Physics I Class 15 Conservation of Angular Momentum Rev. 23-Feb-04 GB 15-1 Angular Momentum of a Particle Review center of rotation (defined) r p m v A n g u la r m o m e n tu m o f a p a r tic le o n c e a c e n te r is d e fin e d : l r p ( W h a t is th e d ir e c tio n o f a n g u la r m o m e n tu m h e r e ? ) Once we define a center (or axis) of rotation, any object with a linear momentum that does not move directly through that point has an angular momentum defined relative to the chosen center. 15-2 Angular Momentum of a Particle Angular Momentum of an Object F o r a s o l i d o b j e c t , e a c h a t o m h a s i t s o w n a n g u l a r m o m e n t u m : l i ri p i ri ( m
i v i) T h e d ire c tio n is th e s a m e a s th e d ire c tio n o f a n g u la r v e lo c ity . T h e m a g n itu d e is | l i | | r i | | p i | sin( so l i m i ri ) m i | ri | | v i | m i ri ri m i ri 2 2 T h e to ta l a n g u la r m o m e n tu m , s u m m in g a ll a to m s , is L
l i m i 2 ri I 15-3 How Does Angular Momentum of a Particle Change with Time? Take the time derivative of angular momentum: d l d d r dp (r p) p r dt dt dt dt Find each term separately: so dr p vp 0 (Why?) dt dp r r Fnet net (Why?) dt
dl net (Newtons 2nd Law for angular momentum.) dt 15-4 Angular Momentum of a Particle: Does It Change if = 0? Y (0,0) r (blue) r (red) X p m v = 1 k g m / s ( + X d i r . )
The figure at the left shows the same particle at two different times. No forces (or torques) act on the particle. Is its angular momentum constant? (Check magnitudes at the two times.) (4,3) B l u e a n g l e : = 9 0 2 l = r p s i n ( ) = ( 3 ) ( 1 ) s i n
( 9 0 ) = 3 k g m / s (0,3) R e d a n g l e : = a r c t a n ( 3 / 4 ) = 3 6 . 8 7
2 l = r p s i n ( ) = ( 5 ) ( 1 ) s i n ( 3 6 . 8 7 ) = 3 k g m / s p [ r
s i n ( ) ] i s t h e c o m p o n e n t o f r a t a r i g h t a n g l e t o . I t i s
c o n s t a n t . I t i s a l s o t h e d i s t a n c e a t c l o s e s t a p p r o a
c h t o t h e c e n t e r . 15-5 Conservation of Angular Momentum Take (for exam ple) two rotating objects that interact. dl1 on1from2 exton1 dt dl 2 on2 from1 exton2 dt The total angular m omentum is the sum of 1 and 2: dL d l 1 d l 2 exton1 exton2 (Why?) dt dt dt If there are
no external torques, then dL 0 dt 15-6 Example 1 A n ic e s k a te r s p in s a t 6 ra d /s e c w ith o u t-s tre tc h e d h a n d s . H e r r o ta tio n a l in e r tia is 1 .5 k g m 2 . S h e th e n p u lls h e r a r m s in , th e r e b y c h a n g in g h e r r o ta tio n a l in e r tia l to 1 .2 k g m 2. W h a t is h e r a n g u la r s p e e d n o w ? N o e x te rn a l to rq u e , s o L re m a in s c o n s ta n t I before after before I before L I after I after before after 1 .5 6 7 .5 ra d /s e c 1 .2 15-7
Example 2 A w h e e l is ro ta tin g fre e ly w ith a n a n g u la r s p e e d o f 3 0 ra d /s e c o n a s h a ft w h o s e ro ta tio n a l in e rtia is n e g lig ib le . A s e c o n d w h e e l, in itia lly a t re s t a n d w ith tw ic e th e ro ta tio n a l in e rtia o f th e firs t is s u d d e n ly c o u p le d to th e s a m e s h a ft. W h a t is th e a n g u la r s p e e d o f th e re s u lta n t c o m b in a tio n o f th e s h a ft a n d tw o w h e e ls ? N o e x te rn a l to rq u e , s o L re m a in s c o n s ta n t I1 after L I1 before after I2 after I 1 before I 1 30 1 before 10 r a d / s e c I1 I 2 I1 2 I1 3 15-8 Class #15 Take-Away Concepts r p.
1. Angular momentumof aparticle(review): l 2. Newtons 2nd Lawfor angular momentum: dl net dt 3. Conservationof angular momentum(noext. torque): dL 0 dt 15-9 Class #15 Problems of the Day ___1. When a woman on a frictionless rotating turntable extends her arms out horizontally, her angular momentum: A. must increase B. must decrease C. must remain the same D. may increase or decrease depending on her initial angular velocity E. changes into kinetic energy 15-10 Answer to Problem 1 for Class #15 The answer is C, angular momentum stays the same. There is no external torque, so there is no way to change the angular momentum. The womans rotational inertia increases, but her angular speed decreases so that the angular momentum remains the same. There is no way to change momentum into energy! 15-11 Class #15 Problems of the Day CHALLENGE PROBLEM
2. Two ice skaters of equal mass perform the following trick: Skater A is at rest on the ice while skater B approaches. As skater B passes by at 10 m/s, his center of mass is 1.8 m from skater As center of mass at the instant of closest approach. At that instant, the skaters reach out and clasp each others hands. Find the rotational speed of the skaters, find the speed of their center of mass, and describe the subsequent path of the center of mass in terms of geometric shape. Treat the skaters as point masses and ignore the friction of the skates on the ice. 15-12 Answer to Problem 2 for Class #15 The system consists of the two skaters. There are no external forces in this system, so we can use conservation of linear and angular momentum. Let the mass of each skater be m. The magnitude of the initial linear momentum is 10 m. The total mass of the system is m+m. Therefore, the speed of the center of mass = (10 m) / (m+m) = 5 meters/s. (before = after) The magnitude of the initial angular momentum about the center of mass at the instant the skaters clasp hands is 0.9 x 10 m = 9 m. After clasping hands, the total rotational inertia of the skaters is 2 m (0.9)2 = 1.62 m. Therefore, = 9 m / 1.62 m = 5.56 rad/s. The path of the center of mass after clasping hands is a straight line, not some kind of cycloid or other curve. 15-13 Activity #15 - Conservation of Angular Momentum Objective of the Activity: 1. 2. 3. Think about conservation of angular momentum. Use conservation of momentum to predict the change in rotational speed in a simple system. Compare measurements with predictions. 15-14
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