k 1 k 2 x (1) x 2 = k 1 x 1 = k 2 k 1 +k 2 x (2) x k series x (3) k 2 x 2 = k 1 k 2 = k 1+k 2 = 1 k k 2 k series

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1 Final Review A Puzzle... Consider wo massless springs wih spring consans k 1 and k and he same equilibrium lengh. 1. If hese springs ac on a mass m in parallel, hey would be equivalen o a single spring wih spring consan k parallel. Find k parallel. k 1 k. If hese springs ac on a mass m in series, hey would be equivalen o a single spring wih spring consan k series. Find k series. k 1 k Soluion 1. For a displacemen x from equilibrium, he force fel by he mass will be -(k 1 + k ) x, and herefore k parallel = k 1 + k.. Consider a massless objec beween he firs and second spring. When he full sysem is displaced by a disance x, he firs spring will be sreched by x 1 and he second spring by x so ha he oal displacemen is given by x 1 + x = x and here is no ne force on he massless objec k 1 x 1 = k x. This implies Therefore, he ne force fel on he mass m equals x 1 = k k 1 +k x (1) x = k 1 k 1 +k x () k x = k 1 k k 1 +k x k series x (3) where we have defined he effecive spring consan k series of he sysem. Noe ha 1 k series = k 1+k k 1 k = 1 k k (4) which is he called he harmonic sum of k 1 and k. Superman Orbis Example Superman sands on op of a mounain and hrows a series of sone horizonally, wih each subsequen sone hrown a an ever greaer velociy. Describe he moions of he sones as he iniial velociy of he hrows increase.

Lecure 18-017-1-01.nb Soluion From Kepler s Firs Law, he sones ravel in an ellipse wih he Earh s cener as one of is foci.

2 Lecure nb Soluion From Kepler s Firs Law, he sones ravel in an ellipse wih he Earh s cener as one of is foci. Le us orien our axes so ha +x poins o he righ and +y poins up away from he cener of he Earh. Given Kepler s Firs Law, he rock mus ravel in an ellipse. Because he iniial velociy is direcly horizonal - perpendicu lar o he line beween yourself and he cener of he Earh - his implies ha: 1. The saring poin of he rock will be he highes poin. The semi-major and semi-minor axes of he ellipse mus be along he y and x direcions, respecively 3. The orbi of he paricle mus be symmeric in he x-direcion Here we see wha he orbis can look like, wih he cener of he ellipse shown in orange and he wo foci shown in dark green: iniial velociy iniial heigh show orbi If we look closely a Newon s drawing, we see ha i does look correc. However, noe ha: 1. If he rock ravels 1 of he way around he world, is final velociy as i crashes o he ground will no be parallel 4

3 Lecure nb 3 o he ground bu insead will poin owards he Earh. If he rock ravels more han 1 way around he world, hen i mus be in an orbi and will no reach land unil i reurns o is saring posiion. In paricular, a rock canno be hrown 3 of he way around he world! 4 Clown Cannon Example During an incredible Cirque Du Soleil performance, wo cannons A and B separaed by a disance d simulaneously fire clowns ino he air. Boh clowns are launched a velociy v bu a differen angles θ A and θ B so ha he wo clown high-five in midair and hen land in he oher cannon! 1. Assume θ B is fixed. Find θ A in erms of θ B so ha he wo clowns cover he same disance d bu do no collide in midair. Will boh clowns reach he maximum heigh of heir rajecory a he same ime? 3. Wha is he minimum disance beween he wo clowns during heir rajecory? In oher words, how far do he clowns have o reach o high five each oher? (Check your answer in he limis θ A = 0, π 4, and π ) θ A θ A θ B d Soluions 1. Boh clowns have acceleraion g downwards. Inegraing his, we find he velociies v A[] = v Cos[θ A ] x + (v Sin[θ A ] - g ) y (5) v B[] = -v Cos[θ B ] x + (v Sin[θ B ] - g ) y (6) (As a quick aside, hese wo equaions enable us o answer Par of he quesion: Will he clowns boh reach heir maximum heighs a he same ime? Absoluely no, since he maximum heigh is deermined by v Sin[θ] - g max heigh = 0, and Sin[θ] will be differen for he wo cannons.) We now inegrae he velociy o obain he posiions of he cupcakes. Cupcake A sared a (0, 0) and cupcake B sared a (d, 0). Thus, we can inegrae he above equaions using v = dr d o obain r A [] = v Cos[θ A ] x + v Sin[θ A ] - 1 g y (7) r B [] = (d - v Cos[θ B ]) x + v Sin[θ B ] - 1 g y (8) Cupcake A his he ground when is y-componen is zero,

4 4 Lecure nb 0 = v Sin[θ A ] - 1 g (9) which solves o = 0 or = v g Sin[θ A]. We wan he second soluion, which allows us o solve for d, d = v v g Sin[θ A] Cos[θ A ] = v g Sin[ θ A] (10) Similarly solving when he y-componen of r B equals 0 allows us o find when cupcake B his he ground, which by he symmery of he equaions mus equal = v g Sin[θ B]. Since cupcake B his cannon A, which simplifies o Subsiuing back ino Equaion (10), or equivalenly 0 = d - v v g Sin[θ B] Cos[θ B ] (11) d = v g Sin[ θ B] (1) v g Sin[ θ A] = v g Sin[ θ B] (13) Sin[ θ A ] = Sin[ θ B ] (14) We would like o solve for θ B in erms of θ A. One obvious soluion is o ake ArcSin of boh sides of he equaion and obain θ B = θ A, bu we specifically don wan his soluion (since we don wan he cupcakes o hi each oher). So wha does his mean? A his poin, i helps o visualize Sin[ x], 1.0 Sin[x] x The answer lies in he fac ha Sin[ x] = Sin π - x. Therefore we can wrie he above relaion for θ A and θ B as Sin[ θ A ] = Sin π - θ B (15) and upon aking he ArcSin of boh sides, θ B = π - θ A (16) Noe ha when θ A = π 4, θ B = π 4 which is he only ime ha boh cupcakes are forced ono he same rajecory.

5 Lecure nb 5 θ A θ A θ B d. As discussed above, he wo clowns will no reach heir maximum heighs a he same ime. The clown ha goes higher will be in he air for longer, bu he horizonal velociy is consan during projecile moion, so he clown ha akes he high road will reach heir maximum heigh laer han he clown ha akes he low road, jus like in he Loch Lomond song. 3. Subsiuing Equaion (16) ino Equaions (7) and (8) and simplifying using rig ideniies, r A = v Cos[θ A ] x + v Sin[θ A ] - 1 g y (17) r B = (d - v Sin[θ A ]) x + v Cos[θ A ] - 1 g y (18) We would like o minimize he disance beween he wo cupcakes over. This disance equals r A - r B = {v (Cos[θ A ] + Sin[θ A ]) - d} + {v (Sin[θ A ] - Cos[θ A ])} (19) To find he minima, we are going o need o differeniae his funcion over and se i equal o 0. However, aking he derivaive of a square roo is incredibly messy. To simplify he calculaion, noice ha he minima of r A - r B occurs a he same ime as he minima of r A - r B, so insead we will differeniae he significanly simpler funcion Differeniaing wih respec o, d r A-r B Simplifying and seing 0 = d r A-r B, r A - r B = {v (Cos[θ A ] + Sin[θ A ]) - d} + {v (Sin[θ A ] - Cos[θ A ])} (0) d = {v (Cos[θ A ] + Sin[θ A ]) - d} {v (Cos[θ A ] + Sin[θ A ])} + {v (Sin[θ A ] - Cos[θ A ])} {v (Sin[θ A ] - Cos[θ A ])} d 0 = v (Cos[θ A ] + Sin[θ A ]) + v (Sin[θ A ] - Cos[θ A ]) - v d (Cos[θ A ] + Sin[θ A ]) = v - v d (Cos[θ A ] + Sin[θ A ]) which we can solve for he ime of minimum disance, min disance = d v (Cos[θ A] + Sin[θ A ]) (3) To find he minimum disance, we subsiue his ime back ino Equaion (19) o obain (1) ()

6 6 Lecure nb r min disance = {v min disance (Cos[θ A ] + Sin[θ A ]) - d} + {v min disance (Sin[θ A ] - Cos[θ A ])} = d (1 + Sin[ θ A]) - d + d Cos[ θ A] = d (Sin[ θ A ] - 1) + Cos[ θ A ] = d (1 - Sin[ θ A ]) = d (Cos[θ A ] - Sin[θ A ]) = d Cos[θ A] - Sin[θ A ] Tha s a prey nea formula! Noe ha min disance does no occur halfway hrough he flighs of eiher cupcake A or cupcake B ( = v g Sin[θ A] and = v g Sin[θ B], respecively) excep for he special cases θ A = 0, π 4, and π. The wo cases θ A = 0 and θ A = π are rivial since d = 0 and he wo clowns sar off a he same spo, so ha r min disance = 0. For θ A π, boh clowns ake roughly he same pah (and if θ 4 A = π exacly, hen boh clowns collide 4 in midair), so we expec r min disance 0, as is indeed seen by he formula above. For any oher rajecory, he poin of closes approach beween he wo clowns will no be when eiher cupcake is a he peak of is fligh. Here is a diagram of he rajecories of he flying cupcakes (shown in purple), wih he poins of closes approach shown in green. θ A θ A θ B d Driven Oscillaions Example Two idenical masses are aached o hree idenical springs as shown below. The sysem sars off a res wih all of he springs a heir unsreched lengh. A ime = 0, he lef mass is subjeced o a driving force F d Cos[ ω ] and he righ mass o a driving force F d Cos[ ω ], where ω = k m 1/. Our goal will be o find he resuling displacemen of he lef mass x L [] and he righ mass x R [].

7 Lecure nb 7 driving m m k k k 1. Wrie down he acceleraions ẍ L [] and ẍ R [] of he lef and righ masses in erms of he forces acing upon each mass. (Check your answer for he special cases x L = x R, x L = 0, and x R = 0 o ensure you do no have any signs flipped). Add your wo equaions from Par 1 o find he sum of acceleraions ẍ L [] + ẍ R [] and he difference in acceleraions ẍ L [] - ẍ R [] 3. Suppose he soluions for he sum and difference of he displacemens have he forms x L [] + x R [] = A Cos[ ω ] x L [] - x R [] = B Cos[ ω ] (5) Subsiue hese equaions ino he differenial equaions from Par and deermine A and B. 4. Find x L [] and x R []. Explain he moion of he lef mass using force analysis Soluion 1. The equaions of moion for he wo masses are m ẍ L [] = -k x L [] + k (x R [] - x L []) + F d Cos[ ω ] m ẍ R [] = -k x R [] - k (x R [] - x L []) + F d Cos[ ω ] I always helps o check hese equaions in various limis. For example, if x L [] = x R [], he middle spring will no exer a force and he wo masses will feel a lefwards force if hey are displaced o he righ (0 < x L [], x R []). If x L [] = 0 and x R [] > 0, he lef mass will feel a spring force o he righ while he righ mass will feel wo spring forces o he lef. If x R [] = 0 and x L [] > 0, he lef mass will feel wo spring forces o he lef and he righ mass will feel a spring force o he righ. You can check ha all of hese saemens are suppored by he equaions above.. The sum and difference of he wo acceleraions is given by (6) ẍ L [] + ẍ R [] = -ω (x L [] + x R []) + 3 F d Cos[ ω ] m ẍ L [] - ẍ R [] = -3 ω (x L [] - x R []) - F d m Cos[ ω ] (7) where we have used ω = k m. Noe ha he op differenial equaion is wrien purely in erms of x L[] + x R [] and is second derivaive, while he boom equaion is wrien purely in erms of x L [] - x R [] and is derivaive. Thus, in erms of he sum and difference of he displacemens, hese wo differenial equaions are uncoupled. Each equaion has he same soluion as a driven harmonic oscillaor. 3. Subsiuing in he forms of Equaion (5) ino Equaion (7), we obain -4 ω A Cos[ ω ] = -ω A Cos[ ω ] + 3 F d Cos[ ω ] m -4 ω B Cos[ ω ] = -3 ω B Cos[ ω ] - F d m Cos[ ω ] (8) Noe ha he Cos[ ω ] erms drop from boh equaions - his mus always happen because A and B canno have ime dependence. Simplify he above equaions,

8 8 Lecure nb 4. Noe ha Subsiuing in Equaions (5) and (9), x L [] = - Fd k x R [] = - Fd k The moion of he sysem is shown below. A = - F d k B = F d k x L [] = (x L[]+x R [])+(x L []-x R []) x R [] = (x L[]+x R [])-(x L []-x R []) Cos[ ω ]+ Fd Cos[ ω ] k = 0 (9) (30) (31) Cos[ ω ]- Fd Cos[ ω ] k = - F d k Cos[ ω ] driving force m m spring force Apparenly, he lef mass is saionary! This mus mean ha he ne forces on he lef mass are always zero, which we can quickly verify. Since x L [] = 0, he only spring force on he lef mass is given by k x R [] = -F d Cos[ ω ] according o Equaion (31). This force is exacly balanced by he driving force exered on he lef mass, hereby resuling in no ne force and hence no ne moion! Leaning Ladder Example A ladder of lengh l and uniform mass densiy sands on a fricionless floor and leans agains a fricionless wall a an iniial angle θ 0 relaive o he verical. I is released from res, whereupon he boom end slides away from he wall, and he op end slides down he wall. When will he ladder lose conac wih he wall?

9 Lecure nb 9 θ l Soluion 1 Le r = l. The ladder has a momen of ineria 1 1 m l = 1 3 m r abou is cener, bu we will use he more general I = η m r. Le θ be he angle beween he wall and he ladder, which also equals he verical angle from he wall o he cener of mass. This symmery implies ha he cener of he ladder (while i says in conac wih he wall) ravels along a circle! Here is wha he moion of he ladder would look like if is ends were forced o remain on he floor and wall (for example, if is ends were clamped ono fricionless rails on he floor and wall). θ (x) cener cener pah v CM v CM Ladder clamped o wall and floor θ r (x) v CM r v CM r θ Before we begin solving his problem, le's gain some insigh ino he moion of he ladder and undersand why i should come away from he wall before i his he ground. The forces acing on he ladder are he normal forces N w and N f from he wall and floor, respecively, as well as graviy acing on he cener of mass.

10 10 Lecure nb The moion of he cener of mass (he cener of he ladder) follows he semi-circle (r Sin[θ], r Cos[θ]). The force N w will accelerae he ladder o he righ, while N f will ry and roae he ladder so ha i says on he wall. However, as θ approaches π, he velociy of he ladder would have o be sraigh down in order for i o say aached o he wall and floor, and his would imply ha here mus be a force acing on i o he lef (i.e. N w would have o be negaive) o achieve his. So somewhere before his poin N w = 0, and since he normal force can be negaive (unless we se up some rails and force he ladder o say aached o he wall), pas his poin he ladder will deaches from he wall. Le s find ou when ha is. By conservaion of energy, m g r Cos[θ 0 ] = m g r Cos[θ] + 1 m r θ + 1 (η m r ) θ (3) where θ appears in he kineic energy erm (because i describes how he cener of mass moves) and he same θ appears in he roaional energy erm (because θ is he angle beween he ladder and he verical). This allows us o solve for he velociy v = r θ as 1 m (1 + η) v = m g r (Cos[θ 0 ] - Cos[θ]) (33) v = g r 1+η (Cos[θ 0] - Cos[θ]) 1/ (34) The velociy has a horizonal componen v x = v Cos[θ] (Cos[θ 0 ] - Cos[θ]) 1/ Cos[θ] (35) We wan o find ou when his is maximized, since a negaive slope of v x (also denoed as x ) implies ha N w poins o he lef. To clarify his poin, here is wha he moion of he ladder would look like if he ladder was clamped ono fricionless rails on he floor and wall. The dashed verical line signifies when Che horizonal velociy x reaches is maximum value, a which poin N w = Ladder clamped o wall and floor θ[] x[] y[] Ladder clamped o wall and floor θ [] x [] y [] Thus, our goal is o find he ime a which v x = x is maximized, assuming ha he ladder remains clamped o he floor and wall, which we do by aking he ime derivaive dv x and seing i equal o zero. However, a very convenien rick ha nearly always works in such siuaions is o noice ha v x is maximized when v x is d maximized (since v x is never negaive), bu his squaring removes he nasy square roos and makes he compuaion significanly easier. We can also ignore all of he exraneous erms ha do no involve θ. Therefore, dv x dθ Sin[θ] Cos[θ] - (Cos[θ 0 ] - Cos[θ]) Cos[θ] Sin[θ] = 3 Sin[θ] Cos[θ] - Cos[θ] Sin[θ] Cos[θ 0 ] = 1 Sin[ θ] (3 Cos[θ] - Cos[θ 0]) Aside from he rivial soluions Sin[ θ] = 0 when he ladder eiher begins fla agains he floor or agains he wall, we see ha v x (and herefore v x ) is maximized when (36) Cos[θ] = 3 Cos[θ 0] (37)

11 Lecure nb 11 If we muliply boh sides of his equaion by r, we see ha he ladder falls when i reaches of is iniial heigh 3 agains he wall! Also, his resul is independen of η, which implies ha for i will hold for any mass disribuion, provided ha he cener of mass is sill a he cener of he ladder. The following simulaion shows he acual moion of he ladder as i deaches from he wall. Ladder full moion Noe ha he deachmen of he ladder from he wall is very suble. I is easier o see his change by looking a he plos of x [] over ime. When he ladder deaches, x [] = consan because here are no longer any horizonal forces on he ladder Ladder full moion θ[] x[] y[] Ladder full moion θ [] x [] y [] Soluion Unil he ladder disconnecs from he wall, where we suppress he ime dependence of θ. Differeniaing wice, we obain x = r Sin[θ] (38) y = r Cos[θ] (39) ẍ = -r θ Sin[θ] + r θ Cos[θ] (40) ÿ = -r θ Cos[θ] - r θ Sin[θ] (41)

12 1 Lecure nb Denoe he normal forces from he wall and floor as N w and N f, respecively. The force equaions in he x and y direcions and he orque equaion (abou he cener of mass of he ladder) are N w = m ẍ (4) N f - m g = m ÿ (43) r (N f Sin[θ] - N w Cos[θ]) = η m r θ (44) where in he las equaion we have used he fac ha θ is he angle abou which he cener of mass roaes. The above 5 equaions have 6 unknowns: N f, N w, ẍ, ÿ, θ, θ. The las equaion comes from conservaion of energy Solve m g r Cos[θ 0 ] = m g r Cos[θ] + 1 m r θ + 1 (η m r ) θ Nw m x ''[], Nf - m g m y ''[], r Nf * Sin[θ[]] - Nw * Cos[θ[]] η m r θ ''[], m g r (Cos[θ[0]] - Cos[θ[]]) == 1 m (1 + η) (r θ '[]) /. x ''[] -r (θ '[]) Sin[θ[]] + r θ ''[] Cos[θ[]], y ''[] -r (θ '[]) Cos[θ[]] - r θ ''[] Sin[θ[]], Nw, Nf, θ '[], θ ''[] [[]] Nw 1 g m (- Cos[θ[0]] + 3 Cos[θ[]]) Sin[θ[]], Nf 1 g m (3 + η - 4 Cos[θ[0]] Cos[θ[]] + 3 Cos[ θ[]]), 1 + η (1 + η) θ [] g (Cos[θ[0]] - Cos[θ[]]), θ [] g Sin[θ[]] r (1 + η) r + r η Looking a he normal force (45) N w = m g 1+η Sin[θ] (3 Cos[θ] - Cos[θ 0]) (46) we see ha i equals zero for he rivial soluions θ = 0 and θ = π (when he ladder eiher begins fla on he floor or fla agains he wall) or he non-rivial soluion for a saring posiion θ 0 0, π when Cos[θ] = 3 Cos[θ 0] (47) Noice ha alhough his mehod was more complicaed, we did gain more insigh abou his problem. In paricular, we now have explici relaions for our unknowns N f, N w, ẍ, ÿ, θ, and θ as a funcion of θ. Mahemaica Iniializaion

1. VELOCITY AND ACCELERATION

1. VELOCITY AND ACCELERATION 1.1 Kinemaics Equaions s = u + 1 a and s = v 1 a s = 1 (u + v) v = u + as 1. Displacemen-Time Graph Gradien = speed 1.3 Velociy-Time Graph Gradien = acceleraion Area under