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pg 75 - circular motion the mégninide of the acceleration...

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Unformatted text preview: circular motion, the mégninide- of the acceleration is .vzlr; So the com.- patients of Newto‘rt‘ 8 law become r? C 11111 x: itsi'nti ="% ' y: itcosfl.-=0 r . where the 0 on the right-hand. side- of the .y. equation reflects the fact that we don‘t want the .car to accelerate in the verticai direction Solv~ ing the )2 equation gives 11: mgicoso Then using this result in the _x_ equation gives mg sine/cos H: mvzfr, or g tanti- -— vzlr The mass can- celed, which is good news because it means our banked road will work for a vehicle of any. maSs Now we can solve for the Banking angle: . merit 5.7: Circtliar Motioré. Looping the Loop ' 7 The “Great American Revolution" roller coaster at Valencia Califorw I nia, includes a loop-thc-iOop section Whose radius 15 6. 3. m at the top (see the chapter opening photo). What is the minimum speed for a ' roller-coaSter car. atthc top of the loop if it’sto stay on the track? - INTERPRET Again, we have. circular motion described by Newton’ 5 sec— ond law. We re asked about the minimum speed for the car to stay on the track. What does it mean to stay on- the track? It means there must he a normal force: bettVeen car anti track; otherwise, the two aren’ t in contactso we can-identify tw0 forces acting on the car; gravity and ' the normal force from the track. Davao? Figure 5.15 shows the phySic‘al situation The situation is es— pecially simple at the top of the track, Where both forces point in the- same direction. We show this in our- free-body diagram, Fig; 5 16. Since that common direction' is downward, it makes sense to choose a ' coordinate system With the y axis downward. The applicable eqi'iation is Newton 1-} second law, and with the. hive forces we ve identified, that becomes" 11 + F= ma EvALuan With both forces in the same direction, we need only the y cemponen't cf NeWton’ s law. With the doanard direction gositive, =1: and F —mg At the' top of the loop, the? car- is in circular motion,lso its acceleration is- toward the center—'doanard—and has At the top, both-ferries point doiimwérd‘andthe I ' - caris‘moiiieniefilyin..___ uniform circular ‘ _ 1 - _ motion ' " ' Gravityis airways downward ' - __ .- but at this pomtthe normal force is horizontal. The net fume ism “Howard the center, and the car isslowiiig as well - as changing direction.- Eliillfilimi ‘ Forces on the roller—coaster car. 5.3 Circuiar Motion 75 ‘ 10- % fin-jig) Z: %n_1( (9s $153313; 1.11).: 180 assess Make sense? At low speed 1} or large radius r, the car 5 motion changes gently and it doesn 1 take a large force to keep it on its circu- lar. path But as v increases or. r decreases, the required force increases and so does the banking angle That’s because the horizontal compo- nent of the dermal force 15 what keréps the car in circular motion, and /' I _ the steeper- the angle. the greater that cemponeiit. ' I Elam-5.]; Our free-body diagram at the tap of theioep. I magnitude v'izlr So. a =_21r,-and the y componcht of Newtoii’ s. law becomes _ - mvz -n+mg.——' r Solving for the speed gives v2 V (iv/m) +gr- 'NoW, the minimum possible speed for contact with the track occurs- when 11 gets arbihfaii- ly small right at the top of. the track, so we find this minimum limit. by setting 11:9. Then the rinswer is - '(9sm132)(63m}—79m/s _A'sse_ss Do you see what’ s happening here? With the minimum Speed I the normal force 'vaniizh‘es at the top of the LOOP, and gravity alone pro- ' vides the force that ice-Legs the object. in its circular path. Since the mo~ tion is circular, that force must have magnitude. owl/r. But the force of - gravity. alone is mg; and rm: 'ng sfollow's directly from equating those two quantities A car moving any slower than. 11mm would lose contact with the track and go into the parabolic; najeetory of a proiee tile. For a car moving faster, there wrinid- be. a nonzero non-rial force contributing to the downward acceleration .at the top of; the loop In the “Great American Revolution,- the acttial speed at the loop 5- top is 9.72111/5 to provide 211 margin of safety. As with many problems invalv- ing gravity, the mass cancels. That’ s'. .a good .thii'rg because it means the safe spreéd doesn’t depend ori the number or mass of the riders I - ...
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