Lecture 31 Angular Momentum; Conservation of Angular Momentum

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Lecture 31: Angular momentum; Conservation of Angular Momentum 1

description

Physics; re-up only; not mine

Transcript of Lecture 31 Angular Momentum; Conservation of Angular Momentum

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Lecture  31:  Angular  momentum;  Conservation  of  Angular  Momentum

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Lecture  Objectives:  1. Define  angular  momentum  for  a  point  particle  and  for  

a  rigid  body.  2. Explain  the  conditions  for  conservation  of  angular  

momentum.    3. Apply  conservation  of  angular  momentum  to  physical  

systems

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(linear)  momentum  p the  larger  the  momentum,  the  harder  to  stop

 Newton’s  First  Law

“an  object  at  rest  remains  at  rest,  at  object  moving  will  keep  on  moving…  until  something  (external  force)  stops  it”

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Anything  that  rotates  keeps  on  rotating  until  something  stops  it.  

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Angular  momentum

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Angular  momentum

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Angular  momentum

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The lightweight wheels on racing bikes have less angular momentum than those on recreational bikes, so it takes less effort to get them turning.

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Conservation  of  angular  momentum

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Sample  Problem  An  acrobatic  physics  professor  stands  at  the  center  of  a  turntable,  holding  his  arms  extended  horizontally  with  a  5.0kg  dumbbell  in  each  hand.  He  is  set  rotating  about  a  vertical  axis,  making  one  revolution  in  2.0s.  Find  the  prof’s  new  angular  velocity  if  he  pulls  the  dumbbells  in  to  his  stomach.  His  moment  of  inertial  (w/o  dumbbells)  is  3.0kgm2  when  his  arms  are  outstretched,  dropping  to  2.2kgm2  when  his  hands  are  in  his  stomach.  The  dumbbells  are  1.0m  from  the  axis  initially  and  0.20m  from  it  at  the  end.

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Using  conservation  of  momentum  and  solving  for  ω2z  :    

 

 

 

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K2  –  K1  =  256J  This  is  extra  energy  is  the  work  the  professor  did  to  pull  his  arms  downward

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Two  disks:  one  (A)  an  engine  flywheel  and  the  other  (B)  a  clutch  plate  attached  to  a  transmission  shaft.  Their  moment  of  inertia  are  IA  and  IB;  initially,  they  are  rotating  with  constant  angular  speeds  ωA  and  ωB  respectively.  We  then  pushed  the  disks  together  with  force  acting  along  the  axis,  so  as  not  to  apply  any  torque  on  either  disk.  The  disk  rub  against  each  other  and  eventually  reach  a  common  final  angular  speed  ω.  Derive  an  expression  for  ω.

Sample  Problem:  Rotational  collision

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Suppose  flywheel  A  has  mass  of  2.0kg,  a  radius  of  0.20m,  and  an  initial  angular  speed  of  50rad/s,  and  the  clutch  plate  B  has  mass  of  4.0kg,  radius  of  0.10m  and  initial  angular  speed  of  200rad/s.    Find  the  common  final  angular  speed  ω  after  the  disks  are  pushed  into  contact.  What  happened  to  the  kinetic  energy  during  this  process?

Sample  Problem:  Rotational  collision  II

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The  moments  of  inertia  of  the  two  disks  are:

From  the  previous  problem,  we  know  the  final  angular  speed,  substituting  the  values:

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Calculating  the  kinetic  energy  before  and  after  collision:

Therefore:  K1  >  K2    kinetic  energy  decreases  (just  as  we  would  expect  from  a  perfectly  inelastic  collision  ☺)

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Seatwork

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Use:  L  =  Iω  For  a  rod  at  rotating  at  its  end,  I  =  1/3  MR2  

2πrad  =  1rev

SW1:  Calculate  the  magnitude  of  L  SW  2:  Use  right  hand  rule  to  get  the  direction  of  L

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SW  3,  4  and  5

Apply  conservation  of  momentum:  Lbefore-­‐jump  =  Lafter-­‐jump  

Ibar  =  1/3Mbarr2

 Hint:  Just  after  the  bug  jumps,  it  has  angular  momentum  in  one  direction  and  the  bar  is  rotating  with  angular  velocity  ωB  in  the  opposite  direction

   

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1)  L  =  Iω  L  =  1/3  MR2  ω  L  =  1/3  (0.006kg)(0.15m)2  (1rev/min)  (2πrad/1rev)(1min/60s)  L  =  4.71x10-­‐6kgm/s2

 

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Apply  conservation  of  momentum  with  the  axis  at  the  nail.  Let  object  A  be  the  bug  and  object  B  be  the  bar.  Initially,  all  the  objects  are  at  rest  and  L1  =  0;  just  after  the  bug  jumps,  it  has  angular  momentum  in  one  direction  and  the  bar  is  rotating  with  angular  velocity  ωB  in  the  opposite  direction:  

 

 

 

 

Increase  in  KE  comes  from  the  work  done  by  the  bug  when  it  pushes  against  the  bar  in  order  to  jump  ☺