Chter ottion o igid Object About Fixed Axis P. () ω ωi. rd s α t. s 4. rd s it+ 4. rd s. s 8. rd (b) θ ω αt P.5 rev min π rd π ωi rd s, ω. min 6. s. rev () ω ωi π / t s α. 5.4 s (b) ω + ωi π π θ ωt t rd s s 7.4 rd 6 6 P.9 ω 5. rev s. π rd s. We will brek the motion into two stges: () eriod during which the tub seeds u nd () eriod during which it slows down. + π While seeding u, θ ωt π. rd s 8. s 4. rd π + While slowing down, θ ωt π. rd s. s 6. rd So, θtotl θ + θ π rd 5. rev P. () v rω ; ω v 45. m s r 5 m.8 rd s (b) P.4 () ( 45. m s) v r r 5 m ω v 5. m s r. m 5. rd s 8. m s towrd the center o trck (b) ω ω + α( Δ θ) i ω ωi ( θ ) ( 5. rd s) ( π ) α Δ.5 rev rd rev 9.8 rd s Δω 5. rd s (c) Δ t α 9.8 rd s.68 s
P. () I mr j j j y (m) 4 In this cse,. kg. kg r r r r 4 r. m +. m. m [ ] I. m. +. +. + 4. kg 4 kg m x (m) 4 kg m 6. rd s (b) K Iω ( ).57 J *P. m 4. kg, r y. m m. kg, r y. m m. kg, r y 4. m ω. rd s bout the x-xis. kg 4. kg FIG. P. () I m r + m r + m r I x x 4.. +.. +. 4. 9. kg m K Ixω ( 9. )(. ) 84 J FIG. P. (b) v rω.. 6. m s K mv 4. 6. 7. J v rω.(.) 4. m s K mv. 4. 6. J v rω 4.(.) 8. m s K mv. 8. 96. J K K + K + K 7. + 6. + 96. 84 J I ω (c) The kinetic energies comuted in rts () nd (b) re the sme. ottionl kinetic energy cn be viewed s the totl trnsltionl kinetic energy o the rticles in the rotting object. x
P. I Mx + m( L x) x di Mx m( L x) (or n extremum) dx ml x M + m di m+ M ; thereore I is t minimum when the xis dx ml o rottion sses through x which is lso the M + m center o mss o the system. The moment o inerti bout n xis ssing through x is M x L L x m CM ml m M m I M m L L M m + M m + + M + m where Mm μ M + m μl FIG. P. P.6 We ssume the rods re thin, with rdius much less thn L. Cll the junction o the rods the origin o coordintes, nd the xis o rottion the z-xis. For the rod long the y-xis, I ml rom the tble. For the rod rllel to the z-xis, the rllel-xis theorem gives L I mr m + ml 4 z y x xis o rottion FIG. P.6 m In the rod long the x-xis, the bit o mteril between x nd x + dx hs mss dx L L nd is t distnce r x + is: rom the xis o rottion. The totl rottionl inerti L L m Itotl ml + ml + x + dx 4 4 L L L L 7 m x ml ml + + x L 4 L L 7 ml ml ml 4 ml + + Note: The moment o inerti o the rod long the x xis cn lso be clculted rom the rllel-xis theorem s L ml m +.
P.7 Tret the tire s consisting o three rts. The two sidewlls re ech treted s hollow cylinder o inner rdius 6.5 cm, outer rdius.5 cm, nd height.65 cm. The tred region is treted s hollow cylinder o inner rdius.5 cm, outer rdius. cm, nd height. cm. Use I m( + ) or the moment o inerti o hollow cylinder. Sidewll: m π (.5 m) (.65 m) ( 6.5 m )(. kg m ).44 kg Iside (.44 kg ) (.65 m ) (.5 m ) 8.68 + kg m Tred: m π (. m ) (.5 m ) (. m )(. kg m ). kg I tred (. kg ) (. m ) + (.5 m ). kg m Entire Tire: I I + I 8.68 kg m +. kg m.8 kg m totl side tred P. We consider the cm s the suerosition o the originl solid disk nd disk o negtive mss cut rom it. With hl the rdius, the cut-wy rt hs one-qurter the ce re nd one-qurter the volume nd one-qurter the mss M o the originl solid cylinder: M 4 M M M M 4 By the rllel-xis theorem, the originl cylinder hd moment o inerti CM + + I M M M M 4 4 The negtive-mss ortion hs M M I 4. The whole cm hs M 4 I M M M M 4 4 ω ω ω 4 48 K I M M nd τ. m. N.5 m 9. N.5 m. N.55 N m P. The thirty-degree ngle is unnecessry inormtion. FIG. P.
P.7 For m, Fy my: + n m g n mg 9.6 N k μkn 7.6 N Fx mx : 7.6 N + T (. kg) () For the ulley, τ Iα : T + T M T+ T (. kg ) T+ T ( 5. kg) For m, + n mgcosθ n 6. kg ( 9.8 m s )( cos. ) 5.9 N FIG. P.7 k μkn 8. N : 8. N T + msin θ m 8. N T + 9.4 N 6. kg () () Add equtions (), (), nd (): (b) () 7.6 N 8. N + 9.4 N. kg 4. N.9 m s. kg T. kg.9 m s + 7.6 N 7.67 N T 7.67 N + 5. kg.9 m s 9. N kg.5 m.5 kg m ωi 5. rev min 5.4 rd s ω ωi 5.4 rd s α.87 rd s t 6. s P.8 I m τ Iα.5 kg m.87 rd s.9 N m The mgnitude o the torque is given by the orce o riction..9 N m, where is FIG. P.8 Thereore,.9 N m nd μkn.5 m yields.8 N μ k n 7. N.
trns. kg. m s 5 J v Krot Iω mr. kg. m s 5 J r 4 P.5 () K mv (b) (c) Ktotl Ktrns + Krot 75 J P.56 I v K mv + Iω m+ v where ω since no sliing occurs. Also, U mgh, U, nd v Thereore, Thus, For disk, So i I m v m + gh gh v + Im I m gh v or disk + gh i v For ring, I m so v or vring Since v > v, the disk reches the bottom irst. disk ring 4gh gh *P.6 () We consider the elevtor-sheve-counterweight-erth system, including n ssengers, s n isolted system nd ly the conservtion o mechnicl energy. We tke the initil conigurtion, t the moment the drive mechnism switches o, s reresenting zero grvittionl otentil energy o the system. Thereore, the initil mechnicl energy o the system is E i K i + U i (/) m e v + (/) m c v + (/)I s ω (/) m e v + (/) m c v + (/)[(/)m s r ](v/r) (/) [m e + m c v + (/)m s ] v The inl mechnicl energy o the system is entirely grvittionl becuse the system is momentrily t rest: E K + U + m e gd m c gd where we hve recognized tht the elevtor cr goes u by the sme distnce d tht the counter- weight goes down. Setting the initil nd inl energies o the system equl to ech other, we hve (/) [m e + m c + (/)m s ] v (m e m c ) gd (/) [8 kg + n 8 kg + 95 kg + 4 kg]( m/s) (8 kg + n 8 kg 95 kg)(9.8
m/s ) d d [89 + 8n](.459 m)/(8n 5) (b) d [89 + 8 ](.459 m)/( 8 5) 94. m (c) d [89 + 8 ](.459 m)/( 8 5).6 m (d) d [89 + 8 ](.459 m)/( 8 5) 5.79 m (e) The rising cr will cost to sto only or n. For n or n, the cr would ccelerte uwrd i relesed. () The grh looks roughly like one brnch o hyerbol. It comes down steely rom 94. m or n, lttens out, nd very slowly roches.459 m s n becomes lrge. (g) The rdius o the sheve is not necessry. It divides out in the exression (/)Iω (/4)m sheve v. (h) In this roblem, s oten in everydy lie, energy conservtion reers to minimizing use o electric energy or uel. In hysicl theory, energy conservtion reers to the constncy o the totl energy o n isolted system, without regrd to the dierent rices o energy in dierent orms. (i) The result o lying F m nd τ Iα to elevtor cr, counterweight, nd sheve, nd dding u the resulting equtions is (8 kg + n 8 kg 95 kg)(9.8 m/s ) [8 kg + n 8 kg + 95 kg + 4 kg] (9.8 m/s )(8n 5)/( 89 + 8n) downwrd P.69 τ will oose the torque due to the hnging object: τ I α T τ : τ T I α () Now ind T, I nd α in given or known terms nd substitute into eqution (). F T mg m y : T m( g ) () t y lso Δ y vt i + () t nd α y t (4) FIG. P.69 8 5 I M + M Substituting (), (), (4), nd (5) into (), we ind (5)
τ ( y) y 5 M y 5 My m g m g t 8 t t 4 t P.75 () Let E reresent the rdius o the Erth. The bse o the building moves est t v ω E where ω is one revolution er dy. The to o the building moves est t v ω ( ) +h. Its estwrd seed reltive to the ground is E h v v ω h. The object s time o ll is given by Δ y + gt, t. g During its ll the object s estwrd motion is unimeded so its delection distnce is h g g Δ x v v t ωh ωh π rd s 5 m.6 cm 86 4 s 9.8 m (b) (c) The delection is only.% o the originl height, so it is negligible in mny rcticl cses. P.77 F T Mg M: τ T Iα M () Combining the bove two equtions we ind nd thus T M( g ) T T M Mg FIG. P.77 (b) T Mg g M M + v + g ( h ) (c) v vi ( x xi) v 4gh For comrison, rom conservtion o energy or the system o the disk nd the Erth we hve
U + K + K U + K + K : gi rot i trns i g rot trns v Mgh+ + + M + Mv v 4gh P.79 () Δ Krot +Δ Ktrns +Δ U m r Note tht initilly the center o mss o the shere is distnce h+ r bove the bottom o the loo; nd s the mss reches the to o the loo, this distnce bove the reerence level is r. The conservtion o energy requirement gives h P mg( h+ r ) mg ( r ) + mv + Iω FIG. P.79 For the shere I 5 mr nd v rω so tht the exression becomes 7 gh gr g v + + () Note tht h h min the condition when the seed o the shere t the to o the loo stisies mv F mg or v g( r) ( r) Substituting this into Eqution () gives ( ) + ( r ) or hmin.7( r).7 hmin r.7 (b) When the shere is initilly t h nd inlly t oint P, the conservtion o energy eqution gives mg( + r ) mg+ mv + mv 5, or v + r g 7 Turning clockwise s it rolls without sliing st oint P, the shere is slowing down with counterclockwise ngulr ccelertion cused by the
torque o n uwrd orce o sttic riction. We hve F y my nd τ Iα becoming mg mα r nd r mr α. 5 5g Eliminting by substitution yields α so tht 7r 5 Fy mg 7 mv / 7 ( + r) mg Fx n mg (since >> r ) r r 7 P.8 () Fx F+ MCM τ F Iα I CM F M F CM CM 4F M F Mg n FIG. P.8 (b) 4F MCM F M F F M (c) v vi + ( x xi) v 8Fd M P.84 Cll the rictionl orce exerted by ech roller t bckwrd on the lnk. Nme s the rolling resistnce exerted bckwrd by the ground on ech roller. Suose the rollers re eqully r rom the ends o the lnk. b M m m F For the lnk, FIG. P.84 Fx mx 6. N t ( 6. kg) The center o ech roller moves orwrd only hl s r s the lnk. Ech roller hs ccelertion nd ngulr ccelertion ( 5. cm ) (. m ) Then or ech, Fx mx + t b. kg
τ Iα 5. cm 5. cm (. kg)( 5. cm ) t + b. cm So t + b kg Add to eliminte b : t (.5 kg) () And 6. N (.5 kg) ( 6. kg) ( 6. N ) ( 7.5 kg).8 m s For ech roller,.4 m s.5 kg.8 m s (b) Substituting bck, t Mg t.6 N 6. N.6 N + b kg.8 m s. N b t n t t n t n t t n t t The negtive sign mens tht the horizontl orce o ground on ech roller is. N orwrd rther thn bckwrd s we ssumed. mg mg b n b b n b FIG. P.84(b) P.85 Fx mx reds + T m. I we tke torques round the center o mss, we cn use τ Iα, which reds + T Iα. For rolling without sliing, α. By substitution, I I T T m m T m IT I ( + ) ( + ) I m T I m mg n FIG. P.85 T I + m T I + m Since the nswer is ositive, the riction orce is conirmed to be to the let.