Page 1Physics 207 – Lecture 17Physics 207: Lecture 17, Pg 1Lecture 17Goals:Goals:••Chapter 12Chapter 12 Define center of mass Analyze rolling motion Use Work Energy relationships Introduce torque Equilibrium of objects in response to forces & torquesAssignment: HW7 due tomorrow Wednesday, Exam ReviewPhysics 207: Lecture 17, Pg 2Combining translation and rotationObjects can have translational energyObjects can have rotational energyObjects can have both K = ½ m v2+ ½ I ω2Page 2Physics 207 – Lecture 17Physics 207: Lecture 17, Pg 31st: A special point for rotationCenter of Mass (CM)A free object will rotate about its center of mass.Center of mass: Where the system is balanced ! A mobile exploits this centers of mass. m1m2+m1m2+mobilePhysics 207: Lecture 17, Pg 4System of Particles: Center of MassHow do we describe the “position” of a system made up of many parts ?Define the Center of Mass (average position): For a collection of N individual point like particles whose masses and positions we know:MmNiii∑=≡1CMrRrr(In this case, N = 2)yxr2r1m1m2RCMPage 3Physics 207 – Lecture 17Physics 207: Lecture 17, Pg 5Sample calculation:Consider the following mass distribution:(24,0)(0,0)(12,12)m2mmRCM= (12,6)kˆjˆiˆ CM CM CM1CMZYXMmNiii++===∑rRrrXCM= (m x 0 + 2m x 12 + m x 24 )/4m metersYCM= (m x 0 + 2m x 12 + m x 0 )/4m metersXCM= 12 metersYCM= 6 metersm at ( 0, 0)2m at (12,12)m at (24, 0)Physics 207: Lecture 17, Pg 6System of Particles: Center of MassFor a continuous solid, one can convert sums to an integral.yxdmrrwhere dm is an infinitesimal mass element but there is no new physics.Mdmrdmdmr∫∫∫== CMrrrRMmRNiii∑=≡1CMrrrPage 4Physics 207 – Lecture 17Physics 207: Lecture 17, Pg 7Connection with motion...An unconstrained rigid object with rotation and translation rotates about its center of mass!Any point p rotating:nTranslatioRotationTOTALKKK+=MmNiii∑=≡1CMrRrr221221)(vKppppRrmmpω==2CM21RotationTOTALmvK+=KVCMωp p p p p p p Physics 207: Lecture 17, Pg 8Work & Kinetic Energy:Work Kinetic-Energy Theorem: ∆K = WNETApplies to both rotational as well as linear motion.What if there is rolling without slipping ?()NET2 CM2 CM212221)vv( I WmKifif=−+−=∆ωωPage 5Physics 207 – Lecture 17Physics 207: Lecture 17, Pg 9Same Example : Rolling, without slipping, MotionA solid disk is about to roll down an inclined plane. What is its speed at the bottom of the plane ? MθhMv ?MMMMMPhysics 207: Lecture 17, Pg 10Rolling without slipping motionAgain consider a cylinder rolling at a constant speed. VCMCM2VCMPage 6Physics 207 – Lecture 17Physics 207: Lecture 17, Pg 11Motion Again consider a cylinder rolling at a constant speed. VCMCM2VCMCMVCMSliding onlyCMRotation onlyVTang= ωRBoth with |VTang| = |VCM |If acceleration acenterof mass= - αRPhysics 207: Lecture 17, Pg 12Example : Rolling MotionA solid cylinder is about to roll down an inclined plane. What is its speed at the bottom of the plane ? Use Work-Energy theoremMθhMv ?Disk has radius RMMMMMMgh = ½ Mv2+ ½ ICMω2 and v =ωRMgh = ½ Mv2+ ½ (½ M R2)(v/R)2 = ¾ Mv2v = 2(gh/3)½Page 7Physics 207 – Lecture 17Physics 207: Lecture 17, Pg 13How do we reconcile force, angular velocity and angular acceleration?Physics 207: Lecture 17, Pg 14From force to spin (i.e., ωωωω) ?τNET = |r| |FTang| ≡ |r| |F| sin θIf a force points at the axis of rotation the wheel won’t turnThus, only the tangential component of the force matters With torque the position & angle of the force mattersFTangentialarFradialFA force applied at a distance from the rotation axis gives a torqueFradialFTangentialθr=|FTang| sin θPage 8Physics 207 – Lecture 17Physics 207: Lecture 17, Pg 15Rotational Dynamics: What makes it spin?τNET = |r| |FTang| ≡ |r| |F| sin θTorque is the rotational equivalent of forceTorque has units of kg m2/s2= (kg m/s2) m = N mFTangentialaarFradialFFτNET= r FTang= r m aTang= r m r α= (m r2) αFor every little part of the wheelPhysics 207: Lecture 17, Pg 16TorqueThis is the rotational version of FNET= ma Moment of inertia, Moment of inertia, IΣΣΣΣimiri2 ,is the rotational is the rotational equivalent of mass.equivalent of mass.If I is big, more torque is required to achieve a given angular acceleration.FTangentialaarFrandialFFThe further a mass is away from this axis the greater the inertia (resistance) to rotationτNET = I αPage 9Physics 207 – Lecture 17Physics 207: Lecture 17, Pg 17Rotational DynamicsτNET = I αααα ≡ |r| |F| sin θA constant torque gives constant angular accelerationif and only if the mass distribution and the axis of rotation remain constant.FTangentialarFradialFPhysics 207: Lecture 17, Pg 18Torque, like ω, has pos./neg. values Magnitude is given by (1) |r| |F| sin θ(2) |Ftangential| |r|(3) |F| |rperpendicular to line of action | Direction is parallel to the axis of rotation with respect to the “right hand rule”And for a rigid objectτ= I αrFF cos(90°−θ) = FTang. rFFradialFarFrθθθθ90°−θr sin θline of actionPage 10Physics 207 – Lecture 17Physics 207: Lecture 17, Pg 19StaticsEquilibrium is established when0 motion nalTranslatioNet=ΣFr0 motion RotationalNet=ΣτrIn 3D this implies SIX expressions (x, y & z) Physics 207: Lecture 17, Pg 20ExampleTwo children (30 kg & 60 kg) sit on a horizontal teeter-totter. The larger child is at the end of the bar and 1.0 m from the pivot point. The smaller child is trying to figure out where to sit so that the teeter-totter remains horizontal and motionless. The teeter-totter is a uniform bar of length 3.0 m and mass 30 kg. Assuming you can treat both children as point like particles, what is the initial angular acceleration of the teeter-totter when the large child lifts up their legs off the ground (the smaller child can’t reach)?The moment of inertia of the bar about the pivot is 30 kg m2.For the static case:0 motion RotationalNet=ΣτrPage 11Physics 207 – Lecture 17Physics 207: Lecture 17, Pg 21Example: Soln.Draw a Free Body diagram (assume g = 10 m/s2)0 = 300 d + 300 x 0.5 + N x 0 – 600 x 1.0 0= 2d + 1 – 4 d = 1.5 m from pivot point0 UseNet=Στr30 kg1 m60 kg30 kg0.5 m300 N300 N600 NNPhysics 207: Lecture 17, Pg 22RecapAssignment: HW7 due tomorrowWednesday: review
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