Chapter 10Spin and Orbital MotionSome DefinitionsOrbital motion: Motion relative to a point, often periodic, but not necessarily so.Spin motion: Motion of an object as it rotates around an axis through its center of mass.Rotational motion: Motion around an axis of rotation. Both orbital and spin motion are examples of rotational motion. Fixed axis of rotation: A single-nonchanging axis around which the object rotates. Rigid bodies: Objects that have a definite unchanging shape. Translational motion: Motion with a fixed direction of net force.Review of TerminolgyθsP1P2θ= s/rrdtdttθθω=∆∆=→∆ 0limdtdttωωα=∆∆=→∆ 0limvT= ds/dt = rdθ/dt = rωaC= vT2/r = ω2ra = aC+ aTaT= dvT/dt = r dω/dt =rαω= 2πνT = 1/νVector (Cross) ProductC = A × BC = AB sinθABθThe direction of C is given by the right hand rule described in the text.We can write the components of the cross product as:Cx= AyBz− AzByCy= AzBx− AxBzCz= AxBy− AyBxThis can be easily remembered by using the determinant form:kjikjiABBA)()()(xyyxxzzxyzzyzyxzyxBABABABABABABBBAAA−+−−−==×−=×Linear momentum of a single particle: pLinear momentum of a system of particles: P = ∑pAngular momentum of a single point particle: lAngular momentum of a system of particles: L = ∑l(Orbital) Angular Momentumpl = r × p+x+yrA 6-kg particle moves to the right at 4 m/s as shown. The magnitude of its angular momentum in kg·m2/s about the point O is:30°4 m/s6 kgO12 mA)ZeroB) 24C) 144D)288E) 249Interactive QuestionProblem: What is the angular momentum of a 4.0 kg object traveling with a velocity of v = 2.3i + 1.5j m/s when it is at (6.8, 5.6) m relative to the point (−3.0, −2.0)?Angular Momentum for Circular Motionl = r × p= r × mvrvThe direction (in this case) for l is up, out of the paper.The magnitude is l = rmv sinθ= rmv= rm (ωr) = mr2ωAnd since ω also points up out of the paper, l = mr2ωFor this case, we define the moment of inertia I = mr2l = IωTorque for Circular Motion∑F = maConsider the tangential motion of a rotating object with a tangential force acting on the object.FT= maTrFT= mr aT= mr rαrFT= mr2ατ= mr2α = Iατis the torquerFTτ = IαSince the angular acceleration has a direction out of the page,∑τ = ∑(mr2)α∑τext+ ∑τint= ∑(mr2)α∑τext= ∑(mr2)α∑τ = I αwhere I = ∑(mr2) is the general form of the rotational inertia, or moment of inertia.This is one form of Newton’s second law for rotation.If there is more than one force with a tangential component,The moment of inertia depends on the object and the axis of rotation. More on this later.General Definition of TorqueIn the previous example we defined τ= rFTand defined the direction of the torque as up out of the page.rFTIt makes sense, then, that the general definition of torque isτ = r × FConsider now a system of particles:L = ∑ili= ∑i(ri× pi)dL/dt = ∑i{(dri/dt × mvi) + (ri× m dvi/dt)} = ∑i{(vi× mvi) + (ri× mai)}= ∑i{0 + (ri× mai)} = ∑i(ri× ∑jFj) = ∑i∑jτj= ∑τnet= ∑τext + ∑τintdL/dt = ∑τextThis is the more general form of Newton’s 2ndlaw for rotation.∑τext= dL/dtSummary∑τext= dL/dtIf the moment of inertia doesn’t change:∑τext= Iατ = r × Fl = r × pGeneral CaseSpecific CaseFor a “point” object in a circular orbit:l = IωI = mr2I = ∑(mr2)Now let’s do some problems with all of these conceptsAxis of rotationFF=F cosθF⊥=F sinθθrr⊥=r sinθτ = r × Fr⊥is often called the “lever arm”TorqueThe direction of the torque is up out of the page and the magnitude is τ= rF sinθ= Fr⊥= rF⊥Problem: Consider the object shown below, and the forces acting on the object. Calculate the torque around an axis perpendicular to the paper through (a) point O(b) point C.45º20º30 N10 NOCProblem: A penny with a mass of 2.5 grams sits on a turntable a distance of 6.6 cm from the center. The turntable starts from rest and spins up to a maximum angular speed of 33 rev/min in 2.2 seconds rotating counterclockwise. The penny does not slide with respect to the turntable.a) Assuming a constant torque what is the torque on the penny with respect to the center of the turntable?b) What is the angular speed of the penny after 1.1 seconds with respect to the center of the turntable?c) What is the angular momentum of the penny after 1.1 seconds with respect to the center of the turntable?d) What is the magnitude of the total force of friction acting on the penny after 1.1 seconds?Conservation of Angular Momentum∑τext= dL/dtWhen there is no net external torque, then the angular momentum does not change with time. It is conserved.dL/dt = 0Lfinal= LinitialThere may be no external torque for many reasons, including the case when the line of force for all forces passes through the axis of rotation. (τ = r × F)Problem: Show how Kepler’s second law, that the area swept out by a planet moving in an ellipse is always the same for equal time intervals, is a result of the conservation of angularMoment of InertiaFor a single “point” object with mass m, rotating at a distance of r around the axis of rotation, I = mr2To calculate the rotational inertia for a more complicated object, we simply sum up the rotational inertia of each part of the object.Problem: Suppose a baton is 1.0 m long with weights on each end weighing 0.3 kg. Neglect the mass of the bar and consider the weights as point objects. What is the moment of inertia for a baton (a) spinning around its center? (b) spinning around one end of the baton?The rotational inertia depends on both the axis of rotation, and how the mass is distributed.Three thin disks with uniform mass density have the same mass, and radius. Rank the objects in order of increasing moment of inertia. The axis of rotation is shown as the dark dot and is perpendicular to the paper.I II IIIA) I, II, III B) III, II, IC) II, III, I D) II, I, IIIE) More information is neededInteractive QuestionFor an object with a continuous mass distribution, the sum becomes an integral and we get I = ∫ r2dmMoments of inertia that have been calculated this way are often found in a table. (See Table 10.1)Calculating the Moment of InertiaYou don’t have to memorize these formulas. They will be given (or I will ask you to derive them.)Problem: What is the moment of inertia of a thin disk with a mass m, radius R, and thickness t, rotating through its center of mass along an axis perpendicular to the plane of the disk?If you know the