Week 11: Chapter 11

Home , Angular momentum, Environment (systems)

The Vector Product

 There are instances where the product of two Week 11: Chapter 11 vectors is another vector  Earlier we saw where the product of two vectors Angular Momentum was a scalar  This was called the dot product  The vector product of two vectors is also called the cross product

The Vector Product and Torque The Vector Product Defined    The torque vector lies in a  Given two vectors, A and B direction perpendicular to   the plane formed by the  The vector (cross) product of A and B is position vector and the   force vector defined as a third vector, CAB      Fr  C is read as “A cross B”  The torque is the vector (or cross) product of the  The magnitude of vector C is AB sin    position vector and the   is the angle between A and B force vector

More About the Vector Product Properties of the Vector Product

 The quantity AB sin  is  The vector product is not commutative. The equal to the area of the order in which the vectors are multiplied is parallelogram  formed important by A and B   To account for order, remember  The direction of C is   A BBA  perpendicular to the    plane formed by A and B  If A is parallel to B ( = 0o or 180o), then    The best way to A B  0 determine this direction   is to use the right-hand  Therefore A  A  0 rule

1 More Properties of the Vector Final Properties of the Vector Product Product     If A is perpendicular to B , then ABAB  The derivative of the cross product with  The vector product obeys the distributive law respect to some variable such as t is      ddA  dB ABCABAC x ( + ) = x + x AB  B A dt dt dt where it is important to preserve the   multiplicative order of A and B

Vector Products of Unit Vector Products of Unit Vectors Vectors, cont

 Signs are interchangeable in cross products ˆˆ ˆˆ ˆ ˆ   ii jjkk  0  A  -BAB   ˆîj ˆˆ jik ˆ  and î   ˆj  î ˆj ˆˆjk k ˆˆˆ j i kiˆˆ ˆîk ˆj

Using Determinants Vector Product Example    The cross product can be expressed as  Given A  23;ˆîjBij ˆˆ 2   ˆˆîjk  Find A B AAAA AA  yzˆˆxz xy ˆ ABAAAxyz  i  j  k  Result BByzBBxz BB xy BBBxyz   AB(2ˆˆ i  3 j )  ( ˆî 2 j )  Expanding the determinants gives  2()223()32ˆˆˆˆˆˆˆîiijji      jj   ˆˆ ˆ ABAByz  AB zy i  AB xz  AB zx j  AB xy  AB yx k 04kkˆˆ  3  07 k ˆ

2 Torque Vector Example Angular Momentum

 Given the force and location  Consider a particle of mass m located at the  ˆˆ vector position r and moving with linear Fij(2.00 3.00 ) N  rij (4.00ˆˆ 5.00 ) m momentum p  Find the net torque  Find the torque produced    dp   rF r  rF [(4.00ˆˆ i  5.00 j )N]  [(2.00 ˆˆi  3.00 j )m]  dt ˆˆ ˆˆ dr [(4.00)(2.00)ii (4.00)(3.00) ij Add the termp  sinceit 0 dt (5.00)(2.00)ˆˆji (5.00)(3.00) ˆˆi j   d()rp ˆ   2.0k N m  dt

Torque and Angular Angular Momentum, cont Momentum

 The instantaneous angular  The torque is related to the angular momentum  momentum L of a particle  Similar to the way force is related to linear momentum relative to the origin O is  defined as the cross dL   product of the particle’s  dt instantaneous position vector r and its  The torque acting on a particle is equal to the time instantaneous linear rate of change of the particle’s angular momentum  momentum p  This is the rotational analog of Newton’s Second    Lrp Law     and L must be measured about the same origin  This is valid for any origin fixed in an inertial frame

More About Angular Angular Momentum of a Momentum Particle, Example     The SI units of angular momentum are  The vector Lrp =  is (kg.m2)/ s pointed out of the diagram  The magnitude is  Both the magnitude and direction of the L = mvr sin 90o = mvr angular momentum depend on the choice of  sin 90o is used since v is origin perpendicular to r  A particle in uniform circular  The magnitude is L = mvr sin  motion has a constant   angular momentum about   is the angle between p and r  an axis through the center  The direction of L is perpendicular to the of its path plane formed by r and p

3 Angular Momentum of a Angular Momentum of a System of Particles System of Particles, cont

 The total angular momentum of a system of  Any torques associated with the internal forces particles is defined as the vector sum of the acting in a system of particles are zero  angular momenta of the individual particles dL   tot    Therefore,  ext   LLLtot12 L n L i dt i  Differentiating with respect to time  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 dLtot dLi   i the system about that origin dtii dt  This is the mathematical representation of the angular momentum version of the nonisolated system model.

Angular Momentum of a System of Particles, final System of Objects, Example

 The resultant torque acting on a system  The masses are about an axis through the center of mass connected by a light equals the time rate of change of angular cord that passes over a momentum of the system regardless of the pulley; find the linear motion of the center of mass acceleration  This applies even if the center of mass is  Conceptualize   accelerating, provided  and L are evaluated  The sphere falls, the relative to the center of mass pulley rotates and the block slides  Use angular momentum approach

Angular Momentum of a Angular Momentum of a Rotating Rigid Object Rotating Rigid Object, cont

 Each particle of the object  To find the angular momentum of the entire rotates in the xy plane about the z axis with an object, add the angular momenta of all the angular speed of  individual particles  The angular momentum of LL  mrI2   an individual particle is Li = zi  ii 2 ii mi ri     L and  are directed along  This also gives the rotational form of the z axis Newton’s Second Law dL d   z II  ext dt dt

4 Angular Momentum of a Angular Momentum of a Rotating Rigid Object, final Bowling Ball

 The rotational form of Newton’s Second Law is also  The momentum of valid for a rigid object rotating about a moving axis inertia of the ball is provided the moving axis: 2/5MR 2 (1) passes through the center of mass  The angular (2) is a symmetry axis momentum of the ball  If a symmetrical object rotates about a fixed axis is Lz = I passing through its center of mass, the vector form  The direction of the holds: L  I  angular momentum is  where L is the total angular momentum measured with in the positive z respect to the axis of rotation direction

Conservation of Angular Conservation of Angular Momentum Momentum, cont

 The total angular momentum of a system is constant  If the mass of an isolated system undergoes in both magnitude and direction if the resultant redistribution, the moment of inertia changes external torque acting on the system is zero  The conservation of angular momentum requires  Net torque = 0 -> means that the system is isolated  a compensating change in the angular velocity  LLLtot = constant or i = f   Ii i = If f = constant  For a system of particles, LL = = constant tot n  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

Clicker Question Conservation Law Summary

A person sitting on a rotating chair is holding two  For an isolated system - heavy dumbbells in his two hands. Initially, the position of dumbbells is in front of his chest. (1) Conservation of Energy: Suddenly, he extends his arms with dumbbells in hands. What will happen to this rotating system  Ei = Ef (chair-man-dumbbells)? Friction force is ignored. (2) Conservation of Linear Momentum: A. It will rotate faster.    ppif B. It will rotate more slowly. C. It will stop rotating. (3) Conservation of Angular Momentum:   D. Rotation speed will not change.  LLif E. Need more conditions to decide.

5 Conservation of Angular Momentum: The Merry-Go-Round Motion of a Top

 The moment of inertia of the  The only external forces system is the moment of acting on the top are the inertia of the platform plus normal force and the the moment of inertia of the gravitational force person  The direction of the angular  Assume the person can be momentum is along the axis treated as a particle of symmetry  As the person moves  The right-hand rule toward the center of the indicates that the torque is rotating platform, the in the xy plane angular speed will increase      To keep the angular  rF rM g momentum constant

Motion of a Top, cont Gyroscope

 The net torque and the angular momentum are  A gyroscope can be used to related:  illustrate precessional dL   motion dt  The gravitational force  A non-zero torque produces a change in the angular produces a torque about the momentum pivot, and this torque is  The result of the change in angular momentum is a perpendicular to the axle precession about the z axis  The normal force produces  The direction of the angular momentum is changing no torque  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

Gyroscope, cont Gyroscope, final

 The torque results in a  To simplify, assume the angular momentum change in angular momentum in a direction due to the motion of the center of mass about perpendicular to the axle. the pivot is zero  The axle sweeps out an angle d in a time interval  Therefore, the total angular momentum is due to dt. its spin  The direction, not the  magnitude, of the angular  This is a good approximation when is large momentum is changing  The gyroscope experiences precessional motion

6 Precessional Frequency Gyroscope in a Spacecraft

 Analyzing the previous vector triangle, the  The angular momentum of the rate at which the axle rotates about the spacecraft about its center of vertical axis can be found mass is zero  A gyroscope is set into dMgh   rotation, giving it a nonzero p dt I angular momentum  The spacecraft rotates in the   is the precessional frequency direction opposite to that of p the gyroscope  This is valid only when  <<  p  So the total momentum of the system remains zero

New Analysis Model 1 New Analysis Model 2

 Nonisolated System (Angular Momentum)  Isolated System (Angular Momentum)  If a system interacts with its environment in the  If a system experiences no external torque from sense that there is an external torque on the the environment, the total angular momentum of system, the net external torque acting on the the system is conserved:   system is equal to the time rate of change of its LL angular momentum: if   Applying this law of conservation of angular dL   tot momentum to a system whose moment of dt inertia changes gives

Iii = Iff = constant