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Question Number 187364 by ajfour last updated on 16/Feb/23
Commented by ajfour last updated on 16/Feb/23
Find the time taken by ball to reach down the y=bx^2 valley. The height =ba^2 (≠ab^2 ⌣^(• •) ) No friction.
$${Find}\:{the}\:{time}\:{taken}\:{by}\:{ball}\:{to} \\ $$$${reach}\:{down}\:{the}\:{y}={bx}^{\mathrm{2}} \:\:{valley}. \\ $$$${The}\:\:{height}\:={ba}^{\mathrm{2}} \:\:\:\left(\neq{ab}^{\mathrm{2}} \:\:\overset{\bullet\:\bullet} {\smile}\right) \\ $$$${No}\:{friction}. \\ $$
Commented by a.lgnaoui last updated on 17/Feb/23
Soit Reference ℜ(ox,oy) oy: a_y =−g v=−gt+v_(0y) at t_0 =0 y_o =ab^2 v_(0y) =0 ⇒y=−(1/2)gt^2 +ab^2 Along axis [ox): a_x =0 v=(√(v_x ^2 +v_y ^2 )) v^2 =2gx x=a =vt y=0 (1/2)gt^2 =ab^2 t=b(√((2a)/g))
$${Soit}\:{Reference}\:\Re\left({ox},{oy}\right) \\ $$$${oy}:\:\:\:{a}_{{y}} =−{g}\:\:\:\:\:{v}=−{gt}+{v}_{\mathrm{0}{y}} \\ $$$$\:\:{at}\:{t}_{\mathrm{0}} =\mathrm{0}\:\:\:\:\:{y}_{{o}} ={ab}^{\mathrm{2}} \:\:\:\:{v}_{\mathrm{0}{y}} =\mathrm{0} \\ $$$$\Rightarrow{y}=−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} +{ab}^{\mathrm{2}} \\ $$$${Along}\:\:{axis}\:\left[{ox}\right):\:\:\:{a}_{{x}} =\mathrm{0} \\ $$$${v}=\sqrt{{v}_{{x}} ^{\mathrm{2}} +{v}_{{y}} ^{\mathrm{2}} } \\ $$$${v}^{\mathrm{2}} =\mathrm{2}{gx}\:\: \\ $$$${x}={a}\:={vt}\:\:\:\:\:{y}=\mathrm{0}\:\: \\ $$$$\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} ={ab}^{\mathrm{2}} \:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{t}}=\boldsymbol{{b}}\sqrt{\frac{\mathrm{2}\boldsymbol{{a}}}{\boldsymbol{{g}}}} \\ $$$$\:\:\: \\ $$
Commented by ajfour last updated on 17/Feb/23
This wont do, wrong, i shall post my solution..soon (hrs..)
$${This}\:{wont}\:{do},\:{wrong},\:{i}\:{shall}\:{post} \\ $$$${my}\:{solution}..{soon}\:\left({hrs}..\right) \\ $$
Answered by ajfour last updated on 17/Feb/23
Commented by ajfour last updated on 17/Feb/23
y=bx^2 (dy/dx)=2bx =tan θ while ((dy/dt))^2 =4b^2 x^2 =4by v^2 =2g(ba^2 −y) v_x ^2 sec^2 θ=2g(ba^2 −y) ((dx/dt))^2 (1+4b^2 x^2 )=2g(ba^2 −y) ((dy/dt))^2 (((1+4b^2 x^2 )/(4b^2 x^2 )))=2g(ba^2 −y) (dy/dt)(√((1+4by)/(4by)))=(√(2g))(√(ba^2 −y)) (1/( (√(2g))))∫_0 ^( t) dt=∫_(ba^2 ) ^( 0) (√((1/(4by))+1))(√(1/(ba^2 −y)))dy .....
$${y}={bx}^{\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{2}{bx}\:=\mathrm{tan}\:\theta \\ $$$${while}\:\:\left(\frac{{dy}}{{dt}}\right)^{\mathrm{2}} =\mathrm{4}{b}^{\mathrm{2}} {x}^{\mathrm{2}} =\mathrm{4}{by} \\ $$$$\:\:{v}^{\mathrm{2}} =\mathrm{2}{g}\left({ba}^{\mathrm{2}} −{y}\right) \\ $$$${v}_{{x}} ^{\mathrm{2}} \mathrm{sec}\:^{\mathrm{2}} \theta=\mathrm{2}{g}\left({ba}^{\mathrm{2}} −{y}\right) \\ $$$$\left(\frac{{dx}}{{dt}}\right)^{\mathrm{2}} \left(\mathrm{1}+\mathrm{4}{b}^{\mathrm{2}} {x}^{\mathrm{2}} \right)=\mathrm{2}{g}\left({ba}^{\mathrm{2}} −{y}\right) \\ $$$$\left(\frac{{dy}}{{dt}}\right)^{\mathrm{2}} \left(\frac{\mathrm{1}+\mathrm{4}{b}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{4}{b}^{\mathrm{2}} {x}^{\mathrm{2}} }\right)=\mathrm{2}{g}\left({ba}^{\mathrm{2}} −{y}\right) \\ $$$$\frac{{dy}}{{dt}}\sqrt{\frac{\mathrm{1}+\mathrm{4}{by}}{\mathrm{4}{by}}}=\sqrt{\mathrm{2}{g}}\sqrt{{ba}^{\mathrm{2}} −{y}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{g}}}\int_{\mathrm{0}} ^{\:{t}} {dt}=\int_{{ba}^{\mathrm{2}} } ^{\:\mathrm{0}} \sqrt{\frac{\mathrm{1}}{\mathrm{4}{by}}+\mathrm{1}}\sqrt{\frac{\mathrm{1}}{{ba}^{\mathrm{2}} −{y}}}{dy} \\ $$$$….. \\ $$
Answered by mr W last updated on 17/Feb/23
Commented by mr W last updated on 17/Feb/23
i replace a with R, ba^2 with H. eqn. of parabola is y=H((x/R))^2 let η=(H/R), ξ=(x/R)∈[−1,1] y′=2H(x/R^2 )=2ηξ y′′=2(H/R^2 )=((2η)/R) radius of curvature r=(([1+(y′)^2 ]^(3/2) )/(∣y′′∣)) ⇒r=(((1+4η^2 ξ)^(3/2) R)/(2η)) (1/2)mv^2 =mg(H−y)=mg(H−Hξ^2 ) ⇒v^2 =2gH(1−ξ^2 ) ⇒v=(√(2gH(1−ξ^2 ))) ds=(√(1+(y′)^2 ))dx=R(√(1+4η^2 ξ^2 ))dξ dt=(ds/v)=(R/( (√(2gH))))(√((1+4η^2 ξ^2 )/(1−ξ^2 )))dξ the time the ball takes to move down the valley: T=(R/( (√(2gH))))∫_0 ^1 (√((1+4η^2 ξ^2 )/(1−ξ^2 )))dξ T=(√((2H)/g))×(1/(2η))∫_0 ^1 (√((1+4η^2 ξ^2 )/(1−ξ^2 )))dξ =k(√((2H)/g)) with k=(1/(2η))∫_0 ^1 (√((1+4η^2 ξ^2 )/(1−ξ^2 )))dξ =(1/(2η))E((π/2)∣−4η^2 ) (√((2H)/g)) is the time for the ball to fall freely from height H. exact formula for k is not possible. examples: η=(H/R)=1 ⇒k≈1.3176 η=(H/R)=2 ⇒k≈1.1018 η=(H/R)=4 ⇒k≈1.0309 η=(H/R)=10 ⇒k≈1.0061
$${i}\:{replace}\:{a}\:{with}\:{R},\:{ba}^{\mathrm{2}} \:{with}\:{H}. \\ $$$${eqn}.\:{of}\:{parabola}\:{is} \\ $$$${y}={H}\left(\frac{{x}}{{R}}\right)^{\mathrm{2}} \\ $$$${let}\:\eta=\frac{{H}}{{R}},\:\xi=\frac{{x}}{{R}}\in\left[−\mathrm{1},\mathrm{1}\right] \\ $$$${y}'=\mathrm{2}{H}\frac{{x}}{{R}^{\mathrm{2}} }=\mathrm{2}\eta\xi \\ $$$${y}''=\mathrm{2}\frac{{H}}{{R}^{\mathrm{2}} }=\frac{\mathrm{2}\eta}{{R}} \\ $$$${radius}\:{of}\:{curvature}\:{r}=\frac{\left[\mathrm{1}+\left({y}'\right)^{\mathrm{2}} \right]^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mid{y}''\mid} \\ $$$$\Rightarrow{r}=\frac{\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi\right)^{\frac{\mathrm{3}}{\mathrm{2}}} {R}}{\mathrm{2}\eta} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} ={mg}\left({H}−{y}\right)={mg}\left({H}−{H}\xi^{\mathrm{2}} \right) \\ $$$$\Rightarrow{v}^{\mathrm{2}} =\mathrm{2}{gH}\left(\mathrm{1}−\xi^{\mathrm{2}} \right) \\ $$$$\Rightarrow{v}=\sqrt{\mathrm{2}{gH}\left(\mathrm{1}−\xi^{\mathrm{2}} \right)} \\ $$$$ \\ $$$${ds}=\sqrt{\mathrm{1}+\left({y}'\right)^{\mathrm{2}} }{dx}={R}\sqrt{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} }{d}\xi \\ $$$${dt}=\frac{{ds}}{{v}}=\frac{{R}}{\:\sqrt{\mathrm{2}{gH}}}\sqrt{\frac{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} }{\mathrm{1}−\xi^{\mathrm{2}} }}{d}\xi \\ $$$${the}\:{time}\:{the}\:{ball}\:{takes}\:{to}\:{move}\:{down} \\ $$$${the}\:{valley}: \\ $$$${T}=\frac{{R}}{\:\sqrt{\mathrm{2}{gH}}}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} }{\mathrm{1}−\xi^{\mathrm{2}} }}{d}\xi \\ $$$${T}=\sqrt{\frac{\mathrm{2}{H}}{{g}}}×\frac{\mathrm{1}}{\mathrm{2}\eta}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} }{\mathrm{1}−\xi^{\mathrm{2}} }}{d}\xi \\ $$$$\:\:\:={k}\sqrt{\frac{\mathrm{2}{H}}{{g}}} \\ $$$${with}\:{k}=\frac{\mathrm{1}}{\mathrm{2}\eta}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} }{\mathrm{1}−\xi^{\mathrm{2}} }}{d}\xi \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}\eta}{E}\left(\frac{\pi}{\mathrm{2}}\mid−\mathrm{4}\eta^{\mathrm{2}} \right) \\ $$$$\:\sqrt{\frac{\mathrm{2}{H}}{{g}}}\:{is}\:{the}\:{time}\:{for}\:{the}\:{ball}\:{to} \\ $$$${fall}\:{freely}\:{from}\:{height}\:{H}.\: \\ $$$$ \\ $$$${exact}\:{formula}\:{for}\:{k}\:{is}\:{not}\:{possible}. \\ $$$${examples}: \\ $$$$\eta=\frac{{H}}{{R}}=\mathrm{1}\:\Rightarrow{k}\approx\mathrm{1}.\mathrm{3176} \\ $$$$\eta=\frac{{H}}{{R}}=\mathrm{2}\:\Rightarrow{k}\approx\mathrm{1}.\mathrm{1018} \\ $$$$\eta=\frac{{H}}{{R}}=\mathrm{4}\:\Rightarrow{k}\approx\mathrm{1}.\mathrm{0309} \\ $$$$\eta=\frac{{H}}{{R}}=\mathrm{10}\:\Rightarrow{k}\approx\mathrm{1}.\mathrm{0061} \\ $$
Commented by mr W last updated on 17/Feb/23
Commented by mr W last updated on 17/Feb/23
when the ball takes the shortest way (blue inclined plane): T=(√(1+(1/η^2 ))) (√((2H)/g)) η=1, k=1.4142 η=2, k=1.1180 η=4, k=1.0308 η=10, k=1.0050 that means the ball may need more time when following the shortest straight way than following the parabola.
$${when}\:{the}\:{ball}\:{takes}\:{the}\:{shortest}\: \\ $$$${way}\:\left({blue}\:{inclined}\:{plane}\right): \\ $$$${T}=\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\eta^{\mathrm{2}} }}\:\sqrt{\frac{\mathrm{2}{H}}{{g}}} \\ $$$$\eta=\mathrm{1},\:{k}=\mathrm{1}.\mathrm{4142} \\ $$$$\eta=\mathrm{2},\:{k}=\mathrm{1}.\mathrm{1180} \\ $$$$\eta=\mathrm{4},\:{k}=\mathrm{1}.\mathrm{0308} \\ $$$$\eta=\mathrm{10},\:{k}=\mathrm{1}.\mathrm{0050} \\ $$$${that}\:{means}\:{the}\:{ball}\:{may}\:{need}\:{more}\: \\ $$$${time}\:{when}\:{following}\:{the}\:{shortest}\: \\ $$$${straight}\:{way}\:{than}\:{following}\:{the}\: \\ $$$${parabola}. \\ $$
Commented by ajfour last updated on 17/Feb/23
Thank you sir, extensive analysis!
$${Thank}\:{you}\:{sir},\:{extensive}\:{analysis}! \\ $$

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