CHAPTER 10 ROTATIONAL MOTION


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1 CHAPTER 0 ROTATONAL MOTON
2 0. ANGULAR VELOCTY Consder argd body rotates about a fxed axs through pont O n xy plane as shown. Any partcle at pont P n ths rgd body rotates n a crcle of radus r about O. The angle between r and the postve xaxs s θ. As the pont P oves around the crcle the angle θ changes whle r reans constant. Thus θ serves as a coordnate to descrbe the rotatonal poston of a partcle. The angle s related to the dsplaceent (arc length) s, through the relaton srθ Here the unt of θ s radan wth o 360 rad. π 57.3 o
3 n analogy to the lnear velocty, the average angular velocty of a partcle oves fro pont A to pont B n the fgure s defned as θ ω t f f θ t θ t The nstantaneous angular velocty s defned as θ dθ ω l t t 0 As t s clear fro the above equatons, the unt of ω s rad./s, or sply s . The drecton of ω s along the axs of rotaton wth ts sense can be deterned fro the rght hand rule: turn the four fngers of your rght hand wth the drecton of rotaton; your thub then gves the drecton of.
4 0. ANGULAR ACCELERATON The average angular acceleraton of a partcle s defned as ω α t f f ω t ω t And the nstantaneous angular acceleraton s defned as ω dω α l The unt of α s rad./s t, or sply s . t ROTATON WTH CONSTANT a dω α dω α dω α ω ω αt o ω ω t o 0
5 dθ Now ω ω o αt dθ ( ωo αt) θ ( ω ) dθ ( ωo αt) d o αt θ θ ω O O t θ θ α t t o 0 Elnatng t fro the equatons we get ω ( θ ) O α θ O ω These three equatons are of the sae fors as those for lnear oton wth the replacng: x θ, v ω, and a α
6 Exaple 0. A wheel rotates wth constant angular acceleraton of 3.5 rad/s. f the angular velocty of the wheel s rad/s at t 0. a) What angle does the wheel rotate through n s? b) What s the angular velocty at t s? Soluton Usng the equaton θ θ θ O ω Now Usng the equaton O t α t rad ω ω o αt rad/s ω 3.5 9
7 0.4 RELATONSHPS BETWEEN ANGULAR AND LNEAR VARABLES t s known that v ω s rθ ds dθ r vrω The tangental acceleraton s defned as a θ dv r dω α a θ rα Whle the radal acceleraton s defned as v a r r rω a a t a r r α r ω 4 a r α ω 4
8 Reark: For argd body rotatng about afxed axs, every partcle on the body has the sae angular velocty and the sae angular acceleraton, whle the lnear velocty and the lnear acceleraton dfferfro pont to pont. Exaple 0. A wheel of radus 30 c rotates about a fxed axs wth ntal angular velocty 50 rev/n and takes 0 s to coe to rest. a) Calculate the angular acceleraton. b) How any rotatons does the wheel ake before cong to rest? c) Calculate the radal and tangental acceleratons at t 0. Soluton a) Notng that each revoluton represents π radans we have for the ntal angular velocty ( π ) 50 ω o 5.7 rad/s 60 ω 5π Usng the equaton ω ω o αt α o 0.79 rad/s t 0
9 b) Now usng the equaton θ θ O ω O t α t π θ 5π 0 (0) 50π 4 θ No. of Rev. 5Rev π c) Usng the equatons a a t r r o rα ω 0.3(5π ) 7.5π π 0.3( ) /s 4 /s
10 0.5 ROTATONAL KNETC ENERGY Consder a rgd body conssts of sall partcles, rotates about fxed axs wth angular velocty ω. The knetc energy of the th partcle of ass and lnear speed v s gven by K But K v v r ω r ω The total knetc energy of the rotatng rgd body s the su of the knetc energes of the ndvdual partcles,.e., K K r ω
11 or K ω wth r s called the oent of nerta. t represents the ass n all rotatonal equatons. For a body of contnuous ass dstrbuton, the suaton n Equaton can be replaced by an ntegraton over the body,.e., r d where r s the perpendcular dstance fro the eleent d to the axs of rotaton.
12
13 Exaple 0.3 Calculate the oent of nerta of a unfor, thn rod of ass M and length L about an axs noral to ts length a) at one end, b) at ts center. Soluton a) Let us dvde the rod nto sall eleents each of ass d. Usng the equaton r d x d Usng the defnton of lnear ass densty λ, we have d M λ d λdx dx L L [ ] λ x dx λ λ x L L x ML 3 L d dx
14 b) Now f the axs of rotaton s at the center of the rod we have agan r d x d L d M λ d λdx dx L L [ ] λ x dx λ x λ L ML L 3 3 L 3 L x d dx
15 0.6 TORQUE Consder a force F actng on the partcle P n a rgd body, as shown. The torque due to the force F s defned by the cross product of the vector r and the force vector F, dsplaceent τ r F O y r F θ x The torque s a vector quantty wth agntude τ Fr snθ r F wth r r snθ s called the oent ar of the force and s defned as the perpendcular dstance fro the pont of rotaton to the lne of acton of the force. τ s postve f t tends to rotate the body counterclockwse, n the drecton of ncreasng θ. t s negatve f t tends to rotate the body clockwse. Fro the st equaton, the drecton of t s perpendcular to the plane contanng r and F.
16 Exaple 0.4 A force F 0 N s appled at pont P, 0.5 fro the center O of a wheel. Op akes an angle of 30 wth xaxs and the force akes an angle of 60 wth xaxs, as shown. Calculate the torque on the wheel. O F 0 N P 30 o 60 o Soluton The ar of the force s gven by r r sn sn The torque of the force s then τ r F N. The negatve sgn eans that the rotaton s clockwse.
17 Exaple 0.5 A rod of length L and weght 5 N s acted on by two forces as shown n Fgure 0.8. Fnd the total torque actng on the rod about pont O sn30 N 30 o 0 N 0 cos30 N 40 N 5 N Soluton Let us frst resolve the 0N force. Takng the ve sense counterclockwse we get τ ( ) ( 5 ) ( 0sn 30) 5 N.
18 0.7 TORQUE AND ANGULAR ACCELERATON Consder a force F actng on a rgd body free to rotate about O. Durng the nfntesal te, the eleent P of ass d wll ove an nfntesal dstance ds along a crcular path of radus r as the body rotates through an nfntesal angle dθ, where dsrdθ The torque actng on the body s τ F t r But F t M( a c ) t and M( ac) at τ ra t d But a t rα t n a d t
19 τ α r d α f there s ore than one force dong torque on the body, the last equaton can be generalzed as τ α The last equaton can be consdered as Newton s second law for rotatonal oton.
20 Exaple 0.6 A wheel of radus R, ass M, and oent of nerta s rotatng on a frctonless, horzontal axle as shown. A lght cord wrapped around the wheel and supports a body of ass. Calculate the lnear acceleraton, a, of the body, the angular acceleraton, α, of the wheel, and the tenson, T, n the cord. Soluton For the wheel we apply τ α TR α () Note that the weght of the wheel and the force of the axle on the wheel don t do any torque on the wheel. Why? And for the ass we apply F a g T a ()
21 The thrd equaton s obtaned fro the relaton between the angular and lnear acceleraton: a Rα (3) Usng Eq. () & Eq.(3) T a R (4) Now addng Eq. () & Eq.(4) g a R Substtutng back n Eq.(4) a g R g T R
22 Exaple 0.7 Two blocks havng asses and are connected to each other by alght cord that passes over afrctonless pulley, wth oent of nerta and radus R, as shown. Fnd the acceleraton of each block and the tensons n the cord. Soluton The freebody dagras of the asses and the pulley are shown. For the wheel we apply α T T R ( τ ( ) ) And for we apply α F x a T () And for we apply a N g T T g T F y a g T a (3) T
23 The fourth equaton s obtaned fro the relaton between the angular and lnear acceleraton: (4) a Rα Subtractng Eq. () & Eq.() ( ) ) (5 a g T T Substtutng Eq. (5) nto Eq.() ( ) R a R a g α R g a Fro Eq. () we get R g T ( ) R g a g T And fro Eq. (3) we get
24 0.8 WORK AND ROTATONAL MOTON Let us consder agan the stuaton n the fgure shown The work dw done by the force F s F dw ds F snφ ds F snφ ( rdθ ) But ( F snφ) r F r τ dw τ dθ t The total work W s then W θ θ τ dθ dw The power s now gven as P Whch s the rotatonal analogue of dθ τ τω P Fv
25 Now τ α dω P dw τω ω dω dw ωdω ω ( ) W ω dω ω o K ω o ω wth K ω s the rotatonal knetc energy. t s the rotatonal analogue of the lnear equaton K v
26 Exaple 0.8 Consder two asses connected by a strng passng over a pulley havng a oent of nerta about ts axs of rotaton, as shown. The strng does not slp on the pulley, and the syste s released fro rest. Fnd the lnear velocty of the asses after descends through a dstance h, and the angular velocty of the pulley at ths te. Soluton Applyng the conservaton of echancal energy prncple we get K U but K 0 Knowng that 0 and v U U U gh v ωr K f v gh ω
27 0 gh gh v R v v ) ( R gh v ) ( R gh v ( ) R R gh R v ) ( ω
28 0.9 ANGULAR MOMENTUM The angular oentu L of the partcle relatve to an orgn s defned as L r p r v where r s the poston vector of the partcle relatve to O. The angular oentu s a vector quantty wth a drecton perpendcular to the plane fored by r and v, and ts sense s deterned by the rght hand rule dscussed n chapter. The S unt of L s kg. /s Consder a rgd body that rotates about the zaxs through the orgn as shown. The agntude of the angular oentu of the partcle s L v r ωr
29 The total angular oentu of the body s, therefore L L r ω L ω Whch s the rotatonal analogue of the lnear equaton 0.0CONSERVATON OF ANGULAR MOMENTUM t s known that but d t ( p) τ r F dr dp r p r d p ( r ) t and F dl dp dp r τ p dp r Whch s the rotatonal analogue of the lnear equaton F v dp
30 Fro the last equaton, f the external torque actng on a syste s zero, then L s constant,.e., L s conserved, or L L f or ω f ω f Exaple 0.9 Two dsks one of ass kg and radus 0., and the other of ass 4 kg and radus 0.5. The lghter dsk s rotatng wth ntal angular velocty of 80rad/s, whle the heaver dsk rotates wth ntal angular velocty of 40 rad/s. f the two dsks are pushed nto contact, fnd the coon angular velocty of the cobnaton. Soluton The oent of nerta of the two dsks are r ( )(0.) 0.0 kg. r ( 4)(0.5) kg. Now ω ( ) ω ω f ω90.8 rad/s ( 0.0)(80) (0.045)(40) (0.065) ω
31 Exaple 0.0 A turntable rotates n a horzontal plane about a fxed, vertcal axs, akng one revoluton n 5 s. The oent of nerta of the table about the axs of rotaton s 000 kg.. A an of ass 70 kg, ntally standng at the center of the turntable, starts to walk a way fro the center along a radus. What s the angular velocty of the table when the an s fro the center? Soluton Let us call the oent of nerta of the table t and the oent of nerta of the an. The ntal oent of nerta of the syste s, therefore ( t ) ( 000 0) 000 kg.
32 When the an has walked to the poston r, the fnal oent of nerta of the syste becoes f ( ) ( ) 80 kg. r t f Applyng the conservaton of angular oentu law: ω fωf ω f ω f Butω π π 0.4 rad/s T ωf rad/s 80
33 Lnear Moton v dx f a s constant W Rotatonal Moton x θ, v ω, a α,, F τ, p L F a f x v Fdx a dv ω dθ α dω v vo at ω ωo αt vot at f α s constant θ ω ot αt v ax ω ωo αθ o W τ α f τ dθ K v K ω p F Fv v dp τω L τ ω dl
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