis problem will help you learn required concepts such as:•Vector Momentum Conservation•Fully Elastic Collisions in Two Dimensionsso please review them before you begin.In the figure above, two identical billiard balls of massmare sitting in a zero gravity vacuum (sothat we can neglect drag forces and gravity). The one on the left is given a push in the -directionxso that it elastically strikes the one on the right (which is at rest) off center at speedv 0. The topball recoils along the direction shown at a speedv tand angleθ trelative to the direction of incidenceof the bottom ball, which is deflected so that it comes out of the collision at speedv bat angleθ brelative to this direction.a) Use conservation of momentum to show that in this special case that the two masses are equal:~v0= ~vu+ ~vband draw this out as a triangle.b) Use the fact that the collision waselasticto show thatv 20= v 2u+ v 2b(where these speeds are the lengths of the vectors in the triangle you just drew).c) Identify this equation and triangle with thepythagorean theoremproving that in this case~vt⊥ ~vb(so thatθ u= θ b+π/2Using these results, one can actually solve for~vuand~vbgiven onlyv 0and either ofθ uorθ b .Reasoning very similar to this is used to analyze the results of e.g. nuclear scattering experimentsat various laboratories around the world (including Duke)!
234Week 5: Torque and Rotation in One Dimension
Week 5: Torque and Rotation inOne DimensionSummary•Rotations in One Dimensionare rotations of a solid object about asingleaxis. Since weare free to choose any arbitrary coordinate system we wish in a problem, we can without lossof generality select a coordinate system where the -axis represents the (positive or negative)zdirection or rotation, so that the rotating object rotates “in” thexyplane. Rotations of arigid body in thexyplane can then be described by asingle angleθ, measured by conventionin the counterclockwise direction from the positive -axis.x•Time-dependent Rotationscan thus be described by:a) Theangular positionas a function of time, ( ).θ tb) Theangular velocityas a function of time,ω t( ) =dθdtc) Theangular accelerationas a function of time,α t( ) =dωdt=d θ 2dt2Hopefully the analogy between these “one dimensional” angular coordinates and their onedimensional linear motion counterparts is obvious.•Forces applied to a rigid object perpendicular to a line drawn from anaxis of rotationexertatorqueon the object. The torque is given by:τ=rFsin( ) =φrF⊥=r F⊥•The torque (as we shall see) is avectorquantity and by convention its direction isperpendicularto the plane containing~rand~Fin the direction given by theright hand rule. Although wenition of the torque is:fiwon’t really work with this until next week, the “proper” de~τ= ~r× ~F•Newton’s Second Law for Rotationin one dimension is:τ=IαwhereIis themoment of inertiaof the rigid body being rotated by the torqe about aedfigiven/speciaxis of rotation. The direction of this (one dimensional) rotation is the right-handed direction of the axis – the direction your right handed thumb points if you grasp thengers curling around the axis in the direction of the rotation or torque.fiaxis with your 235
236Week 5: Torque and Rotation in One Dimension•Themoment of inertia of a point particleof massmlocated a (fixed) distancerfromsome axis of rotation is:I=mr2•The moment of inertia of a rigid collection of point particles is:I=Xim ri i2•the moment of inertia of a continuous solid rigid object is:I= Zr dm 2•The rotational kinetic energy of a rigid body (total kinetic energy of all of the chunks of massthat make it up) is:K rot= 12Iω2•The work done by a torque as it rotates a rigid body through some angledθis:dW=τ dθHence the work-kinetic energy theorem becomes:W = Zτ dθ= ∆K rot•Consequently rotational work, rotational potential energy, and rotational kinetic energy callall be simply added in the appropriate places to our theory of work and energy. The totalmechanical energy includesboththe total translational kinetic energy of the rigid body treatedas if it is a total mass located at its center of massplusthe kinetic energy of rotationaroundits center of mass:K tot= K cm+ K rotThis is a special case of the last theorem we proved last week.•If we know the moment of inertiaIcmof a rigid body about a given axis through its center ofmass, theParallel Axis Theorempermits us to find the moment of inertia of a rigid bodyof massmaround anewaxis parallel to this axis and displaced from it by a distancercm:Inew= Icm+mr2cm•For a distribution of mass with planar symmetry (mirror symmetry about the plane of rotationor distribution only in the plane of rotation), if we letzpoint in the direction of an axis ofrotation perpendicular to this plane andxandybe perpendicular axes in the plane of rotation,then thePerpendicular Axis Theoremstates that:I z= I x+ I y5.1: Rotational Coordinates in One DimensionIn the last week/chapter, you learned how a collection of particles can behave like a “particle” of thesame total mass located at the center of mass as far as Newton’s Second Law is concerned. We alsosaw at least four examples of how problems involving systems of particles can be decomposed intotwo separate problems – one the motionofthe center of mass, which generally obeys Newtonian
Week 5: Torque and Rotation in One Dimension237dynamics as if the whole system is “a particle”, and the other the motioninthe center of masssystem107.This decomposition is useful (as we saw) even if the system has many particles in it and is fluidor non-interacting, but it isveryuseful in helping us to describe themotion of rigid bodies. Thisis because the most general motion of a rigid object is thetranslationof (the center of mass of)the object according to thetotalforce acting on it and Newton’s Second Law (as demonstrated lastweek), plus therotationof that body about its center of mass as unbalanced forces exert atorqueon the object.The first part we are very very familiar with at this point and we’ll take it for granted thatyou can solve for the motion of the center of mass of a rigid object given any reasonable net force.The second we are not familiar with at all, and we will now take the next two weeks to study it indetail and understand it, as rotation isjust as important and commonas translation when itcomes to understanding the motion of nearly everything we see on a daily basis. Doors rotate abouthinges, tires rotate about axles, electrons and protons “just rotate” because of their intrinsic spin,our fingers and toes and head and arms and legs rotate about their joints, our whole bodies rotateabout their center of mass when we get up in the morning, when we do a twirl on ice skates, whenwe summersault on a trampoline, the entire Earth rotates around its axis while revolving around thesun which rotates onitsaxis while revolving around the Galactic center which... just goes to showthat rotation really is ubiquitous, and pretending that it isn’t important or worthy of understandingis not an option, even for future physicians or non-rocket-scientist bio majors.It willtaketwo weeks (and maybe even longer, for some of you) because rotation is a wee bitcomplicated. For many of you, it will be the most difficult single topic we cover this semester, if onlybecause rotation is best described by means of the Evil Cross Product108. Just as we started ourstudy of coordinate motion with motion in only one dimension, so we will start our study or rotationwith “one dimensional rotation” of a rigid body, that is, the rotation of a rigid object through anangleθabout a single fixed axis109.Eventually we want to be able to treat arbitrary rigid objects, ones that have their mass sym-metrically but non-uniformly distributed (e.g. basketballs or ninja stars) or non-uniformly and notparticularly symmetrically distributed (e.g. the human body, automobiles, blobs of putty of arbi-trary shape). But at the moment even the rotation of a basketball on the tip of a player’s fingerseems like too much for us to handleWe therefore start with the simplest possible example – a “rigid” system with all of its massconcentrated in a single point that rotates around some fixed axis. Consider a small “pointlike” ballof massmon arigidmassless unstretchable rod, portrayed in figure 59. The rod itself is pivoted ona frictionless axle in the center so that the mass is constrained to moveonlyon the dashed circle inthe plane of the picture. The mass therefore maintains a constantdistancefrom the pivot –ris aconstant – but theangleθcan vary in time as external forces act on the system.The very first things we need to do are to bring to mind the set ofrotational coordinatesthat107In particular, we solved elastic collisions in the center of mass frame (where they were easy) while the center ofmass of the colliding system obeyed (trivial) Newtonian dynamics, we looked at the exploding rocket where the centerof mass followed the parabolic/Newtonian trajectory, we saw that inelastic collisions turn all of the kinetic energyinthe center of mass frame into heat, and we proved that in general the kinetic energy of a system in the lab is the sumof the kinetic energyofthe system (treated as a particle moving at speedvcm) plus the kinetic energy of all of theparticlesinthe center of mass frame – this latter being the energy lost in a completely inelastic collision or conservedin an elastic one!108Wikipedia: http://www.wikipedia.org/wiki/Cross Product. Something that is covered both in this Wikipediaarticleandin the online Math Review supplement, so now is a really, really great time to pause in reading thischapter and skip off to refresh your memory of it. Itisa memory, we hope, isn’t it? If not, then by all means skip offtolearnit...109The “direction” of a rotation is considered to be along the axis of its rotation in a right handed sense describedlater below. So a “one dimensional rotation” is the rotation of any object about a single axis – it does not imply thattheobject being rotatedis in any sense one dimensional.
238Week 5: Torque and Rotation in One DimensionrrvθsxyFigure 59: A small ball of massmrotates about a frictionless pivot, moving in a circle of radius .rwe have already introduced for doing kinematics of a rotating object. Sinceris fixed, the positionof the particle is uniquely determined by the positive angle ( ), measured by convention as aθ tcounterclockwise rotation about the -axis from the + -axis as drawn in figure 59. We callzxθtheangular positionof the particle.We can easily relaterandθto the real position of the particle. The distance the particle mustmove in the counterclockwise direction from the standard reference position at ( =xr, y= 0) aroundthe circular arc to an arbitrary position on the circle is =srθ s .(the arc length) is aone dimensionalcoordinatethat describes its motion on the arc of the circle itself, and if we knowrands(the lattermeasured from the + -axis) we know exactly where the particle is in the - plane.xx yWe recall that thetangential velocityof the particle on this circle is thenv t=dsdt=d rθ( )dt= rdθdt=rω(461)where we remind you of theangular velocityω=dθdt. Note that for a rigid bodyv r=drdt= 0, thatis, the particle isconstrainedby the rigid rod or solidity of the body to move in circles of constantradiusrabout the center of rotation or pivot so its speed moving towards or away from the circle iszero.Similarly, we can differentiate one more time to find thetangential acceleration:a t=dvtdt= rdωdt= rd θ 2dt2=rα(462)whereα=dωdt=d θ 2dt2is theangular acclerationof the particle.Although themagnitudeofv r= 0, we note well that thedirectionof~vtis constantly changingand we know thata r= −v /r 2= − rω2which we derived in the first couple of weeks and by nowhave used repeatedly to solve many problems.All of this can reasonably be put in a small table that lets us compare and contrast the onedimensional arc coordinates with the associated angular coordinates:5.2: Newton’s Second Law for 1D RotationsWith these coordinates in hand, we can now consider theangular versionof Newton’s Second Lawfor a force~Fapplied to this particle as portrayed in figure 60. This is an example of a “rigid” body
Week 5: Torque and Rotation in One Dimension239AngularArc Lengthθs=rθω=dθdtv t=dsdt=rωα=dωdta t=dvtdt=rαTable 2: Coordinates used for angular/rotational kinemetics in one dimension. Note thatθis therotation anglearounda given fixed axis, in our picture above the -axis, and thatzθmust be givenin (dimensionless)radiansfor these relations to be true. RememberC= 2πris the formula for thecircumference of a circle and is aspecial caseof the general relations=rθ, but only whenθ= 2πradians.yFFr FtxφΤrFigure 60: A force~Fis applied at some angleφ(relative to ) to the ball on the pivoted massless~rrod.rotation, but because we aren’t yet ready to tackle extended objects all of themassis concentratedin the ball at radius . We’ll handle true, extended rigid objects shortly, once we understand a fewrbasic things well.Since the rod is rigid, and pivoted by an unmovable frictionless axle of some sort in the center,the tension in the rodopposes any motion along r. If the particle is moving around the circleat some speedv t(not shown), we expect that:F r− T= Fcos( )φ− T= −mar= − m v 2tr= −mrω2(463)(whereris anoutwarddirected radius, note that the acceleration isintowards the center) as usual.The rotational motion is what we are really interested in. Newton’s Law tangent to the circle isjust:F t= Fsin( ) =φma t=mrα(464)For reasons that will become clear in a bit, we will find it very useful to multiply this whole equationbyrand redefinerFtto be a new quantity called thetorque, given the symbolτ. We will alsocollect the factors ofrand multiply them by themto make a new quantity called themoment ofintertiaand give it the symbol :Iτ=rFt=rFsin( ) =φmr α2=Iα(465)
240Week 5: Torque and Rotation in One DimensionIn particular, this equation contains themoment of inertia of a point massmmoving in acircle of radiusr :Ipoint mass=mr2(466)This looks like, and of course is,Newton’s Second Law for a rigid rotating systeminone dimension, where force is replaced by torque, mass is replaced by moment of inertia, and linearacceleration is replaced by angular acceleration.Although to usso farthis looks just like a trivial algebraic rewrite of something we could haveworked with just as easily as the real thing in thescoordinates, it is actually far more general andpowerful. To completely understand this, we need to understand two things. One of them is howapplying the force~Fto (for example) the rod atdifferent radiir Fchanges the angular accelerationα. The other is how a force~Fapplied at some radiusr Fto the massless rodinternally redistributesto muliple masses attached to the rod at different radii so that all the masses experience thesameangular acceleration. These are the subjects of the next two sections.5.2.1: The -dependence of TorquerLet’s see how the angular acceleration of this mass will scale with the point of application of theforce along the rod, and in the process justify our “inspired decision” to multiplyF tbyrin ourdefinition of the torque in the previous section. To accomplish this we need anewfigure, one wherethe massless rigid rod extends out past/through the massmso it can act as alever armon the massno matter where we choose to apply the force~F .yxFd θφF = Fsin( )tFrrodd = r d lpivotFpmm rrφFFθFigure 61: The force~Fis applied to the pivoted rod at an angleφat the point~r Fwith the massmattached to the rod at radiusrm .This is displayed in figure 61. A massless rod as long or longer than bothrmandr Fis pivotedat one end so it can swing freely (no friction). The massmis attached to the rod at the positionrm. A force~Fis applied to the rod at the position~r F(on the rod) and at an angleφwith respectto the direction of~r F .Turning this into a suitable angular equation of motion is a bit of a puzzle. The force~Fis notapplied directly to themass– it is applied to themassless rigid rodwhich in turntransmitssomeof the force to the mass. However, the external force~Fis not the only force acting on the rod!In the previous example the pivot only exerted aradialforce~F p= − T, and exerted no tangentialforce onmat all. We could even computeT(and hence~F p) if we knew ,θ vtand~Ffrom rotationalkinematics and some vector geometry. In this case, however, if~Fexerts a force on the rod that can
Week 5: Torque and Rotation in One Dimension241be transmitted to and act tangentially upon the massm, it rather seems that the unknown pivotforce~F pcan as well, butwe don’t know~F p !Alas, without knowingallof the forces that act tangentially onm, we cannot use Newton’sSecond Lawdirectly. This motivates us to consider usingwork and energyto obtain a dynamicalprinciple (basically working the derivation of the WKE theorem backwards) becausethe pivot doesnot moveand thereforethe force~F pdoes no work!Consequently, the fact that we do notyetknow~F pwill not matter!So to work. Let us suppose that the force~Fis applied to the rod for a timedt, and that duringthat time the rod rotates through an angledθas shown. In this case we can easily find theworkdone by the force~F. The point on the rod where the force~Fis applied moves a distancedℓ=r dθF.The work is done only by the tangentialcomponentof the force moved through this distanceF tsothat:dW= ~F ·d ~ℓ=F r dθt F(467)The WKE theorem tells us that this work must equal the change in the kinetic energy over thattime:F r dθt F=dK= d12mv2tdtdt=mv dvtt(468)We make a few useful substitutions from table 3 above:F r dθt F=m rmdθdt(r α dtm) =mr αdθ2m(469)and canceldθ(and reorder a bit) to get:τ=r FFt=r FFsin( ) =φmr α2m=Iα(470)This formally proves that my “guess” ofτ=Iαas being the correct form of Newton’s SecondLaw for this simple rotating rigid body holds up pretty well no matter where we apply the force(s)that make up the torque, as long as we define the torque:τ=r FFt=r FFsin( )φ(471)It is left as an exercise for the student to draw a picture like the one above but involvingmanyindependent and arbitraryforces,~F 1acting at~r 1 ,~F 2acting at~r 2, ..., you get:τtot=Xir Fi isin( ) =φ imr α2m=Iα(472)for a single point-like mass on the rod at position~r m. Note well that eachφ iis theanglebetween~r iand~F i, and you should make the (massless, after all) rod long enough for all of the forces to beable to act on it and also pass throughm .In a bit we will pay attention to the fact thatrFsin( ) is the magnitude of theφcross product110of~r Fand~F, and that if we assign thedirectionof the rotation to be parallel to the -axis of azright handed coordinate system whenφis drawn in the sense shown, we can even make this avectorrelation. For the moment, though, we will stick with our simple 1D “scalar” formulation and ask adifferentquestion: what if we have amore complicatedobject than a single mass on a pivoted rigidrod that is being driven by a torque (or sum of torques).The answer is: We have to sum up the object’s totalmoment of inertiaaround the pivot axis.Let’s prove this.110Wikipedia: http://www.wikipedia.org/wiki/Cross Product. Making this a gangbusters good time to go review– or learn – cross products, at least well enough to be able evaluate theirmagnitudeanddirection(using the righthand rule).
242Week 5: Torque and Rotation in One Dimension5.2.2: Summing the Moment of InertiaSuppose we have a massm 1attached to our massless rod pivoted at the origin at the positionr 1 ,and a second massm 2attached at positionr 2. We will then apply the force~Fat an angleφtothe (extended) rod at positionras shown in figure 62, and duplicate our reasoning from the lastchapter (because westilldo not known the unknown force exerted by the pivot, but as long as weconsider work we don’t have to.xFr21 rmm12yrFφFigure 62: A single torqueτ=r FFsin( ) is applied to a rod withφtwomasses,m 1atr 1andm 2atr 2 .The WKE theorem for this picture is now (note thatv 1andv 2are both necessarily tangential):dW=r FFsin( )φ dθ=τ dθ=dK=m v dv1 11+m v dv2 22so as usualτ dθ=m 1r 1dθdt(r α dt1) +m 2r 2dθdt(r α dt2)τ=m r α1 12+m r α2 22τ=m r1 12+m r2 22α=Iα(473)where we have nowdefinedI=m r1 12+m r2 22(474)That is, the total moment of inertia of thetwopoint masses is just the sum of their individualmoments of inertia. From the derivation it should be clear that if we added 3,4,...,N point massesalong the massless rod the total moment of inertia would just be the sum of their individual momentsof inertia.Indeed, as we add more forces acting at different points and directions (in the plane) on therod and add more masses at different points on the rod,everythingwe did above clearlyscales uplinearly– we simply have tosumthe total torque on the right hand side andsumthe total momentof inertia on the left hand side. We therefore conclude that Newton’s Second Law for a systemconstrained to rotate in (one dimension in a) a plane about a fixed pivot is just:τtot=Xir Fi isin( ) =φ iXjm r αj j2= Itotα(475)So much for discrete forces and discrete masses. However, most rigid bodies that we experienceevery day are, on a coarse-grained macroscopic scale, made up of acontinuousdistribution of mass,and instead of a mythical idealized “massless rigid rod” all of this mass is glued together by meansofinternal forces.It is pretty clear that our expressionτ=Iαwill generalize to this case where we will (probably)replace:I=Xjm rj j2→ Zr dm 2(476)but we will need to do just a teensy bit of work to show that this is true and extract any essentialconceptual insight to be found along the way.
Week 5: Torque and Rotation in One Dimension2435.3: The Moment of InertiaWe begin with a specific example to help smooth the way.Example 5.3.1: The Moment of Inertia of a Rod Pivoted at One EndxxypivotdxLdm = dxλFigure 63: A solid rod of lengthLwith a massMuniformly pivoted about one end. One can thinkof such a rod as being the “massless rod” of the previous section with aninfinite numberof massesm iuniformly distributed along its length, that sum to the total massM .In figure 63 above amassiverod pivoted about one end is drawn. We would like to determinehow this particular rod will rotationally accelerate when we (for example) attach a force to it andapply a torque. We therefore must characterize this rod as having a specific massM, a specificlengthL, and we need to say something about thewaythe mass is distributed, because the rodcould be made of aluminum (not very dense) at one end and tungsten (very dense indeed) at theother and still “look” the same. We will assume thatthisrod is uniformly distributed, and that it isvery thin and symmetrical in cross-section – shaped like a piece of wire or perhaps a wooden dowelrod.In a process that should be familiar to you from last week and from the previous section, weknow that the moment of inertia of a sum ofdiscretepoint masses hung on a “massless” rod (thatonly serves to assemble them into a rigid structure) is just:Itot=Xim ri i2(477)the sum of the moments of inertia of the point masses.We can clearlyapproximatethe moment of inertia of the continuous rod by dividing it up intoNpieces, each of length ∆ =xL/Nand mass ∆M =M/N, and treating each small piece as a “pointmass” located atx i= i∗∆ :xIrod≈NXi=1MNi L∗N=NXi=1∆Mx2i(478)As before, the limit of this sum asN→ ∞isby definitionthe integral, and in this limit the sumexactlyrepresents the moment of inertia of the rod.We can easily evaluate this. To do so, we chant our ritual expression: “The mass of the chunkis the mass per unitlengthof the chunk times the length of the chunk”, ordm=λdx=MLdx, so:Irod= ZL0x dm 2=MLZL0x dx 2=ML 23(479)5.3.1: Moment of Inertia of a General Rigid BodyThis specific result can easily be generalized. If we consider a blob-shaped distribution of mass, thedifferentialmoment of inertia of a tiny chunk of the mass in the distribution about some fixed axis
244Week 5: Torque and Rotation in One Dimensionrdmpivot axisFigure 64: A “blob-shaped” chunk of mass, perhaps a piece of modelling clay, constrained to rotateabout an axis through the blob, perhaps a straight piece of nearly massless coat-hanger wire.of rotation is clearly:dI=r dm 2(480)By now you should be getting the idea that summing upall of the little chunksthat make up theobject is just integrating:Iblob= Zblobr dm 2(481)where it is quite one thing to write down this formal expression, quite another to be able to actuallydothe integral over all of the chunks of mass that make up an object.It isn’t too difficult to do this integral for certainsimpledistributions of mass, and we will need acertain “stock repertoire” of moments of inertia in order to solve problems. Which ones you shouldlearn to do depends on the level of the course – math/physics majors should learn to integrateover spheres (and maybe engineers as well), but everybody else can probably get away learning toevaluate the moment of inertia of a disk. In practice, for any reallycomplicatedmass distribution(like the blob of clay pictured above) one would eithermeasurethe moment of inertia or use acomputer to actually break the mass up into a very large number of discrete (but small/point-like)chunks and do thesum.First let’s do an example that is even simpler than the rod.Example 5.3.2: Moment of Inertia of a RingRMd θzds = RdθFigure 65: A ring of massMand radiusRin the - plane rotates freely about the -axis.x yzWe would like to find the moment of inertia of the ring of uniformly distributed massMandradiusRportrayed in figure 65 above. A differential chunk of the ring has lengthds=R dθ. It’smass is thus (say the ritual words!):dm=λds=M2πRR dθ=M2 πdθ(482)
Week 5: Torque and Rotation in One Dimension245and its moment of inertia is very simple:Iring= Zr dm 2= Z2 π0M2 πR dθ 2=MR 2(483)In fact, we could haveguessedthis.Allof the massMin the ring is at the same distanceRfromthe axis of rotation, so its moment of inertia (which only depends on the mass times the distanceand has no “vector” character) is justMR 2just like a point mass at that distance.Because it is so important, we will do the moment of inertia of a disk next. The disk will be manythings to us – a massive pulley, a wheel or tire, a yo-yo, a weight on a grandfather clock (physical)pendulum. Here it is.Example 5.3.3: Moment of Inertia of a DiskdRMdr rddA = rd drθθθrFigure 66: A disk of massMand radiusRis pivoted to spin freely around an axis through its center.In figure 66 a disk of uniformly distributed massMand radiusRis drawn. We would like tofind its moment of inertia. Consider the small chunk of disk that is shaded of areadA. In planepolar coordinates (the only ones we could sanely hope to integrate over) the differential area of thischunk is just its differential heightdrtimes the width of the arc at radiusrsubtended by the angledθ r dθ,. The area is thusdA=r drdθ.This little chunk was selected because the massdmin it moves in a circle of radiusraround thepivot axis. We need to finddmin units we can integrate to cover the disk. We use our litany to set:dm=σdA=MπR2r drdθ(484)and then write down:dI=r dm 2=MπR2r drdθ 3(485)We integrate both sides to get (note that the integrals are independent one dimensional integralsthat precisely cover the disk):Idisk=MπR2ZR0r dr 3!Z2 π0dθ=MπR2R 44(2 )π=12MR 2(486)This is a very important and useful result, so keep it in mind.
246Week 5: Torque and Rotation in One Dimension5.3.2: Table of Useful Moments of InertiaFinally, here is a table of afewuseful moments of inertia of simple uniform objects. In each caseI indicate the value about an axis through the symmetriccenter of massof the object, because wecan use theparallel axis theoremand theperpendicular axis theoremto find the moments ofinertia around at least some alternative axes.ShapeIcmRod from−L/2 toL/2112ML 2RingMR 2Disk12MR 2Sphere25MR 2Spherical Shell23MR 2Generic “Round” Mass of MassMand radiusRβMR2Table 3: A few useful moments of inertia of symmetric objects around an axis of symmetry throughtheir center of mass. You should probably know all of the moments in this table and should be ableto evaluate the first three by direct integration.5.4: Torque as a Cross ProductThis section will be rather abbreviatedthisweek;nextweek we will cover it in gory detail as avectorrelation. For the moment, however, we need to make a number of observations that will helpus solve problems. First, we know that the one-dimensional torque produced by any single forceacting on a rigid object a distancerfrom a pivot axis is just:τ=rF⊥=rFsin( )φ(487)whereF ⊥is just the component of the forceperpendicularto the (shortest) vector~rfrom the pivotaxis to the point of application. This is really just onecomponentof the total torque, mind you,but it is the one we have learned so far and are covering this week.First, let’s make an important observation. Provided that~rand~Flie in a plane (so that theone dimension is the right dimension) the magnitude of the torque is the magnitude of thecrossproduct of~rand~F :τ=| × ~r~F |=rFsin( ) =φrF⊥=r F⊥(488)I’ve used the fact that I can move the sin( ) around to write this in terms of:φr⊥= sin( )rφ(489)which is thecomponent of~rperpendicular toF, also known as themoment arm of thetorque. This is a very useful form of the torque in many problems. It it equally well expressible interms of the familiar:F ⊥= Fsin( )φ ,(490)
Week 5: Torque and Rotation in One Dimension247the component of~Fperpendicular to . This form, too, is often useful. In fact, both forms may be~ruseful (to evaluate different parts of the total torque) in a single problem!If we let thevector torquebe defined by:~τ= ~r× ~F(491)marvelous things will happen. Next week we will learn about them, and will learn about how toevaluate this a variety of ways. For now let’s just learn one.The vector torque~τhas amagnitude| × ~r~F |=rFsin( ) and points in theφdirectiongiven by theright hand rule.The right hand rule, in turn, is the following:The direction of the vector cross product~A × ~Bis in the direction the thumb of yourright handpoints when you begin with the fingers of your right hand lined up with thevector~Aand then curl them naturally through the angleφ < πinto the direction of~B .That is, if you imagine “grasping” the axis in the direction of the torque with your right hand, yourfingers will curl around in the directionfrom~rto~Fthrough the smaller of the two angles in betweenthem (the one less than ).πYou will get lots of practice with this rule, but be sure to practice with yourrightright hand, notyourwrongright hand. Countless students (and physics professors and TAs!) have been embarrassedbe being caught out evaluating the direction of cross products with their left hand111. Don’t be oneof them!The direction of the torquematters, even in one dimension. There is no better problem todemonstrate this than the following one, determine whatdirectiona spool of rope resting on a tablewill roll when one pulls on the rope.Example 5.4.1: Rolling the SpoolI’m not going toquitefinish this one for you, as there are a lot of things one can ask and it is ahomework problem. But I do want you to get a good start.The spool in figure 67 is wrapped many times around with string. It is sitting on a level, roughtable so that for weak forces~Fit will freely roll without slipping (although for a large enough~Fofcourse it will slip or even rise up off of the table altogether).The question is, whichdirectionwill it roll (or will it not roll until it slides) for each of the threedirections in which the string is pulled.The answer to this question depends on thedirection of the total torque, and the relevant pivotis the point thatdoes not movewhen it rolls, where the (unknown!) force of static friction acts.If we choose the pivot to be the point where the spool touches the table, then gravity, the normalforce and static frictionall exert no torque!The only source of torque is~F .So, what is thedirectionof the torque for each of the three forces drawn, and will a torque inthat direction make the mass roll to theleft, theright, orslide(or not move)?Think about it.111Let he or she who is without sin cast the first stone, I always say. As long as it is cast with therighthand...
248Week 5: Torque and Rotation in One DimensionFFFpivotFigure 67: Whatdirectiondoes one expect the spool in each of the figures to roll (or will it roll atall)?5.5: Torque and the Center of GravityWe will often wish to solve problems involving (for example) rods pivoted at one end swinging downunder the influence of near-Earth gravity, or a need to understand the trajectory and motion ofa spinning basketball. To do this, we need the idea of thecenter of gravityof a solid object.Fortunately, this idea is very simple:Thecenter of gravityof a solid object in an (approximately) uniform near-Earthgravitational field is located at the center of mass of the object. For the purpose ofevaluating the torque and angular motion or force and coordinate motion of center ofmass, we can consider that the entire force of gravity acting on the object is equivalentto the the force that would be exerted by the entire mass located as a point mass at thecenter of gravity.The proof for this is very simple. We’ve already done the Newtonian part of it – we know thatthe total force of gravity acting on an object makes the center of mass move like a particle with thesame mass located there. For torque, we recall that:τ=rF⊥=r F⊥(492)If we consider the torque acting on a small chunk of mass in near-Earth gravity, the force (down)acting on that chunk is:dFy= −gdm(493)The torque (relative to the pivot) is just:dτ= −gr dm⊥(494)orτ= − gZL0r dm⊥= −Mgrcm(495)wherercmis (the component of) the position of the center of mass of the object perpendicular togravity. The torque due to gravity acting on the object around the selected pivot axis is the same
Week 5: Torque and Rotation in One Dimension249as the torque that would be produced by the entire weight of the object pulling down at the centerof mass/gravity. Q.E.D.Note that this proof is valid foranyshape or distribution of mass in auniformgravitational field,but in a non-uniform field it would fail and the center of gravity would bedifferentfrom the centerof mass as far as computing torque is concerned. This is the realm oftidesand will be discussedmore in the week where we cover gravity.Example 5.5.1: The Angular Acceleration of a Hanging RodθθL/2LMgMFigure 68: A rod of massMand lengthLis suspended/pivoted from one end. It is pulled out tosome initial angleθ 0and released.This is your first example of what we will later learn to call aphysical pendulum. A rod issuspended from a pivot at one end in near-Earth gravity. We wish to find the angular accelerationαas a function of . This is basically the equation of motion for thisθrotationalsystem – later wewill learn how to (approximately) solve it.As shown above, for the purpose of evaluating the torque, the force due to gravity can beconsidered to beMgstraight down, acting at the center of mass/gravity of the rod (atL/2 in themiddle). The torque exerted at this arbitrary angleθ(positive as drawn, note that it is swung outin the counterclockwise/right-handed direction from the dashed line) is therefore:τ=rFt= −MgL2sin( )θ(496)It isnegativebecause it acts to makeθsmaller; it exerts a “twist” that is clockwise whenθiscounterclockwise and vice versa.From above, we know thatI=ML /23 for a rod pivoted about one end, therefore:τ=−MgL2sin( )θ=ML 23α=Iαor:α=d θ 2dt2=− 3 g2 Lsin( )θ(497)independent of the mass!5.6: Solving Newton’s Second Law Problems Involving RollingOne of the most common applications of one dimensional torque and angular momentum is solvingrolling problems. Rolling problems include things like:
250Week 5: Torque and Rotation in One Dimension•A disk rolling down an inclined plane.•An Atwood’s Machine, but with amassivepulley.•An unwinding spool of line, either falling or being pulled.These problems are all solved by using acombinationof Newton’s Second Law for the motion ofthe center of mass of the rolling object (if appropriate) or other masses involved (in e.g. Atwood’sMachine)andNewton’s Second Law for 1 dimensional rotation,τ=Iα. In general, they will alsoinvolve using therolling constraint:If a round object of radiusrisrolling without slipping, the distancexit travels relativeto the surface it is rolling on equalsrθ, whereθis the angle it rolls through.That is, all three are equivalently the “rolling constraint” for a ball of radiusrrolling on a levelfloor, started from a position atx= 0 where alsoθ= 0:x=rθ(498)v=rω(499)a=rα(500)(501)These are all quite familiar results – they look a lot like our angular coordinate relations – but theyarenot the same thing!These areconstraints, not coordinate relations – for a ball skiddingalong the same floor they will be false, and for certain rolling pulley problems on your homeworkyou’ll have to figure out one appropriate for the particular radius of contact of spool-shaped or yo-yoshaped rolling objects (that may not be the radius of the object!)It is easier to demonstrate how to proceed for specific examples than it is to expound on thetheory any further. So let’s do the simplest one.Example 5.6.1: A Disk Rolling Down an InclinermgNs fmφφFigure 69: A disk of massMand radiusrsits on a plane inclined at an angleφwith respect to thehorizontal. Itrolls without slippingdown the incline.In figure 69 above, a disk of massMand radiusrsits on an inclined plane (at an angle ) asφshown. It rolls without slipping down the incline. We would like to find its accelerationa xdown theincline, because if we know that we know pretty much everything about the disk at all future timesthat it remains on the incline. We’d also like to know whatf s(the force exerted by static friction)
Week 5: Torque and Rotation in One Dimension251when it is so accelerating, so we can check to see if our assumption of rolling without slipping isjustified. Ifφis too large, we are pretty sure intuitively that the disk will slip instead of roll, sinceifφ≥π/2 we know the disk will just fall and not roll at all.As always, since we expect the disk to physically translate down the incline (so~aand~Ftotwillpoint that way) we choose a coordinate system with (say) the -axis directed down the inline andxydirected perpendicular to the incline.Since the diskrolls without slippingwe know two very important things:1) The forcef sexerted by static friction must be less than (or marginally equal to)µ Ns. If, inthe end, itisn’t, then our solution is invalid.2) If itdoesroll, then the distancexit travels down the incline is related to the angleθit rollsthrough byx=rθ. This also means thatv x=rωanda x=rα.We now proceed to write Newton’s Lawsthree times: Once for the -direction, once for theyx-direction and once forone dimensional rotation(the rolling). We start with the:F y= N −mgcos( ) =φmay= 0(502)which leads us to the familarN =mgcos( ).φNext:F x=mgsin( )φ− f s=max(503)τ=rfs=Iα(504)Pay attention here, because we’ll do the following sort of things fairly often in problems. We useI= Idisk= 12mr2andα=a /rxand divide the last equation byron both sides. This gives us:mgsin( )φ− f s=max(505)f s=12max(506)If weadd these two equations, the unknownf scancels out and we get:mgsin( ) =φ32max(507)or:a x= 23 gsin( )φ(508)We can then substitute this back into the equation forf sabove to get:f s= 12max= 13mgsin( )φ(509)In order to roll without slipping, we know thatf s≤µ Nsor13mgsin( )φ≤µ mgscos( )φ(510)orµ s≥ 13tan( )φ(511)Ifµ sis smaller than this (for any given incline angle ) then the disk willφslipas it rolls downthe incline, which is a more difficult problem.We’ll solve this problem again shortly to find out howfastit is going at the bottom of an inclineof lengthLusing energy, but in order to this we have toaddressrotational energy. First, however,we need to do a couple more examples.
252Week 5: Torque and Rotation in One Dimensionmm12m g1m g2TTrMTT1122Figure 70: Atwood’s Machine, but this time with amassivepulley of massMand radiusR. Themassless unstretchable string connecting the two massesrolls without slippingon the pulley,exerting a torque on the pulley as the masses accelerate to match. Assume that the pulley has amoment of inertiaβMr2for some . Writing it this way let’s us useββ≈ 12to approximate the pulleywith a disk, or use observations ofatomeasureβand hence tell something about the distributionof mass in the pulley!Example 5.6.2: Atwood’s Machine with a Massive PulleyOur solution strategy is almost identical to that of our first solution back in week 1 – choose an”around the corner” coordinate system where if we make movingm 2down “positive”, then movingm 1up is also “positive”. To this we add that a positiverotationof thepulleyisclockwise, and thatthe rolling constraint is thereforea=rα.Now we again write Newton’s Second Law once for each mass andonce for the rotating pulley(asτ=Iα):m g2− T 2=m a2(512)T 1−m g1=m a1(513)τ=rT2−rT1=βMr2ar=Iα(514)Divide the last equation byron both sides, then add all three equations to eliminatebothunknown tensionsT 1andT 2. You should get:(m 2−m g1) = (m 1+ m 2+βM a)(515)or:a=(m 2−m g1 )(m 1+ m 2+βM)(516)This is almost like the previous solution – and indeed, in the limitM→ 0isthe previoussolution – but the net force between the two masses now must alsopartiallyaccelerate the mass ofthe pulley. Partially because only the mass near the rim of the pulley is accelerated at the full ratea– most of the mass near the middle of the pulley has a much lower acceleration.Note also that ifM= 0,T 1= T 2! This justifies – very much after the fact – our assertion earlyon that for a massless pulley, the tension in the string is everywhere constant. Here we seewhythatis true – because in order for the tension in the string between two points to be different, there hasto be somemassin between those points for the force difference to act upon! In this problem, thatmass isthe pulley, and to keep the pulley accelerating up with the string, the string has to exert atorqueon the pulley due to the unequal forces.
Week 5: Torque and Rotation in One Dimension253One last thing to note. We are being rather cavalier about thenormalforce exerted by thepulley on the string – all we can easily tell is that the total force acting up on the string must equalT 1+ T 2, the force that the string pulls down on the pulley with. Similarly, since the center of massof the pulley does not move, we have something likeT p− T 1− T 2−Mg= 0. In other words, thereare other questions I could always ask about pictures like this one, and by now you should have agood idea how to answer them.5.7: Rotational Work and Energyrdm vCenter of MassωFigure 71: A blob of mass rotates about an axis through the center of mass, with an angular velocityas shown.We have already laid the groundwork for studying work and energy in rotating systems. Let usconsider the kinetic energy of an object rotating around its center of mass as portrayed in figure 71.The center of mass is at rest in this figure, so this is acenter of mass inertial coordinate system.It is easy for us to write down the kinetic energy of the little chunk of massdmdrawn into thefigure at a distancerfrom the axis of rotation. It is just:dKin CM= 12dmv2= 12dmr ω22(517)To find the total, we integrate over all of the mass of the blob:Kin CM= 12Zblobdmr ω22= 12Iω2(518)which works becauseωis the same for all chunksdmin the bloband is hence a constantthat can be taken out of the integral, leaving us with the integral for .IIf we combine this with the theorem proved at the end of the last chapter we at last can preciselydescribe the kinetic energy of a rotating baseball in rest frame of the ground:K = 12Mv2cm+ 12Iω2(519)That is, the kinetic energy in the lab is the kinetic energyofthe (mass moving as if it is all at the)center of mass plus the kinetic energyinthe center of mass frame,12Iω2. We’ll have a bit more tosay about this when we prove the parallel axis theorem later.5.7.1: Work Done on a Rigid ObjectWe have already done rotational work. Indeed webeganwith rotational work in order to obtainNewton’s Second Law for one dimensional rotations above! However, there is much to be gained byconsidering thetotalwork done by anarbitraryforce acting on an arbitrary extended rigid mass.Consider the force~Fin figure 72, where I drew a regular shape (a disk) only to make it easy to see
254Week 5: Torque and Rotation in One DimensionFRrr FFφMFigure 72: A force~Fis applied in an arbitrary direction at an arbitrary point on an arbitrary rigidobject, decomposed in a center of mass coordinate frame. A disk is portrayed only because it makesit easy to see where the center of mass is.and draw an useful center of mass frame – it could just as easy be a force applied to the blob-shapedmass above in figure 71.I have decomposed the force~Finto two components:F r=Fcos( )φ(520)F ⊥=Fsin( )φ(521)(522)Suppose that this force acts for a short timedt, beginning (for convenience) with the mass at rest.We expect that the work done will consist of two parts:dW=F drr+F ds⊥(523)whereds=r dθ. This is then:dW=W r+W θ=F drr+rF dθ⊥=F drr+τ dθ(524)We know that:dWr=F drr=dKr(525)dWθ=τ dθ=dKθ(526)and if we integrate these independently, we get:W tot=W cm+W θ= ∆K cm+ ∆K θ(527)or thework decomposes into two parts!The work done by the component of the forcethroughthe center of mass accelerates the center of mass and changes the kinetic energyofthe center ofmass of the system as if it is a particle! The work done by the component of the forceperpendicularto the line connecting the center of mass to the point where the force is applied to the rigid objectincreases therotational kinetic energy, the kinetic energyinthe center of mass frame.Hopefully this is all making a certain amount of rather amazing, terrifying,senseto you. Onereason that torque and rotational physics is so important is that we can cleanly decompose thephysics of rotating rigid objectsconsistently, everywhereinto the physics of the motion of the centerof mass androtationabout the center of mass. Note well that we have also written the WKE
Week 5: Torque and Rotation in One Dimension255theorem in rotational terms, and are now justified in usingallof the results of the work and energychapter/week in (fully or partially) rotational problems!Before we start, though, let’s think a teeny bit about the rolling constraint and work, as we willbe solving many rolling problems.5.7.2: The Rolling Constraint and WorkA car is speeding down the highway at 50 meters per second (quite fast!) being chased by the police.Its tires hum as theyrolldown the highway without sliding. Fast as it is going, there arefour pointson this car that arenot moving at all relative to the ground!Where are they?The four places where the tires are in contact with the pavement, of course. Those points aren’tslidingon the pavement, they are rolling, and “rolling” means that they are coming down at restonto the pavement and then lifting up again as the tire rolls on.If the car is travelling at a constant speed (and we neglect or arrange for their to be nodrag/friction) we expect that the road will exertnoforce along the direction of motion of thecar – the force exerted by static friction will be zero. Indeed, that’s why wheels were invented – anobject that is rolling at constant speed on frictionless bearings requires no force to keep it going –wheels are a way to avoid kinetic/sliding friction altogether!More reasonably, the force exerted by static friction willnotbe zero, though, when the carspeeds up, slows down, climbs or descends a hill, goes around a banked turn, overcomes drag forcesto maintain a constant speed.What happens to theenergyin all of these cases, when the only force exerted by the ground isstatic friction at the points where the tire touches the ground? What is the work done by the forceof static friction acting on the tires?Zero!The force of static friction doesnowork on the system.If you think about this for a moment, this result is almost certain to make your head ache. Onthe one hand, it is obvious:dWs=F dxs= 0(528)becausedx= 0 in the frame of the ground – the place where the tires touch the grounddoes notmove, so the force of static friction acts throughzero distanceanddoes no work.Um, but if static friction does no work, how does the car speed up (you might ask)? Whatelsecould be doing work on the car? Oooo, head-starting-to-huuuurrrrrt...Maybe, I dunno,the motor?In fact, the car’s engine exerts atorqueon the wheels that is opposed by thepivot forceat theroad – the point of contact of a rolling object is anatural pivotto use in a problem, because forcesexerted there, in addition to doing no work, exert no torque about that particular pivot. By fixingthat pivot point, the car’s engine creates a net torque that accelerates the wheels and, since theyare fixed at the pivot, propels the car forward. Note well that the actual source of energy, however,is theengine, not theground. This is key.In general, in work/energy problems below, we will treat the force of static friction in rollingproblems (disks, wheels, tires, pulleys) where there is rolling without slipping asdoing no workand hence acting like a normal force or other force of constraint – not exactly a ”conservative force”but one that we can ignore when considering the Generalized Non-Conservative Work – MechanicalEnergy theorem or just the plain old WKE theorem solving problems.Later, when we consider pivots in collisions, we will see that pivot forcesoftencause momentumnot to be conserved – another way of saying that they can cause energy to enter or leave a system
256Week 5: Torque and Rotation in One Dimension– but that they are generally not thesourceof the energy. Like a skilled martial artist, they do notprovide energy themselves, but they are very effective at diverting energy from one form to another.In fact,verymuch like a skilled martial artist, come to think about it.It isn’t really a metaphor...Example 5.7.1: Work and Energy in Atwood’s Machinemm12m g1m g2TTrMTT1122HFigure 73: Atwood’s Machine, but this time with amassivepulley of massMand radiusR(andmoment of inertiaI=βMR2), this time solving a “standard” conservation of mechanical energyproblem.We would like to find the speedvofm 1andm 2(and the angular speedωof massM) when massm > m21falls a heightH, beginning from rest, when the massless unstretchable string connectingthe masses rolls without slipping on the massive pulley. Wecoulddo this problem usingafromthe solution to the example above, finding the timetit takes to reachH, and backsubstituting tofind , but by now we know quite well that it is a lot easier to use energy conservation (since novnon-conservative forces act if the string does not slip) which is already time-independent.Figure??shows the geometry of the problem. Note well that massm 1will goupa distanceHat the same timem 2goesdowna distanceH .Again our solution strategy is almost identical to that of the conservation of mechanical energyproblems of two weeks ago. We simply evaluate the initial and final total mechanical energyin-cluding the kinetic energy of the pulleyandusing the rolling constraintand solve for .vWe can choose the zero of potential energy for the two mass separately, and choose to startm 2aheightHabove its final position, and we start massm 1at zero potential. The final potential energyofm 2will thus be zero and the final potential energy ofm 1will bem gH1. Also, we will need tosubstitute the rolling constraint into the expression for the rotational kinetic energy of the pulley inthe little patch of algebra below:ω= vr(529)Thus:E i=m gH2E f=m gH1+ 12m v12+ 12m v22+ 12βMR ω22orm gH2=m gH1+ 12m v12+ 12m v22+ 12βMR ω22
Week 5: Torque and Rotation in One Dimension257and now we substitute the rolling constraint:(m 2−m gH1 )=12 (m 1+m v2 )2+ 12βMR2v 2R 2(m 2−m gH1 )=12 (m 1+ m 2+βM v)2(530)to arrive atv=s 2(m 2−m gH1 )m 1+ m 2+βM(531)You can do this! It isn’t really that difficult (or thatdifferentfrom what you’ve done before).Note wellthat the pulleybehaves like an extra massβMin the system – all of this mass hasto be accelerated by the actual force difference between the two masses. Ifβ= 1 – a ring of mass– then all of the mass of the pulley ends up moving atvandallof its mass counts. However, fora disk or ball or actual pulley,β <1 becausesomeof the rotating pulley’s mass is moving moreslowly thanvand has less kinetic energy when the pulley is rolling.Also note wellthat the strings do nonetwork in the system. They are internal forces, withT 2doing negative work onm 2but equal and oppositepositivework onM, withT 1doing negative workonM, but doing equal and oppositepositivework onm 1. Ultimately, the tensions in the string serveonly totransfer energybetween the masses and the pulley so that the change in potential energy iscorrectly shared by all of the masses when the string rolls without slipping.Example 5.7.2: Unrolling SpoolMRMgTHFigure 74: A spool of fishing line is tied to a pole and released from rest to fall a heightH, unrollingas it falls.In figure 74 a spool of fishing line that has a total massMand a radiusRand is effectively adisk is tied to a pole and released from rest to fall a heightH. Let’s findeverything: the accelerationof the spool, the tensionTin the fishing line, the speed with which it reachesH .We start by writing Newton’s Second Law for both the translational and rotational motion. We’llmake down -positive. Why not! First the force:yF y=Mg− T=Ma(532)and then the torque:τ=RT=Iα=12MR 2aR(533)
258Week 5: Torque and Rotation in One DimensionWeuse the rolling constraint(as shown) to rewrite the second equation, and divide bothsides byR. Writing the first and second equation together:Mg− T=Ma(534)T=12Ma(535)we add them:Mg= 32Ma(536)and solve for :aa= 23 g(537)We back substitute to findT :T= 13Mg(538)Next, we tackle the energy conservation problem. I’ll do it really fast and easy:E i=MgH= 12Mv2+ 1212MR 2vR2= E f(539)orMgH= 34Mv2(540)andv=r 4gH3(541)Example 5.7.3: A Rolling Ball Loops-the-Loopm,rHRFigure 75: A ball of massmand radiusrrolls without slipping to loop the loop on the circular trackof radiusR .Let’s redo the “Loop-the-Loop” problem, but this time let’s consider asolid ballof massmand radiusrgoing around the track of radiusR. This is a tricky problem to dopreciselybecauseas the normal force decreases (as the ball goes around the track) atsomepoint the static frictionalforce required to “keep the ball rolling” on the track may well become greater thanµ Ns, at whichpoint the ball willslip. Slipping dissipates energy, so one would have to raise the ball slightly at the
Week 5: Torque and Rotation in One Dimension259beginning to accommodate this. Of course, raising the ball slightly at the beginning also increasesNso maybe it doesn’t ever slip. So the best we can solve for is the minimum heightHminit musthave to roll without slipping assuming that it doesn’t ever actually slip, and “reality” is probably abit higher to accommodate or prevent slipping, overcome drag forces, and so on.With that said, the problem’s solution isexactly the same as beforeexcept that in the energyconservation step one has to use:mgHmin= 2mgR+ 12mv2min+ 12Iω2min(542)plus the rolling constraintωmin= vmin/rto get:mgHmin=2mgR+ 12mv2min+ 15mv2min=2mgR+ 710mv2min(543)Combine this with the usual:mg=mv2minR(544)so that the ball “barely” loops the loop and you get:Hmin= 2 7. R(545)only a tiny bit higher than needed for a block sliding on a frictionless track.Really, not all that difficult, right? All it takes is somepractice, both redoing these examples onyour own and doing the homework and it will all make sense.5.8: The Parallel Axis TheoremAs we have seen, the moment of inertia of an object or collection of point-like objects is justI=Xim ri i2(546)wherer iis the distance between the axis of rotation and the point massm iin a rigid system, orI= Zr dm 2(547)whereris the distance from the axis of rotation to “point mass”dmin the rigid object composedof continuously distributed mass.However, in the previous section, we saw that the kinetic energy of a rigid object relative to anarbitraryorigin can be written as the sum of the kinetic energy of the object itself treated as a totalmass located at the (moving) center of mass plus the kinetic energy of the object in the movingcenter of mass reference frame.For the particular case where a rigid object rotatesuniformlyaround an axis that is parallel toan axis through the center of mass of the object, that is, in such a way that the angular velocityofthe center of mass equals the angular velocityaroundthe center of mass we can derive a theorem,called theParallel Axis Theorem, that can greatly simplify problem solving while embodying theprevious result for the kinetic energies. Let’s see how.Suppose we want to find the moment of inertia of the arbitrary “blob shaped” rigid mass dis-tribution pictured above in figure 76 about the axis labelled “New (Parallel) Axis”. This is, by
260Week 5: Torque and Rotation in One DimensionrdmCM AxisNew (Parallel) Axisr’rcmFigure 76: An arbitrary blob of total massMrotates around the axis at the origin as shown. Notewell the geometry of~rcm,~r ′, and~r= ~rcm+ ~r ′ .definition (and using the fact that~r= ~rcm+ ~r ′from the triangle of vectors shown in the figure):I=Zr dm 2=Z~rcm+ ~r ′·~rcm+ ~r ′dm=Zr 2cm+ r ′2+ 2~rcm·~r ′dm=Zr 2cmdm+ Zr ′2dm+ 2~rcm·Z~r ′dm=r 2cmZdm+ Zr ′2dm+ 2M ~rcm·1M Z~r ′dm=Mr2cm+ Icm+ 2M ~rcm·(0)(548)orI= Icm+Mr2cm(549)In case that was a little fast for you, here’s what I did. I substituted~rcm+ ~r ′for . I distributed~rout that product. I used the linearity of integration to write the integral of the sum as the sum ofthe integrals (all integrals over all of the mass of the rigid object, of course). I noted that~rcmisaconstantand pulled it out of the integral, leaving me with the integralM = Rdm. I noted thatRr dm ′2is justIcm, the moment of intertia of the object about an axis through its center of mass. Inoted that (1/M)R~r ′dmis the positionofthe center of mass in center of mass coordinates, whichiszero– by definition the center of mass is at the origin of the center of mass frame.The result, in words, is that the moment of inertia of an object that uniformly rotates aroundany axis is the moment of inertia of the object about an axis parallel to that axis through the centerof mass of the objectplusthe moment of inertia of the total mass of the object treated as a pointmass located at the center of mass as it revolves!This sounds a lot like the kinetic energy theorem; let’s see how the two are related.As long as the object rotates uniformly – that is, the object goes around its own center one timefor every time it goes around the axis of rotation, keeping thesame side pointing in towards thecenter as it goes– then its kinetic energy is just:K = 12Iω2= 12 (Mr2cm)ω 2+ 12 Icmω 2(550)
Week 5: Torque and Rotation in One Dimension261A bit of algebraic legerdemain:K = 12 (Mr2cm)vcmrcm2+ 12 Icmω 2= K (ofcm) +K (incm)(551)as before!Warning!This will not work if an object is revolving many times around its own center ofmass for each time it revolves around the parallel axis.Example 5.8.1: Moon Around Earth, Earth Around SunThis is a conceptual example, not really algebraic. You may have observed that the Moon alwayskeeps the same face towards the Earth – it is said to be “gravitationally locked” by tidal forces sothat this is true. This means that the Moon revolves once on its axis in exactly the same amount oftime that the Moon itself revolves around the Earth. We could therefore compute thetotalangularkinetic energy of the Moon by assuming that it is a solid ball of massM, radius , in an orbit aroundrthe Earth of radiusR, and a period of 28.5 days:Imoon=MR 2+ 25Mr2(552)(from the parallel axis theorem),ω= 2 πT(553)(you’ll need to findTin seconds, 86400×28 5) and then:.K = 12 Imoonω 2(554)All that’s left is the arithmetic.Contrast this with the Earth rotating around the Sun.Itrevolves on its own axis 365.25 timesduring the period in which it revolves around the Sun. To findit’skinetic energy we couldnotusethe parallel axis theorem, but we can still use the theorem at the end of the previous chapter. Herewe would find two different angular velocities:ωday=2 πTday(555)andωyear=2 πTyear(556)(again, 1 day = 86400 seconds is a good number to remember). Then if we letMbe the mass ofthe Earth,rbe its radius, andRbe the radius of its orbit around the Sun (all numbers that arereadily available on Wikipedia112we could find the total kinetic energy (relative to the Sun) as:K = 12MR 2ω 2year+ 1225Mr2ω 2day(557)which is somewhat more complicated, no?Let’s do a more readily evaluable example:Example 5.8.2: Moment of Inertia of a Hoop Pivoted on One SideIn figure??a hoop of massMand radiusRis pivoted at a point on theside, on the hoop itself, notin the middle. We already know the moment of inertia of the hoop about its center of mass. What112Wikipedia: http://www.wikipedia.org/wiki/Earth.
262Week 5: Torque and Rotation in One DimensionpivotMRFigure 77: A hoop of massMand radiusRis pivoted on the side – think of it as being hung on anail from a barn door.is the moment of inertia of the hoop about this new axisparallelto the one through the center ofmass that we used before?It’s so simple:Iside pivot=MR 2+ Icm=MR 2+MR 2= 2MR 2(558)and we’re done!For your homework, you get to evaluate the moment of inertia of a rod about an axis throughits center of mass and about one end of the rod and compare the two, both using direct integrationand using the parallel axis theorem. Good luck!5.9: Perpendicular Axis TheoremIn the last section we saw how a bit of geometry and math allowed us to prove a very useful theorem– useful because we can now learn a short table of moments of inertia about a given axis throughthe center of mass and theneasily extend themto find the moments of inertia of these same shapeswhen they uniformly rotate around a parallel axis.In this section we will similarly derive a theorem that is very useful for relating moments ofinertia ofplanar distributions of mass(only) around axes that areperpendicularto one another– thePerpendicular Axis Theorem. Here’s how it goes.xyxyzrMdmFigure 78: A planar blob of mass and the geometry needed to prove the Perpendicular Axis Theorem
Week 5: Torque and Rotation in One Dimension263Suppose that we wish to evaluateI x, the moment of inertia of the plane mass distributionMshown in figure 78. That’s quite easy:I x= Zy dm 2(559)Similarly,I y= Zx dm 2(560)We add them, and presto chango!I x+ I y= Zy dm 2+ Zx dm 2= Zr dm 2= I z(561)This is it, thePerpendicular Axis Theorem:I z= I x+ I y(562)I’ll give a single example of its use. Let’s find the moment of inertia of a hoop about an axisthrough the centerin the plane of the loop!
264Week 5: Torque and Rotation in One DimensionExample 5.9.1: Moment of Inertia of Hoop for Planar AxisMxyzRFigure 79: A hoop of massMand radiusRis drawn. What is the moment of inertia about thex-axis?This one is really very, very easy. We use the Perpendicular Axis theorembackwardsto get theanswer. In this case we knowI z=MR 2, and want to findI x. We observe thatfrom symmetry,I x= I yso that:I z=MR 2= I x+ I y= 2I x(563)orI x= 12 I z= 12MR 2(564)
Week 5: Torque and Rotation in One Dimension265Homework for Week 5Problem 1.Physics Concepts:Make this week’s physics concepts summary as you work all of the problemsin this week’s assignment. Be sure to cross-reference each concept in the summary to the problem(s)they were key to, and include concepts from previous weeks as necessary. Do the work carefullyenough that you can (after it has been handed in and graded) punch it and add it to a three ringbinder for review and study come finals!Problem 2.This problem will help you learn required concepts such as:•Definition/Evaluation of Moment of Inertia•Parallel Axis Theoremso please review them before you begin.a) Evaluate the moment of inertia of a uniform rod of massMand lengthLabout its center ofmass by direct integration.−L/2L/2xypivotb) Evaluate the moment of inertia of a uniform rod of massMand lengthLabout one end bydirect integration.Alsoevaluate it using the parallel axis theorem and the result you justobtained. Which is easier?Lxypivot
266Week 5: Torque and Rotation in One DimensionProblem 3.This problem will help you learn required concepts such as:•Definition/Evaluation of Moment of Inertia•Parallel Axis Theoremso please review them before you begin.a) Evaluate the moment of inertia of a uniform disk of massMand radiusRabout its axis ofsymmetry by direct integration (this can be set up as a “one dimensional integral” and henceis not too difficult).c)pivotb) Evaluate the moment of inertia of this disk around a pivot at the edge of the disk (for example,a thin nail stuck through a hole at the outer edge) using the parallel axis theorem. Would youcare to do the actual integral to find the moment of inertia of the disk in this case?pivotd)
Week 5: Torque and Rotation in One Dimension267Problem 4.M,RθHThis problem will help you learn required concepts such as:•Newton’s Second Law for Translation and Rotation•Moment of Inertia•Conservation of Mechanical Energy•Static Frictionso please review them before you begin.A round object with massmand radiusRis released from rest to roll without slipping downan inclined plane of heightHat angleθrelative to horizontal. The object has a moment of inertiaI=βmR2(whereβis a dimensionless number such as12or , that might describe a disk or a solid25ball, respectively).a) Begin by relatingv(the speed of the center of mass) to the angular velocity (for the rollingobject). You will use this (and the related two equations forsandθandaand ) repeatedlyαin rolling problems.b) Using Newton’s second law in both its linear and rotational form plus the rolling constraint,show that the acceleration of the object is:a= gsin( )θ1 +βc) Using conservation of mechanical energy, show that it arrives at the bottom of the incline witha velocity:v=s 2gH1 +βd) Show that the condition for the greatest angle for which the object will rollwithout slippingisthat:tan( )θ≤(1 +1β )µ swhereµ sis the coefficient of static friction between the object and the incline.
268Week 5: Torque and Rotation in One DimensionProblem 5.icyroughHm,RH’This problem will help you learn required concepts such as:•Conservation of Mechanical Energy•Rotational Kinetic Energy•Rolling Constraint.so please review them before you begin.A disk of massmand radiusRrolls without slippingdown a rough slope of heightHontoan icy (frictionless) track at the bottom that leads up a secondicy (frictionless)hill as shown.a) How fast is the disk moving at the bottom of the first incline? How fast is it rotating (what isits angular velocity)?b) Does the disk’s angular velocity change as it leaves the rough track and moves onto the ice (inthe middle of the flat stretch in between the hills)?c) How far up the second hill (vertically, findH ′) does the disk go before it stops rising?
Week 5: Torque and Rotation in One Dimension269Problem 6.FFFRrI (about cm)mass mThis problem will help you learn required concepts such as:•Direction of torque•Rolling Constraintso please review them before you begin.In the figure above, a spool of massmis wrapped with string around the inner spool. The spoolis placed on a rough surface (with coefficient of frictionµ s= 0 5) and the string is pulled with force.F≪0 5. mgin the three directions shown.a) For each picture, indicate the direction that static friction will point. Can the spool slip whileit rolls for this magnitude of force?b) For each picture, indicate the direction that the spool will accelerate.c) For each picture, find the magnitude of the force exerted by static friction and the magnitudeof the acceleration of the spool in terms of ,r RorIcm=βmR2 .Note:You can useeitherthe center of massorthe point of contact with the ground (with theparallel axis theorem) as a pivot, the latter beingslightly easierboth algebraically and intuitively.
270Week 5: Torque and Rotation in One DimensionProblem 7.MRm 1m 2HThis problem will help you learn required concepts such as:•Newton’s Second Law•Newton’s Second Law for Rotating Systems (torque and angular acceleration)•Moments of Inertia•The Rolling Constraint•Conservation of Mechanical Energyso please review them before you begin.In the figure above Atwood’s machine is drawn – two massesm 1andm 2hanging over amassivepulley which you can model as a disk of massMand radiusR, connected by a massless unstretchablestring. The string rolls on the pulley without slipping.a) Draw three free body diagrams (isolated diagrams for each object showing just the forcesacting on that object) for the three masses in the figure above.b) Convert each free body diagram into a statement of Newton’s Second Law (linear or rotational)for that object.c) Using therolling constraint(that the pulley rolls without slipping as the masses move up ordown) find the acceleration of the system and the tensions in the string onbothsides of thepulley in terms ofm 1 ,m 2 ,M g, , andR .d) Suppose massm > m21and the system is released from rest with the masses at equal heights.When massm 2has descended a distanceH, use conservation of mechanical energy to findvelocity of each mass and the angular velocity of the pulley.
Week 5: Torque and Rotation in One Dimension271Problem 8.rMRθpivotThis problem will help you learn required concepts such as:•Newton’s Second Law for rotating objects•Moment of Inertia•Parallel Axis Theorem•Work in rotating systems•Rotational Kinetic Energy•Kinetic Energy in Lab versus CMso please review them before you begin.In the picture above, a physical pendulum is constructed by hanging a disk of massMand radiusron the end of a massless rigid rod in such a way that the center of mass of the disk is a distanceRaway from the pivot and so that the whole disk pivots with the rod. The pendulum is pulled toan initial angleθ 0(relative to vertically down) and then released.a) Find thetorqueabout the pivot exerted on the pendulum by gravity at an arbitrary angle .θb) Integrate the torque fromθ= θ 0toθ= 0 to find the total work done by the gravitationaltorque as the pendulum disk falls to its lowest point. Note that your answer should beMgR(1−cos( )) =θ 0MgHwhereHis the initial height above this lowest point.c) Find the moment of inertia of the pendulum about the pivot (using the parallel axis theorem).d) Set the work you evaluated in b) equal to the rotational kinetic energy of the disk12Iω2usingthe moment of inertia you found in c). Solve forω=dθdtwhen the disk is at its lowest point.e) Show that this kinetic energy is equal to the kinetic energy of the moving center of mass of thedisk12Mv2plus the kinetic energy of the disk’s rotation about its own center of mass,12 cm Iω 2 ,at the lowest point.As an optional additional step, write Newton’s Second Law in rotational coordinatesτ=Iα(using the values for the magnitude of the torque and moment of inertia you determined above) andsolve for the angular acceleration as a function of the angle . In a few weeks we will learn to solveθthis equation of motion for small angle oscillations, so it is good to practice obtaining it from thebasic physics now.
272Week 5: Torque and Rotation in One DimensionProblem 9.m,rHRThis problem will help you learn required concepts such as:•Newton’s Second Law•Moments of Inertia•Rotational Kinetic Energy•The Rolling Constraint•Conservation of Mechanical Energy•Centripetal Acceleration•Static Frictionso please review them before you begin.A solid ball of massMand radiusrsits at rest at the top of a hill of heightHleading to acircular loop-the-loop. The center of mass of the ball will move in a circle of radiusRif it goesaround the loop. Recall that the moment of inertia of a solid ball isIball= 25MR 2 .a) Find the minimum heightHminfor which the ballbarelygoes around the loop staying on thetrack at the top, assuming that itrolls without slippingthe entire time independent of thenormal force.b) How does your answer relate to the minimum height for the earlier homework problem whereit was a block that slid around a frictionless track? Does this answer make sense? If it ishigher, where did the extra potential energy go? If it is lower, where did the extra kineticenergy come from?c) Discussion question for recitation: The assumption that the ball will roll around the trackwithout slipping if released from this estimatedminimumheight is not a good one. Why not?
Week 5: Torque and Rotation in One Dimension273Advanced Problem 10.FMm,RθMFThis problem will help you learn required concepts such as:•Newton’s Second Law (linear and rotational)•Rolling Constraint•Static and Kinetic Friction•Changing Coordinate Framesso please review them before you begin.A disk of massmis resting on a slab of massM, which in turn is resting on a frictionlesstable. The coefficients of static and kinetic friction between the disk and the slab areµ sandµ k ,respectively. A small force~Fto the right is applied to the slab as shown, then gradually increased.a) When~Fis small, the slab will accelerate to the right and the disk will roll on the slab withoutslipping. Find the acceleration of the slab, the acceleration of the disk, and theangularacceleration of the disk as this happens, in terms ofm M R,,, and the magnitude of the forceF .b) Find the maximum forceFmaxsuch that it rolls without slipping.c) IfFis greater than this, solve once again for the acceleration and angular acceleration of thedisk and the acceleration of the slab.Hint: The hardest single thing about this problem isn’t the physics (which is really prettystraightfoward). It is visualizing the coordinates as the center of mass of the disk moves with adifferent acceleration as the slab. I have drawntwofigures above to help you with this – the lowerfigure represents a possible position of the disk after the slab has moved some distance to the rightand the disk has rolled back (relative to the slab! It has movedforwardrelative to the ground!Why?) without slipping. Note the dashed radius to help you see the angle through which it hasrolled and the various dashed lines to help you relate the distance the slab has movedx s, the distancethe center of the disk has movedx d, and the angle through which it has rolled . Use this relationθto connect the acceleration of the slab to the acceleration and angular acceleration of the disk.If you can do this one, good job!
274Week 5: Torque and Rotation in One DimensionOptional ProblemsThe following problems arenot required or to be handed in, but are provided to give yousome extra things to work on or test yourself withaftermastering the required problems and conceptsabove and to prepare for quizzes and exams.Optional Problem 11.RFrA cable spool of massM, radiusRand moment of inertiaI=βMR2is wrapped around itsOUTER disk with fishing line and set on a rough rope as shown. The inner spool has a radius .rThe fishing line is then pulled with a forceFto the right so that it rolls down the ropewithoutslipping.a) Find the magnitude and direction of the acceleration of the spool.b) Find the force (magnitude and direction) exerted by the friction of the rope on the spool.c) For one particular value of , the frictional force isrzero!. Find that value. For larger values,which way does friction point? For smaller values, which way does friction point? At thisvalue, if the ropewere not therewould the motion be any different?In analyzing the “walking the spool” problem in class and in the text above, students often askhow they can predict which direction that static friction acts on a rolling spool, and I reply thatthey can’t!I can’t, not always, because in this problem it can pointeither wayand which way itultimately points depends on thedetailsofR r,and ! The best you can do is make a reasonableβguessas to the direction and let the algebra speak – if your answer comes out negative, frictionpoints the other way.
Week 5: Torque and Rotation in One Dimension275Optional Problem 12.MMRMMRThis problem will help you learn required concepts such as:•Moment of Inertia•Newton’s Second Law (translational and rotational)•Conservation of Mechanical Energyso please review them before you begin.A block of massMsits on a smooth (frictionless) table. A massMis suspended from an Acme(massless, unstretchable, unbreakable) rope that is looped around the two pulleys, each with thesame massMand radiusRas shown and attached to the support of the rightmost pulley. Thetwo pulleys each have the moment of inertia of a disk,I= 12MR 2and the rope rolls on the pulleyswithout slipping. At time = 0 the system is released at rest.ta) Draw free body diagrams for each of the four masses. Don’t forget the forces exerted by thebar that attaches the pulley to the mass on the left or the pulley on the right to the table!Note that some of these forces will cancel due to the constraint that the center of mass ofcertain masses moves only in one direction or does not move at all. Note also that the torqueexerted by theweightof the pulley around the near corder of the mass on the table is not notenough to tip it over.b) Write the relevant form(s) of Newton’s Second Law for each mass, translational and rotationalas needed, separately, wherever the forces in some direction donotcancel. Also write theconstraint equations that relate the accelerations (linear and angular) of the masses and pulleys.c) Find the acceleration of the hanging massM(only) in terms of the givens. Note that youcan’tquitejust add all of the equations you get after turning the torque equations into forceequations, but if you solve the equations simultaneously, systematically eliminating all internalforces and tensions, you’ll get a simple enough answer.d) Find the speed of the hanging massMafter it has fallen a heightH, using conservation of totalmechanical energy. Show that it is consistent with the “usual” constant-acceleration-from-restanswerv= √ 2aHfor the acceleration found in c).
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