PHYSICS 06a HOMEWORK #0 SOLUTIONS (Poblem #,, and 3 have been moved to the next aignment.) 3. Supeman want to top the Eath fom pinning. a. Auming a contant angula acceleation, what toque would he have to exet on it to top it in one hou? b. If he exet the toque by puhing on a point on the equato, what foce doe he have to exet? c. Rathe than puhing on a point on the equato, he puhe on a point in Edwadville, IL. Now what foce doe he have to exet? Angula acceleation, by analogy to egula acceleation, i the change in angula velocity divided by the time in which that change occu. Mathematically, thi i witten v v ω α =. Fo thi poblem, we e unconcened with the diection of the vecto, o let t teat them a cala fo implicity. Thi allow u to dicu peed intead of velocity. We e going fom the angula peed found in poblem #9 of aignment #9 θ ( π = = 7.7 0 5 5 ω = ) to zeo in one hou. So ω = 7.7 0 t 86400econd 5 7.7 0 ω 8 and we have α = = =.0 0. Thi i the angula acceleation. t 3600 Now, in the ealm of otating object, toque eve in the ame ole a foce doe in the ealm of moving object: Toque i the thing that make omething tat o top otating (o change the diection of it v otation). We have an analog fo Newton econd law in the v v v otating ealm: Intead of F = ma, we have τ = I α. (Thee i alway ome confuion hee: I eem to have given you two fomula fo the toque, one in which it defined a a foce time a moment am and the othe in which it defined a a moment of inetia time an angula acceleation. What give? Well, thi i pat of the dange of eeing thee mathematical expeion a fomula athe than a what they eally ae: Statement about the elationhip between entitie. The way I think of thee two equation i that one [the one in which toque i expeed a a moment am time a foce] i a caue of the toque while the othe [the one in which toque i elated to a moment of inetia time an angula acceleation] i the effect of the toque. Beak youelf of the habit of thinking of thee mathematical expeion a fomula into which quantitie ae to be plugged and commence thinking of them a deciption of elationhip witten in the language of mathematic and you ll avoid a hot of eo and open up a plethoa of new inight!) We ve jut found the angula acceleation, o all that emain i to multiply thi by the moment of inetia, I. The moment of inetia of an object i the analog of ma: Ma i a meaue of how difficult it i to change the velocity an object. Moment of inetia i a meaue of how
difficult it i to change the angula velocity of an object. It i moe complex than ma ince it depend fit of all on the ma of the object and alo on the geomety of the object and the axi about which the object i otated. A ingle object doe not have a ingle moment of inetia ince we can otate the object about any axi we chooe. (Note that while toque can be detemined fom any oigin we chooe, giving the ame anwe fo any choice, moment of inetia demand a paticula choice of axi. Each one will be diffeent.) The moment of inetia of a ingle blob of ma m itting a ditance fom an axi of otation i I = m. But an extended object, which ha it ma ditibuted, will be diffeent. Fo a olid phee otating about it cente, the moment of inetia i I = m. 5 (Thee i a table of uch value in you text. On an exam, I will povide you with uch a table. You need only memoize that the moment of inetia of a point ma a ditance fom an axi i I = m. Anything ele, you can look up.) If we aume the eath i a olid phee of ma m = 5.97 0 4 kg and adiu 6 = 6.37 0 mete, we find it moment of inetia to be 4 6 37 I = m = 0.4 5.97 0 kg ( 6.37 0 mete) = 9.69 0 kg m. Uing thi and 5 the angula acceleation found peviouly (again, ignoing the diection of the vecto and concentating on it ize only), we find the needed toque to be 37 8 30 τ = Iα = 9.69 0 kg m.0 0 =.96 0 Newton mete. Now, thi link the toque to it effect. Let link it to it caue: The toque we jut found i peumed to be the eult of a foce exeted at the Eath equato. We futhe aume that the foce i exeted peciely pependicula to the Eath adiu (if it ween t, we d have to include the ine of ome angle all we cae about i the component of the 6 foce that i pependicula to the Eath adiu). Taking = 6.37 0 mete (I ve een diffeent value given fo thi, don t be too concened if you ued a lightly diffeent numbe), we have τ = F. (Again, we e ignoing the vecto apect vey conciouly. Thi i a iky thing, a you know! In thi cae, we can get away with it becaue we e not concened with the diection of the eult and we e auming the foce i pependicula to the adiu vecto. Without thee aumption, we d be making a hoible mitake by ignoing the vecto natue of thee quantitie!) Uing thi, and the numbe found above, 30.96 0 Newton mete 3 we have F = τ = = 3.08 0 Newton. 6 6.37 0 mete To find the foce needed at Edwadville, we ue the adiu at ou latitude, a in poblem #9 of aignment #9. Thi amount to imply dividing the numbe above by the coine of 39. Thi give τ.96 0 F = = co(39 ) 6.37 0 30 6 Newton mete = 3.96 0 mete co(39 ) 3 Newton. Notice that Supeman ha to puh hade to top the Eath at Edwadville than he did when puhing at the equato. Thi i fo exactly the ame eaon a why it i eaie to open
a doo by puhing on a knob fa fom the hinge than to open the ame doo by puhing cloe to the hinge.. The popelle on an aiplane ha a diamete of 3 cm and a ma of 6.8 kg. The deied otation ate of the popelle i 700 pm. The engine povide a toque of τ = 550N m. How long doe it take fo the popelle to each opeating peed when the aiplane engine i tated? I intentionally didn t tell you anything about the geomety of the popelle. Thee ae a vaiety of way you can viualize thi. The eaiet i to think of the popelle a a od being otated aound it midpoint. The moment of inetia of uch a od i I = ML. We ae given the toque (the caue of the angula acceleation), o all we need i to find it effect. ω We have τ = Iα. Recalling that α = (again neglecting the vecto apect ince we e t τ unconcened with diection in thi cae), we ee that we can wite α = o I ω I ω t = =. Fom hee it jut plug-and-chug except fo one point: I gave you the α τ popelle final otation ate in evolution pe minute. We eally need it to be in adian pe econd to be meaningful. Since thee ae π adian in each evolution and thee ae 700 π ad 60 econd in each minute, 700 pm = = 8.7. We can now plug thi 60 ad 6.8kg ( 0.3m) 8.7 I ω into the above equation and get t = = = 0. 057. τ 550N m Fankly, that a idiculou numbe! I alway ty to give you ealitic numbe on the poblem I aign. I don t know whee the difficulty cept in, hee: I ued manufactue pecification fo all the numbe. I will do ome moe eeach on thi to ee whee I may have povided an incoect input. Nevethele, the anwe above i conitent with the input. Only the plauibility of the aumption i in quetion. 3. Two childen it on the end of a ee-aw. The ditance between the childen i 3 mete. Child A ha a ma of kg and child B ha a ma of 3 kg. The pivot of the ee-aw i halfway between them. They begin with child B in the ai and child A on the gound. At thi time, the ee-aw make an angle elative to the hoizontal of 0. adian. What i the magnitude (ize) of the net toque on the ee-aw? Thi i jut like poblem #7 and 8 of the peviou aignment, with a twit (pun intentional): Now the foce do not act pependicula to the od. Recall that the definition of toque i NOT ditance time foce. It i ditance time foce in a diection pependicula to the vecto connecting the oigin (which we ae fee to chooe) to the location whee the foce act. That vecto i called the moment am. (We can alo think of thi a ditance in a diection pependicula to the foce time the foce they e equivalent tatement. Peonally, I find it much eaie to viualize the ditance a a fixed quantity and then to conide only the pat of the foce pependicula to that ditance.) So we mut decompoe
ou foce into component that ae paallel to the moment am and component that ae pependicula to the moment am. Fotunately, thi alway give u (teating only the ize of the toque and ignoing it diection fo now we ll deal with the diection in a little while) τ = df in(θ ), whee θ i the angle between the foce and the moment am. Ou ituation i pictued below..5 m F.5 m pivot θ Β W B W A θ Α Fo the ake of ymmety, let pick the pivot a the oigin fo thi poblem. (A a vaiation, I ecommend that you give it a hot with one of the childen a the oigin.) We can now look at the toque due to each of the two childen individually. Thee ae τ A = m A gd in( θ A ) and τ B = m B gd in( θ B ). In both cae, the ditance, d, i.5 mete. Since the ee-aw make an angle of 0. adian with the hoizontal, the angle ae θ = π + 0. A and θ = π 0. B. (Thoe of you who ae paying attention will note that π π in + 0. = in 0.. Let chooe not to ue thi fact in thi cae. Nomally, we hould take advantage of it, howeve.) Ineting numbe, we get m π τ A = magd in( θ A ) = kg 9.8.5m in( + 0.) = 36. 95 Newton mete and m π τ B = mb gd in( θ B ) = 3kg 9.8.5m in( 0.) = 446. 6 Newton mete. Now, although we ae neglecting diection in thi, we cannot completely ignoe it. Note that one of the toque found above hould be poitive and the othe negative. It eally doen t matte which one i which a long a we e neglecting the diection of the net, but we might a well do it ight. (The way to ee that it mut be thi way i to note that both of the foce ae pointed in the ame diection. Howeve, the vecto fom the oigin [the pivot, in thi cae] to child A point in the oppoite diection to the vecto fom the oigin to child B. Thu the toque mut have oppoite ign.) Uing the Right Hand Rule, a demontated in the olution to poblem #, we ee that child A toque point out of the page while child B toque point into the page. Again uing out of the page a poitive, we get a final toque of kg m kg m τ = τ A + τ B = 36.95 446.6 = 9. 67 Newton mete. Thi indicate that the otation induced by thi toque will be clockwie.
4. A yoyo ha a total ma of 85 gam and a adiu of 6 cm. Conide the yoyo to be a pefect olid dik. The hub ha a adiu of 3 millimete. The yoyo begin with it ting completely wound aound the hub. (Neglect the thickening of the hub due to the ting.) When the ting i held and the yoyo allowed to fall by unwinding, what total downwad acceleation will the yoyo expeience? (Hint: Detemine the angula acceleation and then calculate the amount of ting played out due to the otation. Remembe that linea acceleation and angula acceleation ae elated, in thi cae.) R F T Mg Soy fo all the aow in the pictue! We ll need them, o ty to keep them all taight in you mind. The fit thing to ealize in thi poblem i that, a implied by the hint, the downwad acceleation i diectly linked to the angula acceleation. Thi i becaue the yoyo goe down by unwinding ting. The total amount of ting that i unwound fom the hub at any intant i a diect meaue of how fa the yoyo ha taveled (we do need to aume that the ting doen t lip aound the hub; if we wanted to include that lippage, we d need to add fiction to the poblem). So, if we find the ate at which the yoyo i pinning, we can detemine how fat it falling. The yoyo expeience an angula acceleation which eult in a linea acceleation (becaue of the ting). The angula acceleation i caued by a net toque, o we mut ue the angula analog of v v Newton econd law, τ = Iα, to find the angula acceleation.
The next tep i to emembe that toque poblem tat out a foce poblem. (That an impotant entence. Reead it.) To olve a toque poblem, you hould begin with a fee-body diagam, jut a you would with a foce poblem. The cucial diffeence i that with a foce poblem, it eally doen t matte whee you daw the aow. With a toque poblem, the ize of the aow, thei diection, and thei location cay meaning, o you mut be moe caeful. Thee ae two foce acting in thi poblem: The tenion on the ting and the foce of gavity. Newton econd law didn t top being tue all of a udden: If we knew F T we d be done with the poblem ight now. But we don t know F T (it i not the ame a the weight of the yoyo if it wee, the yoyo wouldn t go down). All we know v v v i the ma of the yoyo, o we ll have to wok with thi. Since τ = d F and thee ae two foce but we only know one of them, we d bette put the oigin at the place whee one of the foce act thi i the point whee the ting meet the hub. Thi mean that the only foce ceating a non-zeo toque i the one due to the weight of the yoyo. Since the two foce (the weight and the tenion) point in exactly oppoite diection (if they didn t, the v yoyo v would acceleate away fom the peon holding it, not jut up and down) d and F ae pependicula to each othe and we can jut multiply them without uing the in(θ) tem. Thi give τ = Mg. Time fo a little eview: Newton econd law i a tatement of caue and effect. v v In the otational cae, thi i witten τ = Iα with the caue being the toque and the effect being the angula acceleation. Ty to think about it thi way to avoid confuion. I ve had many conveation with tudent who ae dazed by the fact that we eem to have two equation fo τ. The ouce of the confuion i linked to you deie fo fomula. You ee an equation that ha a τ to the left of the = ign and you think oh, that a fomula fo toque. WRONG! Seeing it witten that way imply mean that it an equation involving toque. It can be tuned into a fomula fo any of the quantitie in the equation. The equation ha meaning that you hould eek to undetand. Do not imply memoize and egugitate thee thing o you ll find youelf confued and in touble! So, we ve found the caue of the angula acceleation: τ = Mg. The caue i linked to the effect by τ = Iα (once again jut concening ouelve with the ize and ignoing the diection fo now). Thee give Iα = Mg. Thi can be olved fo the Mg angula acceleation to give α =. Now we encounte a poblem: We picked the I place whee the ting meet the hub of the yoyo a the oigin. We eally hould ue the moment of inetia about that point. But that i beyond the level of thi cla. So we will jut make the appoximation that the moment of inetia of a dik otated about a point cloe to the cente i appoximately the ame a the moment of inetia of a dik. I ll ue a diffeent technique at the end of thi olution to how you what it look like if we do it exactly. The moment of inetia of a dik otated about it cente i I = MR. We
ue thi to find the total angula acceleation: α = Mg MR g = R a ditubing faction of you latching onto the FALSE definition I =. (Note: I ve noticed MR. Thi i tue fo a olid dik o cylinde otated about it axi. It i not tue fo all hape! Be caeful to ue the coect moment of inetia fo a paticula ituation.) adian If we expe the angula acceleation in, we can wite a = α (notice econd which adiu i being ued that of the hub, not the entie yoyou; make ue you g cm undetand why), o a =. Sticking in the numbe, we have a = 4.9. R Now fo an altenate olution that will yield the exact anwe, without appoximating the moment of inetia. To do thi we need to take the oigin at the cente of the yoyo an axi fo which the moment of inetia i well known exactly. Thi mean we e left with F T a an unknown. But all i not lot. Fom the toque equation, a we have τ = FT and ince a = α (a we dicued above), we can wite α =. Now we wite the otational fom of Newton econd law τ = Iα and put thee thee piece a a togethe to wite F T = I. O, F T = I. Now, Newton econd law (the egula fom) ay Mg F = T Ma, o we can olve fo FT: F T = Mg Ma. Thi can be ubtituted into ou acceleation expeion found fom the toque to give a Mg Ma = I. A bit of algeba on thi to olve fo a give a I Mg = I + Ma = a M + which yield Mg Mg g g a = = = =. Notice that ou two anwe I M + MR R + R M + + ae, a we pedicted, vey imila. If i vey much le than R, the addition of an in the denominato make inignificant diffeence. Howeve, if tat to get big enough that it a ignificant faction of R, we d bette ue thi fom. (Remembe that ou fit anwe wa coect except fo the fact that we ued the wong moment of inetia. If we d ued the accuate moment of inetia fo a dik otated aound a point away fom the cente by a ditance, which i I = M + MR, we would have gotten the pecie anwe uing that method.)
5. Conide again the yoyo decibed in the peviou poblem. Ue conevation of enegy to detemine the angula peed of the yoyo afte it ha decended ½ mete. Enegy i coneved in thi poblem. Thu we can wite the condition fo conevation of enegy P. E. + K. E. = contant. Afte the yoyo ha decended by ½ mete it P.E. ha deceaed by P. E. = Mgh. It K.E. mut have inceaed by the ame amount. But hee thee a diffeence between thi poblem and one we ve done befoe: The K.E. can go to two diffeent place. It can go to the oveall motion of the yoyo moving downwad, called tanlational kinetic enegy, o it can go into the pinning of the yoyo, called otational kinetic enegy. Do NOT peume that it goe into the two 50/50! We ll need to think about how it ditibuted athe than jut auming that it divided up faily. Fotunately fo u, ince the ting equie that the downwad motion i exactly elated to the pinning, we can wite thee two down at the ame time. Remembeing that v = ω, we can wite down the tanlational kinetic enegy a K. E. = Mv = Mω. The otational kinetic enegy i given by tanlational K. E. MR otational = Iω = ( MR ) ω = 4 ω, whee we have ued the moment of inetia of a dik otated about it cente. Setting the change in potential enegy equal to the um of the two kinetic enegie, we gh have Mgh = 4 MR ω + Mω. Thi give ω = R. Putting in the numbe, + adian thi give ω = 73.6. econd If the hub adiu i vey much le than the dik adiu, note that the vat majoity of the enegy i in the otational pat of thi. We d be jutified in appoximating the angula 4gh adian peed, in thi cae, by ω =. Again ineting the numbe, we get ω = 73.8. R econd Clealy the appoximation i quite a good one thi anwe vaie fom the one above by only about 0.3%. 4 6. An ice kate whoe ma i 55 kg i pinning at adian pe econd. She can be appoximated a a cylinde 50 cm in diamete. Someone thow a cat to he and the cat attache itelf to he with it hap claw (neglect the peed with which the cat i thown). The ma of the cat i 7 kg. What will he angula peed be with the cat attached? It i vey impotant that you ecognize the baic fom of thi poblem: Thi i jut anothe inelatic colliion (the cat tick, wheneve omething tick, it inelatic) and o i eentially identical to the othe inelatic colliion poblem that you ve done like the cat in lab. In thoe cae, we wee only concened with egula (i.e., linea) momentum. Hee, we e only concened with angula momentum. Linea momentum and angula momentum ae coneved independently, o you could alo olve thi poblem fo the kate new linea momentum if I d given you the velocity of the cat. You could do thi
uing peciely the ame method a you ve done peviouly. In thi cae, we jut need to woy about angula momentum, howeve. Angula momentum i coneved alway. Jut a in the linea cae, we have v v L befoe = L afte. The addition of the cat to the pinning kate change he moment of inetia. Theefoe, he angula peed mut change in ode fo he angula momentum to emain unchanged. (If we hadn t neglected the thown peed of the cat, thi would not neceaily be tue. We d have to allow fo the cat angula momentum a well.) The ize of the initial angula momentum of the kate i L = Iω. If we teat he a a cylinde, hee moment of inetia i given by I = M. Thu he total angula momentum kg m i L = M ω = 55kg (.5m) = 8.9. Thi will be the final angula momentum of the ytem a well. We ae now left with a bit of a judgment call to make: What hall we ue fo he moment of inetia once the cat attache itelf to he? The cat i a ingle ma which i attached at a ditance fom the kate axi. The moment of inetia of the cat, theefoe, i I cat = M cat. I think the bet technique would be to add thi moment of inetia to that of the kate. Howeve, an altenative method, which i alo valid (although I find it a bit le jutifiable) would be imply to conide the cat to be pat of the kate and jut ue I = M but fo the ma ue the um of the two mae kate and cat. I ll go ahead and ue the method I pefe, but if you ued the othe one, it alight. (Thi i the ot of thing that imply mut be communicated in a poblem olution. We can quibble about how valid an appoximation i, but we eally hould all agee on what appoximation ha been ued.) Thi give I = M kate + M cat o, etting the two angula momenta (befoe the cat and afte the cat) equal to each othe, we have M kate ωinitial = M kate + M cat ω final. Solving fo the final angula peed, we have ω M M ω M = M kate initial kate initial final = initial kate + M cat kate + M cat ω = 0.797ω adian = 8.77. econd
7. Two men ae tanding on a mey-go-ound which i tuning at a contant angula peed, ω. One man i tanding vey nea the cente of the meygo-ound, the othe i tanding vey nea the edge. The man at the cente i tying to thow a baeball diectly to hi fiend at the edge but fo ome eaon the ball alway wind up vey fa fom it intended taget. Pleae explain the mitake the man at the cente i making. (Thi i an example of the "coioli" effect.) Both men ae tanding on a otating object. They vey natually (it eally the way human ae put togethe) peceive the mey-go-ound to be a good fame of efeence. That i, ince they ae not moving elative to the mey-go-ound, they think of the meygo-ound a being motionle, even though they know it pinning aound. Futhe, even when thei intellect ovecome thei intuition ufficiently to convince them that they e moving, they till don t ecognize that they ae acceleating. Any object foced to move in anything othe than a taight path mut acceleate! Thu, thei efeence fame i noninetial. When the man at the cente thow the ball to hi fiend, he doen t allow fo the fact that hi fiend ha a diffeent velocity than he doe. The man at the edge i moving fate than the man at the cente even though they have the ame angula peed. The ball, when thown, will have whateve velocity the man at the cente give it. A een by an obeve in an inetial efeence fame, the ball will have that velocity plu whateve velocity the man at the cente ha (elative to the inetial fame). But the man at the edge will have a diffeent velocity. If the men peume thei efeence fame to be inetial, it will appea that the ball ha ome exta velocity not given to it by the thowe. Since the only thing that can change a velocity i a foce, the men will come to the concluion that thee i ome foce that act on the ball afte it i thown. In eality, no uch foce exit. It i fictitiou. Notice that the coioli foce act tangentially i.e., in a diection tangent to the otation. Thi i pependicula to anothe fictitiou foce that we ve tudied, the centifugal foce, which act adially i.e., along the adiu of the cicle. Don t confue the two! The coioli foce i vey impotant in weathe: A ai tavel fom one latitude to anothe, the gound ha a diffeent peed. Thi eult in ai moving elative to the gound until it get acceleated to the ame peed a the gound. Thi i a majo caue of wind and i the eaon that thing like huicane otate.
8. Conide again the ituation in poblem #7. The angula peed of the adian mey-go-ound i 0. and it i mete in diamete. The baeball econd mete i thown with a velocity in the xˆ diection of 3 (i.e., ignoe the ŷ econd component of the velocity). How fa (meaued along an ac) fom the catche doe the ball wind up? ω v v θ L x Let aume the mey-go-ound i pinning counte-clockwie a een fom above. (The anwe will be the ame which eve way you take the otation ince I only aked fo the ditance fom the catche the ball wind up, not the diection.) Thi i hown in the figue. Now, take the god eye view look down on the otating mey-go-ound fom the pepective of a fixed obeve in an inetial efeence fame diectly above the cente of the mey-go-ound. Thi in t eential, but it make life eaie. Fom thi pepective, the ball will be thown with a velocity v v and will keep thi velocity though the entie poblem it will not change ince thee i no eal foce acting on it afte it i thown. Howeve, in the time that it take fo the ball to make it fom the cente of the mey-goound to it edge, the catche will have moved ome angle θ fom hi tating point. To figue out what θ i, we fit ealize that we know how θ change with time. The angula peed of the mey-go-ound i jut ω = θ. Thi i jut like ou egula peed t except with angle eplacing ditance. We ead thi a the change in angle divided by the time it take to make the change. So, we can conclude that θ = ωt (taking ou initial y
angle to be zeo). Now all we need to do i find the time that it take fo the ball to each the edge of the mey-go-ound. Since the time that it take fo the ball to each the edge of the mey-go-ound doen t depend on the angula peed (note that thi i becaue of the high level of ymmety in a cicle; imagine an elliptical mey-go-ound, o ome othe hape, then the poblem would equie moe wok!), all we need to do i ealize that the peed (that i, the egula peed, not the angula peed) i jut the ditance taveled divided by the time it take to tavel that ditance. In thi cae, the ditance taveled i the adiu of the cicle,. 6 mete We have t =. Subtituting numbe, thi give t = = 0.46 econd. v mete 3 econd Now we ue thi with ou angula peed to find the angle adian θ = ωt = 0. 0.46 econd = 0.046 adian. But wait, we e not done yet. The econd poblem aked how fa fom the catche the ball wind up. That i, it ak fo the ditance. Thi i indicated on the figue a L. (I am oy that we ae uing the ame lette, L, to indicate both the ac length and the angula momentum. Thi ot of thing i maddeningly confuing, but inevitable. Thee ae jut too many concept to eeve a lette fo each of them!) Hee whee uing adian to meaue angle eally come in handy: By definition, L if we meaue ou angle in adian, we have θ. Thu L = θ = 0.046 ad 6 mete = 0.76 mete. (Some folk who ae new to adian will feel uncomfotable with the fact that thi unit eem to have magically vanihed fom the anwe above. In ome ene, thi i becaue it wa neve eally thee. Recall the definition of an angle meaued in adian L. Note that the numeato and the denominato both have the ame unit. So θ neve eally had a unit. By aying the angle i meaued in adian we keep tack of how the numbe wa ceated, but thee not eally a unit thee. I know thi i confuing. Thi i one place whee I ll jut ay to memoize the fact that thi i o without wating too much time on undetanding why it i. Thee won t be too many othe cae like thi.) θ
9. Yet again, conide the mey-go-ound of poblem #7. The catche peceive that a foce i acting on the ball to kew it path. Thi i a fictitiou foce. What ize foce would be needed to caue the motion obeved by the catche if it wee eal? A i o often the cae, thee a had way to do thi and an eay way. I knew that when I wote the poblem. What I lot tack of, howeve, wa jut how had the had way i if you don t know calculu fa too had fo thi coue. (Ty to emembe what it wa like befoe you could ead. It difficult to emembe, in t it? That the way it get with Math alo: Sometime it had to keep hold of the fact that I wan t bon knowing calculu.) So don t feel too bad if you mied it. I m pefectly happy with the eay anwe. But the had way i intuctional, howeve, o I m not oy I thew it at you. I jut hould have given you adequate waning fit. The had way i much moe infomative than the eay way, o I ll do it that way fit. Then, I ll make you lap you head in futation by howing you how eaily it can eally be done. I ll alo include the method uing calculu, fo thoe of you who have that kill. Had way: You mut ecognize that the ball move in a taight line at a contant peed (a equied by Newton fit law) fo it entie tip a een by an obeve in an inetial efeence fame. Thi line i along a adiu of the mey-go-ound, o let call thi peed v adial. Alo, the mey-go-ound move at a contant angula peed fo that entie tip. But, a the ball move out fom the cente of the mey-go-ound, the peed (not angula) of the mey-go-ound inceae teadily. Since egion of the mey-go-ound that ae diffeent ditance fom it cente weep out lage ditance in a given amount of time. L Thi can be een fom the definition of angle θ. If we take θ to be the angle wept out in ome mall inteval of time, a inceae, L mut alo inceae to keep θ contant (a it mut be ince the angula peed i contant). Note that thi peed i in a diection pependicula to v adial. Now, let come up with a tategy fo olving thi poblem: We want an acceleation. (We actually want a foce, but ince we know Newton econd law, we ealize that if we can find the acceleation, we ll have alo found the foce.) We know that acceleation i the change in velocity divided by the time ove which that change occu. We alo know that velocity i the change in the diplacement vecto divided by the time ove which that change occu. So, if we can find the diplacement a a function of time, we can divide by time to find velocity and then divide that by time to find acceleation. Let do it. Now, we can define an angle in tem of the adiu and the ac length it ubtend: L θ. Alo, the angula peed i the angle wept out in ome inteval of time: ω = θ. t L Putting thee togethe, we can wite L ω = θ = =. (You may be toubled that I t t t blithely go between θ and θ. I can do thi by imply taking the angle at the beginning to be zeo. Recall that the change in ome quantity i it value at the end minu it value at the
beginning. So, θ = θ if θ = 0 beginning when it i tue!) Now, figue out what L i a a function of time:. Thi i convenient, but be caeful to ue it only L = ωt. We e almot thee. Now, to find the peed (we e now woking in one dimenion, o we can dop the vecto language), we divide thi by t to get v = L = ω. Pleae be caeful: t Hee i not the total adiu of the mey-go-ound but imply the ditance fom the cente at which the ball happen to be at ome time. If that time happen to be the time at which the ball i at the edge of the mey-go-ound, then i the full adiu. Now, we find the acceleation of the ball. We make the aumption (if you know a bit of calculu you can pove that thi i tue; if you don t, you ll jut have to take my wod fo it) that the acceleation i contant in thi poblem. Thu we can ue the definition of v vend vbegin acceleation (in one dimenion) a = =. Remembe: We e not talking about t t the peed of the ball a it goe fom the cente of the mey-go-ound to it edge. All obeve agee that the ball i not acceleating in that diection. We e talking about the ω acceleation pependicula to thi. In that diection, v begin =0, o we can jut wite a =. t ω We e almot thee. Notice that we can ewite the above equation a a = = ω. But t t t i peciely the initial peed of the ball vadial! So we can wite a = ωvadial. Ineting adian mete mete ome numbe, thi give a =. 3 =.3. Multiply thi by the econd econd econd ma of the ball to find the fictitiou foce. But wait, we e not done yet! I pulled a leight of hand on you. I only included the change in the ball peed (in the non-inetial fame) in calculating the acceleation. Remembe: Acceleation can be a change in the diection of the velocity in addition to (o intead of) a change in it ize! So fa, we ve only calculated the change in the ize of the velocity a een by the catche. He alo ee a change in the diection of the velocity: The ball begin headed diectly towad him, but at the end, it i hooting off in ome othe diection. The math hee get way beyond thi coue, but the eult i a facto of applied mete to the expeion above. So the final anwe i a =.6, but I wouldn t expect econd anyone in the cla to get the exta facto of. (Anyone who han t had calculu, don t ead thi paagaph! Fo thoe of you who have had calculu, it goe like thi: Stat with ( t) = vadialt and θ ( t) = ωt. Thee ae the ize of the component of the poition vecto in a cicula coodinate ytem a a function of time. The deviation of the ball poition a a function of time will lie in the θˆ diection and will be L( t) = ( t) θ ( t) = v t. We take the deivative of thi with epect adialω to time to find the velocity in the θˆ diection. Thi i v dl( t) = = vadialωt. Now, we take dt θ
the deivative with epect to time of the velocity to find the acceleation. Thi i dvθ a = = v adial ω. Man, doe calculu make life eaie!) dt Eay way: That wa petty tough, huh? Epecially that magical facto of! Let do it the eay way: Again ecognizing that the acceleation i in the diection pependicula to v adial, we note the olution found in the peviou poblem: The ditance taveled due to thi foce (in the non-inetial fame) i 0.76 mete. We alo know how long it took fo thi ditance to be taveled: 0.46 econd. Thu, we can wite L = at and olve thi fo L the acceleation. Thi give a =. Plugging in the numbe above, we have t L.76 mete mete a = = =.6. Yup, it that eay! t (.46 econd) econd 0. What i the obital angula momentum of the Eath? That i, the angula momentum due to the Eath evolving aound the un. Thee ae two way to do thi (in t that a hock?). They e actually identical, but they don t look that way unle you ecall v whee the equation you e uing came fom. v v The definition of angula momentum i L = p whee v i the poition vecto of an object and p v i it (egula) momentum. When we wok thi out and allow fo the fact that object fequently don t come a iolated blob v of matte but ae often extended, with ome tuctue, we get a vaiation which i L = I v ω. (Reead that lat entence a couple of time until you ae clea on what it i aying: The two equation ae the ame. One imply take into account all the component mae that make up an object while the othe i the angula momentum of jut one of thoe component mae.) If we teat the Eath going aound the un a an iolated blob of matte moving with ome angula peed ω, we get (conideing only the ize of the vecto we can neglect the vecto co poduct becaue the velocity vecto and the poition vecto ae pependicula to each othe to a vey good appoximation) L = p = mv = m ω, whee I ve ued the fact that v = ω. (If thi final elation in t like econd natue to you by now, concentate on it! It i a vey impotant piece of geomety.) Well, the moment of inetia of an iolated ma a ditance fom the axi of otation i I = m, o we ve eally jut witten L = Iω. I ll ue thi fom ince it moe convenient in thi cae. The Eath goe aound the un one time pe yea. Auming that the ditance fom the Eath to the un and the angula peed both emain contant (and Keple fit and econd law ay that thi i not tue, although it i a vey good appoximation), we can find the angula peed eaily. The Eath tavel an angle of π adian in one yea, 7 which i 3.56 0 econd. Thi give an angula peed of θ π 7 ω = = =.99 0. 7 t 3.56 0 econd
The mean ditance between the Eath and the un i.5 0 Eath ma i at the un i I 5.98 0 4 mete and the kg, o the Eath moment of inetia about a otational axi 47 (.5 0 mete) =.38 0 kilogam mete 4 = m = 5.98 0 kg Putting thee two eult togethe, we get numbe! L = Iω =.38 0 47 kg m.99 0 7 =.75 0 40 kg m. Now that a big.