Homework 7. problems: 9.33, 9.40, 9.65, 9.78

Transcription

1 Hoework 7 probe: 9., 9.4, 9.65, 9.78

2 Probe 9. A biiard ba ovin at 5. / trike a tationar ba of te ae a. After te coiion, te firt ba ove at 4. / and an ane of. wit repect to te oriina ine of otion. Auin an eatic coiion (and inorin friction and rotationa otion), find te truck ba veocit after te coiion. Ti probe contain too uc inforation, wic coud ead to contradiction! In an coiion te tota oentu of te te i conerved v i v f θ φ ) v i + vf + v f v f Hence vf vi vf 5, 4. co,in.5,.65 (One coud ao epre veocit in ter of it poar coordinate [ ] [ ] [ ] (.5 ) + (.65 ) v f vf + vf vf.65 φ arctan arctan 6 vf.5.5 Note tat we found te fina veocit of te econd ba rearde te tpe of te coiion. To verif tat te oution i conitent wit te iven inforation one oud copare te initia and te fina tota kinetic ener vi K + tot,f v ( 5. ) i. K v v v v tot,i f f f + f ( 4. ) + (.5 ) + Te coiion i indeed eatic.

3 Probe 9.4 A rod of ent c a inear denit (a per ent) iven b λ 5 + Were i te ditance fro one end, eaured in eter, and λ i in ra/eter. (a) Wat i te a of te rod? (b) How far fro te end i it center of a. d d c L z a) Conitent wit te fiure and te iven inforation, te a of te indicated fraent of te rod i d λd 5 + d c c In order to find te tota a, te a of a differentia piece ut be added. It require interation of te above function in appropriate iit.. M d 5 + d c rod 5 + Te a of te rod i

4 b) Findin te -coponent of te ocation of center of a require ao interation. Direct fro te definition of center of a c M M d. 5.c Center of a i 5. c fro te c end.

5 Probe 9.8 (a) Conider an etended woe difference portion ave different eevation. Aue te free-fa acceeration i unifor over te. Prove tat te ravitationa potentia ener of te -eart te i iven b U M c, were M i te tota a of te and c i te eevation of it center of a above te coen reference eve. (b) Cacuate te ravitationa potentia ener aociated wit a rap contructed on eve round wit tone wit denit,8 k/ and everwere.6 wide. In a ide view, te rap appear a rit triane wit i 5.7 at te top end and bae d w a) Gravitationa potentia ener of a differentia fraent of an depend on te a of te fraent and it eevation above te reference eve ) du d were i te acceeration due to ravit (free fa acceeration). Te tota ravitationa potentia ener of te require interation of te ravitationa potentia ener over te entire. ) U du d d d Fro te definition of center of a, te -coponent of it poition i ) c d M Subtitutin te intera in equation () 4) U du d d M c Mc z

6 b) Direct fro te definition of center of a, te -coponent of te rap center of a can be found 5) c M w d ρ w w dz dd w d w ρdzdd z d w dd w wdd Fro part (a) 6) U Mc ρ w k J (Ti uc work ut be done b te buider of te rap.)

7 Probe 9.49 Sand fro a tationar opper fa onto a ovin conveor bet at te rate of 5 k/. Te conveer bet i upported b frictione roer. It ove at a contant peed of.75 / under te action of a contant orizonta eterna force F et uppied b te otor tat drive te bet. Find (a) te and rate of cane of oentu in te orizonta direction, (b) te force of friction eerted b te bet on te and, (c) te eterna force F et, and (d) te work done b F et in, and (e) te kinetic ener acquired b te fain and eac econd due to te cane in it orizonta otion. (d) (f) W are te anwer to (d) and (e) different?.75 / F et a) Te a of and fain on te bet in tie dt i d ) d dt dt d were i te iven rate at wic te and fa of te conveor bet. dt Wit a ood approiation, we can aue tat te initia oentu of and faen on te bet in tie dt a a zero vaue. On te and fain on te bet cane it oentu cane it oentu in tat tie. Since te cane i in te aon te bet (-direction), we can iit conideration to one coponent. Te cane in tota oentu of te and i terefore d ) dp v d d v dt dt were v i te peed of te conveor bet. Hence te rate of cane in te tota oentu of te and i dp ) d k v N dt dt

8 b) Tere are tree force eerted on te and but on te frictiona force i not baanced. Te net force i terefore equa to te frictiona force. Hence, fro Newton econd aw of otion (verion ), te -coponent of te (tota) frictiona force eerted on te and i.75 N. c) Fro Newton tird aw of otion, te and eert an oppoite, frictiona force wit te -coponent of -.75 N. Accordin to Newton econd aw of otion, te net force eerted on te conveor bet i zero, wic require tat te otor eert a force wit te -coponent of.75 N. d) One can find te work direct fro it definition reconizin tat te force eerted on te bet b te otor i contant 4) ΔW Fet, d Fet, vδt.75n.75.8j pat e) Te rate of cane in te tota kinetic ener of te and can be found direct fro te definition dk ( ) J 4) tot d v d k v dt dt dt Since te above rate i tie independent, te cane in kinetic ener of te and i dk 5) tot J Δ Ktot Δt.4.4J dt f) Cane in te tota kinetic ener of a te i reated to work done b bot eterna and interna interaction. A te and fa on te conveor bet, kinetic frictiona force are eerted b te bet on te and and b te and on te bet. Unti te and acquire te ae veocit a te bet (ineatic coiion), te ditance traveed b te bet i reater tan te ditance traveed b te and. Terefore, te interna work i neative. Fro te (fourt verion) of te work-ener teore we a concude tat te interna work i 6) Δ W ΔK ΔW.4J.8J.4J int tot et

9 Probe 9.57 An 8 buet i fired into a.5 k bock tat i initia at ret at te ede of a frictione tabe of eit (Fi. P.58). Te buet reain in te bock, and after ipact te bock and fro te botto of te tabe. Deterine te initia peed of te buet. v i M r [, ] v r [, - ] In te proce te buet firt coide wit te bock, ten te entire te ove wit a contant acceeration (free fa). Sueted coordinate te i convenient for conideration bot procee. Coniderin bot, te buet and te bock, a te partice one can reate te veocit v of te te after te coiion wit te initia veocit v i of te buet. Te oentu conervation aw require tat ) v i + ( + M) v fro wic In te free fa otion te poition of te partice i a quadratic function of tie. Conitent wit te arked coordinate te t ) r() t v t + Te ret i at. Fro te iven vertica coponent buet-bock te one can deterine te tie t of te fa. ( ) ) t and te required initia veocit of te te

10 4) v r t t i [, ] [, ] t.45, 4.4, Subtitution in te firt equation ead to 5) v ( + M) i v.8k +.5k 4.4,.8k fro wic te anwer i 6) v i v i 4.4