Angular Momentum

July 19, 2010 Esys  W  Q   Chapter 11, Section 1-3 psys  Fnet t   Lsys  net t Angular Momentum

 Measures rotational motion – Both spin and orbits  Earth orbits the sun and spins on its axis   Is a vector L

funnel.sfsu.edu/courses/gm309/labs/seasons/facts.html Types of Angular Momentum

 Angular momentum is split into 2 types  Translational Angular Momentum – Angular Momentum relative to some point – Earth’s orbit around the sun  Rotational Angular Momentum – An object is spinning around some axis – A spinning top Translational Angular Momentum

 Defined relative to some point    LA  rA  p    Ltrans,A  rA p sin   Center of mass moves around some point  A cross product  Direction determined by Right Hand Rule Right Hand Rule Right Hand Rule 2 Clicker Question #1

 What direction is the angular momentum vector?

– A) Into the board P=mv – B) Out of The board – C) Left r – D) Right Clicker Question #2

 What direction is A x B? – A) Into the board – B) Out of The board – C) Left – D) Right θ B A Cross Products

 Multiply 2 vectors and get another vector  L will always be perpendicular to both r and p – For any cross product    L  r  p  L  ry pz  rz py , rz px  rx pz , rx py  ry px 

 The cross product will give the direction Sample Cross Product

 Put the vectors tail to tail  r  ryˆ  p  mvxˆ    L  r  p  0*(r)  0*(mv) mv*0  0*0 0*0  (r)mv   L  0 0 rmv 

 You get the correct direction for L by doing the cross product Translational Angular Momentum

 An object traveling in a straight line can still have angular momentum – Relative to some point – Important for dealing with collisions Rotational Angular Momentum

 Object spins around an axis through its center of mass – Bike tire, Earth   L  I     Formula derived from: L  r  p

– Find Ltrans for each particle in a multi-particle system and add them up   r  p  m1r1v1 sin1  m2r2v2 sin2  m3r3v3 sin2 

r  r sin 

v  r  2 2 2 Lrot  [m1r1  m2r2  m3r3 ]  I Translation and Moment of Inertia

 The two equations are equivalent    I  r  p  Rotational I is not the same as translational I 2 – Ex: a sphere I  Mr 2 rot 5 2 Itrans  MR

 Can have different ω Example: Ball on a String

 2 ways of solving  Translational    Ltrans  r  mv  mvr  Rotational   L  I   v   r I  mr 2  L  mrv  In both cases the direction is given by the right hand rule

upload.wikimedia.org/wikipedia/commons/9/97/Angular_momentum_circle.png Example: Bike Tire

 All mass is at the edge  Split into 20 parts    M M L  r  p  R vsin90  R2 trans 20 20  M L  20 R 2  MR2 tot 20  Ltot  I  L is into the page  Starting with Translational you can get rotational Total Angular Momentum

 Add the two together    Ltot  Ltrans  Lrot    Ltrans  rA  pcm   Lrot  I Example: The Earth

 The Earth has an orbit and spin  2 2 L  MR2  7.11033 Js rot 5 Earth 24hours  2 L  MR2  2.661040 Js trans Sun,Earth 365days

 Notice that both I and ω are different for each type of L Drawing 3D in 2D

 Angular momentum will always require 3D – It is a cross product  How to draw this easily?  Into the page – Arrow feathers going away  Out of Page – Arrow head coming towards you Clicker Question #3

    Which diagram is the correct representation of L  r  p – A – B C) p – C B) L – D r r L p A) p p D) L r L r