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Chapter 11 Angular Momentum 1 Angular Momentum In analogy to the principle of conservation of linear momentum for an isolated system, the angular momentum of a system is conserved if no external torques act on the system. 2 The Vector Product There are instances where the product of two vectors is another vector Earlier we saw where the product of two vectors was a scalar This was called the dot product The vector product of two vectors is also called the cross product 3 The Vector Product and Torque The torque vector lies in a direction perpendicular to the plane formed by the position vector and the force vector rF The torque is the vector (or cross) product of the position vector and the force vector 4 The Vector Product Defined Given two vectors, A and B The vector (cross) product of A and B is defined as a third vector, C A B C is read as “A cross B” The magnitude of vector C is AB sin q q is the angle between A and B 5 More About the Vector Product The quantity AB sin q is equal to the area of the parallelogram formed by A and B The direction of C is perpendicular to the plane formed by A and B The best way to determine this direction is to use the right-hand rule 6 Properties of the Vector Product The vector product is not commutative. The order in which the vectors are multiplied is important To account for order, remember A B B A If A is parallel to B (q = 0o or 180o), then A B 0 Therefore A A 0 7 More Properties of the Vector Product If A is perpendicular to B , then A B AB The vector product obeys the distributive law A x (B + C) = A x B + A x C 8 Final Properties of the Vector Product The derivative of the cross product with respect to some variable such as t is d dA dB A B B A dt dt dt where it is important to preserve the multiplicative order of A and B 9 Vector Products of Unit Vectors ˆi ˆi ˆj ˆj kˆ kˆ 0 ˆi ˆj ˆj ˆi kˆ ˆj kˆ kˆ ˆj ˆi kˆ ˆi ˆi kˆ ˆj 10 Vector Products of Unit Vectors, cont Signs are interchangeable in cross products A -B A B and ˆi ˆj ˆi ˆj 11 Using Determinants The cross product can be expressed as ˆi A B Ax Bx ˆj Ay By kˆ Ay Az By Bz Az ˆi Ax Bz Bx Az ˆ Ax j Bx Bz Ay kˆ By Expanding the determinants gives A B Ay Bz Az By ˆi Ax Bz AzBx ˆj Ax By Ay Bx kˆ 12 Vector Product Example Given A 2ˆi 3ˆj; B ˆi 2ˆj Find A B Result A B (2ˆi 3ˆj) ( ˆi 2ˆj) 2ˆi ( ˆi ) 2ˆi 2ˆj 3ˆj ( ˆi ) 3ˆj 2ˆj 0 4kˆ 3kˆ 0 7kˆ 13 Torque Vector Example Given the force and location F (2.00 ˆi 3.00 ˆj) N r (4.00 ˆi 5.00 ˆj) m Find the torque produced r F [(4.00ˆi 5.00ˆj)N] [(2.00ˆi 3.00ˆj)m] [(4.00)(2.00)ˆi ˆi (4.00)(3.00)ˆi ˆj (5.00)(2.00)ˆj ˆi (5.00)(3.00)ˆi ˆj 2.0 kˆ N m 14 Angular Momentum Consider a particle of mass m located at the vector position r and moving with linear momentum p Find the net torque dp r F r dt dr Add the term p sinceit 0 dt d (r p ) dt Similar to the equation: dp F dt Why? 15 Angular Momentum, cont The instantaneous angular momentum L of a particle relative to the origin O is defined as the cross product of the particle’s instantaneous position vector r and its instantaneous linear momentum p L r p 16 Torque and Angular Momentum The torque is related to the angular momentum Similar to the way force is related to linear momentum dL dt The torque acting on a particle is equal to the time rate of change of the particle’s angular momentum This is the rotational analog of Newton’s Second Law and L must be measured about the same origin This is valid for any origin fixed in an inertial frame 17 More About Angular Momentum The SI units of angular momentum are (kg.m2)/ s Both the magnitude and direction of the angular momentum depend on the choice of origin The magnitude is L = mvr sin f f is the angle between p and r The direction of L is perpendicular to the plane formed by r and p 18 Angular Momentum of a Particle, Example A particle moves in the xy plane in a circular path of radius r Find the magnitude and direction of its angular momentum relative to an axis through O when its velocity is v 19 Angular Momentum of a Particle, Example The linear momentum of the particle is changing in direction but not in magnitude. Is the angular momentum of the particle always changing? 20 Angular Momentum of a Particle, Example The vector L = r p is pointed out of the diagram The magnitude is L = mvr sin 90o = mvr sin 90o is used since v is perpendicular to r A particle in uniform circular motion has a constant angular momentum about an axis through the center of its path 21 Angular Momentum of a System of Particles The total angular momentum of a system of particles is defined as the vector sum of the angular momenta of the individual particles Ltot L1 L2 Ln Li i Differentiating with respect to time dLtot dLi i dt dt i i 22 Angular Momentum of a System of Particles, cont Any torques associated with the internal forces acting in a system of particles are zero Therefore, ext dL tot dt The net external torque acting on a system about some axis passing through an origin in an inertial frame equals the time rate of change of the total angular momentum of the system about that origin This is the mathematical representation of the angular momentum version of the nonisolated system model. 23 Angular Momentum of a System of Particles, final The resultant torque acting on a system about an axis through the center of mass equals the time rate of change of angular momentum of the system regardless of the motion of the center of mass This applies even if the center of mass is accelerating, provided and L are evaluated relative to the center of mass 24 System of Objects, Example A sphere of mass m1 and a block m2 are connected by a light cord that passes over a pulley M The mass of the thin rim is M. 25 System of Objects, Example The masses are connected by a light cord that passes over a pulley; find the linear acceleration Conceptualize M The sphere falls, the pulley rotates and the block slides Use angular momentum approach 26 System of Objects, Example cont Categorize The block, pulley and sphere are a nonisolated system Use an axis that corresponds to the axle of the pulley M Analyze At any instant of time, the sphere and the block have a common velocity v Write expressions for the total angular momentum and the net external torque L = m1vR + m2vR + MvR and ext dL m1 gR dt 27 System of Objects, Example final Analyze, cont Solve the expression for the linear acceleration M dL ext m1 gR dt d (m1 m2 M )vR dt (m1 m2 M ) Ra Solve a! 28 Angular Momentum of a Rotating Rigid Object Each particle of the object rotates in the xy plane about the z axis with an angular speed of w The angular momentum of an individual particle is Li = mi ri2 w L and w are directed along the z axis 29 Angular Momentum of a Rotating Rigid Object, cont To find the angular momentum of the entire object, add the angular momenta of all the individual particles Lz Li mi ri 2 w Iw i i This also gives the rotational form of Newton’s Second Law ext dLz dw I I dt dt 30 Angular Momentum of a Rotating Rigid Object, final The rotational form of Newton’s Second Law is also valid for a rigid object rotating about a moving axis provided the moving axis: (1) passes through the center of mass (2) is a symmetry axis If a symmetrical object rotates about a fixed axis passing through its center of mass, the vector form holds: L Iw where L is the total angular momentum measured with respect to the axis of rotation 31 Angular Momentum of a Bowling Ball Estimate the magnitude of the angular momentum of a bowling ball spinning at 10 rev/s. 32 Angular Momentum of a Bowling Ball The momentum of inertia of the ball is 2/5MR 2 The angular momentum of the ball is Lz = Iw The direction of the angular momentum is in the positive z direction 33 Conservation of Angular Momentum The total angular momentum of a system is constant in both magnitude and direction if the resultant external torque acting on the system is zero Net torque = 0 -> means that the system is isolated Ltot = constant or Li = Lf For a system of particles, L tot = n L n = constant 34 Conservation of Angular Momentum, cont If the mass of an isolated system undergoes redistribution, the moment of inertia changes The conservation of angular momentum requires a compensating change in the angular velocity Ii wi = If wf = constant This holds for rotation about a fixed axis and for rotation about an axis through the center of mass of a moving system The net torque must be zero in any case 35 Conservation Law Summary For an isolated system (1) Conservation of Energy: Ei = Ef (2) Conservation of Linear Momentum: p i pf (3) Conservation of Angular Momentum: L i Lf 36 Conservation of Angular Momentum: The Merry-Go-Round A horizontal platform in the shape of a circular disk rotates freely in a horizontal plane about a frictionless vertical axle. The platform has a mass M=100kg and a radius R=2m. A student with mass 60kg walks slowly from the rim of the disk toward its center. If the angular speed of the system is 2 rad/s when the student is at the rim, what is the angular speed when she reaches a point r=0.5m from the center? 37 Conservation of Angular Momentum: The Merry-Go-Round The moment of inertia of the system is the moment of inertia of the platform plus the moment of inertia of the person Assume the person can be treated as a particle As the person moves toward the center of the rotating platform, the angular speed will increase Why? To keep the angular momentum constant 38 Conservation of Angular Momentum: The Merry-Go-Round 1 I i I pi I si MR 2 mR 2 2 I f I pf I sf I iwi I f w f 1 MR 2 mr 2 2 1 1 MR 2 mR 2 wi MR 2 mr 2 w f 2 2 Solve ωf 39 Conservation of Angular Momentum: The Merry-Go-Round What if you measured the kinetic energy of the system before and after the student walks inward? Are the initial kinetic energy and the final kinetic energy the 40 Conservation of Angular Momentum: The Merry-Go-Round 1 K i I iwi2 880 J 2 1 K f I f w 2f 1810 J 2 The kinetic energy increases. The student must do work to move herself closer to the center of rotation. This extra kinetic energy comes from chemical potential energy in the student’s body. 41 Motion of a Top The only external forces acting on the top are the normal force and the gravitational force The direction of the angular momentum is along the axis of symmetry The right-hand rule indicates that the torque is in the xy plane r F r Mg 42 Motion of a Top The direction of L parallel to that of is L f L Li the top precesses about the z-axis 43 Motion of a Top, cont The net torque and the angular momentum are related: dL dt A non-zero torque produces a change in the angular momentum The result of the change in angular momentum is a precession about the z axis The direction of the angular momentum is changing The precessional motion is the motion of the symmetry axis about the vertical The precession is usually slow relative to the spinning motion of the top 44 Gyroscope A gyroscope can be used to illustrate precessional motion The gravitational force produces a torque about the pivot, and this torque is perpendicular to the axle The normal force produces no torque dL dt 45 Gyroscope, cont The torque results in a change in angular momentum in a direction perpendicular to the axle. The axle sweeps out an angle df in a time interval dt. The direction, not the magnitude, of the angular momentum is changing dL dt Mghdt sin( df ) df L L L df Mgh wp dt Iw wp: precessional frequency The gyroscope experiences precessional motion 46 Gyroscope, final To simplify, assume the angular momentum due to the motion of the center of mass about the pivot is zero Therefore, the total angular momentum is due to its spin This is a good approximation when w is large 47 Precessional Frequency Analyzing the previous vector triangle, the rate at which the axle rotates about the vertical axis can be found df Mgh wp dt Iw wp is the precessional frequency This is valid only when wp << w 48 Gyroscope in a Spacecraft The angular momentum of the spacecraft about its center of mass is zero A gyroscope is set into rotation, giving it a nonzero angular momentum The spacecraft rotates in the direction opposite to that of the gyroscope So the total momentum of the system remains zero 49 New Analysis Model 1 Nonisolated System (Angular Momentum) If a system interacts with its environment in the sense that there is an external torque on the system, the net external torque acting on the system is equal to the time rate of change of its angular momentum: dLtot dt 50 New Analysis Model 2 Isolated System (Angular Momentum) If a system experiences no external torque from the environment, the total angular momentum of the system is conserved: L i Lf Applying this law of conservation of angular momentum to a system whose moment of inertia changes gives Iiwi = Ifwf = constant 51