4-6 ROTATIONAL MOTION

Transcription

1 Chpter 4 Motions in Spce 51 Reinforce the ide tht net force is needed for orbitl motion Content We discuss the trnsition from projectile motion to orbitl motion when bll is thrown horizontlly with eer incresing speed. Teching Tips Students re often surprised to lern tht projectile motion nd orbitl motion re similr. We lso conert km/s to 1,000 mph to impress the students with the gret speed required. Flwed Resoning Exmines the concepts of centripetl nd centrifugl forces for stellite motion. 4-6 ROTATIONAL MOTION Gols Introduce the notion of center of mss. Argue tht rottionl nd trnsltionl motions re independent. Content When n object moes through spce, its motion cn be thought of s combintion of rottion bout its center of mss nd trnsltion of its center of mss. Teching Tips This section llows instructors to skip Chpter on Rottions without lter consequences. Physics on Your Own A ping-pong or styrofom bll is weighted off-center nd used in gme of ctch. The chnge in the bll's center of mss mkes the rottionl motion quite obious. Video Encyclopedi 4 #1 Center of Mss Disc #7 Air Tble Center of Mss Computer Animtions Actie Figure Animtions re ilble on the Multimedi Mnger Instructor s Resource CD. They re orgnized by textbook chpter, nd ech nimtion comes within shell tht proides informtion on how to use the nimtion, explortion ctiities, nd short quiz. Answers to the Conceptul Questions 1... The cr would trel in the direction of its elocity becuse there would be no net force cting on it. 3. The grittionl force proides the necessry centripetl force.

2 5 Chpter 4 Motions in Spce 4. The friction force with the ground proides the necessry centripetl force. If the frictionl force is not strong enough, the skter will skid to the outside of the corner. 5. The elocity points to the left, which is stright hed to the drier. The chnge in elocity nd therefore the ccelertion point stright down becuse the speed is not chnging but the direction is. The net force must point in the direction of the ccelertion, which is down. 6. ) The elocity is tngent to the pth. b-d) The chnge in elocity, the ccelertion, nd the net force ll point rdilly inwrd. 7. The elocity ectors point in opposite directions but he the sme size.. The ccelertions ectors point in opposite directions, but they he the sme size nd both point towrd the center of the circulr pth. 9. The speed chnges: If the force hs component in the direction of the motion, the speed increses, nd if the force hs component in the direction opposite the motion, the speed decreses. 10. As long s the force stys perpendiculr to the elocity there will only be chnge in the direction of the object, its speed does not chnge. Your sketch should show pth which continully spirls inwrd. 11. The ccelertion is in the sme direction s the chnge in elocity. ) b) 1. The ccelertion is in the sme direction s the chnge in elocity. ) b) 13. ) Speeding up in stright line. b) Speeding up nd turning left. 14. ) Slowing down in stright line. b) Slowing down nd turning right. 15. ) Exmples of objects moing down nd speeding up would include flling object nd downwrd moing eletor s it begins to moe. b) Exmples of objects moing up nd slowing down would include ny object thrown upwrd nd n upwrd moing eletor s it reches its destintion. 16. As you strt moing, you re moing down nd speeding up. Therefore your ccelertion must be downwrd. As you pproch the lobby, you re moing down but slowing down. Your ccelertion ector must be directed up. 17. The friction force with the ground proides the necessry centripetl force. 1. The friction force with the ground proides the necessry centripetl force. 19.The tension force upwrd must be greter thn the weight of the monkey to yield net force upwrd, towrd the center of the circle. 0. The horizontl prt of the norml force exerted by the rod on the cr proides the mjority of the centripetl force. The horizontl prt of the friction force cn either increse or decrese the centripetl force, depending on the speed of the cr. 1. You must exert extr force t the bottom to counterct the force of grity pointing wy from the center of the circle. Remember tht the net force must be equl to the centripetl force.. In ddition to supporting Trzn's weight the ine must exert extr force to mke Trzn moe in circle. 3. Erth s speed must be nerly uniform. The net force on erth must be perpendiculr to its elocity, so it does not chnge speed. 4. The Sun does follow circulr pth, but becuse the mss of the Sun is so much greter thn the mss of Erth, it hs much smller orbit bout their common center of mss. This slight wobble could indicte plnets orbiting distnt strs.

3 Chpter 4 Motions in Spce 53 5.Doubling your speed would qudruple your centripetl ccelertion while reducing your rdius to hlf would only double your centripetl ccelertion. 6. Doubling the speed qudruples the centripetl ccelertion while doubling the rdius cuts it by hlf. The net effect is tht the centripetl ccelertion on the lrge circle is twice wht it is on the smll one. 7. The only force is the force of grity cting erticlly downwrd.. 9. The bll's elocity is horizontl towrd home plte. The net force nd the ccelertion point erticlly downwrd. 30. A B C 31. You get the sme ccelertion ech time. Becuse the horizontl nd erticl motions re independent, the sidewrd motion hs no effect on the downwrd ccelertion. 3. The independence of the erticl nd horizontl motions mens tht they will hit t the sme time. 33. Since both blls he the sme erticl speed t the beginning, they will hit the floor t the sme time. The horizontl motions do not mtter. 34. Since free-fll ccelertion is independent of mss, both blls will hit t the sme time. 35. The demonstrtion works becuse both the freely flling gorill nd the incoming projectile fll t the sme rte nd thus meet t some point long the gorill's downwrd pth. 36. If the two sides of the cnyon re t the sme height, he fils becuse he flls during the time required to cross the cnyon. 37. Assuming tht the initil speed of the bll doesn't chnge, kicking the bll t lrger ngle reltie to the ground results in more hng time. 3. As the ngle of the fce of the iron increses, the bll strts off with lrger ngle reltie to the ground nd therefore goes higher nd shorter. 39. The pple exerts grittionl force of 5 newtons on Erth, by Newton s 3 rd lw. 40. The forces re equl in mgnitude nd opposite in direction by Newton s 3 rd lw. 41. There would only be one force, the grittionl force exerted by the Sun, which points towrd the Sun. 4. There would only be one force, the grittionl force exerted by Erth, which points towrd the center of Erth. 43. The object s center of mss is the only point tht will follow simple prbolic pth. Point C is most likely nerest the object s center of mss.

4 54 Chpter 4 Motions in Spce 44. Since the softbll hs more mss thn the tennis bll, the center of mss will be locted nerer the softbll. Therefore, point A is the most likely to follow prbolic pth. Answers to the Exercises 1. ) 5 m/s west b) 5 m/s est 10 m/s west 5 m/s west -5 m/s west 5 m/s est 5 m/s west -10 m/s west c) 15 m/s est -5 m/s west 10 m/s est 15 m/s est. ) 5 km/h right b) 175 km/h left 5 km/h 100 km/h right -75 km/h right -75 km/h 100 km/h left 175 km/h 3. = ( 40 m/s ) + ( 30 m/s ) = 50 m/s t 37 north of west 4. = ( 10 m/s ) + ( 50 m/s ) = 130 m/s t 3 est of north 550 m/s m/s 50 km/h west m/s m/s ssouth km/h north km/h 404 m/s west

5 Chpter 4 Motions in Spce m/s - m/s south 53 o 6 m/s est = = t = 5m s ( m s) + ( 6m s) s ( 53 north of est) 1 m/s est 4 m/s west 5 m/s west = = t = ( 4ms) ( 1ms) ( ) 10 m s west 0.5 s ( 10 m/s ) 7. ) = m/s = = b) F = m = ( 10 kg ) ( m/ s ) = 40 N r 50 m ( m/s ) 4. ) = = = 3 m/s = 1.3 m/s r 3 m b) F = m = ( 60 kg ) ( m/ s ) = 0 N ; smller. 9. ) b) c) 10. ) C T h 3600 s = = 1dy 1h dys s ( ) = π = π = r m m 11 C m = = = 7 T s ( m s) m s = = = 11 r m m s d) F m ( ) ( ) = = = kg m s N πr π ( m ) = = = 10 T 4 h 3600 s 7 dys 1 dy 1 h ( m s) m/s = = = m s r m b) F m ( )( ) 3 = = = kg m s N 11. The horizontl speed remins 45 m/s; the erticl speed decreses by 10 m/s to m/s. 1. The horizontl speed remins 45 m/s; the erticl speed decreses by (10 m/s) to m/s (tht is, downwrd t m/s). 13. d = t = ( 40 m/s ) ( 5 s ) = 00 m (constnt )

6 56 Chpter 4 Motions in Spce ( )( ) ( 1 1 d = t = 10 m s 5 s = 15 m free fll) d 50 m 14. = = = 3.57 m s t 14 s 10 m/s 15. tup = = = 1s t = s 10 m/s ( )( ) R = t = 30 m s s = 60 m 30 m/s 16. tup = = = 3s t =6s R = t = ( 6m s) ( 6s) = 36m 10 m/s ( kg) 6 ( ) 1 11 N m m s F M = = G = = M r kg m ( 10 3 m s) = = = m s r m 3 Answers to the Problems in Problem Soling 1. ) 3 m/s west b) 3 m/s est c) 9 m/s est. ) 60 km/h forwrd b) 140 km/h bckwrd 3. = ( 100 mph ) + ( 100 mph ) = 141 mph southest 4. = ( ) ( ) 10 mph + 10 mph = 170 mph southwest

7 Chpter 4 Motions in Spce = ( ) ( ) 3 m/ s + 6 m/ s = 6.71 m/ s 3 m/s tn θ= = 0.5 θ= 6.6 north of west 6 m/s θ 6. = ( ) ( ) 50 km/h km/h = 11 km/h 100 km/h tn θ= = θ= 63.4 north of west 50 km/h 7. ν t ( )( ) = = 4 m/s west s = m s west ) = + = 3 m/s west + m/s west = 11 m/s west f i b) = + = 6 m/s est + m/s west = m/s west f i. ν t ( )( ) 9. = = 3 m/s est s = 6 m s est ) = = m/s est 6 m/s est = m/s est i f b) = = 4 m/s west 6 m/s est = 10 m/s west i f ( ) π r π m 1h = 17,300 m/s 4.5 h 3600 s = = T ( 17,300 m s) = = = 0.71 m s r m 10. ) r = r + h = 6370 km km = 690 km erth ( ) 6 πr π m 1min b) ν = = 7630 m/s T 95 min = 60 s c) ( 7630 m s) = = = 6 r m 11. ( ) ( ).41 m s 1 m/s Fc = m = 0 kg = 115 N r 10 m F = F F = F mg = 115 N 0 kg 9. m/s = 36 N set c g c 1. ( ) ( ) ( )( ) 50 m/s Fc = m = 70 kg = 75 N r 00 m F = F + F = F + mg = 75 N + 70 kg 9. m/s = 1560 N set c g c ( )( ) θ

8 5 Chpter 4 Motions in Spce 13. The horizontl speed remins t 15 m/s becuse there is no force in this direction s long s ir resistnce is neglected. f = i + t = 1 m/s + ( 9. m/s ) ( 1 s ) =. m/s (upwrd) 14. h = 15 m/s; t ( ) ( ) 15. ) h t = = g = + = 1 m s + 9. m s s = 7.6 m s (down) f i (. ) 196m = 4 s 9. m/s b) x = t = ( 0 m/s )( s ) = 40 m = s R 45 m 16. ) R = ht = t = = 3s 15 m s 1 b) d t h 1 = i + 0 ( )( ) 9.m s 3s 17. t ( )( ) = + = 44.1 m = + = 0+ 9.m/s 3s = 9.4m f i f h ( ) ( ) 1. t 0 ( 9.m/s )( s) 19.6m/ = + = 9.4 m s + 40 m s = 49.6 m s = + = + = s f i ( ) 19.6 m/s + ( 30 m/s ) = 35. m/s = f + h = 19. It tkes 1 s for the footbll to rech the top of its pth nd 1s dditionl for it to return to its originl height. Therefore, x = t = ( 15 m s)( s) = 30 m 0. ) 1 s b) x = h t = ( 9. m/s )( s ) = 19.6 m