# Chapter 10 Angular Momentum

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2 96 Chapte 0 (b) The nose o the plane tends to vee downwad. The change in angula momentum Δ pop o the popelle is to the ight, so the net toque τ on the popelle is towad the ight as well. The popelle must exet an equal but opposite toque on the plane. This letwad diected toque exeted by the popelle on the plane tends to cause a letwad-diected change in angula momentum o the plane. This means the plane tends to otate clockwise as viewed om the ight. 7 You ae sitting on a spinning piano stool with you ams olded. (a) When you extend you ams out to the side, what happens to you kinetic enegy? What is the cause o this change? (b) Explain what happens to you moment o inetia, angula speed and angula momentum as you extend you ams. Detemine the Concept The otational kinetic enegy o the you-stool system is given by K ot ω. Because the net toque acting on the you-stool system is zeo, its angula momentum is conseved. (a) You kinetic enegy deceases. nceasing you moment o inetia while conseving you angula momentum deceases you kinetic enegy K. (b) Extending you ams out to the side inceases you moment o inetia and deceases you angula speed. The angula momentum o the system is unchanged. Estimation and Appoximation 9 An ice-skate stats he piouette with ams outstetched, otating at.5 ev/s. Estimate he otational speed (in evolutions pe second) when she bings he ams lat against he body. Pictue the Poblem Because we have no inomation egading the mass o the skate, we ll assume that he body mass (not including he ams) is 50 kg and that each am has a mass o 4.0 kg. et s also assume that he ams ae.0 m long and that he body is cylindical with a adius o 0. Because the net extenal toque acting on he is zeo, he angula momentum will emain constant duing he piouette. Because the net extenal toque acting on he is zeo: Δ i 0 o ω ω 0 () ams in amsin amsout amsout

3 Angula Momentum 97 Expess he total moment o inetia with he ams outstetched: Teating he body as though it is cylindical, calculate the moment o inetia o he body, minus he ams: Modeling he ams as though they ae ods, calculate thei moment o inetia when they ae outstetched: + ams out body ams m body.00 kg m ams ( 50kg)( 0.0m) [ ( 4kg)(.0m) ].67 kg m Substitute to detemine he total moment o inetia with he ams outstetched: Expess he total moment o inetia with he ams lat against he body: ams out ams in.00 kg m.67 kg m + body + ams.00 kg m. kg m +.67 kg m [( 4.0kg)( 0.0m) ] Solve equation () o ω ams in to obtain: amsout ω ams in ω amsin amsout Substitute numeical values and evaluate ω : ams in ω ams in.67 kg m. kg m 4ev/s (.5ev/s) A.0-g paticle moves at a constant speed o.0 mm/s aound a cicle o adius 4.0 mm. (a) Find the magnitude o the angula momentum o the paticle. (b) l( l + )h, whee l is an intege, ind the value o l( l+) and the appoximate value o l. (c) By how much does l change i the paticle s speed inceases by one-millionth o a pecent, and nothing else changing? Use you esult to explain why the quantization o angula momentum is not noticed in macoscopic physics. Pictue the Poblem We can use mv to ind the angula momentum o the paticle. n (b) we can solve the equation l( l + )h o l( l +) and the appoximate value o l.

4 98 Chapte 0 (a) Use the deinition o angula momentum to obtain: mv.40 0 (.0 0 kg)(.0 0 m/s)( m) kg m /s.4 0 kg m /s (b) Solve the equation l( l + )h o l ( l +) : l + () h ( l ) Substitute numeical values and l + evaluate l ( ): l( l ).40 0 kg m /s.05 0 J s Because l >>, appoximate its value with the squae oot o l l + : ( ) l. 0 6 (c) The change in l is: the paticle s speed inceases by one-millionth o a pecent while nothing else changes: Δ l l l () new v v + 0 and + 0 v ( + 0 )v ( + 0 ) Equation () becomes: l new and l new ( ) [( + 0 ) ] l + new ( + 0 ) Substituting in equation () yields: ( + 0 ) h Δl l new l 0 h h h h Substitute numeical values and evaluate Δ l : Δl.40 0 kg m /s.05 0 J s and 8 l l. 0 Δ 6 %

5 Angula Momentum 99 The quantization o angula momentum is not noticed in macoscopic physics because no expeiment can detect a actional change in l o 0 6 %. The Coss Poduct and the Vecto Natue o Toque and Rotation 7 A oce o magnitude F is applied hoizontally in the negative x diection to the im o a disk o adius R as shown in Figue 0-4. Wite F and in tems o the unit vectos ˆ i, j ˆ, and k ˆ, and compute the toque poduced by this oce about the oigin at the cente o the disk. Pictue the Poblem We can expess F and in tems o the unit vectos î and ĵ and then use the deinition o the coss poduct to ind τ. Expess F in tems o F and the unit vecto : î F Fiˆ Expess in tems o R and the unit vecto : ĵ Rˆj ( ) ( ) F τ F FR ˆj iˆ FR iˆ ˆj Calculate the coss poduct o and : Toque and Angula Momentum FR kˆ 7 A.0-kg paticle moves diectly eastwad at a constant speed o 4.5 m/s along an east-west line. (a) What is its angula momentum (including diection) about a point that lies 6.0 m noth o the line? (b) What is its angula momentum (including diection) about a point that lies 6.0 m south o the line? (c) What is its angula momentum (including diection) about a point that lies 6.0 m diectly east o the paticle? Pictue the Poblem The angula momentum o the paticle is p whee is the vecto locating the paticle elative to the eeence point and p is the paticle s linea momentum. (a) The magnitude o the paticle s angula momentum is given by: Substitute numeical values and evaluate : psinφ mvsinφ mv (.0kg)( 4.5m/s)( 6.0m) 54kg m /s ( sinφ)

6 00 Chapte 0 Use a ight-hand ule to establish the diection o : 54kg m /s, upwad (b) Because the distance to the line along which the paticle is moving is the same, only the diection o dies: (c) Because p 0 o a point on the line along which the paticle is moving: 54kg m /s, downwad 0 45 n Figue 0-46, the incline is ictionless and the sting passes though the cente o mass o each block. The pulley has a moment o inetia and adius R. (a) Find the net toque acting on the system (the two masses, sting, and pulley) about the cente o the pulley. (b) Wite an expession o the total angula momentum o the system about the cente o the pulley. Assume the masses ae moving with a speed v. (c) Find the acceleation o the masses by using you esults o Pats (a) and (b) and by setting the net toque equal to the ate o change o the system s angula momentum. Pictue the Poblem et the system include the pulley, sting, and the blocks and assume that the mass o the sting is negligible. The angula momentum o this system changes because a net toque acts on it. (a) Expess the net toque about the cente o mass o the pulley: Rg( m sinθ m ) (b) Expess the total angula momentum o the system about an axis though the cente o the pulley: τ net Rm g sinθ Rm g whee we have taken clockwise to be pos to be consistent with a positive upwad velocity o the block whose mass is m as indicated in the igue. ω + m vr + m vr vr R + m + m (c) Expess τ as the time deivative o the angula momentum: d τ dt ar R d dt vr R + m + m + m + m

7 Equate this esult to that o Pat (a) and solve o a to obtain: a Angula Momentum 0 ( m θ m ) g R sin + m + m Consevation o Angula Momentum 49 You stand on a ictionless platom that is otating at an angula speed o.5 ev/s. You ams ae outstetched, and you hold a heavy weight in each hand. The moment o inetia o you, the extended weights, and the platom is 6.0 kg m. When you pull the weights in towad you body, the moment o inetia deceases to.8 kg m. (a) What is the esulting angula speed o the platom? (b) What is the change in kinetic enegy o the system? (c) Whee did this incease in enegy come om? Pictue the Poblem et the system consist o you, the extended weights, and the platom. Because the net extenal toque acting on this system is zeo, its angula momentum emains constant duing the pulling in o the weights. (a) Using consevation o angula momentum, elate the initial and inal angula speeds o the system to its initial and inal moments o inetia: Substitute numeical values and evaluate ω : (b) Expess the change in the kinetic enegy o the system: i i ωi ω 0 ω ωi 6.0kg m ω.8kg m Δ K K K i (.5ev/s) 5.0ev/s ω ω i i Substitute numeical values and evaluate ΔK: ΔK ev π ad (.8kg m ) 5.0 ( 6.0kg m ) 0.6 kj s ev ev π ad.5 s ev (c) Because no extenal agent does wok on the system, the enegy comes om you intenal enegy.

8 0 Chapte 0 5 A lazy Susan consists o a heavy plastic disk mounted on a ictionless beaing esting on a vetical shat though its cente. The cylinde has a adius R 5 and mass M 0.5 kg. A cockoach (mass m 0.05 kg) is on the lazy Susan, at a distance o 8.0 om the cente. Both the cockoach and the lazy Susan ae initially at est. The cockoach then walks along a cicula path concentic with the cente o the azy Susan at a constant distance o 8.0 om the axis o the shat. the speed o the cockoach with espect to the lazy Susan is 0.00 m/s, what is the speed o the cockoach with espect to the oom? Pictue the Poblem Because the net extenal toque acting on the lazy Susancockoach system is zeo, the net angula momentum o the system is constant (equal to zeo because the lazy Susan is initially at est) and we can use consevation o angula momentum to ind the angula velocity ω o the lazy Susan. The speed o the cockoach elative to the loo v is the dieence between its speed with espect to the lazy Susan and the speed o the lazy Susan at the location o the cockoach with espect to the loo. Relate the speed o the cockoach with espect to the loo v to the speed o the lazy Susan at the location o the cockoach: Use consevation o angula momentum to obtain: Expess the angula momentum o the lazy Susan: Expess the angula momentum o the cockoach: Substitute o S and C in equation () to obtain: Solving o ω yields: v v ω () 0 () S C ω S C S MR ω v CωC m ω v MR ω m ω ω mv MR + m 0 Substitute o ω in equation () to m v obtain: v v MR + m Substitute numeical values and evaluate v : ( 0.05kg)( m) ( 0.00 m/s) ( 0.5m)( 0.5m) + ( 0.05kg)( m) v 0.00 m/s 0 mm/s

9 Angula Momentum 0 Remaks: Because the moment o inetia o the lazy Susan is so much lage than the moment o inetia o the cockoach, ate the cockoach begins moving, the angula speed o the lazy Susan is vey small. Theeoe, the speed o the cockoach elative to the loo is almost the same as the speed elative to the lazy Susan. *Quantization o Angula Momentum 55 The z component o the spin o an electon is h, but the magnitude o the spin vecto is 0.75h. What is the angle between the electon s spin angula momentum vecto and the positive z-axis? Pictue the Poblem The electon s spin angula momentum vecto is elated to its z component as shown in the diagam. The angle between s and the positive z-axis is φ. z h φ θ 0.75h s h Expess φ in tems o θ to obtain: φ 80 θ Using tigonomety, elate the magnitude o s to its z component: h θ cos 0.75h Substitute o θ in the expession o φ to obtain: h θ 80 cos 0.75h 5 57 You wok in a bio-chemical eseach lab, whee you ae investigating the otational enegy levels o the HB molecule. Ate consulting the peiodic chat, you know that the mass o the bomine atom is 80 times that o the hydogen atom. Consequently, in calculating the otational motion o the molecule, you assume, to a good appoximation, that the B nucleus emains stationay as the H atom (mass kg) evolves aound it. You also know that the sepaation between the H atom and bomine nucleus is 0.44 nm. Calculate (a) the moment o inetia o the HB molecule about the bomine nucleus, and (b) the otational enegies o the bomine nucleus s gound state (lowest enegy) l 0, and the next two states o highe enegy (called the ist and second excited states) descibed by l, and l.

10 04 Chapte 0 Pictue the Poblem The otational enegies o HB molecule ae elated to l and E0 accoding to K l l( l +) E 0 whee E0 h. (a) Neglecting the motion o the bomine molecule: Substitute numeical values and evaluate HB : HB mp HB m H 7 9 (.67 0 kg)( m) kg m kg m (b) Relate the otational enegies to l and E : 0 ( ) K l l l + E 0 whee E 0 h HB Substitute numeical values and evaluate E : 0 E 0 h meV 4 ( J s) 47 ( 0 kg m ).46 ev J J Evaluate E 0 to obtain: E K meV Evaluate E to obtain: E K ( + )(.00meV).0meV Evaluate E to obtain: E K ( + )(.00meV) Collisions with Rotations 6.0meV 6 Figue 0-5 shows a thin uniom ba o length and mass M and a small blob o putty o mass m. The system is suppoted by a ictionless hoizontal suace. The putty moves to the ight with velocity v, stikes the ba at a distance d om the cente o the ba, and sticks to the ba at the point o contact. Obtain expessions o the velocity o the system s cente o mass and o the angula speed ollowing the collision. Pictue the Poblem The velocity o the cente o mass o the ba-blob system does not change duing the collision and so we can calculate it beoe the collision

11 Angula Momentum 05 using its deinition. Because thee ae no extenal oces o toques acting on the ba-blob system, both linea and angula momentum ae conseved in the collision. et the diection the blob o putty is moving initially be the +x diection. et lowe-case lettes ee to the blob o putty and uppe-case lettes ee to the ba. The diagam to the let shows the blob o putty appoaching the ba and the diagam to the ight shows the ba-blob system otating about its cente o mass and tanslating ate the peectly inelastic collision. M M d y d v m v h ω The velocity o the cente o mass beoe the collision is given by: Using its deinition, expess the location o the cente o mass elative to the cente o the ba: Expess the angula momentum, elative to the cente o mass, o the ba-blob system: Expess the angula momentum about the cente o mass: ( M m) v mv + MV + o, because V 0, m v v M + m md ( M + m) y md y M + m below the cente o the ba. ω ω () mv ( d y ) md mmvd mv d M + m M + m

12 06 Chapte 0 Using the paallel axis theoem, expess the moment o inetia o the system elative to its cente o mass: M + My + m d ( y ) Substitute o y and simpliy to obtain: md md M + M + m d M + m M + m Substitute o and in equation ω mmd M + M + m mmvd M M + m + Mmd () and simpliy to obtain: ( ) Remaks: You can veiy the expession o by letting m 0 to obtain M and letting M 0 to obtain A uniom od o length and mass M equal to 0.75 kg is attached to a hinge o negligible mass at one end and is ee to otate in the vetical plane (Figue 0-55). The od is eleased om est in the position shown. A paticle o mass m 0.50 kg is suppoted by a thin sting o length om the hinge. The paticle sticks to the od on contact. What should the atio / be so that θ max 60 ate the collision? Pictue the Poblem Assume that thee is no iction between the od and the hinge. Because the net extenal toque acting on the system is zeo, angula momentum is conseved in this peectly inelastic collision. The od, on its downwad swing, acquies otational kinetic enegy. Angula momentum is conseved in the peectly inelastic collision with the paticle and the otational kinetic o the ate-collision system is then tansomed into gavitational potential enegy as the od-plus-paticle swing upwad. et the zeo o gavitational potential enegy be at a distance below the pivot and use both angula momentum and mechanical enegy consevation to elate the distances and and the masses M and m. Use consevation o enegy to elate the initial and inal potential enegy o the od to its otational kinetic enegy just beoe it collides with the paticle: K Ki + U U i o, because K i 0, K + U U 0 i 0

13 Angula Momentum 07 Substitute o K, U, and U i to obtain: ( M ) ω + Mg Mg 0 Solving o ω yields: ω g etting ω epesent the angula speed o the od-and-paticle system just ate impact, use consevation o angula momentum to elate the angula momenta beoe and ate the collision: Solve o ω to obtain: Use consevation o enegy to elate the otational kinetic enegy o the od-plus-paticle just ate thei collision to thei potential enegy when they have swung though an angle θ max : Δ i 0 o M + m ω' M ω ( ) ( ) 0 ω' M + m K M ω Ki + U U i 0 Because K 0: ( )( cosθ ) + mg ( cosθ ) 0 ω' + Mg () max max Expess the moment o inetia o the system with espect to the pivot: M + m Substitute o θ max, and ω in equation (): g ( M ) M + m Mg ( ) + mg Simpliy to obtain: m m m M M M

14 08 Chapte 0 Dividing both sides o the equation by yields: m 6 + m m + M M M et α m/m and β / to obtain: 6α β + αβ + αβ 0 Substitute o α and simpliy to obtain the cubic equation in β: 8β + 6β + 4β 0 Use the solve unction o you calculato to ind the only eal value o β: Pecession β A bicycle wheel that has a adius equal to 8 is mounted at the middle o an axle 50 long. The tie and im weigh 0 N. The wheel is spun at ev/s, and the axle is then placed in a hoizontal position with one end esting on a pivot. (a) What is the angula momentum due to the spinning o the wheel? (Teat the wheel as a hoop.) (b) What is the angula velocity o pecession? (c) How long does it take o the axle to swing though 60 aound the pivot? (d) What is the angula momentum associated with the motion o the cente o mass, that is, due to the pecession? n what diection is this angula momentum? Pictue the Poblem We can detemine the angula momentum o the wheel and the angula velocity o its pecession om thei deinitions. The peiod o the pecessional motion can be ound om its angula velocity and the angula momentum associated with the motion o the cente o mass om its deinition. (a) Using the deinition o angula momentum, expess the angula momentum o the spinning wheel: w ω MR ω R ω g Substitute numeical values and 0 N 9.8m/s ev π ad s ev 8.J s 8J s ( 0.8m) evaluate :

15 Angula Momentum 09 (b) Using its deinition, expess the angula velocity o pecession: Substitute numeical values and evaluate ω p : d φ MgD ω p dt ω p ( 0 N)( 0.5m) 8.J s 0.4ad/s 0.44 ad/s (c) Expess the peiod o the pecessional motion as a unction o the angula velocity o pecession: (d) Expess the angula momentum o the cente o mass due to the pecession: π π T ω 0.44 ad/s p p ω MD ω p p 5s Substitute numeical values and evaluate p 0 N 9.8m/s J s : p ( 0.5m) ( 0.44 ad/s) Geneal Poblems The diection o p is eithe up o down, depending on the diection o. 7 A paticle whose mass is.0 kg moves in the xy plane with velocity v (.0 m / s)ˆ i along the line y 5. m. (a) Find the angula momentum about the oigin when the paticle is at ( m, 5. m). (b) A oce F (.9 N) ˆ i is applied to the paticle. Find the toque about the oigin due to this oce as the paticle passes though the point ( m, 5. m). Pictue the Poblem While the -kg paticle is moving in a staight line, it has angula momentum given by p whee is its position vecto and p is its linea momentum. The toque due to the applied oce is given by τ F. (a) The angula momentum o the paticle is given by: p

16 0 Chapte 0 m ˆ + 5.m Expess the vectos and p : ( ) i ( )j and p mviˆ.0kg ( 9.0kg m/s)iˆ ( )(.0 m/s) ˆ iˆ Substitute o and p :and simpliy to ind : (b) Using its deinition, expess the toque due to the oce: Substitute o and F and simpliy to ind τ : [( m) iˆ + ( 5.m) ˆj ] ( 9.0kg m/s) ( 47.7 kg m /s)( ˆj iˆ ) τ F τ ( 48kg m /s)kˆ [( m) iˆ + ( 5.m) ˆj ] (.9 N) ( 5.9 N m)( ˆj iˆ ) ( N m)kˆ iˆ iˆ 77 Repeat Poblem 76, this time iction between the disks and the walls o the cylinde is not negligible. Howeve, the coeicient o iction is not geat enough to pevent the disks om eaching the ends o the cylinde. Can the inal kinetic enegy o the system be detemined without knowing the coeicient o kinetic iction? Detemine the Concept Yes. The solution depends only upon consevation o angula momentum o the system, so it depends only upon the initial and inal moments o inetia. 79 Keple s second law states: The line om the cente o the Sun to the cente o a planet sweeps out equal aeas in equal times. Show that this law ollows diectly om the law o consevation o angula momentum and the act that the oce o gavitational attaction between a planet and the Sun acts along the line joining the centes o the two celestial objects. Pictue the Poblem The pictoial epesentation shows an elliptical obit. The tiangula element o the aea is da ( dθ ) d θ.

17 Angula Momentum dθ da θ Dieentiate da with espect to t to obtain: Because the gavitational oce acts along the line joining the two objects, τ 0. Hence: θ da d ω () dt dt m ω constant () Eliminate ω between equations () and () to obtain: da dt m constant 8 The tem pecession o the equinoxes ees to the act that Eath s spin axis does not stay ixed but sweeps out a cone once evey 6,000 y. (This explains why ou pole sta, Polais, will not emain the pole sta oeve.) The eason o this instability is that Eath is a giant gyoscope. The spin axis o Eath pecesses because o the toques exeted on it by the gavitational oces o the Sun and moon. The angle between the diection o Eath s spin axis and the nomal to the ecliptic plane (the plane o Eath s obit) is.5 degees. Calculate an appoximate value o this toque, given that the peiod o otation o Eath is.00 d and its moment o inetia is kg m. Deleted: the eath Pictue the Poblem et ω P be the angula velocity o pecession o Eath-asgyoscope, ω s its angula velocity about its spin axis, and its moment o inetia with espect to an axis though its poles, and elate ω P to ω s and using its deinition. Use its deinition to expess the pecession ate o Eath as a giant gyoscope: Substitute o and solve o τ to obtain: τ ω P τ ω P ωω P

18 Chapte 0 The angula velocity ω s o Eath about its spin axis is given by: Substitute o ω to obtain: ω π whee T is the peiod o T otation o Eath. π ωp τ T Substitute numeical values and evaluateτ: τ π 7 ( kg m )( s ) N m 4h 600s d d h

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