3. No. The relationship between the parts of a non-rigid object can change. Different parts of the object may have different values of ω.

Transcription

1 CHAPTE : ainal Min Answers Quesins. The deer will regiser a disance greaer han he disance acually raveled. The deer cuns he nuber f revluins and he calibrain gives he disance raveled per revluin (πr). The saller ire will have a saller radius, and a saller acual disance raveled per revluin.. A pin n he ri f a disk raing wih cnsan angular velciy has n angenial accelerain since he angenial speed is cnsan. I des have radial accelerain. Alhugh he pin s speed is n changing, is velciy is, since he velciy vecr is changing direcin. The pin has a cenripeal accelerain, which is direced radially inward. If he disk s angular velciy increases unifrly, he pin n he ri will have bh radial and angenial accelerain, since i is bh ving in a circle and speeding up. The agniude f he radial cpnen f accelerain will increase in he case f he disk wih a unifrly increasing angular velciy, alhugh he angenial cpnen will be cnsan. In he case f he disk raing wih cnsan angular velciy, neiher cpnen f linear accelerain will change... The relainship beween he pars f a nn-rigid bjec can change. Differen pars f he bjec ay have differen values f ω. 4. Yes. The agniude f he rque eered depends n nly n he agniude f he frce bu als n he lever ar, which invlves bh he disance fr he frce he ais f rain and he angle a which he frce is applied. A sall frce applied wih a large lever ar culd creae a greaer rque han a larger frce wih a saller lever ar. 5. When yu d a si-up, yu are raing yur runk abu a hriznal ais hrugh yur hips. When yur hands are behind yur head, yur en f ineria is larger han when yur hands are sreched u in frn f yu. The si-up wih yur hands behind yur head will require re rque, and herefre will be harder d. 6. unning invlves raing he leg abu he pin where i is aached he res f he bdy. Therefre, running fas requires he abiliy change he leg s rain easily. The saller he en f ineria f an bjec, he saller he resisance a change in is rainal in. The clser he ass is he ais f rain, he saller he en f ineria. Cncenraing flesh and uscle high and clse he bdy iniizes he en f ineria and increases he angular accelerain pssible fr a given rque, iprving he abiliy run fas. 7.. If w equal and ppsie frces ac n an bjec, he ne frce will be zer. If he frces are n c-linear, he w frces will prduce a rque.. If an unbalanced frce acs hrugh he ais f rain, here will be a ne frce n he bjec, bu n ne rque. 8. The speed f he ball will be he sae n bh inclines. A he p f he incline, he ball has graviainal penial energy. This energy beces cnvered ranslainal and rainal kineic energy as he ball rlls dwn he incline. Since he inclines have he sae heigh, he ball will have he sae iniial penial energy and herefre he sae final kineic energy and he sae speed in bh cases. 9. ll he spheres dwn an incline. The hllw sphere will have a grea en f ineria and will ake lnger reach he b f he incline. 8 Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey 8

2 Giancli Physics fr Scieniss & Engineers, 4 h Ediin. The w spheres will reach he b a he sae ie wih he sae speed. The larger, re assive sphere will have he greaer al kineic energy a he b, since he al kineic energy can be saed in ers f ass and speed.. A ighrpe walker carries a lng, narrw bea in rder increase his r her en f ineria, aking rain (and falling ff he wire) re difficul. The greaer en f ineria increases he resisance change in angular in, giving he walker re ie cpensae fr sall shifs in psiin.. The en f ineria f a slid sphere is given by by M and ha f a slid cylinder is given 5 M. The slid sphere, wih a saller en f ineria and herefre a saller resisance change in rainal in, will reach he b f he incline firs and have he greaes speed. Since bh bjecs begin a he sae heigh and have he sae ass, hey have he sae iniial penial energy. Since he penial energy is cpleely cnvered kineic energy a he b f he incline, he w bjecs will have he sae al kineic energy. Hwever, he cylinder will have a greaer rainal kineic energy because is greaer en f ineria re han cpensaes 4 fr is lwer velciy. A he b, v = gh and v = gh Since rainal kineic energy is K Iω, r sphere 7 = hen K = gh and K = gh. r 7 sphere r cylinder cylinder.. The en f ineria will be leas abu an ais parallel he spine f he bk, passing hrugh he cener f he bk. Fr his chice, he ass disribuin fr he bk will be clses he ais. 4. Larger. The en f ineria depends n he disribuin f ass. Iagine he disk as a cllecin f any lile bis f ass. Mving he ais f rain he edge f he disk increases he average disance f he bis f ass he ais, and herefre increases he en f ineria. (See he Parallel Ais here.) 5. If he angular velciy vecr f a wheel n an ale pins wes, he wheel is raing such ha he linear velciy vecr f a pin a he p f he wheel pins nrh. If he angular accelerain vecr pins eas (ppsie he angular velciy vecr), hen he wheel is slwing dwn and he linear accelerain vecr fr he pin n he p f he wheel pins suh. The angular speed f he wheel is decreasing. Sluins Prbles. (a) ( π ) 45. rad 6 = π 4 rad =.785rad (b) ( π ) 6. rad 6 = π rad =.5rad (c) ( π ) 9. rad 6 = π rad =.57 rad (d) ( π ) 6. rad 6 = π rad = 6.8rad (e) ( π ) 445 rad 6 = 89π 6 rad = 7.77 rad 8 Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey 9

3 Chaper ainal Min. The subended angle (in radians) is he diaeer f he Sun divided by he Earh Sun disance. diaeer f Sun θ = r Earh Sun π rad radius f Sun = θ r =.5.5 = Earh Sun. We find he diaeer f he sp fr he definiin f radian angle easure. diaeer 5 8 θ = diaeer = θ r = Earh Mn (.4 rad)(.8 ) = 5 r Earh Mn rev π rad in 4. The iniial angular velciy is ω = 65 68rad s. = in rev 6 sec definiin f angular accelerain. ω 68rad s α = = = 7 rad s 4. s Use he 5. (a) 5 rev π rad in ω = = 6.8rad sec 6 rad sec in rev 6s (b) v ωr = = 6.8 rad sec.75 = 46 s a = ω r = 6.8 rad sec.75 =. s 4 6. In each revluin, he wheel ves frward a disance equal is circuference, π d. 7 = rev ( π d) = = = 4 rev πd π The angular velciy is epressed in radians per secnd. The secnd hand akes revluin every 6 secnds, he inue hand akes revluin every 6 inues, and he hur hand akes revluin every hurs. (a) Secnd hand: (b) Minue hand: rev π rad π rad ω = = rad sec.5 6sec rev sec rev π rad in π rad rad ω = =.75 6 in rev 6 s 8 sec sec rev π rad h π rad 4 rad (c) Hur hand: ω = =.45 h rev 6 s,6 sec sec (d) The angular accelerain in each case is, since he angular velciy is cnsan. 8. The angular speed f he erry-g-rund is π rad 4.s =.57rad s. (a) v ωr = =.57 rad sec. =.9 s (b) The accelerain is radial. There is n angenial accelerain. a = ω r =.57 rad sec. =. s wards he cener 8 Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey

4 Giancli Physics fr Scieniss & Engineers, 4 h Ediin 9. Each lcain will have he sae angular velciy ( revluin per day), bu he radius f he circular pah varies wih he lcain. Fr he diagra, we see r = cs θ, where is he radius f he Earh, and r is he radius a laiude θ. π π rad day = = = 6.8 = 464 s T day 864s π π rad day 6 v = ωr = r = 6.8 cs66.5 = 85 s T day 864s π π rad day 6 v = ωr = r = 6.8 cs 45. = 8 s T day 864s 6 (a) v ωr r (b) (c) r θ θ. (a) The Earh akes ne rbi arund he Sun in ne year. θ π rad year 7 ω = = =.99 rad s rbi 7 year.6 s (b) The Earh akes ne revluin abu is ais in ne day. θ π rad day 5 ω = = = 7.7 rad s rain day 86,4s. The cenripeal accelerain is given by a = ω r. Slve fr he angular velciy. a, 9.8 s rad rev 6s ω = r =.7 = s π rad in =. Cnver he rp values angular velciies. rev π rad in ω =.6 rad s = in rev 6 sec rp rev π rad in ω = 8 = 9.rad s in rev 6 sec (a) The angular accelerain is fund fr Eq. -a. ω ω 9.rad s.6 rad s α = = =.9rad s.9 rad s 4. s (b) T find he cpnens f he accelerain, he insananeus angular velciy is needed. ω = ω + α =.6 rad s +.9rad s.s =.5rad s The insananeus radial accelerain is given by a a = ω r =.5rad s.5 = 6 s The angenial accelerain is given by aan = α r. a = αr =.9rad s.5 =.4 s an = ω r. 8 Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey

5 Chaper ainal Min. (a) The angular rain can be fund fr Eq. -a. The iniial angular frequency is and he final frequency is rp. rev π rad. in. ω ω in rev 6 s α = = =.454 rad s.5 rad s 7 s (b) Afer 7. in (4 s), he angular speed is as fllws. ω = ω + α = + = rad s 4s 6.7 rad s 4 4 Find he cpnens f he accelerain f a pin n he uer skin fr he angular speed and he radius. a a an rad = α =.454 rad s 4.5 = 6. s 4 4 = ω = 6.7 rad s 4.5 =.6 s 4. The angenial speed f he urnable us be equal he angenial speed f he rller, if here is n slippage. v = v ω = ω ω ω = 5. (a) The direcin f ω is alng he ale f he wheel, he lef. Tha is he direcin f ω is als alng is ais f rain, s i is sraigh up. Tha is he Tha is als he angular velciy f he ais f he wheel. (b) A he insan shwn in he ebk, we have he vecr relainship as shwn in he diagra. ω = ω + ω = 44. rad s + 5. rad s = 56. rad s ˆ i direcin. The ˆ +k direcin. ω 5. θ θ = an = an = 8.5 ω 44. ω d (c) Angular accelerain is given by α = ω. Since d ω = ω + ω, and ω is a cnsan 5.k ˆ rad s, α = dω. ω is raing cunerclckwise abu he z ais wih he angular d velciy f, ω = ω csω ˆi sin ω ˆj. ω and s if he figure is a =, hen ( ) ω d ( ω + ω ) ( cs ˆ sin ˆ ) dω d ω ω i ω j ωω ( sinω ˆ csω ˆ ) d α = = = = = i j d d d d ˆ ˆ α = = j = 44.rad s 5.rad s j= 54rad s ˆj ( ) ωω ω ω z 6. (a) Fr cnsan angular accelerain: ω ω rev in 5 rev in rev in rad in π α = = =.5 s.5s rev 6 s = 96.4 rad s 96 rad s 8 Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey

6 Giancli Physics fr Scieniss & Engineers, 4 h Ediin (b) Fr he angular displaceen, given cnsan angular accelerain: in θ = ( ω + ω) = ( 5 rev in + rev in)(.5 s) = 98rev 6 s 7. The angular displaceen can be fund fr Eq. -9d. θ = ω = ω + ω = + = 4 5rev in s in 6s.8 rev 8. (a) The angular accelerain can be fund fr Eq. -9b wih ω =. θ ( rev) α = = = 4. rev in (. in) (b) The final angular speed can be fund fr θ ( ω ω ) θ ( rev) ω = ω = = 4. rp. in = + wih ω =., 9. (a) The angular accelerain can be fund fr Eq. -9c. α θ (b) The ie ce a sp can be fund fr θ = ( ω + ω ) θ 5rev 6s ω ω 85rev in rev π rad in rad = = = = 5 rev in rev 6 s s = = = ω + ω 85 rev in in 9s. dω. We sar wih α =. We als assue ha α is cnsan, ha he angular speed a ie = is d ω and ha he angular displaceen a ie = is., ω dω α = dω = αd dω = αd ω ω = α ω = ω + α d ω θ dθ ω = ω + α = dθ = ( ω + α ) d dθ = ( ω + α ) d θ = ω + α d. Since here is n slipping beween he wheels, he angenial cpnen f he linear accelerain f each wheel us be he sae. a = a α r = α r (a) an an sall sall large large sall large rsall. c α = α = large sall ( 7. rad s ) =.6857 rad s.69 rad s r. c large (b) Assue he pery wheel sars fr res. Cnver he speed an angular speed, and hen use Eq. -9a. rev π rad in ω = 65 = 6.87 rad s in rev 6 s 8 Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey

7 Chaper ainal Min ω ω 6.87 rad s ω = ω + α = = = α.6857 rad s 9.9s 4. We are given ha θ = (a) dθ ω = = , where ω is in rad/sec and is in sec. d (b) dω α = =. + 9., where α is in rad sec and is in sec. d (c) ω (.) = 8.5.(.) + 6.4(.) = 9rad s α. = = 4 rad s (d) The average angular velciy is he angular displaceen divided by he elapsed ie. θ θ(.) θ(.) ω = = avg.s.s =.s = 8rad s (e) The average angular accelerain is he change in angular velciy divided by he elapsed ie. ω ω(.) ω(.) α = = avg.s.s = = 9 rad s.s. (a) The angular velciy is fund by inegraing he angular accelerain funcin. ω dω α = dω = αd dω αd ( ) d ω d = = = (b) The angular psiin is fund by inegraing he angular velciy funcin. θ = θ dθ ω = dθ = ωd dθ = ωd = d d (c) ω. s = =.7 rad s 4 rad s 6 4 θ. s = = 4.67 rad 5 rad 4. (a) The aiu rque will be eered by he frce f her weigh, pushing angenial he circle in which he pedal ves. τ = rf= rg=.7 6 kg 9.8 s =. i (b) She culd eer re rque by pushing dwn harder wih her legs, raising her cener f ass. She culd als pull upwards n he handle bars as she pedals, which will increase he dwnward frce f her legs. 8 Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey 4

8 Giancli Physics fr Scieniss & Engineers, 4 h Ediin 5. Each frce is riened s ha i is perpendicular is lever ar. Call cunerclckwise rques psiive. The rque due he hree applied frces is given by he fllwing. τ = =.8 i applied frces Since his rque is clckwise, we assue he wheel is raing clckwise, and s he fricinal rque is cunerclckwise. Thus he ne rque is as fllws. τ = ( 8 )(.4 ne ) ( 8 )(.4 ) ( 5 )(. ) +.4 i =.4 i =.4 i, clckwise 6. The rque is calculaed by τ = rf sin θ. See he diagra, fr he p view. (a) Fr he firs case, θ = 9. τ = rf sinθ =.96 sin 9 = i (b) Fr he secnd case, θ = 6.. τ = rf sinθ =.96 sin 6. = 7 i r θ F 7. There is a cunerclckwise rque due he frce f graviy n he lef blck, and a clckwise rque due he frce f graviy n he righ blck. Call clckwise he psiive direcin. τ = gl gl = g ( l l ), clckwise 8. The lever ar he pin f applicain f he frce is alng he ais. Thus he perpendicular par f he frce is he y cpnen. Use Eq. -b. τ = F = = 5.86 i, cunerclckwise 9. The frce required prduce he rque can be fund fr τ = rf sin θ. The frce is applied perpendicularly he wrench, s θ = 9. τ 75i F = = = 7 r.8 The ne rque sill us be 75i. This is prduced by 6 frces, ne a each f he 6 pins. We assue ha hse frces are als perpendicular heir lever ars. τ 75 i τ = ne ( 6 Fpin ) r F = = = 7 pin pin 6r Fr each rque, use Eq. -c. Take cunerclckwise rques be psiive. (a) Each frce has a lever ar f.. abu C τ =. 56 sin +. 5 sin 6 = 7i (b) The frce a C has a lever ar f., and he frce a he p has a lever ar f.. abu P τ =. 56 sin sin 45 = i ( sig fig) The negaive sign indicaes a clckwise rque.. Fr a sphere raing abu an ais hrugh is cener, he en f ineria is as fllws. I = M =.8 kg.648 =.8 kgi Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey 5

9 Chaper ainal Min. Since all f he significan ass is lcaed a he sae disance fr he ais f rain, he en f ineria is given by I = M. ( ) I = M =.kg.67 =. kgi The hub ass can be ignred because is disance fr he ais f rain is very sall, and s i has a very sall rainal ineria.. (a) The rque eered by he fricinal frce is τ = rf sin θ. The frce f fr fricin is assued be angenial he clay, and s θ = 9. τ al fr ( ) = rf sinθ =..5 sin 9 =.9 i (b) The ie sp is fund fr ω = ω + α, wih a final angular velciy f. The angular accelerain can be fund fr τ = Iα. al The ne rque (and angular accelerain) is negaive since he bjec is slwing. ω ω ω ω (.6rev s)( π rad rev ) = = = = s α τ I.9 i.kgi direcin f rain F fr 4. The ygen lecule has a dubbell geery, raing abu he dashed line, as shwn in he diagra. If he al ass is M, hen each a has a ass f M/. If he disance beween he is d, hen he disance fr he ais f rain each a is d/. Trea each a as a paricle fr calculaing he en f ineria. ( ) 4 I = M d + M d = M d = Md ( i ) d = = = I M 4.9 kg 5. kg. 5. The rque can be calculaed fr τ = Iα. The rainal ineria f a rd abu is end is given by I = ML. (.7 rev s )( π rad rev ) ω τ = Iα = ML =. kg.95 = 56 i.s 6. (a) The en f ineria f a cylinder is M. I = M =.8 kg.85 =.7 kgi.7 kgi (b) The wheel slws dwn n is wn fr 5 rp res in 55.s. This is used calculae he fricinal rque. ω ( 5rev in)( π rad rev)( in 6 s) τ = Iα = I = fr fr (.7 kgi ) 55. s =.9 i The ne rque causing he angular accelerain is he applied rque plus he (negaive) fricinal rque. τ = τ + τ = Iα applied ω τ = Iα τ = I τ applied fr fr fr 8 Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey 6

10 Giancli Physics fr Scieniss & Engineers, 4 h Ediin 75 rev in π rad rev in 6 s = 5. s.7 kgi (.9 i) 5.4 = i 7. (a) The sall ball can be reaed as a paricle fr calculaing is en f ineria. I = M =.65 kg. =.94 kgi (b) T keep a cnsan angular velciy, he ne rque us be zer, and s he rque needed is he sae agniude as he rque caused by fricin. τ = τ τ = τ = τ = Fr= applied fr applied fr fr (. )(.) =.4 i 8. (a) The rque gives angular accelerain he ball nly, since he ar is cnsidered assless. The angular accelerain f he ball is fund fr he given angenial accelerain. aan τ = Iα = M α = M = Ma = an (.6 kg)(.)( 7. s ) = 7.8 i 7.8 i (b) The riceps uscle us prduce he rque required, bu wih a lever ar f nly.5 c, perpendicular he riceps uscle frce. τ = Fr F = τ r = 7.8 i.5 = 9. (a) The angular accelerain can be fund fr he fllwing. ω ω vr ( 8.5 s) (.) α = = = = = 78.4 rad s 78 rad s.5 s (b) The frce required can be fund fr he rque, since τ = Fr sin θ. In his siuain he frce is perpendicular he lever ar, and s θ = 9. The rque is als given by τ = Iα, where I is he en f ineria f he ar-ball cbinain. Equae he w epressins fr he rque, and slve fr he frce. Fr sinθ = Iα Iα d + L F = = rsinθ rsin 9 ball ball ar ar + (.5) α. kg..7 kg. = 78.4 rad s = (a) T calculae he en f ineria abu he y ais (verical), use he fllwing. i i (.5 ) (.5 ) (. ) (. ) I = M = + M + + M ( M) = = 5.kg.5 +. = 6.6kgi (b) T calculae he en f ineria abu he -ais (hriznal), use he fllwing. I = M = + M.5 =.66kgi i iy (c) Because f he larger I value, i is en ies harder accelerae he array abu he verical ais. 8 Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey 7

11 Chaper ainal Min 4. The rque required is equal he angular accelerain ies he en f ineria. The angular accelerain is fund using Eq. -9a. Use he en f ineria f a slid cylinder. ω ω = ω + α α = ( kg)( 7. ) (.68 rad s) ω M ω 4 τ = Iα = ( M ) = = =. i 4s 4. The rque supplied is equal he angular accelerain ies he en f ineria. The angular accelerain is fund using Eq. -9b, wih ω =. Use he en f ineria f a sphere. θ θ θ = ω + α α = ; τ = Iα = ( Mr 5 ) ( i ) ( π ) 5τ s M = = = kg 4r θ rad 4. The applied frce causes rque, which gives he pulley an angular accelerain. Since he applied frce varies wih ie, s will he angular accelerain. The variable accelerain will be inegraed find he angular velciy. Finally, he speed f a pin n he ri is he angenial velciy f he ri f he wheel. ω F dω F F T T T τ = F = Iα α = = dω = d dω = d T I d I I v ω Fd T T ω = = + = Fd I I. vt = ω = F d T (.. ) d ( ) s I = = I I i (. ) (.85kgi ). ( ) v = 8.s = 8.s 8.s is = s 7 s 44. The rque needed is he en f ineria f he syse (erry-g-rund and children) ies he angular accelerain f he syse. Le he subscrip gr represen he erry-g-rund. ω ω ω τ = Iα = ( I + I gr children ) = ( M + gr child ) ( 5 rev in)( π rad rev)( in 6 s) = [ ( 76kg) + ( 5kg) ](.5 ).s = 4.5i 4i The frce needed is calculaed fr he rque and he radius. We are ld ha he frce is direced perpendicularly he radius. τ = Fsin θ F = τ = 4.5i.5 = 7 ω 8 Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey 8

12 Giancli Physics fr Scieniss & Engineers, 4 h Ediin 45. Each ass is reaed as a pin paricle. The firs ass is a he ais f rain; he secnd ass is a disance l fr he ais f rain; he hird ass is l fr he ais, and he furh ass is l fr he ais. (a) I = Ml + M l + M l = 4Ml (b) The rque rae he rd is he perpendicular cpnen f frce ies he lever ar, and is als he en f ineria ies he angular accelerain. Iα 4Ml α 4 τ = Iα = F r F = = = Ml α r l (c) The frce us be perpendicular he rd cnnecing he asses, and perpendicular he ais f rain. An apprpriae direcin is shwn in he diagra. 46. (a) The free bdy diagras are shwn. e ha nly he frces prducing rque are shwn n he pulley. There wuld als be a graviy frce n he pulley (since i has ass) and a nral frce fr he pulley s suspensin, bu hey are n shwn. (b) Wrie ewn s secnd law fr he w blcks, aking he psiive direcin as shwn in he free bdy diagras. : F = F gsin θ = a A TA A A A ( sinθ ) F = g + a TA A A = + = 8.kg 9.8 s sin. s sig fig : F = gsin θ F = a B B B TB B ( sinθ ) F = g a TB B B = (.kg) ( 9.8 s ) sin 6. s = (c) The ne rque n he pulley is caused by he w ensins. We ake clckwise rques as psiive. τ = ( F F TB TB) = ( )(.5) =.94i.9i Use ewn s secnd law find he rainal ineria f he pulley. The angenial accelerain f he pulley s ri is he sae as he linear accelerain f he blcks, assuing ha he sring desn slip. a τ = Iα = I = ( F F TB TB ) I ( F F ) ( )(.5) TB TB = = = a. s 47. (a) The en f ineria f a hin rd, raing abu is end, is add geher. al ( l ) l.59 kgi. l l l ML There are hree blades I = M = M = 5kg.75 = 898kgi.9 kgi y F TA F A F TB θ B θ A Ag F B Bg θ B F TA θ A y F F TB 8 Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey 9

13 Chaper ainal Min (b) The rque required is he rainal ineria ies he angular accelerain, assued cnsan. ω ω 5. rev/sec π rad rev τ = I α = I = 898 kgi = 75 i al al 8.s 48. The rque n he rr will cause an angular accelerain given by α = τ I. The rque and angular accelerain will have he ppsie sign f he iniial angular velciy because he rr is being brugh res. The rainal ineria is ha f a slid cylinder. Subsiue he epressins fr angular accelerain and rainal ineria in he equain ω = ω + αθ, and slve fr he angular displaceen. ω ω ω ω M ω ω = ω + αθ θ = = = = α τ τ 4τ ο ο ο ο ( I) ( M ) rev π rad in (.8 kg)(.7 ), in rev 6s rev = = 464rad 4(.i) π rad = 79 rev The ie can be fund fr θ = ( ω + ω ) θ ( 79rev). 6s = = = 8.6s ω + ω, rev in in 49. (a) Thin hp, radius (b) Thin hp, radius, widh w I = Mk = M k = I = Mk = M + Mw k = + w (c) Slid cylinder I = Mk = M k = (d) Hllw cylinder I = Mk = M ( + ) k = ( + ) (e) Unifr sphere I = Mk = Mr k = r 5 5 (f) Lng rd, hrugh cener I = Mk = Ml k = l (g) Lng rd, hrugh end I = Mk = Ml k = (h) ecangular hin plae I = Mk = M l + w k = l + w l 5. The firing frce f he rckes will creae a ne rque, bu n ne frce. Since each rcke fires angenially, each frce has a lever ar equal he radius f he saellie, and each frce is perpendicular he lever ar. Thus τ = 4 F. This rque will cause an angular accelerain ne accrding τ = Iα, where I = M + cbining a cylinder f ass M and radius wih 4, 4 pin asses f ass and lever ar each. The angular accelerain can be fund fr he 8 Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey

14 Giancli Physics fr Scieniss & Engineers, 4 h Ediin ω kineaics by α =. Equaing he w epressins fr he rque and subsiuing enables us slve fr he frce. ω ( M + 4 ) ω 4F = Iα = ( M + 4 ) F = τ 4 ( ( 6kg) + 4( 5 kg ))( 4. )( rev in)( π rad rev)( in 6 s) = = in 6s in 5. We assue ha >, and s will accelerae dwn, will B A B A accelerae up, and he pulley will accelerae clckwise. Call he direcin f accelerain he psiive direcin fr each bjec. The asses will have he sae accelerain since hey are cnneced by a crd. The ri f he pulley will have ha sae accelerain since he crd is aking i rae, and s α = ar Fr he free-bdy diagras fr each bjec, we have he pulley. fllwing. F = F g = a F = g + a ya TA A A TA A A F = g F = a F = g a yb B TB B TB B B τ = = = a F r F r Iα I TB TA r Subsiue he epressins fr he ensins in he rque equain, and slve fr he accelerain. a ( ) B A F r F r = I g a r g + a r = I a = g TB TA B B A A ( + + A B ) r r I r If he en f ineria is ignred, hen fr he rque equain we see ha F = F, and he TB TA accelerain will be a I = = ( B A) ( + ) A B g. included will be saller han if he en f ineria is ignred. 8 Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey a We see ha he accelerain wih he en f ineria 5. (a) The free bdy diagra and analysis fr prble 5 are applicable here, fr he n-fricin case. ( B A) ( B A) ( B A) a = g = g = g + + I r + + r r + + (.8 kg.5kg).8 kg +.5kg +.4 kg A B A B P A B P = 9.8 s =.8667 s.87 s (b) Wih a fricinal rque presen, he rque equain fr prble 5 wuld be dified, and he analysis prceeds as fllws. a a τ = F r F r τ = Iα = I TB TA fr ( g a B B ) r ( g + a A A ) r τ = I fr r r I τ = r fr ( B A ) g + + a r B A ( B A ) g = ( + + B A p ) a r +θ F TA F TA I r F TB + y A B + y Ag Bg F TB

15 Chaper ainal Min The accelerain can be fund fr he kineaical daa and Eq. -a. v v. s v = v + a a = = =.6 s 6.s τ r ( ) g ( ) a = + + fr B A B A p =.4.65kg 9.8 s 7.5kg.6 s =.6 i 5. A p view diagra f he haer is shwn, jus a he insan f release, alng wih he accelerain vecrs. (a) The angular accelerain is fund fr Eq. -9c. [( 6.5 s) (. ) ] ( 8π rad) ( vr) ω ω ω = ω + α θ α = = θ θ = = 9.7 rad s 9.7 rad s (b) The angenial accelerain is fund fr he angular accelerain and he radius. a = αr = 9.7 rad s. =.64 s.6 s an (c) The cenripeal accelerain is fund fr he speed and he radius. a = v r = 6.5 s. = 585. s 585 s rad (d) The ne frce is he ass ies he ne accelerain. I is in he sae direcin as he ne accelerain. F a a a = = + = 7. kg.64 s s = 47 ne ne an rad (e) Find he angle fr he w accelerain vecrs. a.64 s θ = = = an an an.4 a 585. s rad a ne a θ rad a an 54. (a) See he free bdy diagra fr he falling rd. The ais f rain wuld be cing u f he paper a he pin f cnac wih he flr. There are cnac frces beween he rd and he able (he fricin frce and he nral frce), bu hey ac hrugh he ais f rain and s cause n rque. Thus nly graviy causes rque. Wrie ewn s secnd law fr he rain f he rd. Take cunerclckwise be he psiive direcin fr rainal quaniies. Thus in he diagra, he angle is psiive, bu he rque is negaive. dω τ = Iα = g( l cs φ ) = l d g dω dω dφ dω g cs cs d ωdω φ = = = d dφ d dφ φ φ φ = l l F φ F fr l g φ ω g g g cs d d sin sin φ φ = ω ω φ = ω ω = φ l l l π / 8 Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey

16 Giancli Physics fr Scieniss & Engineers, 4 h Ediin (b) The speed f he ip is he angenial speed f he ip, since he rd is raing. A he ablep, φ =. ( φ) v = ωl = gl sin v = gl 55. The parallel ais here is given in Eq. -7. The disance fr he cener f ass f he rd he end f he rd is h = l. I = I + Mh = Ml + M l = + Ml = Ml CM We can cnsider he dr be ade f a large nuber f hin hriznal rds, each f lengh l =., and raing abu ne end. Tw such rds are shwn in he diagra. The en f ineria f ne f hese rds is l where i, i is he ass f a single rd. Fr a cllecin f idenical rds, hen, he en f ineria wuld be I = il = Ml. The heigh f he dr des n ener in he calculain direcly. I = Ml = 9. kg. = 6.kgi i l 57. (a) The parallel ais here (Eq. -7) is be applied each sphere. The disance fr he cener f ass f each sphere he ais f rain is h =.5 r. I = I + Mh = Mr + M.5r =.65 Mr I = 5.Mr fr ne CM 5 al sphere (b) Treaing each ass as a pin ass, he pin ass wuld be a disance f.5r fr he ais f rain. I = M appr (.5r ) = 4.5Mr I I appr eac 4.5Mr 5.Mr %errr= = = I 5.Mr 5. eac = 5% The negaive sign eans ha he appriain is saller han he eac value, by abu 5%. 58. (a) Treaing he ball as a pin ass, he en f ineria abu AB is I = M. (b) The parallel ais here is given in Eq. -7. The disance fr he cener f ass f he ball he ais f rain is h =. I = I + Mh = Mr + M CM 5 (c) 5 I I appr eac M Mr + M Mr 5 %errr= = = I Mr + M Mr + M eac 5 5 = = = r +..9 The negaive sign eans ha he appriain is saller han he eac value, by abu.%. 8 Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey

17 Chaper ainal Min 59. The.5-kg weigh is reaed as a pin ass. The rigin is placed a he cener f he wheel, wih he direcin he righ. Le A represen he wheel and B represen he weigh. + ( 7. kg A A B B )( ) + (.5 kg)(. ) (a) = = CM kg A B =.88.9 (b) The en f ineria f he wheel is fund fr he parallel ais here. I = I + I = I + M + M wheel weigh wheel wheel CM weigh weigh CM CM = M + M + M wheel wheel CM weigh weigh CM = 7. kg kg..88 =.4 kgi 6. We calculae he en f ineria abu ne end, and hen use he parallel ais here find he en f ineria abu he cener. Le he ass f he rd be M, and use Eq. -6. A sall ass dm can be fund as a sall lengh d ies he ass per uni lengh f he rd. M M l I dm d M l l l = = = = l end = + l = l = l l = l end CM CM end 4 I I M I I M M M M d.. 6. (a) We chse crdinaes s ha he cener f he plae is a he rigin. Divide he plae up in differenial recangular eleens, each wih an area f da = ddy. The ass f an eleen is M d = ddy. l w rain is The disance f ha eleen fr he ais f = + y. Use Eq. -6 calculae he en f ineria. w/ l / w/ l / M 4M I = cener dm = ( + y ) ddy = ( + y ) ddy l w l w l w / / w/ w/ 4M M = ( ) ( ) y dy y dy w + l l = + w l l M = ( w ) + ( w ) = M ( + w ) w l l (b) Fr he ais f rain parallel he w diensin (s he rain ais is in he y direcin), we can cnsider he plae be ade f a large nuber f hin rds, each f lengh l, raing abu an ais hrugh heir cener. The en f ineria f ne f hese rds is l i, 8 Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey 4 w dy y y l d w

18 Giancli Physics fr Scieniss & Engineers, 4 h Ediin where i is he ass f a single rd. Fr a cllecin f idenical rds, hen, he en f ineria wuld be I = M. y l = l A siilar arguen wuld give I = Mw. i i This illusraes he perpendicular ais here, Eq. -8, I = I + I. z y 6. Wrk can be epressed in rainal quaniies as W = τ θ, and s pwer can be epressed in W θ rainal quaniies as P = = τ = τω. rev π rad in hp P = τω = ( 55i ) 75 = 4 hp in rev 6 s 746 W 6. The energy required bring he rr up speed fr res is equal he final rainal kineic energy f he rr. K rev π rad in = Iω = ( 4.5 kgi ) 975 =. J in rev 6 s 4 r 64. T ainain a cnsan angular speed ω will require a rque τ seady r ppse he fricinal rque. The pwer required by he r is P = τ ω = τ ω. r seady fricin seady τ f = = fricin fricin P r Iα M ω ω π rad (.8 rev s) ω ω f rev 5 = M ω = seady ( kg)( 5.5) =.86 W 6s hp = = 746 W 5.86 W 58.9 hp 6 hp 65. The wrk required is he change in rainal kineic energy. The iniial angular velciy is. π rad 4 ω ω ω r f i f 4 W = K = I I = M = 64 kg 7.5 =.4 J 8. s 66. Mechanical energy will be cnserved. The rain is abu a fied ais, s K = K = Iω Fr graviainal penial r. energy, we can rea he bjec as if all f is ass were a is cener f ass. Take he lwes pin f he cener f ass as he zer lcain fr graviainal penial energy. E = E U = K iniial final iniial final ( cs ) b ( l ) Mg l θ = Iω = M ω b l θ l csθ l g ω = θ = ω = θ b b b l ( cs ) ; v l gl ( cs ) 8 Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey 5

19 Chaper ainal Min 67. The nly frce ding wrk in his syse is graviy, s echanical energy is cnserved. The iniial sae f he syse is he cnfigurain wih n he grund and all bjecs a res. The final sae f he syse has B jus reaching he grund, and all bjecs in in. Call he zer level f graviainal penial energy be he grund level. Bh asses will have he sae speed since hey are cnneced by he rpe. Assuing ha he rpe des n slip n he pulley, he angular speed f he pulley is relaed he speed f he asses by ω = v. All bjecs have an iniial speed f. E = E i f v + v + Iω + gy + gy = v + v + Iω + gy + gy A i B i i A i B i A f B f f A f B f A A M B h v f gh= v + v + B A f B f ( M ) + gh A v f ( B A) gh ( M) A B ( )( ) ( 8. kg 5. kg. kg) 8.kg 5.kg 9.8 s.5 = = = s 68. (a) The kineic energy f he syse is he kineic energy f he w asses, since he rd is reaed as assless. Le A represen he heavier ass, and B he ligher ass. K = I ω + I ω = r ω + r ω = r ω + A A B B A A A B B A A B = =. 5.6 rad s 7. kg 4.84 J (b) The ne frce n each bjec prduces cenripeal in, and s can be epressed as F F A A A A B B B B = r ω = 4. kg. 5.6 rad s = 6. = rω =. kg. 5.6 rad s = Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey 6. rω These frces are eered by he rd. Since hey are unequal, here wuld be a ne hriznal frce n he rd (and hence he ale) due he asses. This hriznal frce wuld have be cuneraced by he uning fr he rd and ale in rder fr he rd n ve hriznally. There is als a graviy frce n each ass, balanced by a verical frce fr he rd, s ha here is n ne verical frce n eiher ass. (c) Take he 4. kg ass be he rigin f crdinaes fr deerining he cener f ass. + A A B B ( 4. kg)( ) + (. kg)(.4 ) = = =.8 fr ass A CM + 7. kg A B S he disance fr ass A he ais f rain is nw.8, and he disance fr ass B he ais f rain is nw.4. e-d he abve calculains wih hese values. K = I ω + I ω = r ω + r ω = ω r + r F F A A B B A A A B B A A A B B = 5.6 rad s 4. kg.8 +. kg.4 = 4.74 J A A A A B B B B = r ω = 4. kg rad s =.6 = rω =. kg rad s =.6 e ha he hriznal frces are nw equal, and s here will be n hriznal frce n he rd r ale.

20 Giancli Physics fr Scieniss & Engineers, 4 h Ediin 69. Since he lwer end f he ple des n slip n he grund, he fricin des n wrk, and s echanical energy is cnserved. The iniial energy is he penial energy, reaing all he ass as if i were a he CM. The final energy is rainal kineic energy, fr rain abu he pin f cnac wih he grund. The linear velciy f he falling ip f he rd is is angular velciy divided by he lengh. E = E U = K gh = Iω gl = L v L v inial final iniial final end end = gl = 9.8 s. = 8. s 7. Apply cnservain f echanical energy. Take he b f he incline be he zer lcain fr graviainal penial energy. The energy a he p f he incline is hen all graviainal penial energy, and a he b f he incline, here is bh rainal and ranslainal kineic energy. Since he cylinder rlls wihu slipping, he angular velciy is given by ω = v. v E = E Mgh = Mv + I ω = Mv + M = Mv p b CM 4 v 4 4 = = = gh 9.8 s s 7. The al kineic energy is he su f he ranslainal and rainal kineic energies. Since he ball is rlling wihu slipping, he angular velciy is given by ω = v. The rainal ineria f a sphere abu an ais hrugh is cener is I =. 5 v K K K v I v v 7 = + = + ω = + = al rans r 5 =.7 7. kg.7 s = 7. J 7. (a) Fr he daily rain abu is ais, rea he Earh as a unifr sphere, wih an angular frequency f ne revluin per day. K = Iω = M ω daily daily 5 Earh daily 4 6 π rad day 9 = 6. kg J 5 = day 86,4s (b) Fr he yearly revluin abu he Sun, rea he Earh as a paricle, wih an angular frequency f ne revluin per year. K = Iω = M ω yearly yearly Sun- yearly Earh 4 π rad day = 6. kg.5.7 J = 65day 86,4s Thus he al kineic energy is K 9 K.6 J.7 J.7 J. daily yearly + = + = The kineic energy due he daily in is abu, ies saller han ha due he yearly in. 8 Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey 7

21 Chaper ainal Min 7. (a) Mechanical energy is cnserved as he sphere rlls wihu slipping dwn he plane. Take he zer level f graviainal penial energy he level f l sinθ he cener f ass f he sphere when i is n he level surface a he b f he plane. All f he energy is penial energy a he p, and all is kineic energy (f bh ranslain and rain) a he b. E = E U = K = K + K inial final iniial final CM r v b gh = gl sin θ = v + Iω = v + b b b ( r 5 ) r b b b v = gh = gl sinθ = 9.8 s. sin. = 8.67 s ω 8.7 s v = r 8.67 s = =.9 rad s.54 K v v 5 CM b b (b) = = = K Iω r b v b ( r 5 ) r (c) The ranslainal speed a he b, and he rai f kineic energies, are bh independen f he radius and he ass. The rainal speed a he b depends n he radius. 74. (a) Since he cener f ass f he spl is sainary, he ne frce us be. Thus he frce n he hread us be equal he weigh f he spl and s F = Mg. hread (b) By he wrk energy here, he wrk dne is he change in kineic energy f he spl The spl has rainal kineic energy. W = K K = Iω = M ω = M ω final iniial 4 l θ 75. Use cnservain f echanical energy equae he energy a pins A and B. Call he zer level fr graviainal penial energy be he lwes pin n which he ball rlls. Since he ball rlls wihu slipping, ω = vr. E = E U = U + K = U + K + K A B A B B final B B CM B r g = gr + v + Iω B B A y = B C v = gr + v + r v = g r r B B 5 B 7 8 Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey 8

22 Giancli Physics fr Scieniss & Engineers, 4 h Ediin 76. (a) We wrk in he acceleraing reference frae f he car. In he acceleraing frae, we us add a ficiius frce f agniude Ma, in he ppsie direcin he accelerain f he rain. This rain rel grund is discussed in deail in secin -8 f he ebk. Since he ball is rlling wihu slipping, α = a. See he free-bdy diagra fr ball rel rain he ball in he acceleraing reference frae. Wrie ewn s secnd law fr he hriznal direcin and fr rques, wih clckwise rques as psiive. Cbine hese relainships find a ball rel, he accelerain f he ball in he acceleraed frae. a a ball rel rain rain ball rel rain α fr fr 5 ball rel rain τ = F= I = I F = Ma F = F Ma = Ma Ma Ma = Ma = fr rain rel ball rel 5 ball rel rain rel ball rel grund rain rain grund rain 5 7 a rain rel grund And s as seen fr inside he rain, he ball is acceleraing backwards. (b) Use he relaive accelerain relainship. a = a + a = a + a = a 5 ball rel ball rel rain rel 7 rain rel rain rel 7 rain grund rain grund grund grund And s as seen fr uside he rain, he ball is acceleraing frwards, bu wih a saller accelerain han he rain. 77. (a) Use cnservain f echanical energy. Call he zer level fr graviainal penial energy be he lwes pin n which he pipe rlls. Since he pipe rlls wihu slipping, ω = v. See he aached diagra. E = E U = K = K + K iniial final iniial final CM r gd sinθ = v + Iω b b vb = v + b ( ) = v b v = gdsinθ = 9.8 s 5.6 sin7.5 = 4.6 s b (b) The al kineic energy a he base f he incline is he sae as he iniial penial energy. K = U = gdsinθ =.545kg 9.8 s 5.6 sin7.5 = 8.99 J final iniial (c) The fricinal frce supplies he rque fr he bjec rll wihu slipping, and he fricinal frce has a aiu value. Since he bjec rlls wihu slipping, α = a. Use ewn s secnd law fr he direcins parallel and perpendicular he plane, and fr he rque, slve fr he cefficien f fricin. a τ = F = Iα = = a F = a fr fr F = F gcs θ F = gcsθ y = F fr g y Ma rain F θ F Mg D F fr 8 Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey 9

23 Chaper ainal Min F = gsin θ F = a F = gsinθ fr fr F F gsinθ µ F = µ gcs θ µ an θ fr saic s s s a µ = anθ = an7.5 =.58 s in 78. (a) While he ball is slipping, he accelerain f he cener f ass is cnsan, and s cnsan accelerain relainships ay be used. Use Eq. -b wih resuls fr Eaple -. v v v ( µ k ) = v + a = v g 7µ g + 7µ g = 49µ g k k k (b) Again ake use f he fac ha he accelerain is cnsan. Once he final speed is reached, he angular velciy is given by ω = vr. v 5 v = v + a = v + ( µ g k ) = v ; ω = 7 7µ k g 79. (a) The al kineic energy included he ranslainal kineic energy f he car s al ass, and he rainal kineic energy f he car s wheels. The wheels can be reaed as ne cylinder. We assue he wheels are rlling wihu slipping, s ha v = ω. CM wheels 5 7 v r CM = + = + ω = + CM r CM wheels CM ( wheels wheels ) wheels K K K M v I M v M s 5 = ( M + M wheels ) v = CM ( 7 kg) ( 95k h) = 4.74 J.6 k h 5 4. J (b) The fracin f kineic energy in he ires and wheels is K I ω + M v K r ( + ) + K K rans wheels M M v M. v r wheels wheels CM wheels wheels CM wheels = = = + ω + + CM wheels wheels CM wheels K M v I M M v M M kg = =.8 7 kg (c) A free bdy diagra fr he car is shwn, wih he fricinal frce f F fr a each wheel cause he wheels rll. A separae diagra f ne wheel is als shwn. Wrie ewn s secnd law fr he hriznal in f he car as a whle, and he rainal in f ne wheel. Take clckwise rques as psiive. Since he wheels are rlling wihu slipping, a α. CM wheels = a CM F I α M fr wheels wheels wheels τ = 4 = = F = M a fr 8 wheels CM wheels 4F fricin F fricin F g F w 8 Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey

24 Giancli Physics fr Scieniss & Engineers, 4 h Ediin F = F 4 F = M a w fr CM F 4 M a = M a a w 8 wheels CM CM w = = =.8 s. s CM ( M + M ) ( 7 kg) F wheels 5 (d) If he rainal ineria were ignred, we wuld have he fllwing. F 5 w F = F = M a a = = = w CM CM.64 s M kg a CM %errr= = = 6% a.8 s CM wihu slipping, ω = vr. All lcains y = B D = 8 Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey.64 s.8 s 8. (a) The fricin frce acceleraes he cener f ass f he wheel. If he wheel is spinning (and slipping) clckwise in he diagra, hen he surface f he wheel ha uches he grund is ving he lef, and he fricin frce is he righ r frward. I acs in he direcin f in f he velciy f he cener f ass f he wheel. (b) Wrie ewn s secnd law fr he direcin, he y direcin, and he rain. Take clckwise rques (abu he cener f ass) as psiive. F = F Mg = F = Mg y F µ F µ Mg fr k k F = F = Ma a = = = = µ g F fr fr k M M M F F Mg µ g fr k τ = F= Iα α = = = fr M M Bh he accelerain and angular accelerain are cnsan, and s cnsan accelerain kineaics ay be used epress he velciy and angular velciy. µ g k v = v + a = µ g ; ω = ω + α = ω k e ha he velciy sars a and increases, while he angular velciy sars a ω and decreases. Thus a se specific ie T, he velciy and angular velciy will be ω = v, and he ball will rll wihu slipping. Slve fr he value f T needed ake ha rue. µ g ω k ω = v ω T = µ gt T = k µ g (c) Once he ball sars rlling wihu slipping, here is n re fricinal sliding frce, and s he velciy will reain cnsan. ω v = µ gt = µ g = ω final k k µ g 8. (a) Use cnservain f echanical energy equae he energy a pin A he energy a pin C. Call he zer level fr graviainal penial energy be he lwes pin n which he ball rlls. Since he ball rlls k A k θ y C Mg

25 Chaper ainal Min given fr he ball are fr is cener f ass. E = E A C U = U + K = U + K + K A C C C C C CM r [ csθ] g = g r + v + Iω C C v = g r + v + r C 5 r [ cs θ ] C ( ) θ v = g r cs = 9.8 s.45 cs 45 =.557 s.6 s C 7 7 (b) Once he ball leaves he rap, i will ve as a prjecile under he influence f graviy, and he cnsan accelerain equains ay be used find he disance. The iniial lcain f he ball is given by = ( r ) sin 45 and y = ( r ) cs 45. The iniial velciy f he ball is given by v = v cs 45 and v = v sin 45. The ball lands when y = r =.5. Find C y C he ie f fligh fr he verical in, and hen find D fr he hriznal in. Take he upward direcin as psiive fr he verical in. y = y + v + a = r cs 45 + v sin 45 g y C = =.77s,.58s We use he psiive ie. D = = + v = r sin 45 + v cs 45 C =.45 sin s cs 45.77s = Wrie he rainal versin f ewn s secnd law, wih cunerclckwise rques as psiive. τ = τ τ = F l F = I α = M α ne fr C CM 5 CM ewn s secnd law fr he ranslainal in, wih lef as he psiive direcin, gives he fllwing. F F = F = a a = ne If he sphere is rlling wihu slipping, we have α = a. Cbine hese relainships CM analyze he relainship beween he rques. a 7 F l = F + M α = F + M = F+ Ma = F+ F = F 5 CM τ = 7 5 τ fr And since he rque due he nral frce is larger han he rque due fricin, he sphere has a cunerclckwise angular accelerain, and hus he rainal velciy will decrease. 8. Since he spl rlls wihu slipping, each pin n he edge f he spl ves wih a speed f v = rω = v CM relaive he cener f he spl, where v is he speed f he cener f he spl CM relaive he grund. Since he spl is ving he righ relaive he grund, and he p f he spl is ving he righ relaive he cener f he spl, he p f he spl is ving wih a speed f v relaive he grund. This is he speed f he rpe, assuing i is unrlling CM wihu slipping and is a he uer edge f he spl. The speed f he rpe is he sae as he speed f he persn, since he persn is hlding he rpe. S he persn is walking wih a speed f wice 8 Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey

26 Giancli Physics fr Scieniss & Engineers, 4 h Ediin ha f he cener f he spl. Thus if he persn ves frward a disance l, in he sae ie he cener f he spl, raveling wih half he speed, ves frward a disance l. The rpe, say cnneced bh he persn and he spl, us herefre unwind by an aun l als. 84. The linear speed is relaed he angular velciy by v = ω, and he angular velciy (rad / sec) is relaed he frequency (rev / sec) by Eq. -7, ω = π f. Cbine hese relainships find values fr he frequency. v v v.5 s 6s ω = π f = f = ; f = = 48rp = π π π (.5) in v.5 s 6s f = = = rp π π (.58 ) in 85. (a) There are w frces n he y-y: graviy and sring ensin. If he p f he sring is held fied, hen he ensin F T des n wrk, and s echanical energy is cnserved. The iniial graviainal penial energy is cnvered in rainal and ranslainal kineic energy. Since he y-y rlls wihu slipping a he pin f cnac f he sring, he g velciy f he CM is relaed he angular velciy f he y-y by v = rω, where r is he radius f he inner hub. Le be he ass f he inner hub, CM and M and be he ass and radius f each uer disk. Calculae he rainal ineria f he y-y abu is CM, and hen use cnservain f energy find he linear speed f he CM. We ake he f graviainal penial energy be a he b f is fall. 5. kg ( 5. kg).5 kg I = r + M = r + M CM al 5 = 5. kg kg.75 = 7.8 kgi = + M = + = U iniial = K final I CM I CM gh = v + I ω = v + v = v al al CM CM al CM CM + al CM r r v gh (.5 kg)( 9.8 s )(. ) al = = = = CM ICM + al (.5 kg) + (b) Calculae he rai K K. r r I v K K I I v CM CM ω r r CM CM CM = = = r = iniial al al al K U gh gh r gh 5 ( 7.8 kgi )(.895 s) ( ) ( ) 5 ( 7.8 kgi ) ( 5. ) = =.96 = 96% 5..5 kg 9.8 s s 8 Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey

27 Chaper ainal Min 86. As discussed in he e, fr he reference frae f he ale f he wheel, he pins n he wheel are all ving wih he sae speed f v = rω, where v is he speed f he ale f he wheel relaive he grund. The p f he ire has a velciy f v he righ relaive he ale, s i has a velciy f v he righ relaive he grund. v = v + v = v he righ + v he righ = v he righ p rel p rel cener rel grund cener grund p rel grund v = v = v + a = a =. s.5s = 5. s 87. Assue ha he angular accelerain is unifr. Then he rque required whirl he rck is he en f ineria f he rck (reaed as a paricle) ies he angular accelerain. ω ω.5 kg.5 rev rad in π τ = Iα = r = 85 =. 5.s in rev 6s i Tha rque ces fr he ar swinging he sling, and s ces fr he ar uscles. 88. The rque is fund fr τ = Iα. The angular accelerain can be fund fr ω = ω + α, wih an iniial angular velciy f. The rainal ineria is ha f a cylinder. ω ω 8 rev s rad rev π τ = Iα = M =.5.4 kg. = 5 i 6.s 89. (a) The linear speed f he chain us be he sae as i passes ver bh sprckes. The linear speed is relaed he angular speed by v = ω, and s ω = ω. If he spacing f he F F eeh n he sprckes is a disance d, hen he nuber f eeh n a sprcke ies he spacing disance us give he circuference f he sprcke. d = π and s (b) ω ω = 5 = 4. F (c) ω ω = 4 8 =.5 F d =. Thus π d d ω ω = ω = F π π ω F F F 9. The ass f a hydrgen a is. aic ass unis. The aic ass uni 7 is.66 kg. Since he ais passes hrugh he ygen a, i will have n rainal ineria. (a) If he ais is perpendicular he plane f he lecule, hen each hydrgen a is a disance l fr he ais f rain. perp I = H = 7 9 l..66 kg.96 O θ l H l y =. kgi 45 (b) If he ais is in he plane f he lecule, bisecing he H-O-H bnds, each hydrgen a is a disance f l = l θ = y ( ) sin 9.6 sin 5 = Thus he en f ineria is as fllws. I =..66 kg kg plane H l y = = i 7 45 H 8 Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey 4

28 Giancli Physics fr Scieniss & Engineers, 4 h Ediin 9. (a) The iniial energy f he flywheel is used fr w purpses give he car ranslainal kineic energy ies, and replace he energy ls due fricin, fr air resisance and fr braking. The saeen f he prble leads us ignre any graviainal penial energy changes. (b) K W = K K F cs8 = M v K fr final iniial fr car car flywheel K = F + M v flywheel flywheel fr car car = 5 ( 45 )(.5 ) ( ) ( kg) ( 95k h) = J.7 J = Iω flywheel flywheel 8 ( ) ( 4 kg)(.75) s.6 k h KE KE.65 J ω = = = = rad s I M wrk (c) T find he ie, use he relainship ha pwer =, where he wrk dne by he r ie will be equal he kineic energy f he flywheel. 8 (.65 J) W W P = = = = P ( 5 hp)( 746 W hp).476 s 5in 9. (a) Assuing ha here are n dissipaive frces ding wrk, cnservain f echanical energy ay be used find he final heigh h f he hp. Take he b f he incline be he zer level f graviainal penial energy. We assue ha he h hp is rlling wihu sliding, s ha ω = v. elae he θ cndiins a he b f he incline he cndiins a he p by cnservain f energy. The hp has bh ranslainal and rainal kineic energy a he b, and he rainal ineria f he hp is given by I =. v E = E v + Iω = gh v + = gh b p (. s) v h = = =. g 9.8 s h. The disance alng he plane is given by d = = = sinθ sin5 (b) The ie can be fund fr he cnsan accelerain linear in. ( 4.9 ) = ( v+ v ) = = =.6 s v+ v +. s This is he ie g up he plane. The ie ce back dwn he plane is he sae, and s he al ie is 5.s. 8 Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey 5

29 Chaper ainal Min 9. The wheel is rlling abu he pin f cnac wih he sep, and s all rques are be aken abu ha pin. As sn as he wheel is ff he flr, here will be nly w frces ha can eer rques n he wheel he pulling frce and he frce f graviy. There will n be a nral frce f cnac beween he wheel and he flr nce he wheel is ff he flr, and any frce n he wheel fr he pin f he sep cann eer a rque abu ha very pin. Calculae he ne rque n he wheel, wih clckwise rques psiive. The iniu frce ccurs when he ne rque is. τ = F h g h = h h g F ( ) h F Mg h Mg h h = = h h 94. Since fricinal lsses can be ignred, energy will be cnserved fr he arble. Define he psiin f graviainal penial energy be he b f he rack, s ha he b f he ball is iniially a heigh h abve he psiin f graviainal penial energy. We als assue ha he arble is rlling wihu slipping, s ω = vr, and ha he arble is released fr res. The arble has bh ranslainal and rainal kineic energy. (a) Since r, he arble s CM is very clse he surface f he rack. While he arble is n he lp, we hen appriae ha is CM will be ving in a circle f radius. When he arble is a he p f he lp, we appriae ha is CM is a disance f abve he psiin f graviainal penial energy. Fr he arble jus be n he verge f leaving he rack eans he nral frce beween he arble and he rack is zer, and s he cenripeal frce a he p us be equal he graviainal frce n he arble. v p f lp = g v = g p f lp Use energy cnservain relae he release pin he pin a he p f he lp. E = E K + U = K + U release p f release release p f p f lp lp lp vp f lp + gh = v + Iω + g = v + p f p f p f ( r 5 ) + g lp lp lp r gh = v + g = g + g =.7 g h = p f lp (b) Since we are n assue ha r, hen while he arble is n he lp prin f he rack, i is ving in a circle f radius r, and when a he p f he lp, he b f he arble is a heigh f ( r) abve he psiin f graviainal penial energy (see he diagra). Fr he arble jus be n he verge f leaving he rack eans he nral frce beween he arble and he rack is zer, and s he cenripeal frce a he p us be equal he graviainal frce n he arble. v p f lp p f lp g v g r r = = Use energy cnservain equae he energy a he release pin he energy a he p f he lp. y = 8 Pearsn Educain, Inc., Upper Saddle iver, J. All righs reserved. This aerial is preced under all cpyrigh laws as hey 6 r