Curvilinear Motion: Normal and Tangential Components

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1 15 Crviliear Moio: Noral ad Tageial Copoe Ref: Hibbeler 1.7, Bedford & Fowler: Dyaic.3 Whe he pah of a paricle i kow, a - coordiae ye wih a origi a he locaio of he paricle (a a ia i ie) ca be helpfl i decribig he oio of he paricle. Hibbeler give a cocie procedre for aalyi i ecio 1.7, which we will apply o he followig exaple. Exaple: A kaeboarder i a qarer pipe. A kaeboarder ha dropped io a 3 eer (radi) qarer pipe. Whe he dropped a diace of 1 eer her peed i 3. / ad i icreaig by 7.8 /. Deerie he oral ad ageial copoe of he kaeboarder acceleraio. Solio The eqaio of he kaeboarder pah i ha of a qarer circle: y = R R x where R i he radi, 3. Fir, we eed o deerie he acal locaio of he kaeboarder whe he ha dropped 1. y θ 1 v A = 3. / A x She i obvioly R 1 = off he grod, ad her x-poiio ca be deeried fro he pah eqaio ig eiher MATLAB ieraive roo fider (fzero), or ybolic ah capabiliy. Here, we fir creae a fcio (a -file) whoe roo we wih o fid ad he call he fzero ieraive olver o he ewly creaed fcio.

2 %Defie he fcio o which o e fzero. fcio F = FidMii(x) R=3; Y=; %3 eer radi % eer off he grod %The eqaio decribig he caeboarder pah i ha of % ad qarer circle => x^ y^ = R^ % or iilarly => y = R - qr(x^ y^) % % The eig he eqaio eqal o oe arbirary variable % ad olvig for he roo F = R - qr(r^ - x^) - Y; >> X = fzero('fidmii',) %Ge: X =.884 So, a he ia of iere, he kaeboarder i locaed a X =.88, Y =. Nex, he eqaio of he lope of he pah a hi poi ca be deeried ig MATLAB ybolic derivaive operaor a follow: >> y x y r %Defie he x, y, ad R a ybolic variable %Fid ybolic rel o he fir derivaive >> y = r - qr( r^ - x^ ); %Defie ybolic eqaio >> dydx = diff(y,x) %Fid fir derivaive dydx = 1/(r^-x^)^(1/)*x Noe: lowercae variable are he ybolic variable while ppercae variable are aiged a eric vale. By aigig he derivaive o a ilie fcio ad X ad R o heir repecive vale, he eric vale of he derivaive ca be calclaed. >> R = 3; >> X =.884; >> DYDX = ilie('1/(r^-x^)^(1/)*x','x','r') DYDX = Ilie fcio: DYDX(x,r) = 1/(r^-x^)^(1/)*x >> DYDX(X,R) a =.88 The agle i degree a poi A ca he be deeried a: >> aa(.88) * 180/pi a =

3 The kaeboarder acceleraio ca be wrie i er of oral ad ageial velociie, a a = v& = 7.8 v ( 3. ) The radi of crvare,, of he pah a he poi of iere (X=.88, Y = ) ca be calclaed a: = dy 1 dx d dx y 3 I MATLAB, hi eqaio i evalaed a follow: >> y = r - qr( r^ - x^ ); %Defie Sybolic Eqaio >> dydx = diff(y,x) %Fir Derivaive dydx = 1/(r^-x^)^(1/)*x >> DYDX = ilie('1/(r^-x^)^(1/)*x','x','r'); %Creae ilie fcio DYDX >> dydx = diff(diff(y,x),x) %Secod Derivaive dydx = 1/(r^-x^)^(3/)*x^1/(r^-x^)^(1/) %Creae ilie fcio DYDX >> DYDX = ilie('1/(r^-x^)^(3/)*x^1/(r^-x^)^(1/)','x','r'); >> RHO = (1 DYDX(X,R)^)^(3/) / ab(dydx(x,r)) RHO = The calclaed radi of crvare i 3 which hold coe a o rprie ice he pah i a circle of radi 3. The eqaio for he acceleraio of he kaeboarder i er of ageial ad oral copoe i ow: a = v& = 7.8 v ( 3. ) 3 a φ A θ The agide of he acceleraio i fod wih he followig calclaio:

4 >> a = qr(7.8^ ( 3.^ / 3 )^) %eer per ecod qared a = While he agle (i degree), φ, i fod ig he aa() fcio ad liplyig by he coverio facor 180 o /π. >> PHI = aa(v_do / ( v^/rho )) *180/pi PHI = The agle of he acceleraio i degree fro he poiive x axi i: >> ANGLE = THETA 90 PHI ANGLE = Aoaed MATLAB Scrip Solio %Defie he fcio o which o e fzero. fcio F = FidMii(x) R=3; Y=; %3 eer radi % eer off he grod %The eqaio decribig he caeboarder pah i ha of % ad qarer circle => x^ y^ = R^ % or iilarly => y = R - qr(x^ y^) % % The eig he eqaio eqal o oe arbirary variable % ad olvig for he roo F = R - qr(r^ - x^) - Y; %Call fzero o he FidMii fcio wih a iiial ge of X = fzero('fidmii',); %Ge: fprif('x = %1.4f.\', X) %Defie all oher variable ig ppercae leer o idicae he variable ha bee aiged a vale. V_do = 7.8; V = 3.; R = 3; Y = ; %Defie he x, y, ad r a ybolic variable. %Noe: Lowercae variable are all ybolic variable while % ppercae variable are aiged a eric vale. y x y r; %Fid ybolic rel o fir ad ecod derivaive y = r - qr( r^ - x^ ); %Defie Sybolic Eqaio dydx = diff(y,x) %Fir Derivaive

5 DYDX = ilie('1/(r^-x^)^(1/)*x','x','r'); dydx = diff(diff(y,x),x) %Secod Derivaive DYDX = ilie('1/(r^-x^)^(3/)*x^1/(r^-x^)^(1/)','x','r'); %Calclae he agle a poi A. THETA = aa(x) *180/pi; fprif('the agle a poi a i %1.4f degree.\', THETA) %Calclae he radi of crvare a X =.88 ad y =. RHO = (1 DYDX(X,R)^)^(3/) / ab(dydx(x,r)); fprif('the radi of crvare a X =.88 ad y = i %1.4f.\', RHO) %Calclae he agide of acceleraio. A = qr(v_do^ (V^/RHO)^); fprif('the agide of acceleraio i %1.4f /^.\', A) %Calclae he algle of acceleraio relaive o he, coordiae ye. PHI = aa(v_do / ( V^/RHO )) *180/pi; %Calclae he agle of acceleraio relaive o he poiive x-axi. ANGLE = THETA 90 PHI; fprif('the agle of acceleraio relaive o he poiive x-axi i %1.4f degree.\', ANGLE)

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