Rotation and Conservation of Angular Momentum

Transcription

1 Chapter 4. Rotaton and Conservaton of Angular Momentum Notes: Most of the materal n ths chapter s taken from Young and Freedman, Chaps. 9 and Angular Velocty and Acceleraton We have already brefly dscussed rotatonal moton n Chapter when we sought to derve an expresson for the centrpetal acceleraton n cases nvolvng crcular moton (see Secton.4 and equaton (.46)). We wll revst these notons here but wth a somewhat broader scope. We rentroduce some basc relatons between an angle of rotaton θ about some fxed axs, the radus, and the arc traced by the radus over the angle θ. Fgure shows these relatonshps. Frst, the natural angular unt s the radan, not the degree as one mght have expected. The defnton of the radan s such that t s the angle for whch the radus r and the arc s have the same length (see Fgure a). The crcumference of a crcle equals π tmes the radus; t therefore follows that rad = 360 π = (4.) Second, as we prevously saw n Chap., the angle s expressed wth θ = s r (4.) and can be seen from Fgure b. We can defne an average angular velocty as the rato of an angular change Δθ over a Fgure The relatons between an angle of rotaton about some fxed axs, the radus, and the arc traced by the radus over the angle

2 tme nterval Δt ω ave,z = Δθ z Δt. (4.3) For example, an object that accomplshes one complete rotaton n one second has an average angular velocty (also sometmes called average angular frequency) of π rad/s. If we make these ntervals nfntesmal, then we can defne the nstantaneous angular velocty (or frequency) wth Δθ ω z = lm z Δt 0 Δt = dθ z dt. (4.4) That s, the nstantaneous angular velocty s the tme dervatve of the angular dsplacement. The reason for the presence of the subscrpt z n equatons (4.3) and (4.4) wll soon be made clearer. It should be noted that an angular dsplacement Δθ can ether be postve or negatve; t s a matter of conventon how the sgn s defned. We wll defne an angular dsplacement as postve when t s effected n a counter-clockwse drecton, as seen from an observer, when the fxed about whch the rotaton s done s pontng n the drecton of the observer. Ths s perhaps more easly vsualzed wth Fgure. 4.. Vector Notaton Snce a rotaton s defned n relaton to some fxed axs, t should perhaps not be too surprsng that we can use a vector notaton for angular dsplacements. That s, just as we can defne a vector Δr composed of lnear dsplacements along the three ndependent Fgure Conventon for the sgn of an angle

3 axes n Cartesan coordnates wth Δr = Δx e x + Δye y + Δze z, (4.5) we can do the same for an angular dsplacement vector Δθ wth Δθ = Δθ x e x + Δθ y e y + Δθ z e z. (4.6) It s understood that n equaton (4.6) Δθ x s an angular dsplacement about the fxed x-axs, etc. The notaton used n equatons (4.3) and (4.4) s now understood as meanng that the angular dsplacement and velocty are about the fxed z-axs. An example s shown n Fgure 3, along wth the so-called rght-hand rule, for the angular velocty vector ω = dθ dt. (4.7) The ntroducton of a vector notaton has many benefts and smplfes the form of several relatons that we wll encounter. A frst example s that of the nfntesmal arc vector dr that results for an nfntesmal rotaton vector dθ of a rgd body (please note that we have ntentonally replaced s for the fnte arc n equaton (4.) wth dr and not ds ). Let us consder the specal case shown n Fgure 4 where an nfntesmal rotaton dθ = dθ z e z about the z-axs s effected on a vector r = r e x algned along the x-axs. As can be seen from the fgure, the resultng nfntesmal arc dr = dr e y wll be orented along the y-axs. We know from equaton (4.) that dr = rdθ, (4.8) Fgure 3 Shown s the vector representaton for an angular velocty about the, along wth the socalled rght-hand rule

4 y (+z) r dθz dr x Fgure 4 - Infntesmal rotaton of a rgd body about the. but how can we mathematcally determne the orentaton of the nfntesmal arc from that of the rotaton and radus? To do so, we must ntroduce the cross product between two vectors. Let a and b two vectors such that a = a x e x + a y e y + a z e z b = b x e x + b y e y + b z e z. (4.9) Then we defne the cross product It s mportant to note that a b = ( a y b z a z b y )e x + ( a z b x a x b z )e y + ( a x b y a y b x )e z. (4.0) It s then straghtforward to establsh the followng a b = b a. (4.) e x e y = e z e y e z = e x e z e x = e y, (4.) and e e = 0, (4.3) where = x, y, or z. Comng back to our smple example of Fgure 4, and consderng equatons (4.8) and (4.) we fnd that dr e y = dθ z e z r e x. (4.4) Although equaton (4.4) results from a specal case where the orentaton of the dfferent vectors was specfed a pror, ths relaton can be generalzed wth

5 dr = dθ r, (4.5) as could readly verfed by changng the orentaton of r and dθ n Fgure 4. We therefore realze that the nfntesmal arc dr represents the change n the radus vector r under a rotaton dθ ; hence the chosen notaton. Moreover, we can fnd a vector generalzaton of equaton (.44) n Chapter that establshed the relatonshp between the lnear and angular veloctes by dvdng by an nfntesmal tme nterval dt on both sdes of equaton (4.5). We then fnd where v = dr dt and ω = dθ dt. v = ω r, (4.6) Just as we dd for the angular velocty n equatons (4.3) and (4.4) (but usng a vector notaton), we can defne an average angular acceleraton over a tme nterval Δt wth α ave = Δω Δt (4.7) and an nstantaneous angular acceleraton wth Δω α = lm Δt 0 Δt = dω dt. (4.8) Combnng equatons (4.7) and (4.8), we can also express the nstantaneous angular acceleraton as the second order tme dervatve of the angular dsplacement α = dω dt = d dθ dt dt = d θ dt. (4.9) 4.. Constant Angular Acceleraton We have so far observed a perfect correspondence between the angular dsplacement dθ, velocty ω, and acceleraton α wth ther lnear counterparts dr, v, and a. We also studed n Secton. of Chapter the case of a constant lnear acceleraton and found that

6 ( ) = a (constant) ( ) = v 0 + at a t v t (4.0) r( t) = r 0 + v 0 t + at, wth r 0 and v 0 the ntal poston and velocty, respectvely. We further combned these equatons to derve the followng relaton v t ( ) v 0 = a r( t) r 0. (4.) Because the relatonshp between θ, ω, and α s the same as that between r, v, and a, we can wrte down smlar equatons for the case of constant angular acceleraton ( ) = α (constant) ( ) = ω 0 + αt α t ω t (4.) θ ( t) = θ 0 + ω 0 t + αt and ω t ( ) ω 0 = α θ ( t) θ 0 (4.3) wthout dervng them, snce the process would be dentcal to the one we went through for the constant lnear acceleraton case Lnear Acceleraton of a Rotatng Rgd Body We prevously derved equaton (4.6) for the lnear velocty of a rotatng rgd body. We could thnk, for example, of a sold, rotatng dsk and focus on the trajectory of a pont on ts surface. Snce ths pont, whch at a gven nstant has the velocty v, does not move lnearly but rotates, there must be a force pullng t toward the centre pont of the dsk. Ths leads us to consder, one more tme, the concept of the centrpetal acceleraton dscussed n Chapter (see Secton.4). It s, however, possble to use equaton (4.6) to combne two types of accelerated motons. To do so, we take ts tme dervatve dv dt = d ( ω r) dt = dω dt r + ω dr dt = α r + ω v. (4.4)

7 On the second lne of equaton (4.4) we use the fact that the dervatve of a product of functons, say, f and g, yelds d dt ( fg) = df dt g + f dg dt. (4.5) The same result holds for the scalar or cross products between two vectors a and b. We can verfy ths as follows d dt ( a b) = d dt ( a x b x + a y b y + a z b z ) = d ( a x b x ) + d dt dt ( a y b y ) + d dt ( a z b z ) = da x dt b db x + a x x + da y dt dt b db y + a y y + da z dt dt b db z + a z z dt (4.6) = da x dt b x + da y dt b y + da z dt b db z + a x db x + a y db y + a z z dt dt dt = da db b + a dt dt and f we consder the x component of a b (see equaton (4.0)) d dt ( a b) x = d dt ( a y b z a z b y ) = d ( a y b z ) d dt dt = da y dt b + a z y ( a z b y ) db z dt = da y dt b da z z dt b y + a y = da dt b x + a db dt da z dt b y + a z x db z dt a z. db y dt db y dt (4.7) If we also consder smlar solutons for the y and z, then we have d dt ( a b) = da dt b + a db dt. (4.8) Returnng to equaton (4.4), we nsert equaton (4.6) back nto t to get a = α r + ω ( ω r). (4.9)

8 The term wthn square brackets can be further expanded usng the followng dentty (whch could be proven combnng equaton (4.0) and the expanson for the scalar product) We then fnd that a ( b c) = ( a c)b ( a b)c. (4.30) a = α r + ( ω r)ω ( ω ω )r = α r + ( ω r)ω ω r. (4.3) Let us now examne the last two terms on the rght-hand sde of second of equatons (4.3). Frst, we break down the poston vector nto two parts r = r + r, (4.3) where r and r are the parts of r that are parallel and perpendcular to the orentaton of the angular velocty vector ω, respectvely. The perpendcular component r s smply the dstance of the pont under consderaton from the axs of rotaton. It follows that we can wrte for the second term ( ω r)ω = ( ωr )ω = ω r. (4.33) If we know combne ths equaton wth the thrd term of equaton (4.3), then we fnd that ( ω r)ω ω r = ω r ω r ( ) = ω r r = ω r. (4.34) Equaton (4.3) for the lnear acceleraton of a pont on a rgd body becomes a = α r ω r. (4.35) The frst term on the rght-hand sde of ths equaton s the tangental acceleraton due to the angular acceleraton a tan = α r. (4.36)

9 We should note that only the perpendcular component r contrbutes to the magntude of the tangental acceleraton, because of the presence of the cross product. We can verfy ths as follows α r = α ( r + r ) = α r + α r, (4.37) but α r = 0 snce α and r are parallel to one another. For our rgd body, we then rewrte equaton (4.35) as whle the tangental acceleraton becomes and a = α r ω r, (4.38) a tan = α r (4.39) a tan = αr. (4.40) The second term on the rght-hand sde of equaton (4.35) s the radal acceleraton due to the angular acceleraton a rad = ω r. (4.4) Ths acceleraton s nothng more than the vector form of the centrpetal acceleraton dscussed n Chapter for crcular motons. The mnus sgn n equaton (4.4) ndcated that the acceleraton s drected toward the orgn (or the axs of rotaton). The magntude of the radal acceleraton s a rad = ω r, (4.4) whch s the same result obtaned wth equaton (.46). Accordngly, we could have worked out ths analyss by frst notcng that, for rgd body, equaton (4.6) smplfes to v = ω r = ω r + r = ω r. ( ) (4.43) It would have then become clear from the onset that only the perpendcular component r partakes n the analyss

10 4..4 Exercses. (Prob. 9.7 n Young and Freedman.) A safety devce brngs the blade of a power mower from an ntal angular speed of ω to rest n.00 revoluton. At the same constant angular acceleraton, how many revolutons would t take the blade to come to rest from an ntal angular speed ω that was three tmes as great (ω = 3ω ). Soluton. We use equaton (4.3) (n one dmenson) to relate the necessary quanttes. That s, ω = α π, (4.44) where the π corresponds to one revoluton (.e., θ = π ). We therefore have For the second angular speed we have α = ω 4π. (4.45) ω = αθ = ω 4π θ, (4.46) or θ = π ω ω = 8π. (4.47) It therefore takes 9.00 revolutons to stop the blade.. (Prob. 9. n Young and Freedman.) You are to desgn a rotatng cylndrcal axle to lft 800-N buckets of cement from the ground to a rooftop 78.0 m above the ground. The buckets wll be attached to the free end of a cable that wraps around the rm of the axle; as the axle turns the buckets wll rse. (a) What should the dameter of the axle be n order to rase the buckets at a steady.00 cm/s when t s turnng at 7.5 rpm? (b) If nstead the axle must gve the buckets an upward acceleraton of m/s, what should the angular acceleraton of the axle be? Soluton

11 (a) We know that the magntude of the tangental velocty at rm s v = ωr (4.48) and wll also be the speed at whch the buckets wll rse. We therefore have r = v ω = 0.0 m/s π rad/s =.55 cm. (4.49) (b) The angular acceleraton can be evaluated wth equaton (4.40) α = a tan r 0.4 m/s = m = 5.7 rad/s. (4.50) 4. Moment of Inerta and Rotatonal Knetc Energy We wll once agan concentrate on a gven pont on or n our rotatng rgd body located at poston r, the subscrpt dentfes the partcle located at that pont. We now calculate the knetc energy assocated wth ths partcle wth K = m v = m ( ω r ), (4.5) where we used equaton (4.6) for the velocty. We wll now make use of the followng ( a b) ( c d) = ( a c) ( b d) ( a d) ( b c), (4.5) and equaton (4.5) becomes (wth a = c = ω and b = d = r ) K = m ( ω r ) ω r ( ) ( ) = m ω r ω r ( ) = m ω r ωr,, (4.53) - 7 -

12 and snce, as before, whle because r and r are perpendcular to one another r = r, + r, (4.54) we fnally fnd r = r, + r,, (4.55) K = m r, ω. (4.56) We now defne a new quantty I = m r, (4.57) and we rewrte equaton (4.53) as K = I ω. (4.58) and we see that I serves the same role for the rotatonal knetc energy of the partcle as the mass does for the knetc energy due to lnear moton. If we now sum over all partcles that compose the rgd body, we fnd for the total rotatonal knetc energy K = = Iω, I ω (4.59) where we ntroduced the moment of nerta of the rgd body I = I = m r,. (4.60) The moment of nerta s a functon of the geometry of the rgd body as well as the dstrbuton of the matter wthn t. And as was mentoned above, ts role for rotatonal motons s smlar to that of the mass when dealng wth lnear motons. Ths mples, - 7 -

13 among other thngs, that the greater the moment of nerta, the harder t s to start the body rotatng from rest (or slowng t down when already rotatng). The last of equatons (4.60) would only be useful n calculatng the moment of nerta for cases where the rgd body s made of a dscrete arrangement of partcles. For a body made from a contnuous dstrbuton of matter, the summaton n equaton (4.60) s replaced by an ntegral I = r dm, (4.6) where dm s an nfntesmal element of mass located a dstance r away from the axs of rotaton. If we defne ρ has the mass densty of the body (n kg/m 3 ) and dv as the elemental volume where dm s located, then we can wrte I = r ρ dv. (4.6) The calculaton of ths ntegral for dfferent shapes and geometres of objects s beyond the scope of our study. But one mportant aspect that comes out from such analyses s that, for a gven body (e.g., a sphere or a cube), the poston of the axes about whch the moment of nerta s calculated (.e., the axs of rotaton n our case) wll affect the value of the ntegral n equaton (4.6). Examples of moments of nerta for a few rgd bodes and axs postons are shown n Fgure 5. Fgure 5 The moment nerta for dfferent geometres of rgd bodes

14 4.3 Gravtatonal Potental Energy of a Rgd Body We can proceed n a smlar manner to determne the gravtatonal potental energy of a rgd body as we dd for the rotatonal knetc energy. That s, we once agan concentrate on a gven pont on or n our rotatng rgd body located at poston r, the subscrpt dentfes the partcle located at that pont whose gravtatonal potental energy s where as usual U = m gy, (4.63) r = x e x + y e y + z e z, (4.64) or y = r e y. We can determne the total gravtatonal potental energy by summng over all the partcles that make the rgd body U = U = g m y. (4.65) Referrng to equaton (3.56) n Chapter 3, where we defned the centre of mass of a body, we can transform equaton (4.65) to U = Mgy cm, (4.66) where M = m s the total mass of the body. The total gravtatonal potental energy of an extended, rgd body s calculated as f all ts mass was concentrated at ts centre of mass. 4.4 Exercses 3. (Prob n Young and Freedman.) Four small spheres, each of whch you can a = rω =.. = regard as a pont of mass 0.00 kg, are arranged n a square m on a sde and connected by extremely lght rods. Fnd the moment of anerta rad 060 of m/s the system 08; about an axs (a) through the centre of the square, perpendcular to ts g plane; = 980 (b) m/sbsectng = a = 08g. two opposte sdes of the square; (c) that passes through the centres of the upper left and lower rght rad/s. In part (c), v and spheres. rad Soluton. Then v= rω = (0. 35 m)(67. 0 rad/s) = 5. 7 m/s. rad (0 35 m)(67 0 rad/s) 060 m/s EVALUATE: In parts (a) and (b), snce a rato s used the unts cance a are calculated from ω, and ω must be IDENTIFY and SET UP: Use Eq. (9.6). Treat the spheres as pont m EXECUTE: The object s shown n Fgure 9.30a. (a) (a) The dstance from the centre s the same for the four spheres; denotng t by r, we have r = (0. 00 m I = mr = 4 I = kg Fgure 9.30a (b) The object s shown n Fgure 9.30b.

15 r = ( 0.00) + ( 0.00) = 0.83 m. The moment of nerta s then I = r m ( ) ( kg) = 0.83 m = kg m. rad = = ( ; m)(67 a. = 008 rad/s) g = 060 m/s a g = rω 980. m/s = arad 060 m/s EVALUATE: In parts 08; (a) and (b), snce a rato s used the unts cance rad/s. g = In 980 part (c), m/s = a = 08g. v and a rad are calculated from ω, and ω must be EVALUATE: IDENTIFY and In parts SET UP: (a) Use and Eq. (b), (9.6). snce Treat a rato the s used spheres the as unts pont canc m EXECUTE: rad/s. In part The (c), object v and s a rad shown are n calculated Fgure 9.30a. from ω, and ω must be (a) IDENTIFY and SET UP: Use Eq. (9.6). Treat the spheres as pont m EXECUTE: The object s shown n Fgure 9.30a. (a) (4.67) r = (0. 00 m Fgure 9.30a (4.68) (b) Fgure The object 9.30a s shown n Fgure 9.30b. I = mr = 4 r = (0. 00 I = kg I = mr = I = k (b) In ths case agan the dstance of the four masses to the axs s the same wth r = 0.00 m, whch mples that I = r m ( ) ( kg) = 0.00 m = 0.03 kg m. It s half of the value obtaned n (a). (4.69) (c) Now, two of the masses are located on the axs and have r = 0 and do not contrbute to the moment of nerta. The other two have r = 0.00 moment of nerta then becomes Fgure 9.30b ( ) + ( 0.00) = 0.83 m. The (b) The object s shown n Fgure 9.30b. (c) Fgure The object 9.30b s shown n Fgure 9.30c. (c) The object s shown n Fgure 9.30c. Fgure 9.30c r = m I = mr = 4 r = m I = kg I = mr = I = k r = m I = mr = ( r = m I = kg I = mr = I = kg I = mr ( ) ( 0.00 kg) = 0.83 m = 0.03 kg m. 4. (Prob n Young and Freedman.) A frctonless pulley has the shape of a unform EXECUTE: (a) () sold dsk of mass.50 kg and radus 0.0 cm. A.50 kg stone m. s attached I 3 ML to a 3 very lght wre that s wrapped around the rm of the pulley, and the EXECUTE: system (a) s released () from rest. (a) How far must the stone fall so that the pulley has 4.50 J of knetc I 3 energy? ML 3 (b) What percent of total knetc energy does the pulley have? Soluton. EVALUATE: Fgure 9.30c In general I depends on the axs and our answer for par It just happens that I s the same n parts (b) and (c) IDENTIFY: EVALUATE: Use In Table general 9.. I depends The correct on (4.70) the expresson axs and to our use answer n each for cas pa and It just the happens locaton that of the I s axs. the same n parts (b) and (c) SET IDENTIFY: UP: In each Use case Table express 9.. The the correct mass n expresson kg and the to length use n n each m, ca so kg and m the. locaton of the axs. SET UP: In each case = express = the (. mass n kg 50 kg)(0. and the 750 m) = length 0. n m, 469 kg s m Copyrght 0 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected un No porton of ths materal may be reproduced, n any form or by any means, wthout permss Copyrght 0 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected u No porton of ths materal may be reproduced, n any form or by any means, wthout perms (a) The velocty of the stone v = ve y (the y-axs s postve-vertcal) s related to the angular velocty ω = ωe z of the dsk and ts radus r (located n the xy-plane ) wth = = (. 50 kg)( m) = kg m

16 v = ω r = ωr e y, (4.7) or v = ωr. The conservaton of energy before and after the stone starts fallng tells us that mgy 0 = mgy + mv + Iω, (4.7) where m s the mass of the stone and I = MR s the moment of nerta of the dsk, whch we determned usng Fgure 5f. We can then wrte mg( y 0 y ) = mω R + 4 MR ω. (4.73) But the knetc energy of the pulley s gven by the last term on the rght hand-sde of equaton (4.73), and ω = 4K dsk MR J =.50 kg 0.00 cm = 80 rad /s. ( ) (4.74) We fnally wrte y 0 y = ω R + M g m = 80 rad /s 0.04 m.50 kg m/s.50 kg = m. (4.75) The stone need to fall m for 4.50 J of rotatonal knetc energy to be stored n the dsk. (b) The percent of knetc energy stored n the pulley s

17 K dsk K tot = = MR ω 4 mω R + MR ω 4 + m M M = M + m = 45.5%. (4.76) It s nterestng to note that ths fgure a constant of tme and s only a functon of the masses of the pulley and stone. 4.5 Parallel-Axs Theorem We already stated n Secton 4. that the moment of nerta of a rgd body depends on the choce of the axs about whch t s calculated. There s, however, a useful theorem that lnks the moment of nerta obtaned when usng an axs that goes through ts centre of mass, we denote t by I cm, and another that s parallel to t but located some dstance d away, let us call t I p. We can convenently, but n all generalty, locate the orgn of the system of axes at the poston of the centre of mass. It follows from ths that r cm = 0. (4.77) We can also, wthout any loss of generalty, choose the z-axs as that gong through the centre of mass and about whch the moment of nerta s I cm. Wth ths choce, we can wrte for d = d ( ) the locaton of the parallel axs d = ae x + be y. (4.78) We now use the second of equatons (4.60) for the defnton of the moment of nerta I cm = m r, ( ) = m x + y. (4.79) We can also use the same defnton for I p, but the dstance of a pont located at r, (relatve to the z-axs ) s r, d from the parallel axs. We therefore have, wth M =, m

18 I p = m x a = m x + y = m x + y ( ) + ( y b) ( ) ax + by ( ) ax + by ( ) ( ) + a + b ( ) + d = I cm am x b m y + d m = I cm Max cm Mby cm + Md, (4.80) or alternatvely I p = I cm Md r cm, + Md. (4.8) We know from equaton (4.77) that r cm, = 0, whch then yelds the fnal result for the parallel-axs theorem I p = I cm + Md. (4.8) For example, let us use ths equaton to calculate the moment of nerta for the slender rod of Fgure 5b, where the axs used for I p s located at one end of the rod, startng wth the result of Fgure 5a, where the axs used for I cm goes through the centre of mass. In ths case d = L, then I p = I cm + ML 4 = ML + ML 4 = ML 3, (4.83) whch s the result expected Exercses 5. (Prob n Young and Freedman.) About what axs wll a unform, balsa-wood sphere have the same moment of nerta as does a thn-walled, lead sphere of the same mass and radus, wth the axs along a dameter? Soluton. If we the subscrpts and for the unform and thn-walled spheres, respectvely, then from Fgure 5h and for an axs along a dameter

19 I,cm = 5 MR I,cm = 3 MR. (4.84) We need to determne d for I,p = I,cm + Md = I,cm. (4.85) We therefore have d = ( M I,cm I,cm ) = R 3 5 = 4 5 R, (4.86) or 4.6 Angular Momentum and Torque d = 0.56R. (4.87) We now seek to further expand on the correspondence between lnear and rotatonal motons for a rgd body. In what wll follow, we wll put a further restrcton that the axs rotaton about whch the rotaton takes place s a symmetry axs. For example, one can thnk of the long axs of a cylnder, as n Fgure 5f. Whle ths smplfcaton wll ease our analyss, the results can be shown to be of greater generalty, but ths demonstraton s beyond the scope of our study (see footnote on a later page). We already know of the correspondences between the followng quanttes when consderng translatonal and rotatonal motons: x θ v ω a α m I. (4.88)

20 But we have yet to fnd or defne analogs to the lnear momentum p and the force F. To do so, we start by consderng a mass element m located at r n the rgd body, and for whch v = ω r. (4.89) We now frst multply by m (to get the lnear momentum) and then cross-multply wth r. That s, usng equatons (4.30) and (4.3), ( ) r m v = r m ω r = m r ω m ( ω r )r = m r ω ( ω r )r, m ( ω r )r, ( ) ωm r, r, = m r ω r, ω = m r, ω ωm r, r,. (4.90) (Note that r, can be postve or negatve n ths equaton.) We ntroduce a new vector quantty L r m v r p, (4.9) whch we sum all over the volume spanned by the rgd body (.e., over ). Equaton (4.90) then becomes L = ωm r, ( ωm r, r, ) ω m r, r,. = ω m r, (4.9) We know from equaton (4.60) that the frst term on the rght-hand sde s the product of the moment of nerta and the angular velocty Iω, but the last term wll cancel out when the axs of rotaton s an axs of symmetry. That s, m r, r, = 0. (4.93) Ths s because as one effects the sum about the axs of symmetry, for each term m r, r, there wll exst another one such that m j r j, r j, = m r, r,. It follows that the total angular momentum L s defned as

21 L = = Iω L (4.94) when the rgd body s rotatng about an axs of symmetry. We then fnd a new correspondence between lnear and angular momentum that s perfectly consstent wth equatons (4.88),.e., p = mv L = Iω. (4.95) Pushng the analogy further we could postulate that the analog of the force for rotatonal moton must be the tme dervatve of the angular momentum snce accordng to Newton s Second Law We would then defne the torque wth whch accordng to equaton (4.94) would yeld F net = dp dt. (4.96) τ = dl dt, (4.97) τ = d ( Iω ) dt = I dω dt = Iα, (4.98) snce the moment of nerta I s constant for a rgd body. Equatons (4.97) and (4.98) are the correct relatons that lnk the torque and angular momentum for a rgd body rotatng about an axs of symmetry (although equaton (4.97) s true n general). But just as the angular momentum s also related to the lnear momentum through equaton (4.9), there s a smlar connecton between the torque and the force. We thus proceed as follows τ = d dt L, (4.99) Although the result L = Iω was derved for rotaton about symmetry axes and that the more general relaton s mathematcally dfferent, t s always possble to brng ths more general relaton nto that smpler form through a judcous orentaton of the system of axes chosen as a bass for the rgd body (commonly called the prncpal axes)

22 whch, snce v m v = 0, becomes τ == d dt ( ) r p dr = dt p + r dp dt = v m v + r dp dt = r F net, = τ. (4.00) The total torque on the rgd body s therefore the sum of the all the torques actng on the ndvdual partcles that form the rgd body. We note that the torques on the ndvdual partcles result from the net force appled on these partcles, whch s the sum of the correspondng external and nternal forces. But f the nternal forces that characterze the nteracton between partcles are central forces,.e., f they act along the lne that jons a gven par of nteractng partcles, then the torques resultng from these forces wll cancel each other (because of Newton s Thrd Law; see Fgure 6). It follows that only external forces are nvolved n the determnaton of the total torque actng on a rgd body. We can therefore wrte τ = τ ext,. r F ext, (4.0) We then fnd our last correspondence between translatonal and rotatonal motons Fgure 6 The cancellaton of torques resultng from the nternal forces on a par of partcles due to ther nteracton

23 0- Chapter 0 (d) SET UP: Consder Fgure 0.d. F = ma τ = Iα. (4.0) EXECUTE: τ = Fl It s also mportant to note that the second and thrd of equatons (4.00) are vald n τ = (0.0 N)(.73 m) = 7. 3 N m general for a sngle partcle wth Fgure 0.d 4.6. Exercses τ = r F net. (4.03) 6. (Prob. (e) SET 0.3 UP: Consder n Young Fgure and 0.e. Freedman.) A square metal plate 0.80 m on each sde s pvoted about on axs through pont O at ts centre and perpendcular to the plate. EXECUTE: τ = Fl Calculate the net torque about ths axs due to the r = three 0 so l forces = 0 and shown τ = 0 n Fgure 7 f the magntude of the forces are F = 8.0 N, F = 6.0 N, and F 3 = 4.0 N. The plate and all forces Fgure are 0.e n the plane of the page. Soluton. Usng equaton (4.03) for the torque w can wrte τ = r F net = rf net sn( φ), where SET φ UP: s the Let counterclockwse angle between torques r and be Fpostve. net. We then have =.6 N, l = rsn φ = (.00 m)sn 60 =.73 m Ths force tends to produce a clockwse rotaton about the axs; by the rght-hand rule the vector τ s drected nto the plane of the fgure. (f) SET UP: Consder Fgure 0.f. Fgure 0.f τ τ τ ( )sn 35 EXECUTE: Fl l = rsn φ, φ = 80, ( ) (4.04) (4.05) and snce the torque s negatve, t s drected nto the page (.e., the plate s movng τ = so l = 0 and τ = 0 EVALUATE: The torque s zero n parts (e) and (f) because the moment arm s zero; the lne of acton of the force passes through the axs. 0.. IDENTIFY: Fl τ = wth l = rsn φ. Add the two torques to calculate the net torque. EXECUTE: τ = Fl = (8. 00 N)(5. 00 m) = N m. τ =+ Fl = (. 0 N)(. 00 m)sn30. 0 =+. 0 N m. = + = 8. 0 N m. The net torque s τ = ( 8.0 N) m 8. 0 N m, clockwse. EVALUATE: Even though F < F, the magntude of τ s greater than the magntude of τ, because F has a larger moment arm IDENTIFY and SET UP: Use Eq. (0.) to calculate the magntude of each torque and use the rght-hand rule (Fgure 0.4 n the textbook) to determne the drecton. Consder Fgure 0.3. Fgure 0.3 Fgure 7 Torques on the metal plate of Prob. 6. Let counterclockwse be the postve sense of rotaton. Copyrght 0 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst. No porton of ths materal may be reproduced, n any form or by any means, wthout permsson n wrtng from the publsher

24 clockwse. For the second force τ = ( 6.0 N) ( m)sn( 35 ) =.34 N, (4.06) the torque s comng out of the page. Fnally, for the last torque τ 3 = ( 4.0 N) ( m)sn( 90 ) =.78 N, (4.07) also comng out of the page. 7. (Prob. 0.7 n Young and Freedman.) A.0-kg box restng on a horzontal, frctonless surface s attached to a 5.00-kg weght by a thn, lght wre that passes over a frctonless pulley. The pulley has the shape of a unform sold dsk of mass.00 kg and dameter m. After the system s released, fnd (a) the tenson on the wre on both sdes of the pulley and the acceleraton of the box, and (b) the horzontal and vertcal components of the force that the axle exerts on the pulley. Soluton. (a) From the frst two free-body dagrams shown n Fgure 8 we have T = m a m g T = m a, (4.08) or when combnng these two equatons T T = m g ( m + m )a. (4.09) From the thrd free-body dagram of the fgure we can also wrte for the torque actng on the pulley 0-8 Chapter 0 Fgure I DENTIFY: The tumbler has knetc energy due to the lnear moton of hs center of mass plus knetc energy due to hs rotatonal moton about hs center of mass. SET UP: vcm = Rω. ω = 0.50 rev/s = 3.4 rad/s. rot cm K = I ω. Fgure 8 The free-body dagrams for Prob. 7. EXECUTE: (a) K = K + K wth K - 84 I - = MR wth R = m. K = Mv and K = I ω. cm cm = Mv and

25 τ = r( T T ) = Iα (4.0) = I a r, snce the tangental acceleraton s a = αr, wth r the radus of the pulley (see equaton (4.40)). We can transform equaton (4.0) to T T = Mr a r = Ma, (4.) wth I = Mr then have from Fgure 5f. If we now subtract equatons (4.09) and (4.), we a = = m g m + m + M 5.00 kg 9.80 m/s ( )kg =.7 m/s. (4.) Insertng the frst of equatons (4.) nto equatons (4.08) we fnd T = m m g m + m + M.0 kg 5.00 kg 9.80 m/s = ( )kg = 3.6 N (4.3) and m T = + M m + m + M m g ( )kg = 5.00 kg 9.80 m/s ( )kg = 35.4 N. (4.4) It s mportant to note that the reason for havng T T s that we account the moment of nerta of the pulley wth equaton (4.)

26 (b) The last free-body dagram of Fgure 8 shows that three forces are actng on the pulley, whch mples that t must react wth a force F that equals mnus the resultant of the three forces. That s, ( ) F = T + T + m g = ( T )e x ( T m g)e y = T e x + ( T + m g)e y = ( 3.6e x e y )N. (4.5) 4.7 Combned Translatonal and Rotatonal Motons Consder two frames of reference: one nertal frame of axes x, y, and z and a rotatng frame wth axes x, y, and z ted to a rotatng rgd body; ths s shown n Fgure 9. If we choose a pont P n the rgd body located at r from the orgn of the rotatng axes and we further located ths orgn to that of the nertal frame wth R, then we can wrte r = R + r (4.6) for the poston of P n the nertal frame. We now nqure about the moton of P as seen by an observer located at the orgn of the nertal system wth z Rgd Body z P r y r R x y x Fgure 9 A fxed nertal frame wth axes, and a rotatng frame wth axes. The vector locates the centre of mass of the rotatng rgd body

27 d r dt = d dt ( ) R + r = dr dt + dr dt. (4.7) Upon usng equaton (4.6) we fnd that v = V + v = V + ω r, (4.8) where v dr dt v d r dt V dr dt. (4.9) We therefore fnd that the moton of a pont on the rgd body can be broken n ts rotaton moton about a gven centre of rotaton and the translatonal moton of that centre relatve to an nertal frame of reference Wth Rotaton about the Centre of Mass We now constran the rotaton to be about an axs that passes through the centre of mass of the rgd body. That s, r s now the poston of P from the centre of mass and R the locaton of the centre of mass to the orgn of the nertal frame of reference. We have already shown n Secton 3.4 of Chapter 3 that the total lnear momentum P of the rgd body measured n the nertal frame, obtaned by summng over all ponts such as P, equals the total mass M of the rgd body tmes the velocty V of ts centre of mass. Ths s agan readly verfed wth equaton (4.8) P = m v ( ) = m V + ω r = V m + ω m r = MV + ω m r, (4.0) whch, snce by defnton m r = 0, becomes

28 P = MV. (4.) The next obvous nqury to make at ths pont concerns the total angular moment L measured n the nertal frame. We do so as follows L = L = r m v ( ) m ( V + v ) = R + r ( ) = m R V + R v + r V + r v = R P + m R v + r V. ( ) + r m v (4.) However, the second and thrd terms can be shown to cancel snce d m ( R v + r V) = m dt = d R dt = 0, ( ) V r + r V R r m r V m r (4.3) because, once agan, m r = 0 by vrtue of the defnton for the centre of mass. Equaton (4.) then yelds the mportant result L = R P + r p. (4.4) That s, the total angular momentum of a rgd body about the orgn of an nertal frame s the sum of the angular momentum of the centre of mass about that orgn and the angular momentum of the rgd body about ts centre of mass Energy Relatons We just saw that the overall moton of a rgd body can be advantageously broken down nto motons about ts centre of mass and of ts centre of mass relatve to the orgn of an nertal frame. Does the same type of relaton exst for the knetc energy? To answer ths queston we calculate for the pont P

29 K = m v = m ( V + v ) (4.5) = m V ( + V v + v ), where we used equaton (4.8). Summng the energy over the entre rgd body we get K = m V ( + V v + v ) = MV + V m v + m v. (4.6) However, the second term on the rght-hand sde can be shown to vansh snce m v = d dt = 0 m r (4.7) from the defnton of the centre of mass, and K = MV + m v. (4.8) Usng the result already establshed wth equatons (4.5) through (4.59) we can fnally wrte K = MV + I cmω, (4.9) where the moment of nerta relatve to (an axs passng through) the centre of mass of the rgd body I cm s gven n equaton (4.79). Equaton (4.9) answers our earler queston and mples that the total knetc energy of the rgd body as seen n an nertal frame s the sum of the knetc energy of a partcle of mass M movng wth the velocty of the centre of mass and the knetc energy of moton (or rotaton) of the rgd body about ts centre of mass. Fnally, the potental gravtatonal energy of a rgd body of total mass M that exhbt translatonal and rotatonal moton (about any axs of rotaton) remans as was descrbed n Secton 4.3 wth

30 4.7.3 Exercses U = Mgy cm. (4.30) 8. (Prob. 0. n Young and Freedman.) A hollow, sphercal shell wth mass.00 kg rolls wthout slppng down a 38.0 slope. (a) Fnd the acceleraton, the frcton force, and the mnmum coeffcent of frcton needed to prevent slppng. (b) How would your answers to part (a) change f the mass were doubled to 4.00 kg? Soluton. We wll set the x-axs to be down the nclne and let the shell be turnng n the postve drecton. The free-body dagram s shown n Fgure 0. (a) Accordng to the fgure we have for the forces actng on the centre of mass and the torque actng on the sphere ma cm = mgsn( β ) f s I cm α = f s R, (4.3) where we must use the statc (and not dynamc) force of frcton snce the pont of the sphere contactng the surface s not movng relatve to t. But snce the sphere s not slppng we also have that a cm = α R and from the second of equatons (4.3) whch we can nsert n the frst of equatons (4.3) to get Alternatvely, we can evaluate a cm = f s = a cmi cm R, (4.3) gsn( β ) + I cm mr. (4.33) f s = mgsn( β ). (4.34) + mr I cm For a hollow sphere we have I cm = 3mR (4.35) whch leads to

31 n N (b) The acceleraton s ndependent of m and doesn t change. The frcton force s proportonal to m so wll double; f = N. The normal force wll also double, so the mnmum µ s requred for no slppng wouldn t change. EVALUATE: If there s no frcton and the object sldes wthout rollng, the acceleraton s g sn β. Frcton and rollng wthout slppng reduce a to 0.60 tmes ths value. Fgure 0. Fgure 0 Free-body dagram for the rollng sphere n Prob. 8. Copyrght 0 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst. No porton of ths materal may be reproduced, n any form or by any means, wthout permsson n wrtng from the publsher. ( ) a cm = 3 5 gsn β = 3.6 m/s f s = 5 mgsn( β ) = 4.83 N. (4.36) The coeffcent of statc frcton s gven by µ s = f s n = f s mgcos β ( ) = 5 tan( β ) = (4.37) Ths coeffcent s requred to ensure that the sphere does not slp down the nclne. (b) The only part of (a) that would change f the mass of the sphere was doubled s the magntude of the statc frcton force, whch would also double to 9.66 N. 9. (Prob. 0.4 n Young and Freedman.) A unform marble rolls down a symmetrcal bowl, startng from rest at the top of the left sde. The top of each sde s a dstance h above the bottom of the bowl. The left half of the bowl s rough enough to cause the marble to roll wthout slppng, but the rght half has no frcton because t s coated wth ol. (a) How far up the smooth sde wll the marble go, measured vertcally from the bottom? (b) How hgh would the marble go f both sdes were as rough as the left sde? (c) How do you account for the fact that the marble goes hgher wth frcton on the rght than wthout frcton? - 9 -

32 torque on the marble, mv rot rot + K = mgh + K. h = = = h g g 7 (b) mgh = mgh so h = h. EVALUATE: (c) Wth frcton on both halves, all the ntal potental energy gets converted back to potental energy. Wthout frcton on the rght half some of the energy s stll n rotatonal knetc energy when the marble s at ts maxmum heght. Fgure 0.4 Fgure A marble rollng nsde a bowl, n Prob IDENTIFY: Apply conservaton of energy to the moton of the wheel. SET UP: The wheel at ponts and of ts moton s shown n Fgure 0.5. Soluton. Take y = 0 at the center of the wheel when t s at the bottom of the hll. (a) We wll let y = 0 at the bottom of the bowl, shown n Fgure. Snce the marble of mass m starts from rest, ts knetc energy at the bottom of the bowl wll be gven by equaton (4.9) K = mv cm+ I cmω = mv cm+ I cm v cm R, (4.38) K = (0. 800) MR ω + MR ω = MR ω where R s the radus of the marble and we mposed the no-slppng condton v cm = ω R. K = 0, U = 0, Because of the prncple U = Mgh of conservaton of energy, ths (change n) knetc energy equals Thus 0900MR. ω the change n potental + W frc = Mgh gravtatonal energy wth and, snce I cm = 5 mr, mv cm+ I cm v cm = 0 7 v cm R = mgh, (4.39) gh. (4.40) But the at hll. the bottom of the bowl the no-slppng condton does not hold anymore and the rotatonal knetc energy wll be conserved, as no torque (due to frcton) wll be actng on the marble. Conservaton of energy then dctates that from the bottom of the bowl to the heght h the marble wll reach on the rght sde of the bowl we must have Copyrght 0 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst. No porton of ths materal may be reproduced, n any form or by any means, wthout permsson n wrtng from the publsher. or Fgure 0.5 cm cm The wheel has both translatonal and rotatonal moton so ts knetc energy s K = I ω + Mv K + U + W = K + U EXECUTE: other Wother = Wfrc = 3500 J (the frcton work s negatve) ω K = I + Mv ; v= Rω and I = MR so M = w/ g = 39 N/(980. m/s ) = 400. kg W frc MR ω + h = Mg (0. 900)(40. 0 kg)( m) (5. 0 rad/s) 3500 J h = =. 7 m (40. 0 kg)(9. 80 m/s ) EVALUATE: Frcton does negatve work and reduces h IDENTIFY: Apply τz = Iαz and F = ma to the moton of the bowlng ball. SET UP: acm = Rα. f s = µ s n. Let + x be drected down the nclne. EXECUTE: (a) The free-body dagram s sketched n Fgure 0.6. The angular speed of the ball must decrease, and so the torque s provded by a frcton force that acts up mv cm+ I cmω = I cmω + mg h, (4.4)

33 h = v cm g = 5 7 h, (4.4) where equaton (4.40) was used. (b) If the rght sde of the bowl had the same roughness as the left sde, then the knetc energy stored n the rotaton of the marble at the bottom of the bowl would be transferred back to potental gravtatonal energy and equaton (4.4) would be replaced by we would fnd that h = h from equaton (4.39). mv cm+ I cmω = mg h (4.43) (c) As was stated above, the marble goes hgher when the whole bowl has a rough surface because the knetc energy stored n the rotaton marble at the bottom of the bowl s transferred back to potental gravtatonal energy. 4.8 Work and Power n Rotatonal Moton We know from our treatment n Chapter that the nfntesmal work dw done by a force F actng through an nfntesmal dstance dr s gven by But for a rotatonal moton we know from equaton (4.5) that dw = F dr. (4.44) dr = dθ r, (4.45) where dr s now understood to be the nfntesmal arc subtended by the radus r over an nfntesmal angular dsplacement vector dθ. Insertng equaton (4.45) n equaton (4.44) we have but snce ( a b) c = ( c a) b = ( b c) a we can also wrte dw = ( dθ r) F, (4.46) dw = ( r F) dθ = τ dθ, (4.47)

34 where τ = r F s the torque on the system. If we now ntegrate equaton (4.47) through a fnte angular dsplacement θ θ we fnd that the total work done s W = θ θ τ dθ θ = ( r F) dθ. θ (4.48) Ths relaton s the analog of equaton (.39) n Chapter for translatonal motons r W = F dr. (4.49) r It follows from equaton (4.48) that for rotatonal moton a force can only do work f t has a component perpendcular to the radus, as expected. Of course, the work-energy theorem derved n Secton. of Chapter also apples here. For example, f rotaton happens about a symmetry axs we have τ dθ = ( Iα ) dθ = I dω dt dθ = I dω dt ωdt = I ω dω, (4.50) and W = = I I ω dω d( ω ) = Iω Iω (4.5) = ΔK rot. It should be noted that ths result could also be obtaned for the specal case of a constant angular acceleraton usng equaton (4.3), n a manner smlar to what was done n Chapter for translatonal motons. The power assocated wth the work done by the torque (or the force) can readly be determned wth

35 P = dw dt = d ( τ dθ ) dt = τ dθ dt (4.5) and fnally P = τ ω. (4.53) Agan ths result s analogous to the one obtan wth equaton (.47) n Chapter for translatonal motons 4.8. Exercses P = F v. (4.54) 0. (Prob n Young and Freedman.) You connect a lght strng to a pont on the edge of a unform vertcal dsk wth radus R and mass M. The dsk s free to rotate wthout frcton about a statonary horzontal axs through ts centre. Intally, the dsk s at rest wth the strng connecton at the hghest pont on the dsk. You pull the strng wth a constant horzontal force F untl the wheel has made exactly one-quarter revoluton about the horzontal axs, and then you let go. (a) Use equaton (4.48) to fnd the work done by the strng. (b) Use equaton (4.49) to fnd the work done by the strng. Do you obtan the same result as n part (a)? (c) Fnd the angular speed of the dsk. (d) Fnd the maxmum tangental acceleraton of a pont on the dsk. (e) Fnd the maxmum radal (centrpetal) acceleraton of a pont on the dsk. Soluton. From Fgure we can verfy that F = Fe x R = R cos( θ )e x + sn( θ )e y. (4.55) (a) Snce the force and the torque are dependent on the angle θ we wrte

36 y F R θ (+z) x W = Fgure The strng-dsk arrangement of Prob. 0. τ dθ = ( R F) dθ = { R sn( θ )e x + cos( θ )e y ( Fe x )} e z dθ = FR( e y e x ) e z cos( θ )dθ, ( ) (4.56) but snce cos( θ )dθ = sn( θ ) and e y e x = e z, we have (wth θ = 0 and θ = π ) W = FR sn θ = FR. ( ) sn( θ ) (4.57) (b) Usng equaton (4.49) wth the arc dr = dθ R we have W = F ( dθ R) = ( R F) dθ, (4.58) whch equals the second of equatons (4.56) (we also used F ( dθ R) = ( R F) dθ ). It follows that the two approaches,.e., ths one and the one of part (a), wll gve the same result. (c) We know from equaton (4.5) that

37 ω = W I = 4FR MR = F MR. (4.59) (d) The tangental acceleraton s gven by a tan = α R = τ I R (4.60) and snce τ and R are perpendcular to one another a tan = τ R I = FRcos( θ )R MR (4.6) = F M cos( θ ), whch wll be maxmum wth F M at θ = 0. (e) The radal acceleraton s gven by a rad = ω R ( ) MR 4F sn θ = R, (4.6) where we used the value of W from equaton (4.57) at an arbtrary angle θ. The centrpetal acceleraton wll be at a maxmum at θ = π wth a rad = 4F M. 4.9 The Conservaton of Angular Momentum When dervng the prncple of conservaton of lnear momentum n Chapter 3, we consdered an solated system of partcles; ths prncple only apples for such system. Our dervaton rested entrely on Newton s Thrd Law. We showed that f nteractons between pars of partcles happen through nternal forces n such a way that F j = F j, (4.63)

38 where and j denote a par of partcles, then the total lnear momentum of the system was conserved dp tot dt = 0. (4.64) On the other hand, f the system was not solated but was subjected to a net external force, then the total lnear momentum of the system was allowed to change accordng to Newton s Second Law F ext = dp tot dt. (4.65) For the purpose of nvestgatng a smlar conservaton of angular momentum we must also consder an solated system; certanly a rgd body satsfes ths requrement. From our prevous study on the relaton between the torque and angular momentum, we have a perfect correspondence between the force-lnear momentum and torque-angular momentum pars. For example, we know from equaton (4.97) that τ = dl dt. (4.66) For a rgd body not subjected to a net external torque, and therefore solated from any agent that could change ts state of rotaton, equaton (4.66) tells us that dl dt = 0. (4.67) In other words, when the net torque appled to a system s zero, then the total angular momentum of the system s conserved and remans unchanged. It s mportant to note that ths prncple s a unversal conservaton law, to the same fundamental level as the prncples of conservaton of energy and lnear momentum. It s possble that nternal torques arse between the components of a system (just as nternal forces could be present when nvestgatng the total lnear momentum of an solated system of partcles). But f we agan call upon Newton s Thrd Law such that these nternal torques arse from nteracton forces that satsfy equaton (4.63), the sum of all such nternal torques can be wrtten as τ nt, = r F j j ( ) j pars = r F j + r j F j, (4.68)

39 where the last summaton s on unque pars of partcles and j (.e., f we consder and j then we shouldn t nclude j and to avod double-countng). But from equaton (4.63) we can wrte ( ) F j τ nt, = r r j j pars. (4.69) If we now enforce the further constrant that the nternal forces be central,.e., that they are drected along the straght lne jonng the two nteractng partcles (.e., F j s parallel or ant-parallel to r r j ), then and ( r r j ) F j = 0 (4.70) τ nt, = 0. (4.7) Ths mples that parts of the system can experence a change n ther angular momentum, but the total angular momentum must be conserved when no external torque s appled. Equatons (4.68) to (4.7) put on a frm mathematcal bass what was dscussed n Secton 4.6 and llustrated n Fgure 6. When an external torque s appled, then the total angular momentum wll change n accordance wth equaton (4.66) Exercses. (Prob. 0.4 n Young and Freedman.) Under some crcumstances, a star can collapse nto an extremely dense object made mostly of neutrons and called a neutron star. The densty of a neutron star s approxmately 0 4 tmes that of ordnary sold matter. Suppose we represent the star as a unform, sold, rgd sphere, both before and after the collapse. The star s ntal radus was km (comparable to our sun); ts fnal radus s 6 km. If the orgnal star rotated once n 30 days, fnd the angular speed of the neutron star. Soluton. The angular momentum must be conserved wth I star ω = I neutron ω 5 MR starω = 5 MR neutronω, (4.7) or

40 ω = ω R star R neutron π = rad/sec = 4640 rad/s (or 738 rev/s). (4.73). (Prob n Young and Freedman.) The outstretched hands and arms of a fgure skater preparng for a spn can be consdered a slender rod pvotng about an axs through ts centre. When the skater s hands and arms are brought n and wrapped around hs body to execute the spn, the hands and arms can be consdered a thn-walled, hollow cylnder. Hs hands and arms have a combned mass of 8.00 kg. When outstretched, they span.80 m; when wrapped, they form a cylnder of radus 5.0 cm. The moment of nerta about the rotaton axs or the remander of the body s constant and equal to kg m. If hs orgnal angular speed s rev/s, what s hs fnal angular speed? Soluton. The angular momentum must be conserved through ths maneuver,.e., or From Fgure 5 we have that ( I arms, + I body )ω = ( I arms, + I body )ω, (4.74) I ω = ω arms, + I body I arms, + I body. (4.75) I arms, = ML I arms, = MR, (4.76) where L and R are the length and radus of the hands and arms when outstretched and brought n, respectvely. It follows that

41 ML + I ω = ω body MR + I body = ( π 0.4)rad/s = 7.5 rad/s (or.4 rev/s). (4.77) 4.0 Precesson We have all experenced the remarkable dynamcal behavor of a spnnng top. One ntrgung aspect s the observaton of an ncreasng rotatonal-wobble of the spnnng axs as the angular speed of the top (about the spnnng axs) s slowng down. Ths type of moton s called precesson. We have now developed all the tools necessary to understand ths behavor. Let us consder a flywheel wth ts symmetry, and spn, axs postoned horzontally as depcted n Fgure 3. The flywheel s spnnng wth an angular velocty ω about ts symmetry axs, ntally drected along the x drecton (.e., ω = ωe x ), and s smultaneously subjected to gravty wth ts weght w = we z pontng downward. The presence of ths force located at the centre of mass of flywheel-axs system brngs a torque ntally pontng along the y-axs (.e., pontng nto the page) τ = r w = τe y. (4.78) It would perhaps be ntutve to thnk that the presence of ths torque would start the flywheel rotatng about the y-axs and eventually brng t n contact wth the ground. Ths s ndeed what would be observed f the flywheel were not spnnng about ts symmetry axs. But the flywheel s dynamcs are much more nterestng because of ts rotatonal moton Fgure 3 Precesson of a flywheel as t spns about ts symmetry axs