1 Chapter 10 Spin and orbital motion Rotation: All around us: wheels, skaters, ballet, gymnasts, helicopter, rotors, mobile engines, CD disks, … Atomic world: electrons— “spin”, “orbit”. Universe: planets spin and orbiting the sun, galaxies spin, … Chapter 4 kinematics Chapter 10 dynamics 2 §10.1 Some concepts about rotation 1. Spin—describe rotational motion of a system about an axis through its center of mass. 2. Rigid body—a system composed of many pointlike particles that maintain fixed distances from each other at all time. each particle of the spinning rigid body system executes circular motion about the axis through the center of mass. 3. Orbital motion—the center of the mass of the system is moving in space from a perspective of a particular reference frame. The motion of the center of mass must not be circular. §10.1 Some concepts about rotation 3 4. The orbital angular momentum of a particle Define: vmrprL rrrr r ×=×= x y z m θ r r p r o ⊥ r L r ⊥ p Magnitude: prrprmvL ⊥⊥ === θsin Direction: pr rr and Perpendicular to the plane containing the §10.1 Some concepts about rotation Notice: 1 is measured with respect to the origin at O; 2 unit: kg·m 2 /s; 3 whatever the path or trajectory of a particle is straightline , curved path, closed orbital path, …. L r 5. The angular momentum of the circular orbital motion of a particle (a) Angular momentum §10.1 Some concepts about rotation 4 mvrL vmrprL = ×=×= rrrr r rv rv ω ω = ×= rrr r r p r o m ω r ωω 22 mrrmL ==then (b) Moment of inertia of a particle ω r r 2 mrL = §10.1 Some concepts about rotation 2 mrI = ωω rr r ImrL == 2 thendefine §10.2 The time rate of change of angular momentum and torque 1. The time rate of change of angular momentum for a single particle t p rp t r pr tt L d d d d )( d d d d r rr r rr r ×+×=×= prL rr r ×= total d d d d 0 d d Fr t p r t L vmvpvp t r r r r r r rrrrr r Q ×=×=∴ =×=×=× r r F r θ o m 5 Define: totaltotal Fr r rr ×=τ 2. torque is the position vector of the point of application of the force with respect to the chosen origin. r r Unit of the torque: N·m Magnitude: ⊥⊥ === rFFrrF θτ sin Direction: Perpendicular to the plane containing the Fr r r and r r F r θ o m dr = ⊥ ⊥ F §10.2 The time rate of change of angular momentum and torque r r F r θ o m dr = ⊥ ⊥ F If ,thecross ,0 ,or0 total OF F r r = = πθ 0 total =τ r §10.2 The time rate of change of angular momentum and torque Discussion: 6 3. Dynamics of circular orbital motion of a single particle ωω rr r Q ImrL == 2 totaltotal d d d d τ r r r r r r =×=×= Fr t p r t L Example 1: P 433 10.5 Example 2: P 433 10.6 α ω ωτ r r rr I t II t ===∴ d d )( d d total Can not be used in noncircular orbital motion. §10.2 The time rate of change of angular momentum and torque §10.3 The angular momentum of a system of particles and moment of inertia of rigid body 1. The angular momentum of a system of particles ∑ ∑ ∑ ×=×== ii i iiiiii vmrprLL rrrr rr ? ? ? ′ += ′ += iCMi iCMi vvv rrr rrr rrr Q θ i p r o CM r r i r r i m i r r ′ C () () ii i iCMi ii iiiCM iCMi ii iiiCM ii i iCM vmrvmrvmr vvmrvmr vmrrL ′ × ′ +× ′ +×= ′ +× ′ +×= × ′ += ∑∑∑ ∑∑ ∑ rrrrrr rrrrr rrr r ∴ 7 ∑ ×=× i CMCMiiCM vMrvmr rrrr First term: 0=×′=×′=×′ ∑∑ CMCMCM i iiCMi i i vrMvrmvmr rrrrrr Second term: The position vector of center of mass with respect to the center of mass ii i i vmr rr ′ × ′ ∑ Third term: is the vector sum of angular momentum of all particles with respect to the center of mass. §10.3 The angular momentum of a system of particles and moment of inertia of rigid body spinorbital LL vmrvMrL ii i iCMCM rr rrrr r += ′ × ′ +×= ∑ then 2. Spin angular momentum of a rigid body about a axis through the center of mass i v ′ r o ω r z o′ i m i r ′ r ⊥ ′ i r r //i r′ r ii i i vmrL rr r Q ′ × ′ = ∑spin ⊥⊥ ′ + ′ = ′′ ×= ′ iiiii rrrrv rrrrrr // ω )()( //spin ⊥⊥ ′ ×× ′ + ′ =∴ ∑ i i iii rrrmL rrrr r ω §10.3 The angular momentum of a system of particles and moment of inertia of rigid body 8 §10.3 The angular momentum of a system of particles and moment of inertia of rigid body ∑ ∑ ∑ ∑ ⊥ ⊥ ⊥⊥ ⊥ ′ + ′ ?= ′ ×× ′ + ′ ×× ′ = i ii i iii ii i i ii i i rm rrm rrm rrmL ω ω ω ω r r rrr rrr r 2 // //spin )( )( i v ′ r o ω r z o′ i m i r ′ r ⊥ ′ i r r //i r ′ r The vector is an involved vector summation. The rotation of an oddly shaped object about any axis of rotation is beyond the scope of this course. i v ′ r o ω r z o′ i m i r ′ r ⊥ ′ i r r //i r′ r 3. The moment of inertia or rotational inertia of a rigid body about a fixed axis through center of mass ∑∑ ⊥⊥ ′ ?= ′ ×× ′ i iiiii i i rrmrrm rrrr ωω )( // If 1the rigid body is symmetry about the axis; 2the axis is fixed. This term has no effect. Then ∑ ∑ ⊥ ⊥⊥ ′ = ′ ×× ′ = i ii ii i i rm rrmL ω ω r rrr r 2 spin )( §10.3 The angular momentum of a system of particles and moment of inertia of rigid body 9 i v ′ r o ω r z o′ i m i r ′ r ⊥ ′ i r r //i r′ r ∑ ⊥ ′ = i iiCM rmI 2 Define: This is the moment of inertia or rotational inertia of a rigid body about a fixed axis through center of mass The spin angular momentum of a rigid body ωω rr r )( 2 spin ∑ ⊥ == i iiCM rmIL §10.3 The angular momentum of a system of particles and moment of inertia of rigid body 4. The moment of inertia of various rigid bodies (a) Point particle 2 mrI = r –a distance from the axis of rotation (b) Collection of point particles ∑ ⊥ = i i i rmI 2 --the perpendicular distance of each mass m i from the axis of rotation ⊥i r §10.3 The angular momentum of a system of particles and moment of inertia of rigid body (c) Rigid body of distributive mass mrI d 2 ∫ ⊥ = 10 --the perpendicular distance of each mass dm from the axis of rotation ⊥ r =md λλ：densitylineardl σσ：densitysurfacedS ρρ：densityvolumdV The element of mass: §10.3 The angular momentum of a system of particles and moment of inertia of rigid body l ll l A m m2 m3 m4 m5 2 2 22 32 )2)(54( )2(32 ml lmm lmmlI = ++ += Example1: 5 particles are connected by 4 light staffs as shown in figure. Find the moment of the system with respect to the axis through point A, and perpendicular to the paper plane. Solution: 2 ⊥∑ = i i i rmI §10.3 The angular momentum of a system of particles and moment of inertia of rigid body 11 Example 2: There is a light thin staff of mass m and length L. Find the moment with respect to different axis. mx mrI d d 2 2 ∫ ∫ = = o x md x 2 L 2 L ? 2 33 3 2 2 2 12 1 883 1 2 2 3 1 d mL LL L m L L x L m x L m x L L = ? ? ? ? ? ? ? ? += ? == ∫ ? Solution: §10.3 The angular momentum of a system of particles and moment of inertia of rigid body L md o x x 23 0 222 3 1 03 1 ddd mL L x L m x L m xmxmrI L == === ∫∫∫ §10.3 The angular momentum of a system of particles and moment of inertia of rigid body 12 §10.3 The angular momentum of a system of particles and moment of inertia of rigid body Some rotational inertias Summary of the previous sections 1. The orbital angular momentum of a particle 2. The time rate of angular momentum for a single particle For a circular orbital motion of a particle ωω rr r ImrL == 2 totaltotal d d d d τ r r r r r r =×=×= Fr t p r t L Define: totaltotal Fr r rr ×=τ x y z m θ r r p r o ⊥ r L r ⊥ p r r total F r θ o m vmrprL rrrr r ×=×=Define: Rotational inertia r r p r o m ω r 13 For a circular orbital motion of a particle α ω ωτ r r rr I t II t === d d )( d d total 3. The angular momentum of a system of particles spinorbital LL vmrvMrL ii i iCMCM rr rrrr r += ′ × ′ +×= ∑ Summary of the previous sections ( ) ( ) iCMi i iCMii i i vvmrrvmrL ′ +× ′ +=×= ∑∑ rrrrrr r θ i p r o CM r r i r r i m i r r ′ C 4. The angular momentum of a rigid body spinning about a fixed axis through center of mass ωω rr r spin 2 spin )( IrmL i ii == ∑ ⊥ Summary of the previous sections i v r o ω r z o′ i m i r r ⊥i r r //i r r ∫ = mrI d 2 spin ∑ ⊥ = i ii rmI 2 spin The rotational inertia It represents the inertia of rotating body. 14 Some rotational inertias Summary of the previous sections 1. The time rate of change of the angular momentum of a system of particles A group of particles with masses , n mmm ,,, 21 L and angular momentums n LLL r L rr ,,, 21 ∑ =+++= i i LLLLL rr L rrr 321 The total angular momentum of the system is totaltotal d d d d τ r r r rr =×== ∑∑ i ii i i Fr t L t L Then §10.4 The dynamics of rigid body with a fixed axis totali,total d d τ r r r r =×= ii i Fr t L Recall 15 1 m 2 m 3 m 4 m 5 m 6 m 2. The torques due to internal and external forces 2 F r 1 F r 1 r r 2 r r o ∑∑∑ +== i i i i i i exintotal ττττ rrrr A system bounded with orange colour 0 )( )( in121 212121in = ×?= ×+×= Frr FrFr r rr r r r rr τ For any two element mass of the system m 1 and m 2 ∑ ≡ i in 0 i τ r therefore 12 F r 21 F r §10.4 The dynamics of rigid body with a fixed axis ∑∑∑ =×==×= i extexttotaltotal d d ii i ii i i FrFr t L ττ r r rr r r r 3. The dynamics of rigid body with a fixed axis--The rotational counterpart of Newton’s second law of a rigid body with fixed axis t L i d d i exttotal r rr == ∑ ττ ω r rr spinspin ILL == α ω τ r r r r spinspinexttotal d d d d I t I t L === §10.4 The dynamics of rigid body with a fixed axis The torque due to the internal forces do not change the angular momentum of a system. z zz I tt L α ω τ spinz d d d d === am t v m t p F r rr r === d d d d exttotal 16 Example 1: As shown in figure a uniform disk, with mass M=2.5 kg and radius R=20 cm, mounted on a fixed horizontal axle. A block with mass m=1.2 kg hangs from a massless cord that is wrapped around the rim of the disk. Find the acceleration of the falling block, the angular acceleration of the disk, and the tension in the cord. The cord does not slip, and there is no friction at the axle. §10.4 The dynamics of rigid body with a fixed axis gm r gM r N r Solution: 2 2 rad/s24 2.0 8.4 N0.6 2 1 m/s8.4 2 2 ?= ? == =?= ?= + ?= R a MaT mM m ga α )3( )2( 2 1 )1( 2 α α Ra MRRT mamgT = =? =? + + §10.4 The dynamics of rigid body with a fixed axis 17 Example 2: Find the acceleration of the falling block by direct application of . tL dd total r r =τ Solution: RmvIL z )(+= ω mRa R a MR mRaI t v mR t I mvRI t Rmg += += += += 2 2 1 d d d d )( d d )( α ω ω mM mg a 2 2 + = §10.4 The dynamics of rigid body with a fixed axis gm r N r g r Example 3: As shown in figure, one block has mass M=500 g, the other has mass m=460 g, and the pulley, which is mounted in horizontal frictionless bearings, has a radius of 5.00 cm. When released from rest, the heavier block falls h=75.0 cm in 5.00 s ( without the cord slipping ). (a) What is the magnitude of the blocks? (b) What are the tensions of the cord in both sides? (c) What is the magnitude of the pulley’s angular acceleration? (e) What is its rotational inertia? §10.4 The time rate of change of the spin angular momentum and dynamics of rigid body 18 Solution: gM r gm r 1 T r 1 T r 2 T r 2 T r + + 2 21 2 1 2 1 )( ath Ra IRTT mamgT MaTMg = = =? =? =? α α 22 2 2 1 22 mkg1038.1 1.2rad/s N54.4 N87.4 m/s100.6 ?×= = = = ×= ? ? I T T a α §10.4 The time rate of change of the spin angular momentum and dynamics of rigid body 1. The angular impulse-momentum theorem totaltotal d d d d τ r r r rr =×== ∑∑ i ii i i Fr t L t L Recall tL dd total τ r r = ∫∫ ==? f i f i t t L L tLL dd total τ r rr r r The angular impulse-momentum theorem Compare with the impulse-momentum theorem total d d F t p r r = ItFpp f i f i t t p p rr rr r r ===? ∫∫ dd total §10.5 The angular impulse-momentum theorem and the procession of a rapidly spinning top 19 §10.5 The angular impulse-momentum theorem and The procession of a rapidly spinning top Example 4:Two cylinder having radii R 1 and R 2 and rotational inertias I 1 and I 2 respectively, are supported by axes perpendicular to the plane of the screen. The two cylinders are initially rotating with angular speeds , respectively. The small cylinder is moved to the right until it touches the large one and the two cylinders rotate at constant rates in opposite directions. Find the final angular speeds of the two cylinders in terms of I 1 , I 2 , R 1 , R 2 , . 2010 andωω 2010 andωω 2 R 2 I 10 ω 1 R 1 I 1 O2 O 20 ω 2 R 2 I 1 ω 1 R 1 I 1 O 2 O 2 ω )3( )2(d )1(d 2211 2022222 1011111 RR IItRf IItRf ωω ωω ωω = ?= ?=? ∫ ∫ Solution: According to the angular impulse-momentum theorem Free body diagram: §10.5 The angular impulse-momentum theorem and The procession of a rapidly spinning top We can get 1 2 22 2 1 20 2 1210211 2 IRIR RIRRI + + = ωω ω 1 2 22 2 1 2021210 2 21 1 IRIR RRIRI + + = ωω ω 12 ff rr ?= 1 f r 222 111 Rf Rf = ?= τ τ 20 2. The procession of a rapidly spinning top 0 spin =L r If the spinning angular momentum is zero, i.e. jMgrgMr c ? sin total φτ =×= rrr The torque of the gravity with respect to the point O §10.5 The angular impulse-momentum theorem and The procession of a rapidly spinning top It will make the top fall down. If the spinning angular momentum is not zero, i.e. 0 spin ≠L r tL t L dd, d d totaltotal ττ r r r r == According to the angular impulse-momentum theorem C r r gM r spin L r x z §10.5 The angular impulse-momentum theorem and The procession of a rapidly spinning top ttL ?==? ∫ totaltotal d ττ rr r The change of the angular momentum is 21 The result is that the top moves in the direction perpendicular to the plane of symmetry axis and the local vertical direction. LgMr c r rrr ⊥×= total τbecause LLLLL rrrrr =′+=′ ? The total angular momentum §10.5 The angular impulse-momentum theorem and The procession of a rapidly spinning top φ φ C r r C r r φ φ θd L r ? L r ? 3. The angular speed of the procession φ θ θφ sin d d dsind spin spin L L LL = = §10.5 The angular impulse-momentum theorem and The procession of a rapidly spinning top φ φ θd spinspin spin procs sin sin sin d/d d d L Mgr L Mgr L tL t == == φ φ φ θ ω 22 4. The application of the procession of top ω o c ω r r r v r f r gm r §10.5 The angular impulse-momentum theorem and The procession of a rapidly spinning top Think! why does the bicycle wheel behave as shown in the following video film? §10.6 Simultaneous spin and orbital motion 2. Rotational kinetic energy of a system 1. The kinetic energy of a spinning system ∑∑ ⊥ ×== i iiii i rmvmKE 22 total )( 2 1 2 1 rr ω for spinning motion ⊥ = ii rv ω 222 rot 2 1 2 1 ωω CM i ii IrmKE == ∑ ⊥ for orbital motion of the center of mass 2 2 1 CM MvKE = 23 Total kinetic energy 22 spin 2 1 2 1 CMCM MvIKE += ω §10.6 Simultaneous spin and orbital motion Example 4: A uniform spherical shell of mass M and radius R rotates about a vertical axis on frictionless bearings. A massless cord passes around the equator of the shell, over a pulley of rotational inertial I and radius r, and is attached to a small object of mass m. There is no friction on pulley’s axle; the cord does not slip on the pulley. What is the speed of the object after it falls a distance h from rest? §10.6 Simultaneous spin and orbital motion 24 Solution: mghPE r v R v mvIMRKE = = = ++= pull sph 22 pull 2 sph 2 2 1 2 1 ) 3 2 ( 2 1 ω ω ωω According to the conservation law of mechanical energy PE=KE mMmrI gh M r I m mgh v 3/2/1 2 32 1 2 1 2 2 ++ = ++ = §10.6 Simultaneous spin and orbital motion 3. The total angular momentum (a) Spinning motion spinspin ω r r CM IL = (b) Orbital motion of center of mass orbit 2 orbit ω rrrrr r ⊥ =×=×= MrvMrprL CMCMCMCM (c) Total angular momentum of the system orbit 2 spinspinorbital ωω rr rrr ⊥ +=+= MrILLL CM Example : P450 10.10 §10.6 Simultaneous spin and orbital motion 25 4. Synchronous rotation If the spin angular velocity and the orbital angular velocity are parallel, and have the same magnitude, then, the motion is called synchronous rotation. ωωω rrr == orbitspin 5. Parallel axis theorem The total angular momentum for a synchronous rotation 2 orbit 2 spintotal MdII IMdIL CM CM += =+= ωωω rrr r §10.6 Simultaneous spin and orbital motion The parallel axis theorem: 2 MdII CM += The moment of inertia I of the system about an axis parallel to the symmetry axis of the system and separated from it by a perpendicular distance d is I. §10.6 Simultaneous spin and orbital motion 6. Rolling motion without slipping (a) The spin and orbital axes are parallel and separated by the radius of the circularly shaped system undergoing rolling motion. without slipping—ABS system? 26 (b) Rolling constrains: α ω ω θ θ R t Raa t v Rv t R t s Rs t ==== == = d d d d or d d d d center center center (c) Velocity rrr ′ += rrr center vvv ′ += rrr center r r center r r r ′ r o §10.6 Simultaneous spin and orbital motion (d) Total kinetic energy CM CM CMCM ImRI IImRKE IKEandmvKE += =+= == 2 2222 total 2 rot 2 trans 2 1 2 1 2 1 2 1 2 1 ωωω ω CM v CM v ωR ωR CM v2 ? C C C TTT BBB CM v CM v §10.6 Simultaneous spin and orbital motion + = 27 Example 1: P459 10.12, find the acceleration and the tension in the cord. R Solution: method 1: choose the center of mass of the body as the reference point. α α Ra ITR maTmg CM = = =? CM CM CM ImR mgI Tg ImR mR a + = + = 22 2 §10.6 Simultaneous spin and orbital motion T r gm r o R + method 2: choose the point P on the rim of the body as the reference point. R R a mRII ImgR CM = += = α α 2 CM CM CM ImR mgI Tg ImR mR a + = + = 22 2 The same results can be obtained §10.6 Simultaneous spin and orbital motion T r gm r o R + P 28 6. A general discussion about the angular momentum and the torque(§10.17) (a) Angular momentum about a point P (When both spin and orbital motion exist) i r ′ r P o i m P r ri r r PiiiPi rrrrrr rrrrrr ?= ′′ += or i i iiP vrmL ′ × ′ = ∑ rr r Pii Pii vvv t r t r t r rrr rrr ?= ′ ??= ′ d d d d d d )()( PiP i iiP vvrrmL rrrr r ?×?= ∑ §10.6 Simultaneous spin and orbital motion P i iiP P i ii i i i P P i ii ii P i i P P i ii i P i ii PiP i ii PiP i ii P arm arm t p r t v rrm t vm rr t v rrm t v rrm vv t rrm vvrr t m t L rrr rr r r r rr r rr r rr r rr rrrr rrrr r ×′?= ×′?×′= ×??×?= ×??×?= ?×?+ ?×?= ∑ ∑∑ ∑∑ ∑∑ ∑ ∑ )( )( d d d d )( d )d( )( d d )( d d )( )]( d d [)( )()]( d d [ d d τ (b) The time rate of change of the angular momentum about point P §10.6 Simultaneous spin and orbital motion 29 if 1P is in an inertial reference frame; 2P is the center of mass of the system; 3P has an acceleration parallel or antiparallel to the vector locating the center of mass. then t L P P d d r r =τ For a rigid body rotating about the axis through the center of mass αττ rr r r I t L P P === total d d §10.6 Simultaneous spin and orbital motion §10.7 Conservation of angular momentum ∑ = i LL itotal rr 1. The total angular momentum of a system of particles 2. The time rate of change of the angular momentum exttotal total d d τ r r = t L 3. Conservation of angular momentum If 0 exttotal =τ r then vectorconstant total =L r 30 §10.7 Conservation of angular momentum The diver’s angular momentum is constant The student decreasing his inertia to increase his angular speed. The angular momentum is constant. Examples: 2211 ωω II = §10.7 Conservation of angular momentum An idealized spacecraft containing a flywheel. If the flywheel is made to rotate clockwise as shown, the spacecraft itself will rotate counterclockwise. When the flywheel is braked to a stop, the spacecraft will also stop rotating but will be reoriented by the angle ?θ sc . 0 2211 =+ ωω rr II 31 §10.7 Conservation of angular momentum As shown in figure, the bicycle wheel whose rotational inertia about its central axis is I wh , the wheel is rotating at an angular speed ω wh counterclockwise. When the wheel is inverted, the student, the stool and the wheel’s center rotate together as a composite body about the stool’s rotation axis, with rotational inertia I b . With what angular speed and in what direction does the composite body rotate after the inversion of the wheel. §10.7 Conservation of angular momentum Exercise 1: 32 Solution: The angular momentum of the system is conserved b whwh b bbwhwhbf whwhwhwhbf whibiwhfbf I I IkIL kIkIL LLLL ω ω ωω ωω 2 ? 2 ? 0 ? = == +=? +=+ r r rrrr §10.7 Conservation of angular momentum Exercise 2: In the overhead view of the figure, four thin, uniform rods, each of mass M and length d= 0.50 m, are rigidly connected to a vertical axle to form a turnstile. The turnstile rotates clockwise about the axle, which is attached to a floor, with initial angular velocity ω i =2.0 rad/s. A mud ball of mass m=M/3 and initial speed v i =12 m/s is thrown along the path shown and sticks to the end of one rod. What is the final angular velocity ω f of the ball -turnstile system? §10.7 Conservation of angular momentum 33 Solution: Mechanical energy, linear momentum and angular momentum, which is conserved? iball,its,fball,fts, LLLL rrrr +=+ 2 ts 22 ts tsfts,tsits, 3 4 or ]) 2 ( 12 1 [4 MdI d MMdI ILIL fi = += == ωω o oo 60cos )60180sin( iball, dmv dmvL i i ?= ??= + ff mdIL ωω 2 ballfball, == §10.7 Conservation of angular momentum o 60cos 3 4 3 4 222 iiff mdvMdMdMd ?=+ ωωω rad/s80.0)60cos4( 5 1 ?=?= o iif vd d ωω iball,its,fball,fts, LLLL rrrr +=+ + §10.7 Conservation of angular momentum 34 Example 3:The particle of mass m in figure slides down the frictionless surface through height h and collides with the uniform vertical rod of mass M and length d, sticking to it. The rod pivots about point O through the angle θ before momentarily stopping. Find θ. §10.7 Conservation of angular momentum Solution: 22 3 1 2 mdMd ghmd + =ω §10.7 Conservation of angular momentum )cos 22 ()cos( )( 2 1 ) 2 ( 12 1 )( 2 1 22 rod 22 rod 2 rod 2 θθ ω ω dd Mgddmg mdI d MMdI mdImvd mvmgh ?+? =+ += += = ] )3)(2( 6 1[cos 2 1 MmMmd hm ++ ?= ? θ 35 Exercise 4:during a jump to his partner, an aerialist is to make a quadruple somersault lasting a time t=1.87 s. For the first and last quarter revolution, he is in the extended orientation shown in figure, with rotational inertia I 1 =19.9 kgnullm 2 around his center of mass. During the rest of the flight he is in a tight tuck, with rotational inertia I 2 =3.93kg null m 2 . What must be his angular speed around his center of mass during the tuck. §10.7 Conservation of angular momentum §10.7 Conservation of angular momentum Solution: There is no net external torque about his center of mass, his angular momentum about his center of mass is conserved. 1 22 12211 I I II ω ωωω == First and last quarter -revolution rev5.0 111 == tωθ rev5.3 222 == tωθ The rest of time 36 )( 1 s87.1 2 2 11 22 2 22 11 2 2 1 1 21 θ θ ωω θ ω θ ω θ ω θ +=+= =+=+= I I I I t ttt Then we have 10.14rad/srev/s23.3 2 ==ω §10.7 Conservation of angular momentum