Roller-Coaster Coordinate System

Transcription

1 Winer 200 MECH 220: Mechanics 2 Roller-Coaser Coordinae Sysem Imagine you are riding on a roller-coaer in which he rack goes up and down, wiss and urns. Your velociy and acceleraion will change (quie abruply), as you rise along he rack. I would be ineresing o describe he moion in erms of a coordinae sysem relaive o you ravelling along he roller-coaser rack. The equaion for he roller-coser is, of course specified, i.e. S(x, y, z) = 0. If you are ravelling on his rack, hen your posiion will be deermined if you know he disance s along he rack from some reference poin, say he origin s = 0. The magniude of your velociy isheraeofchangeofs wih respec o ime, d or ṡ. The direcion of he velociy is angen o he rack. If we define a uni vecor e (s) angen o he curve S(x, y, z), hen he velociy is given v = ṡ e. The angen vecor can be expressed in erms of he posiion vecor of he car r relaive o some coordinae sysem ha defines he curve S(x, y, z). In Figure, we assume a Caresian coordinae sysem: y S(x,y,z) r() Δr r( + Δ) z x Figure : Curve s(x, y, z) in Caresian space The magniude of Δ r is Δs and he direcion of Δ r is e. We can herefore express e as: Δ r e = lim Δs 0 Δs = d r From he velociy v =ṡ e, we can obain he acceleraion by simply differeniaing v wih respec o ime, i.e.: a = d v d = s e +ṡ e Noe ha e is a funcion of he posiion s only. Thus, we wrie e = d e acceleraion vecor is: a = s e +ṡ 2 d e d = d e ṡ and he

2 Winer 200 MECH 220: Mechanics 2 2. We nex mus deermine how e changes wih posiion along he prescribed pah as in Figure y 5 s 6 7 s 6 s 5 4 s s 3 s 2 2 s z x Figure 2: Uni vecors along curve s Figure 2 illusraes he change of e as we move along he curve S(x, y, z) = 0. If we consider he change of e when we move from s o s +Δs, hen Δ e = Δθ, whereδθ is he angle beween e (s) and e (s +Δs) as shown in Figure 3. e (s) Δe (s) -e (s) Δθ e (s + Δs) Figure 3: Uni vecor change from s o Δs The direcion of Δ e is normal o e.thus,wemaywrie: Δ e =Δθ e N where e N is a uni vecor normal o e.sinceδ e (s) is a funcion of s only, we divide he above equaion by Δs and ake he limi as Δs 0, i.e.: Δ e lim Δs 0 Δs = d e = lim Δθ e N Δs 0 Δs Noe ha Δθ and Δs can be relaed o he local curvaure of he curve. The local curvaure is he deflecion of he angen vecor e per uni lengh of ravel Δs along he curve. 2

3 Winer 200 MECH 220: Mechanics 2 e (s) Δs Δθ e (s + Δs) Figure 4: Δs and Δθ relaionship From Figure 4, we ge Δθ =Δs and hence Δθ curvaure. Thus we wrie: Δs = where is defined as he local radius of d e The acceleraion vecor can now be wrien as: = e N a = s e + ṡ2 e N Consider a simple example of a poin moving in a circle of radius R as in Figure 5. e θ e r R θ Figure 5: Poin moving in a circle The space curve s(x, y) = 0 is jus he equaion for a circle y = R 2 x 2. The vecor e is e θ in polar coordinaes and ṡ e = R θ e θ is he velociy, always angen o he circle. The acceleraion s e is he linear acceleraion R θ e θ. For a planar circle, he radius of curvaure is = R. The second erm in he acceleraion can hus be wrien as: ṡ 2 e N = R2 θ 2 R e r = R θ 2 e r 3

4 Winer 200 MECH 220: Mechanics 2 where e N is defined o be in he negaive e r direcion. This simple example illusraes he equivalence of he roller coaser coordinaes o he polar coordinaes wih e θ = e and e N = e r. A hree-dimensional space curve can also be wised like a coiled spring for example. In his case, he plane conaining he vecor e and e N (i.e. he osculaing plane) will roae abou e as we move along he curve. To define he degree of wis or orsion as we move along he curve, we define a hird vecor normal o he osculaing plane, i.e. e b = e e N as shown in Figure 6. Thus, as he osculaing plane roaes when he curve wiss, he vecor e b will be defleced by an angle Δφ. e N e e b e b e e N Figure 6: Three dimensional coordinae sysem along a curve If we move from s o s +Δs, le he roaion of he osculaing plane be Δφ. Looking sraigh a he vecor e,wege: e e b Δφ Δe b= -Δφ en e N Δφ Δe N = Δφ e b Figure 7: Uni vecor changes in he b and N-direcions due o Δφ Thus Δ e b = Δφ e N and he magniude is Δφ and Δ e b is poining in he negaive e N direcion. Similar o he curvaure,wedefineheorsion τ of he curve as he wis per uni lengh of ravel Δs, i.e. τ = dφ. Dividing Δ e b by Δs, wege Δ e b Δs = Δφ Δs e N and aking he limi as Δs 0, we wrie: 4

5 Winer 200 MECH 220: Mechanics 2 d e b = e N τ To find he orsion, we noe ha e b = e e N, hence: d e b = d e e N + e d e N Since d e = e N, he firs erm is e N e N = 0 and hence: Wih e N = e b e,wege: d e b = e d e N d e N = d e b e + e b d e = e N e + e b e N τ and hence: d e N = e b e τ The se of equaions giving he change of he hree vecors wih respec o s is referred o as Frene s equaions: d e = e N d e N d e b = e b e τ = τ e N Given he equaion for he space curve S(x, y, z) = 0, we can define he posiion vecor r from some origin o a poin on he curve. Then, we may obain he curvaure and orsion of he curve in erms of he posiion vecor r. For example, e = d r. Since: d e = e N = d2 r 2 we see ha = 2. Thus he curvaure is given by he magniude of d2 r 2. To find he orsion, we differeniae he laer equaion once more wih respec o s, we ge: d 3 r 3 = d e N + d e N 5

6 Winer 200 MECH 220: Mechanics 2 If we cross produc he above wih d2 r = e 2 N, we see ha he firs erm on he righ-hand side vanishes and we ge: We obain: 2 3 = e N d e N 2 = ( ( e 2 N e b )) e τ 3 = e 2 + e τ 3 b If we do produc he above equaion wih e = d r,wege: d r d2 r 2 Hence, he orsion is given by he expression: τ = 2 3 = 2 τ ( d r d2 r 2 Since =, wemayreplace 2, we obain: 2 τ = d r ( 2 d 2 r 2 d3 ) r 3 ) 3 Thus, when he equaion for he space curve is given, hence r is defined, he curvaure and he orsion can be obained by he above formulae. An alernae expression for he curvaure can be obained if we know he velociy and acceleraion vecors, v and a. Taking he cross produc of V and a,wege: v a = ṡ e = ṡ3 e b d2 r 2 ( s ) e + ṡ2 e N Since e b is a uni vecor, aking he magniude of he above yiel: or he curvaure is given by v a = ṡ3 = v a ṡ 3 = v a v 3 We shall illusrae he finding of he curvaure and orsion of a space curve. Consider a spiral wih radius R and pich P in Figure 8. The posiion vecor for his spiral is given by: r = R sin θ e x + R cos θ e z + Pθ e y 6

7 Winer 200 MECH 220: Mechanics 2 y P = P 2π z s=0 r θ R Rcosθ Pθ Rsinθ x Figure 8: Spiral wih a radius R and pich P where P = P 2π is he pich. The lengh of he curve is given by d r = r r. Wih he posiion vecor given, we can find: Thus: d r =(R cos θ e x R sin θ e z + P e y ) dθ = d r d r = R 2 + P 2 dθ = h dθ wherewehavedefined h = R 2 + P 2. We can now evaluae d r Similarly, we find: d r = d r dθ as: dθ = h (R cos θ e x R sin θ e z + P e y ) 2 = h2 (R sin θ e x + R cos θ e z ) d 3 r 3 = h3 ( R cos θ e x + R sin θ e z ) The curvaure of he spiral can now be readily obained since all he derivaives of r wih respec o s have been found: and he orsion: = 2 = 2 d2 r 2 = R Rh2 = R 2 + P 2 7

8 Winer 200 MECH 220: Mechanics 2 τ = ( d r 2 d 2 r 2 ) 3 d2 r 2 τ = P Ph2 = R 2 + P 2 If we are given he condiion for he moion along his spiral, like sliding down he spiral saircase a some prescribed acceleraion, hen we can readily obain he velociy and he acceleraion a any posiion along his spiral. 8