CHAPTER LAGRANGIAN MECHANICS. Solution ( t (, t + αη ( t ( t (, t +αη ( t where (, t sinωt nd (, cos t ω ωt T V k ω t J α ω dt so: ( ( t t + + ( ω cosωt αη ω ( sinωt αη dt t t t α t J ( α ω ( cos ωt sin ωt dt αω ( ηcosωt ωηsinωt dt ( η ω η dt + + t t Eine the ter liner in α : t t t t ( ( ηcosωt ωηsinωt dt η t cosωt + ωηsinωtdt ωηsinωtdt t t t t η t η t ( st ter vnishes t both endpoints: ( ( t so J( α ω cos ωtdt+ α ( η ω η t t t ω ωt ωt + α η ω η 4 [ sin sin ] ( which is iniu t α t t dt dt t. V z T ( + y + z L T V ( + y + z z L L L, y, z y z d d, y dt, dt y d z dt z
, y L z d Fro equtions.4.5 qi dt q i, const y, y const z. Choosin enerlized coordinte s liner displceent down the inclined plne (See Fiure 8.6., for rollin without slippin ω 7 T + Iω + 5 For V t the initil position of the sphere, V sin 7 L T V + sin L 7 d, 5 dt L sin 7 sin 5 5 sin 7 Fro equtions 8.6. - 8.6. 7 5 5 c sin 7.4 ( For the distnce of the hnin block below the ede of the tble: T + nd V L T V + L d, dt L
(b T + + nd V l l L T V + + + l d ( +, ( + dt L + l ( + + l l+ l +.5 The four sses hve positions: : + : l + : l + 4 : l + l ( + + ( ( ( V l + + l + + l + l 4 T ( + + ( + + ( ( + + 4 L T V ( + + ( + + ( + + 4( ( ( 4 ( 4 cons. + + + + + t L ( + ( + ( ( + + d L ( + + ( dt L ( ( + + ( (
( ( ( ( + + + + 4 ( + 4 ( + ( + + + + ( ( + 4 4 4 L ( + ( L ( 4 4 ( + ( + ( 4 4 4 For, 4,, nd 4 : 5, 5 + 8 ( ( + +, Substitutin into the second eqution: 9 9 + + 8 5 5 88 8, 5 5 6 5 4 Accelertions: 5 : + 7 : + : + 5 4 :.6 See fiure.5., replcin the block with bll. The squre of the speed of the bll is clculted in the se wy s for the block in Eple.5.6. v + + cos The bll lso rottes with nulr velocity ω so 4
T v + Iω + M For rollin without slippin, ω. I 5 T ( + + cos + + M 5 V sin, for V t the initil position of the bll. 7 L T V + + cos sin 5 + M + 7 d 7 + cos, cos + 5 dt 5 L sin 7 cos sin 5 + 5 ( sin cos 7 L d + cos + M, ( M cos + + dt L + M + ( cos + + 5 sin cos 5 7 7 cos 5 sin cos 5cos 7 + M ( M (.7 Let be the slnt heiht of the prticle v + ω T v ( + ω V sin sinωt L T V ( + ω sinωt ωt L d, dt L ω sinω t ω sinωt ω sinωt 5
The solution to the hooeneous eqution ωt ωt ω, is Ae + Be Assuin prticulr solution to hve the for C ω ω Csinωt ω Csinωt sinωt ω ω Ae + Be + sin ωt ω At tie t, nd A+ B t t ωa ωb+ ω Csinωt, A ω B + ω ωt ωt ωt ωt ( e + e ( e e + ω ω sinωt Fro Appendi B, we use the identities for hyperbolic sine nd cosine to obtin coshωt sinhωt+ sinωt ω ω.8 In order tht prticle continues to ove in plne in rottin coordinte syste, it is necessry tht the is of rottion be perpendiculr to the plne of otion. For otion in the y plne, ω ˆk ω. v v + ω r ( i ˆ+ ˆjy + kˆ ω ( i ˆ + ˆjy v iˆ( ω y + ˆj( y + ω ( T v v ωy+ ω y + y + yω+ ω L T V L d ( ω y, ( ω y, dt V ( y ω+ ω V ( ωy ( y ω + ω F ωy ω ( L y +ω y d dt y (, ( y+ ω p, 6
V ( ω+ ω y y y ( ω ( ω ω V y+ + y y ( ω ω Fy y+ y For coprison, fro eqution 5.. ( A nd ω F + ω v + ω ( ω r F i ˆ + ˆjy + ωkˆ i ˆ + ˆjy + ωkˆ ωkˆ i ˆ + ˆjy F ωy ω y ( ( ( ( ( ω ω F y+ y.9 Choosin the is of rottion s the z is v v + ω r i ˆ+ ˆjy+ kz ˆ + ωkˆ i ˆ + ˆjy+ kz ˆ v iˆ ωy + ˆj y+ ω + kz ( ( ( ( ˆ T v v y+ y + y + y + + z L T V L d ( ω y ( ω y, dt V ( y ω+ ω V ( ωy ( y ω + ω F ωy ω ( ( ω ω ω ω L d y +ω y dt y V ( ω+ ω y y y V ( y+ ω ( ω+ ω y y y (, ( y+ ω ( ω ω F y+ y L d z, z z, dt z V z Fz z Fro eqution 5.. ( A nd ω V z z 7
F + ω v + ω ( ω r F ( i ˆ + ˆ jy + kz ˆ + ωk ˆ ( i ˆ + ˆ jy + kz ˆ + ωk ˆ ωk ˆ ( i ˆ + ˆ jy + kz ˆ F ωy ω y ( ( ω ω F y y Fz z. T ( r + r V k( r l rcos r k L T V ( r + r ( r l + rcos L d r, r r dt r L r k ( r l cos + r r r k r l + cos ( L r L r sin d ( r r sin dt 4 4 t & y T ( + y V y ( + cos ( + cos y sin L T V ( + cos + sin ( cos 8 4 [ + cos + ] ( cos 8 4 cos sin where we used the trionoetric identies cos + cos nd sin cos. (See Eple 4.6. ( + sin y ( cos 8
Let s sin so s cos s L s s ks where k The eqution of otion is thus k s+ s or s+ s - siple hronic oscilltor. Coordintes: cosωt+ bsin y sinωt bcos ω sinωt+ b cos y ω cosωt+ b sin L T V ( + y y ω b + b ω sin( ω t ( sin ω t b cos + d L b + bω ( ω cos( ωt dt L b ωcos( ωt bsin d The eqution of otion dt is ω cos( t sin b ω + b Note the eqution reduces to eqution of siple pendulu if ω.. Coordintes: lcosωt+ lcos( + ωt y l t+ l + t sinω sin ( ω ( sin ( ( cos( ωlsinωt + ω l + ωt y ωlcosωt+ + ω l + ωt L T + y l ω + + ω + ( (... ( + ( + + cos ωt ( cos cos ωt sin ωt sin l ω ω ωt ωt ωt... sin sin cos sin cos 9
( l ( l ( cos ω + + ω + ω + ω d dt l ω sin + l ω + ω sin ( ( + ω sin (b The bed eecutes siple hronic otion ( > opposite the point of ttchent. (c The effective lenth is "" l ω.4 v ˆj t+ l ( iˆcos + ˆjsin il ˆ cos + ˆ( j t + l sin T v v ( l cos + t + tl sin + l sin bout point dietriclly V t lcos t L T V ( l + t + tl sin lcos L l d + tl sin, l + l sin + tl cos dt L tl cos l sin l + l sin + tl cos tl cos l sin + + sin l For sll oscilltions, sin + + l π l T π ω +.5 ( T + I nd T + 5 V I 5
L T V + + T 5 The eqution of constrint is f(, The Lrne equtions with ultipliers re d f + λ t dt L d L f + λ dt L d L L f λ λ dt + λ nd fro the -eqution 5 λ Differentitin the eqution of constrint nd substitutin into the bove 5 nd 7 λ 7 (b The enerlized force tht is equivlent to the tension T is f Q λ λ 7.6 For v the velocity of differentil ss eleent, d, of the sprin t distnce below the support v v, d d T v + ( v d d + T + ( V k l d ( k l d ( V k l ( L T V k l + + + + d, + dt,
k( l + + k( l + + + For y l + k, + y+ ky The block oscilltes bout the point l+ + k with period π + T π ω k.7 Note: 4 objects ove their coordintes re lbeled i : The coordinte of the ovble, ssless pulley is lbeled p. Two equtions of constrint: f, l ( p p ( ( p ( ( p f,, + + l L T V + + d f j + λ j qi dt qi j qi Thus: ( + λ ( + λ ( + λ Now pply Lrne s equtions to the ovble pulley note p So: f f (4 λ + λ or λ λ p p Now - p cn be eliinted between f nd f ( f f + + l+ l (5 Thus + + With little lebr we cn solve ( (5 for the 5 unknowns,,, λ, nd λ
λ λ + + + + 4 4. As the check, let nd Thus, there is no ccelertion nd λ.8 (See Eple 5.. ( r reˆr r re ˆ ˆ r + r e Constrint: f ( ωt so ω nd T ( r + r L d r dt r nd d + λ f dt r ω r rr r + λ ωt ωt r Ae Be r pply constrint + ( ωt ωt r ωae ωbe r ( ωl so A+ B λ rr ω ω A ωb ωl l l A l A B l ωt ωt thus.. r ( e e r lsinhωt λ ω l sinhωtcoshω t r ωlcoshωt.88 now r l t t T so T sinh ω ω (b There re wys to clculte F (i See Eple 5.. F ω l ωt ωt ωl ωt ωt ( e e nd ( e + e F ω lcoshωt f (ii λ is the enerlized force, in this cse torque T, ctin on the bed f T rf λ
λ ω ω F r lsinhωt ω lcoshωt f l sinh tcoshωt.9 (See Eple 4.6. The eqution of constrint is ( f r, r T ( r + r V rcos r L ( r + r rcos d f + λ r dt r r L d L f λ f f + dt r Thus r cos r + λ r sin r rr Now r, r r so cos + λ sin sin nd d d so d sind or cos + λ cos hence, ( nd when λ prticle flls off heisphere t cos. Let rk the loction of the center of curvture of the ovble surfce reltive to fied oriin. This point defines the position of. r rks the position of the prticle of ss. r is the position of the prticle reltive to the ovble center of curvture of. Kinetic enery T + r r r + r + r + r + r ( ( but r r eˆ ˆ r + r e T + + r + r + ( re ˆ ˆ r + r e + + r + r + r sin r cos 4
Potentil enery L T V Eqution of constrint ( V y rsin f f r, r r f f d f Lrne s equtions + λ i dt i i d r sin rcos dt d r sin + r sin + r cos rcos + r sin dt usin constrint r constnt ; r r ( sin + cos ( + r rr rsin rsin r cos + r sin + r cos + rcos usin constrint r constnt ; r r sin cos + ( λ r r+ cos sin + r + sin + sin + Apply constrint λ cos + + sin + ( Now, plu solution for ( into eqution (: ( sin cos sin cos + + ( f sin f sin cos cos where f + d d d d d Now dt d dt d d So let d f sin f sin cos cos d Let y f sin d dy d ( sin y + d d d 5
( Hence: d( y d( sin nd d( y y y ( but t so y( sin sin ( f sin Now we cn solve for nd plu d d sin After soe lebr yields ( + f sin cos f sin d d f sin ( sin d ( sin, into ( to obtin λ ( The solution for λ ( is thus deterined fro eqution ( f sin cos + cos + sin collectin ters: ( f cos f sin cos sin + λ λ Plu, into the bove --- plus --- lot of lebr yields λ ( f sin f sin cos sin f + sin f sin where f + As check, let so it is ioveble then f nd f nd we hve λ ( sin which checks L T V ( R + R φ + z V L R d, R R, dt R. ( T v ( R + R φ + z R φ R V R 6
V R R φ R R R φ Q R L R d φ, R φ + RR V φ φ dt, φ φ φ V R φ + RR φ Q φ φ L d V z, z, z dt z z z V z Qz z For F, usin the coponents of fro eqution..: FR R Rφ Fφ R φ + R φ, F z z (, ( Fro Section., since R nd z re distnces, QR nd Qz re forces. However, since φ is n nle, Q φ is torque. Since F φ is coplnr with nd perpendiculr to R, Q φ (b RF nd ll equtions ree. φ v r + r + r φ sin L T V ( r + r + r φ sin V L d r, V r r, r + r φ sin dt r r r V r r r φ sin Qr r L r d, r + rr V dt, r φ sin cos V r + rr r φ sin cos Q L r d φ sin, r φ sin + rr φsin + r φ sin cos φ dt, φ V φ φ V r φsin + rr φsin + r φ sin cos Q φ φ is force F. Q nd Q φ re torques. Qr r Since φ is in the y plne, the oent r for φ is r sin ; i.e. Q rsin F. Q rf. Fro eqution..4 φ φ 7
( φ sin r ( + φ sin cos ( φsin + φsin + φ cos r r r F r r r F r r r F φ The equtions ree. V V V V V, so nd F. φ r For sphericl coordintes, usin eqution..: v r + r φ sin + r L T V ( r + r φ sin + r V ( r L d r, V r r, r φ sin + r dt r r r r r φ sin r F. For centrl field, ( r r L r d, r + rr L dt, r φ sin cos r + rr r φ sin cos L r φ sin, φ d r φ sin + rr φsin + r φ sin cos dt φ L φ r φsin + rr φsin + r φ sin cos. Since α constnt, there re two derees of freedo, r nd. vr r, vφ r φ sinα T v ( r + r φ sinα V rcosα L T V ( r + r φ sin α rcosα L d r, r r, L r φ sin α cosα dt r r r r φ sin α cosα L r L φ sin α, φ φ 8
Sy d ( r φsinα dt r φsinα constnt r cosα r d dr d r r r r dt dr dr dr d( r cosαdr r r r cosα + C r The constnt of intertion C is the totl enery of the prticle: kinetic enery due to the coponent of otion in the rdil direction, kinetic enery due to the coponent of otion in the nulr direction, nd the potentil enery. U( r + rcosα r For φ,, nd turnin points occur t r Then r cos α + C r ( cosα r Cr + The bove eqution is qudrtic in r ( r 4 nd hs two roots. l.4 Note tht the reltion obtined in Proble., ( cosα r Cr +, defines the turnin points. For the prticle to rein on sinle horizontl circle, there ust be two roots with r r. Thus ( r r divided into the bove epression leves ter tht is liner in r. ( r+ r c r rr + r r Cr + r r r + rr ( r C r r r ( r C r r ( r C r+ r ( r C ( + ( r r C r r r C For the reinder to vnish, both ters ust equl zero. 4r rc r C r 9
r Cr + r r And with r φ sinα φ r sin α Fro Prob.., r cosα r For sll oscilltions bout r, r r + ε. r ε ε ε + r r r r r ε ε cosα r r ε + ε cosα 4 r r 4 π r π ω φ π T sinα L d da + qa q + dt dt da A A A Usin the hint, + y + z dt y z.5 L v + qv A ( + y + z + q ( A + ya y + za z, A A y A z q + y + z A A A A A y A z + q + y + z q + y + z y z A A A A y z q y ( A z ( A z y q v A y z q y z (
( q v B Due to the cyclic nture of the Crtesin coordintes, i.e., iˆ ˆj kˆ y q ( v B y z q ( v B z r q v B Altoether, (.6 V z T ( + y + z T p p p, siilrly, y pz y nd z H T + V ( p + py + pz + z H p p H p, p constnt d p ( siilrly, py constnt, or y dt H py y p y H pz z pz H d pz, p z ( z z z dt These ree with the differentil equtions for projectile otion in Section 4...7 ( Siple pendulu V lcos T l T p l, p l p H T + V lcos l H p p l H l sin p
(b Atwood s chine V ( l T + + I [Includes the pulley] T I p + +, p I + + p H T + V ( I l + + H p p I + + H ( p, p ( (c Prticle slidin down sooth inclined plne V sin T T p p, p H T + V sin H p p H sin p, p sin.8 ( F r L T( q, q T( q, t note, potentil enery is tie dependent. i i i V k F so V Fdr e r r T v v ( r + r βt
pr r r r p r p r pr p k H pi qi L prr + p e r r Substitutin for r nd pr p k βt H + e r r (b H T + V which is tie-dependent (c E is not conserved p r.9 Locte center of coordinte syste t C.M. The potentil is independent of the center of ss coordintes. Therefore, they re inorble. (r, (r, p i q L i p r p pr p r r H βt L T V r + r + r + r k r+ r l where l is the lenth of the reled sprin nd k is the sprin constnt. p pr r r r r p r p r nd siilrly for ( ( ( H + + + k ( r+ r l Equtions of otion First,, re inorble coordintes, so p, p re ech conserved. H p r k( r+ r l r r H pr k( r+ r l r r p p r r p r pr constnt. The rdil oent re equl nd opposite.
. L k t t t δ Ldt t δldt δ k dt t t t ( δ kδ dt t d δ δ dt t t d t δ dt ( dt d ( t δ δ t dt t Intertin by prts: t t t t t t t t nd t ( δ dt δ δ d δ d d( ( dt dt dt t t δ d δ dt t t t t ( δ δ k dt + k. ( L c v V c Let γ v c L γ p This is the enerlized oentu for prt (b. Thus, Lrne s equtions for the -coponent re d d V p dt dt nd so on for the y nd z coponents. (b H v p L but So pi vi γ i i p c c H + + V γc H pc i i i γ c γc + γ + V 4
4 H ( p c + c + V γ c (c Now, if T γ c then we hve 4 4 p c pc + c c + 4 c γ v c c + 4 c c + γ v c ( 4 4 c v c + c v c v c c Thus H γ c + V T + V 4 4 4 γ c v (d T c + +... v c c T v + c ------------------------------------------------------------------------------------------------------------ 5