Page 1 of 6Conservation of Angular Momentum = "Spin"We can assign a direction to the angular velocity: direction of = direction of axis + right hand rule (with right hand, curl fingers in direction of rotation, thumb pointsin direction of .) The direction of is abstract. There is nothingactually moving in the direction of the vector. We want to choose the direction of so that this gives us information about how the object is spinning. When a wheel is spinning on an axis, there is no one direction of motion (different parts of the wheel are moving in different directions). But there is always only one direction of the axis of rotation. So by giving both the direction of the axis (direction of and how fast the object is spinning around that axis (magnitude of ddtqw = w =r r) , we are describing its spinning motion exactly.Similarly, we can assign a direction to angular acceleration :2 12 1ddt t t tw - ww Dwa = =D -v vv vr;Direction of is direction of , not the direction of (likedirection of a is direction of v, not the direction of v)Can also assign direction to torque with cross product.The cross-product of two vectors A and B is a third vector A B C� =v vv defined like this: The magnitude is| A B| A Bsin� = qvv. The direction of C A B= �v vvis the direction perpendicular to the plane defined by the vectors A and B , plus right-hand-rule. (Curl fingers from first vector A to second vector B, thumb points in direction of A B�vv3/25/2009 © M.Dubson, University of Colorado at BoulderABABspinning, and slowingxyzi^j^k^Page 2 of 6ˆ ˆ ˆ ˆ ˆi j k i i 0ˆ ˆ ˆ ˆ ˆj k i j j 0 etc.� = � =� = � =Vector torque defined as ˆr Ft = �vvThe vector torque is defined with respect to an origin (which is usually, but not always, the axis of rotation). So, if you change the origin, you change the torque (since changing the origin changes the position vector r).With these definitions of vector angular acceleration and vector torque, the fixed-axis equationIt = a becomes dI Idtwt = a =vvv ( like dvF m a mdt= =vvv)Now, a new concept: angular momentum L = "spin". Angular momentum, avector, is the rotational analogue of linear momentum. So, based on ouranalogy between translation and rotation, we expect L I= wvv ( likep m v=vv) . Note that this equation implies that the direction of L is thedirection of . Definition of angular momentum of a particle with momentum p = mv at position r relative to an origin is L r p� �vv v3/25/2009 © M.Dubson, University of Colorado at BoulderroriginFpoint of application of force (out of page)= vector out of page= vector into pageLroriginpparticle L (out of page)Page 3 of 6Like torque , the angular momentum L is defined w.r.t. an origin, often the axis of rotation.We now show that the total angular momentum of a object spinning about a fixed axis isL I= wvv. Consider a object spinning about an axis pointing along the +z direction. We place the orgin at the axis.tot i i i i i ii i i2i i i i ii itotL L r p r (m v )ˆ ˆ ˆz r m v z m r z IL I= = � = �= = w = w= w� � �� �v vv v vvvvIf something has a big moment of inertia I and is spinning fast (big ), then it has a big "spin", big angular momentum. Angular momentum is a very useful concept, because angular momentum is conserved.Important fact: the angular momentum of a object spinning about an axis that passes through the center of mass is given by L = Icmindependent of the location of the origin; that is, even if the origin is chosen to be outside the spinning object, the angular momentum has the same value as if the origin was chosen to be at the axis. (Proof not given here).Conservation of Angular Momentum: If a system is isolated from external torques, then its total angular momentum L is constant.ext = 0 Ltot = I = constant ( like Fext = 0 ptot = constant )Here is a proof of conservation of angular momentum: First, we argue that netd Ld tt =vv( this is like netd pFd t=vv ) :ii i i ii iid pd L d d rr p p r .d t d t d t d t� �� �= � = � + �� �� �� �� �� �rrrr r r r 3/25/2009 © M.Dubson, University of Colorado at Boulderrimivixyzvi = riaxisPage 4 of 6Now, the first term in the last expression is zero:( ) ( )i i i i i i ii i iid rp v m v m v v 0d t� �� = � = � =� �� �� � �rrr r r r , since any vector crossed into itself is zero. So, we have ( )ii i ii id pd Lr r Fd t d t� �= � = �� �� �� �rrrr r (since netd pFd t=rr) . Finally,( )i i i neti ir F� = t = t� �rrr r, so we have netd Ld t= trr. So now we have, netLtDt =Drr if net = 0 , then L0 L constanttD= � =Drr. Done.It turns out that only 4 things are conserved: Energy Linear momentum p Angular momentum L Charge qConservation of Angular Momentum is very useful for analyzing the motion of spinning objects isolated from external torques — like a skater or a spinning star.If ext = 0 , then L = I = constant. If I decreases, must increase, to keep L = constant.Example: spinning skater.Example: rotation of collapsing star. A star shines by converting hydrogen (H) into helium (He) in a nuclear reaction. When the H is used up, the nuclear fire stops, and gravity causes the star to collapse inward.3/25/2009 © M.Dubson, University of Colorado at BoulderIi i = If f( I big, small ) ( I small, big )Page 5 of 6As the star collapses (pulls its arms in), the star rotates faster and faster. Star's radius can get much smaller: Ri 1 million miles Rf 30 miles( ) ( )22i i f f52 22 2i i f f5 52 2i i f f2i f i2f i fI I (Sphere I = M R )M R M RR RR T2( using = 2 f )R T Tw = ww = ww = wwp= = w p =wIf Ri >> Rf, then Ti >>>Tf . The sun rotates once every 27 days. "Neutron stars" with diameter of about 30 miles typically rotates 100 time per second.Let's review the correspondence between translational and rotational motionTranslation Rotation x xvtD=D = tDqwD vatD=D = tDwaD F = r F M I = m r2Fnet = M a net = I KEtrans = (1/2)M v2 KErot = (1/2 ) I 2p = …
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