Page 1Physics 207 – Lecture 16Physics 207: Lecture 16, Pg 1Lecture 16Goals:Goals:••Chapter 12Chapter 12 Extend the particle model to rigid-bodies Understand the equilibrium of an extended object. Analyze rolling motion Understand rotation about a fixed axis. Employ “conservation of angular momentum” conceptAssignment: HW8 due March 17thPhysics 207: Lecture 16, Pg 2Exercise Rotational DefinitionsA. The wheel is spinning counter-clockwise and slowing down.B. The wheel is spinning counter-clockwise and speeding up.C. The wheel is spinning clockwise and slowing down.D. The wheel is spinning clockwise and speeding up A friend at a party (perhaps a little tipsy) sees a disk spinning and says “Ooh, look! There’s a wheel with a negative ω and positive α!” Which of the following is a true statement about the wheel??Page 2Physics 207 – Lecture 16Physics 207: Lecture 16, Pg 3Home Exercise (review) : Wheel And Rope A wheel with radius r = 0.4 m rotates freely about a fixed axle. There is a rope wound around the wheel. Starting from rest at t = 0, the rope is pulled such that it has a constant acceleration aT= 4 m/s2. How many revolutions has the wheel made after 10 seconds? (One revolution = 2π radians)aTrPhysics 207: Lecture 16, Pg 4Home exercise: Wheel And Rope A wheel with radius r = 0.4 m rotates freely about a fixed axle. There is a rope wound around the wheel. Starting from rest at t = 0, the rope is pulled such that it has a constant acceleration aT= 4 m/s2. How many revolutions has the wheel made after 10 seconds? (One revolution = 2π radians) Revolutions = R = (θ − θ0) / 2π and aT= α rθ = θ0+ ω0∆t + ½ α ∆t2 R = (θ − θ0) / 2π = 0 + ½ (aT/r) ∆t2 / 2π R = (0.5 x 10 x 100) / 6.28aTrPage 3Physics 207 – Lecture 16Physics 207: Lecture 16, Pg 5System of Particles (Distributed Mass): Until now, we have considered the behavior of very simple systems (one or two masses). But real objects have distributed mass ! For example, consider a simple rotating disk and 2 equal mass m plugs at distances r and 2r. Compare the vtangentialand kinetic energies at these two points.12ωPhysics 207: Lecture 16, Pg 6System of Particles (Distributed Mass): The rotation axis matters too! Twice the radius, four times the kinetic energy KEY POINT: It matters where you put your mass!1 K= ½ m v2= ½ m (ω r)22 K= ½ m (2v)2= ½ m (ω 2r)2ω221221Rotational)(K rmmvω==v2v1Page 4Physics 207 – Lecture 16Physics 207: Lecture 16, Pg 7Exercise Rotational Kinetic EnergyA. 1/9 B. 1/3C. 1D. 3E. 9 We have two balls of the same mass. Ball 1 is attached to a 0.1 m long rope. It spins around at 2 revolutions per second. Ball 2 is on a 0.3 m long rope. It spins around at 2 revolutions per second. What is the ratio of the kinetic energyof Ball 2 to that of Ball 1 ?2221ωmrK=Ball 1Ball 2Physics 207: Lecture 16, Pg 8Exercise Rotational Kinetic Energy K2/K1= ½ m ωr22/ ½ m ωr12= 0.22/ 0.12= 4 What is the ratio of the kinetic energy of Ball 2 to that of Ball 1 ?(A) 1/9 (B) 1/3 (C) 1 (D) 3 (E) 9Ball 1Ball 2Page 5Physics 207 – Lecture 16Physics 207: Lecture 16, Pg 9A special point for rotationSystem of Particles: Center of Mass (CM) A supported object will rotate about its center of mass. Center of mass: Where the system is balanced ! Building a mobile is an exercise in finding centers of mass. m1m2+m1m2+mobilePhysics 207: Lecture 16, Pg 10System of Particles: Center of Mass How 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 6Physics 207 – Lecture 16Physics 207: Lecture 16, Pg 11Sample 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 16, Pg 12System of Particles: Center of Mass For a continuous solid, one can convert sums to an integral.yxdmrrwhere dm is an infinitesimal mass element but there is no new physicsMdmrdmdmr∫∫∫== CMrrrRMmRNiii∑=≡1CMrrrPage 7Physics 207 – Lecture 16Physics 207: Lecture 16, Pg 13Connection with motion... So, for a rigid rotating object whose CM is moving, it rotates about its center of mass!For a point p rotating:nTranslatioRotationTOTALKKK+=MmNiii∑=≡1CMrRrr221221)(KppppRrmvmpω==2CM21RotationTOTALVK MK+=VCMωp p p p p p p Physics 207: Lecture 16, Pg 14Rotational Kinetic Energy Consider the simple rotating system shown below. (Assume the masses are attached to the rotation axis by massless rigid rods). The kinetic energy of this system will be the sum of the kinetic energy of each piece: K = ½ m1v12 + ½ m2v22 + ½ m3v32 + ½ m4v42rr1rr2rr3rr4m4m1m2m3ω∑==41221viiimKPage 8Physics 207 – Lecture 16Physics 207: Lecture 16, Pg 15Rotation & Kinetic Energy Notice that v1= ω r1 , v2= ω r2 , v3= ω r3 , v4= ω r4 So we can rewrite the summation: We recognize & define a new quantity, moment of inertia or I, and write:rr1rr2rr3rr4m4m1m2m3ω24122141222141221][rrvωω∑∑∑======iiiiiiiiimmmK221RotationalIω=K∑=≡Niiirm12IPhysics 207: Lecture 16, Pg 16Calculating Moment of Inertiawhere r is the distance from the mass to the axis of rotation.∑=≡Niiirm12IExample: Calculate the moment of inertia of four point masses(m) on the corners of a square whose sides have length L, about a perpendicular axis through the center of the square:mmmmLPage 9Physics 207 – Lecture 16Physics 207: Lecture 16, Pg 172/2LR=Calculating Moment of Inertia... For a single object, I depends on the rotation axis! Example: I1= 4 m R2= 4 m (21/2L / 2)2LI = 2mL2I2= mL2mmmmI1= 2mL22/LR=LR=Physics 207: Lecture 16, Pg 18Home Exercise Moment of Inertia A triangular shape is made from identical balls and identical rigid, massless rods as shown. The moment of inertia about the a, b, and c axes is Ia, Ib, and Icrespectively. Which of the following is correct:((A)A) Ia > Ib> Ic(B)(B) Ia > Ic> Ib(C)(C) Ib> Ia> Icabc∑=iiirm2IPage 10Physics 207 – Lecture 16Physics 207: Lecture 16, Pg 19Home Exercise Moment of Inertia Ia= 2 m (2L)2 Ib= 3 m L2 Ic= m
View Full Document
Unlocking...