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Physics 207, Lecture 16, Oct. 29 Agenda: Chapter 13 Center of Mass Torque Moment of Inertia Rotational Energy Rotational Momentum Assignment: Wednesday is an exam review session, Exam will be held in rooms B102 & B130 in Van Vleck at 7:15 PM MP Homework 7, Ch. 11, 5 problems, NOTE: Due Wednesday at 4 PM MP Homework 7A, Ch. 13, 5 problems, available soon Physics 207: Lecture 16, Pg 1 Chap. 13: Rotational Dynamics Up until now rotation has been only in terms of circular motion with ac = v2 / R and | aT | = d| v | / dt Rotation is common in the world around us. Many ideas developed for translational motion are transferable. Physics 207: Lecture 16, Pg 2 Conservation of angular momentum has consequences How does one describe rotation (magnitude and direction)? Physics 207: Lecture 16, Pg 3 Rotational Dynamics: A child’s toy, a physics playground or a student’s nightmare A merry-go-round is spinning and we run and jump on it. What does it do? We are standing on the rim and our “friends” spin it faster. What happens to us? We are standing on the rim a walk towards the center. Does anything change? Physics 207: Lecture 16, Pg 4 Summary (with comparison to 1-D kinematics) Angular constant Linear a constant 0 t v v0 at 1 2 0 0 t t 2 1 2 x x0 v 0t at 2 And for a point at a distance R from the rotation axis: x = R v = R a=R Physics 207: Lecture 16, Pg 7 System 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. 1 2 Compare the velocities and kinetic energies at these two points. Physics 207: Lecture 16, Pg 12 System of Particles (Distributed Mass): An extended solid object (like a disk) can be thought of as a collection of parts. The motion of each little part depends on where it is in the object! 1 K= ½ m v2 = ½ m (r)2 The rotation axis matters too! 2 K= ½ m (2v)2 = ½ m (2r)2 Physics 207: Lecture 16, Pg 13 System of Particles: Center of Mass If an object is not held then it rotates about the center of mass. Center of mass: Where the system is balanced ! Building a mobile is an exercise in finding centers of mass. m1 + m2 m1 + m2 mobile Physics 207: Lecture 16, Pg 14 System 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 pointlike particles whose masses and positions we know: mi ri N RCM i 1 M RCM m1 r1 m2 r2 y x (In this case, N = 2) Physics 207: Lecture 16, Pg 15 Sample calculation: Consider the following mass distribution: mi ri N RCM i 1 M XCM î YCM ĵ ZCM k̂ XCM = (m x 0 + 2m x 12 + m x 24 )/4m meters RCM = (12,6) YCM = (m x 0 + 2m x 12 + m x 0 )/4m meters (12,12) 2m XCM = 12 meters YCM = 6 meters m (0,0) m (24,0) Physics 207: Lecture 16, Pg 16 System of Particles: Center of Mass For a continuous solid, convert sums to an integral. dm y r dm r dm RCM M dm r x where dm is an infinitesimal mass element. Physics 207: Lecture 16, Pg 17 Rotational Dynamics: What makes it spin? a A force applied at a distance FTangential from the rotation axis F TOT = |r| |FTang| ≡ |r| |F| sin f Frandial r Torque is the rotational equivalent of force Torque has units of kg m2/s2 = (kg m/s2) m = N m A constant torque gives constant angular acceleration iff the mass distribution and the axis of rotation remain constant. Physics 207: Lecture 16, Pg 18 Lecture 16, Exercise 1 Torque In which of the cases shown below is the torque provided by the applied force about the rotation axis biggest? In both cases the magnitude and direction of the applied force is the same. Remember torque requires F, r and sin or the tangential force component times perpendicular distance L F F A. Case 1 B. Case 2 L axis C. Same case 1 case 2 Physics 207: Lecture 16, Pg 19 Lecture 16, Exercise 1 Torque In which of the cases shown below is the torque provided by the applied force about the rotation axis biggest? In both cases the magnitude and direction of the applied force is the same. Remember torque requires F, r and sin f or the tangential force component times perpendicular distance (A) case 1 (B) case 2 (C) same L F F L axis case 1 case 2 Physics 207: Lecture 16, Pg 20 Rotational Dynamics: What makes it spin? a A force applied at a distance FTangential from the rotation axis F TOT = |r| |FTang| ≡ |r| |F| sin f Frandial r Torque is the rotational equivalent of force Torque has units of kg m2/s2 = (kg m/s2) m = N m TOT = r FTang = r m a =rmr = m r2 For every little part of the wheel Physics 207: Lecture 16, Pg 21 TOT = m r2 and inertia a The further a mass is away from this axis the greater the inertia (resistance) to rotation FTangential F Frandial r This is the rotational version of FTOT = ma Moment of inertia, I ≡ m r2 , (here I is just a point on the wheel) is the rotational equivalent of mass. If I is big, more torque is required to achieve a given angular acceleration. Physics 207: Lecture 16, Pg 22 Calculating Moment of Inertia N I mi ri 2 where r is the distance from the mass to the axis of rotation. i 1 Example: 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: m m m m L Physics 207: Lecture 16, Pg 23 Calculating Moment of Inertia... For a single object, I depends on the rotation axis! Example: I1 = 4 m R2 = 4 m (21/2 L / 2)2 I1 = 2mL2 m m m m I2 = mL2 I = 2mL2 L Physics 207: Lecture 16, Pg 24 Lecture 16, Exercise 2 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 Ic respectively. Which of the following is correct: I mi ri 2 i (A) Ia > Ib > I c a (B) Ia > Ic > I b b (C) Ib > Ia > Ic c Physics 207: Lecture 16, Pg 25 Calculating Moment of Inertia... For a discrete collection of point masses we found: N I mi ri 2 i 1 For a continuous solid object we have to add up the mr2 contribution for every infinitesimal mass element dm. dm An integral is required to find I : I r dm 2 r Physics 207: Lecture 16, Pg 27 Moments of Inertia... Some examples of I for solid objects: R Solid sphere of mass M and radius R, about an axis through its center. I = 2/5 M R2 R Thin spherical shell of mass M and radius R, about an axis through its center. Use the table… See Table 13.3, Moments of Inertia Physics 207: Lecture 16, Pg 29 Rotation & 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: 4 K mi v 1 2 i 1 2 i K = ½ m1v12+ ½ m2v22+ ½ m3v32+ ½ m4v42 m4 r4 m3 r3 r1 m1 r2 m2 Physics 207: Lecture 16, Pg 31 Rotation & Kinetic Energy Notice that v1 = r1 , v2 = r2 , v3 = r3 , v4 = r4 So we can rewrite the summation: 4 4 K mi v mi r 1 2 i 1 2 i 1 2 2 2 i i 1 1 2 4 [ m r ] i 1 2 i i 2 We recognize the quantity, moment of inertia or I, and write: m4 K I 1 2 2 r4 m3 r3 r1 r2 m1 m2 Physics 207: Lecture 16, Pg 32 Lecture 16, Exercise 2 Rotational Kinetic Energy 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 0.2 m long rope. It spins around at 2 revolutions per second. 2 K 12 I What is the ratio of the kinetic energy of Ball 2 to that of Ball 1 ? A. ¼ B. ½ C. 1 D. 2 Ball 1 E. 4 I mi ri 2 i Ball 2 Physics 207: Lecture 16, Pg 33 Lecture 16, Exercise 2 Rotational Kinetic Energy K2/K1 = ½ m r22 / ½ m r12 = 0.22 / 0.12 = 4 What is the ratio of the kinetic energy of Ball 2 to that of Ball 1 ? (A) 1/4 (B) 1/2 Ball 1 (C) 1 (D) 2 (E) 4 Ball 2 Physics 207: Lecture 16, Pg 34 Rotation & Kinetic Energy... The kinetic energy of a rotating system looks similar to that of a point particle: Point Particle K mv 1 2 Rotating System 2 v is “linear” velocity m is the mass. K I 1 2 2 is angular velocity I is the moment of inertia about the rotation axis. I mi ri 2 i Physics 207: Lecture 16, Pg 35 Moment of Inertia and Rotational Energy So K I 1 2 2 where I mi ri 2 i Notice that the moment of inertia I depends on the distribution of mass in the system. The further the mass is from the rotation axis, the bigger the moment of inertia. For a given object, the moment of inertia depends on where we choose the rotation axis (unlike the center of mass). In rotational dynamics, the moment of inertia I appears in the same way that mass m does in linear dynamics ! Physics 207: Lecture 16, Pg 36 Work & Kinetic Energy: Recall the Work Kinetic-Energy Theorem: K = WNET This is true in general, and hence applies to rotational motion as well as linear motion. So for an object that rotates about a fixed axis: K I WNET 1 2 2 f 2 i Physics 207: Lecture 16, Pg 38 Connection with CM motion If an object of mass M is moving linearly at velocity VCM without rotating then its kinetic energy is 2 K T 12 MVCM If an object of moment of inertia ICM is rotating in place about its center of mass at angular velocity then its kinetic energy is K R I CM 1 2 2 What if the object is both moving linearly and rotating? K I CM MV 1 2 2 1 2 2 CM Physics 207: Lecture 16, Pg 45 Angular Momentum: We have shown that for a system of particles, momentum p mv dp F 0 dt is conserved if What is the rotational equivalent of this? angular momentum is conserved if L I dL dt 0 Physics 207: Lecture 16, Pg 59 Example: Two Disks A disk of mass M and radius R rotates around the z axis with angular velocity 0. A second identical disk, initially not rotating, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity F. z z 0 F Physics 207: Lecture 16, Pg 60 Example: Two Disks A disk of mass M and radius R rotates around the z axis with initial angular velocity 0. A second identical disk, at rest, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity F. No External Torque so Lz is constant Li = Lf I i i = I f ½ mR2 0 = ½ 2mR2 f z z 0 F Physics 207: Lecture 16, Pg 61 Lecture 16, Oct. 29 Assignment: Wednesday is an exam review session, Exam will be held in rooms B102 & B130 in Van Vleck at 7:15 PM MP Homework 7, Ch. 11, 5 problems, NOTE: Due Wednesday at 4 PM MP Homework 7A, Ch. 13, 5 problems, available soon Physics 207: Lecture 16, Pg 62 p = mv Angular Momentum: Definitions & Derivations We have shown that for a system of particles FEXT dp dt Momentum is conserved if FEXT 0 What is the rotational equivalent of this? r F The rotational analog of force F is torque Define the rotational analog of momentum p to be angular momentum, L or L r p Physics 207: Lecture 16, Pg 63 Recall from Chapter 9: Linear Momentum Definition: For a single particle, the momentum p is defined as: p ≡ mv (p is a vector since v is a vector) So px = mvx etc. Newton’s 2nd Law: F = ma dv d m (mv) dt dt dp F dt Units of linear momentum are kg m/s. Physics 207: Lecture 16, Pg 64 Linear Momentum and Angular Momentum So from: p mv Newton’s 2nd Law: F = ma dv d m (mv) dt dt dp F dt L I dL dt dL r F dt Units of angular momentum are kg m2/s. Physics 207: Lecture 16, Pg 65 Putting it all together dL r F dt dL r F dt Lrp EXT dL dt L r F dt r p L r p EXT r FEXT In the absence of external torques Total angular momentum is conserved EXT dL 0 dt Physics 207: Lecture 16, Pg 66 Angular momentum of a rigid body about a fixed axis: Consider a rigid distribution of point particles rotating in the x-y plane around the z axis, as shown below. The total angular momentum around the origin is the sum of the angular momentum of each particle: (since ri , vi , are perpendicular) Lz ri pi mi ri v i mi ri vi k̂ i i We see that L is in the z direction. m2 Using vi = ri , we get L z mi ri k̂ 2 i L I v1 i v2 r2 m3 j r3 m1 i r1 v3 Physics 207: Lecture 16, Pg 67 Example: Two Disks A disk of mass M and radius R rotates around the z axis with angular velocity 0. A second identical disk, initially not rotating, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity F. z z 0 F Physics 207: Lecture 16, Pg 68 Example: Two Disks A disk of mass M and radius R rotates around the z axis with initial angular velocity 0. A second identical disk, at rest, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity F. No External Torque so Lz is constant Li = Lf I i i = I f ½ mR2 0 = ½ 2mR2 f z z 0 F Physics 207: Lecture 16, Pg 69 Demonstration: Conservation of Angular Momentum Figure Skating : z z Arm A LA = LB IA IB A B Arm B No External Torque so Lz is constant even if internal work done. Physics 207: Lecture 16, Pg 70 Demonstration: Conservation of Angular Momentum Figure Skating : z z Arm A IAA= LA = LB = IBB Arm B IA < I B A > B ½ IAA2 > ½ IB B2 (work needs to be done) No External Torque so Lz is constant even if internal work done. Physics 207: Lecture 16, Pg 71 Angular Momentum Conservation A freely moving particle has a well defined angular momentum about any given axis. If no torques are acting on the particle, its angular momentum remains constant (i.e., will be conserved). In the example below, the direction of L is along the z axis, and its magnitude is given by LZ = pd = mvd. y x d m v Physics 207: Lecture 16, Pg 72 Example: Bullet hitting stick A uniform stick of mass M and length D is pivoted at the center. A bullet of mass m is shot through the stick at a point halfway between the pivot and the end. The initial speed of the bullet is v1, and the final speed is v2. What is the angular speed F of the stick after the collision? (Ignore gravity) M m F D D/4 v1 v2 before after Physics 207: Lecture 16, Pg 73 Example: Bullet hitting stick What is the angular speed F of the stick after the collision? (Ignore gravity). Process: (1) Define system (2) Identify Conditions (1) System: bullet and stick (No Ext. torque, L is constant) (2) Momentum is conserved (Istick = I = MD2/12 ) Linit = Lbullet + Lstick = m v1 D/4 + 0 = Lfinal = m v2 D/4 + I f M m F D D/4 v1 v2 before after Physics 207: Lecture 16, Pg 74 Example: Throwing ball from stool A student sits on a stool, initially at rest, but which is free to rotate. The moment of inertia of the student plus the stool is I. They throw a heavy ball of mass M with speed v such that its velocity vector moves a distance d from the axis of rotation. What is the angular speed F of the student-stool system after they throw the ball ? M v F I Top view: before d I after Physics 207: Lecture 16, Pg 75 Example: Throwing ball from stool What is the angular speed F of the student-stool system after they throw the ball ? Process: (1) Define system (2) Identify Conditions (1) System: student, stool and ball (No Ext. torque, L is constant) (2) Momentum is conserved Linit = 0 = Lfinal = m v d + I f M v F I Top view: before d I after Physics 207: Lecture 16, Pg 76 An example: Neutron Star rotation Neutron star with a mass of 1.5 solar masses has a diameter of ~11 km. Our sun rotates about once every 37 days f / i = Ii / If = ri2 / rf2 = (7x105 km)2/(11 km)2 = 4 x 109 gives millisecond periods! period of pulsar is 1.187911164 s Physics 207: Lecture 16, Pg 77 Angular Momentum as a Fundamental Quantity The concept of angular momentum is also valid on a submicroscopic scale Angular momentum has been used in the development of modern theories of atomic, molecular and nuclear physics In these systems, the angular momentum has been found to be a fundamental quantity Fundamental here means that it is an intrinsic property of these objects Physics 207: Lecture 16, Pg 78 Fundamental Angular Momentum Angular momentum has discrete values These discrete values are multiples of a fundamental unit of angular momentum The fundamental unit of angular momentum is h-bar Where h is called Planck’s constant h 34 kg m 1.054 10 2 s L n (n 1,2,3,...) Physics 207: Lecture 16, Pg 79 Intrinsic Angular Momentum Physics 207: Lecture 16, Pg 80 Angular Momentum of a Molecule Consider the molecule as a rigid rotor, with the two atoms separated by a fixed distance The rotation occurs about the center of mass in the plane of the page with a speed of Physics 207: Lecture 16, Pg 81 Angular Momentum of a Molecule (It heats the water in a microwave over) L / I CM E = h2/(82I) [ J (J+1) ] J = 0, 1, 2, …. Physics 207: Lecture 16, Pg 82 Center of Mass Example: Astronauts & Rope Two astronauts are initially at rest in outer space and 20 meters apart. The one on the right has 1.5 times the mass of the other (as shown). The 1.5 m astronaut wants to get back to the ship but his jet pack is broken. There happens to be a rope connected between the two. The heavier astronaut starts pulling in the rope. (1) Does he/she get back to the ship ? (2) Does he/she meet the other astronaut ? m M = 1.5m Physics 207: Lecture 16, Pg 83 Example: Astronauts & Rope (1) There is no external force so if the larger astronaut pulls on the rope he will create an impulse that accelerates him/her to the left and the small astronaut to the right. The larger one’s velocity will be less than the smaller one’s so he/she doesn’t let go of the rope they will either collide (elastically or inelastically) and thus never make it. m M = 1.5m Physics 207: Lecture 16, Pg 84 Example: Astronauts & Rope (2) However if the larger astronaut lets go of the rope he will get to the ship. (Too bad for the smaller astronaut!) In all cases the center of mass will remain fixed because they are initially at rest and there is no external force. To find the position where they meet all we need do is find the Center of Mass m M = 1.5m Physics 207: Lecture 16, Pg 85