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Question Number 225941 by mr W last updated on 16/Nov/25
Commented by mr W last updated on 16/Nov/25
Q222520 at least dropped from some points, the ball will return back to its starting position A.
$${Q}\mathrm{222520} \\ $$$${at}\:{least}\:{dropped}\:{from}\:{some}\:{points}, \\ $$$${the}\:{ball}\:{will}\:{return}\:{back}\:{to}\:{its} \\ $$$${starting}\:{position}\:{A}. \\ $$
Answered by mr W last updated on 16/Nov/25
let μ=(b/a), λ=(d/a) −(x^2 /a^2 )+(((y+b)^2 )/b^2 )=1 −(d^2 /a^2 )+(((k+b)^2 )/b^2 )=1 ⇒k=b((√(1+(d^2 /a^2 )))−1)=μa((√(1+λ^2 ))−1) −((2x)/a^2 )+((2(y+b)y′)/b^2 )=0 y′=tan θ=((b^2 x)/(a^2 (y+b)))=((b^2 d)/(a^2 b(√(1+(d^2 /a^2 )))))=((μλ)/( (√(1+λ^2 )))) say velocity at C is u. t=(d/u) 2(y_C −k)=d tan ((π/2)−2θ)=(d/(tan 2θ)) y_C −k=((gt^2 )/2)=(g/2)((d/u))^2 =(d/(2 tan 2θ)) ((gd)/u^2 )=(1/(tan 2θ))=((1−tan^2 θ)/(2 tan θ)) =((1−((μ^2 λ^2 )/(1+λ^2 )))/((2μλ)/( (√(1+λ^2 )))))=((1+(1−μ^2 )λ^2 )/(2μλ(√(1+λ^2 )))) ((gλa)/u^2 )=((1+(1−μ^2 )λ^2 )/(2μλ(√(1+λ^2 )))) ⇒(u^2 /(2ga))=((μλ^2 (√(1+λ^2 )))/(1+(1−μ^2 )λ^2 )) (y_C /a)=(k/a)+(λ/(2 tan 2θ)) =μ((√(1+λ^2 ))−1)+((1+(1−μ^2 )λ^2 )/(4μ(√(1+λ^2 )))) mgy_C +((mu^2 )/2)=mgh (h/a)=(y_C /2)+(u^2 /(2ga))=(k/a)+(λ/(2 tan 2θ))+((μλ^2 (√(1+λ^2 )))/(1+(1−μ^2 )λ^2 )) ⇒(h/a)=μ((√(1+λ^2 ))−1)+((1+(1−μ^2 )λ^2 )/(4μ(√(1+λ^2 ))))+((μλ^2 (√(1+λ^2 )))/(1+(1−μ^2 )λ^2 ))
$${let}\:\mu=\frac{{b}}{{a}},\:\lambda=\frac{{d}}{{a}} \\ $$$$−\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left({y}+{b}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$−\frac{{d}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left({k}+{b}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Rightarrow{k}={b}\left(\sqrt{\mathrm{1}+\frac{{d}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}−\mathrm{1}\right)=\mu{a}\left(\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }−\mathrm{1}\right) \\ $$$$−\frac{\mathrm{2}{x}}{{a}^{\mathrm{2}} }+\frac{\mathrm{2}\left({y}+{b}\right){y}'}{{b}^{\mathrm{2}} }=\mathrm{0} \\ $$$${y}'=\mathrm{tan}\:\theta=\frac{{b}^{\mathrm{2}} {x}}{{a}^{\mathrm{2}} \left({y}+{b}\right)}=\frac{{b}^{\mathrm{2}} {d}}{{a}^{\mathrm{2}} {b}\sqrt{\mathrm{1}+\frac{{d}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}}=\frac{\mu\lambda}{\:\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }} \\ $$$${say}\:{velocity}\:{at}\:{C}\:{is}\:{u}. \\ $$$${t}=\frac{{d}}{{u}} \\ $$$$\mathrm{2}\left({y}_{{C}} −{k}\right)={d}\:\mathrm{tan}\:\left(\frac{\pi}{\mathrm{2}}−\mathrm{2}\theta\right)=\frac{{d}}{\mathrm{tan}\:\mathrm{2}\theta} \\ $$$${y}_{{C}} −{k}=\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}=\frac{{g}}{\mathrm{2}}\left(\frac{{d}}{{u}}\right)^{\mathrm{2}} =\frac{{d}}{\mathrm{2}\:\mathrm{tan}\:\mathrm{2}\theta} \\ $$$$\frac{{gd}}{{u}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{2}\theta}=\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\theta}{\mathrm{2}\:\mathrm{tan}\:\theta} \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}−\frac{\mu^{\mathrm{2}} \lambda^{\mathrm{2}} }{\mathrm{1}+\lambda^{\mathrm{2}} }}{\frac{\mathrm{2}\mu\lambda}{\:\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }}}=\frac{\mathrm{1}+\left(\mathrm{1}−\mu^{\mathrm{2}} \right)\lambda^{\mathrm{2}} }{\mathrm{2}\mu\lambda\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }} \\ $$$$\frac{{g}\lambda{a}}{{u}^{\mathrm{2}} }=\frac{\mathrm{1}+\left(\mathrm{1}−\mu^{\mathrm{2}} \right)\lambda^{\mathrm{2}} }{\mathrm{2}\mu\lambda\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }} \\ $$$$\Rightarrow\frac{{u}^{\mathrm{2}} }{\mathrm{2}{ga}}=\frac{\mu\lambda^{\mathrm{2}} \sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }}{\mathrm{1}+\left(\mathrm{1}−\mu^{\mathrm{2}} \right)\lambda^{\mathrm{2}} } \\ $$$$\frac{{y}_{{C}} }{{a}}=\frac{{k}}{{a}}+\frac{\lambda}{\mathrm{2}\:\mathrm{tan}\:\mathrm{2}\theta} \\ $$$$\:\:\:\:=\mu\left(\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }−\mathrm{1}\right)+\frac{\mathrm{1}+\left(\mathrm{1}−\mu^{\mathrm{2}} \right)\lambda^{\mathrm{2}} }{\mathrm{4}\mu\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }} \\ $$$$ \\ $$$${mgy}_{{C}} +\frac{{mu}^{\mathrm{2}} }{\mathrm{2}}={mgh} \\ $$$$\frac{{h}}{{a}}=\frac{{y}_{{C}} }{\mathrm{2}}+\frac{{u}^{\mathrm{2}} }{\mathrm{2}{ga}}=\frac{{k}}{{a}}+\frac{\lambda}{\mathrm{2}\:\mathrm{tan}\:\mathrm{2}\theta}+\frac{\mu\lambda^{\mathrm{2}} \sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }}{\mathrm{1}+\left(\mathrm{1}−\mu^{\mathrm{2}} \right)\lambda^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{h}}{{a}}=\mu\left(\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }−\mathrm{1}\right)+\frac{\mathrm{1}+\left(\mathrm{1}−\mu^{\mathrm{2}} \right)\lambda^{\mathrm{2}} }{\mathrm{4}\mu\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }}+\frac{\mu\lambda^{\mathrm{2}} \sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }}{\mathrm{1}+\left(\mathrm{1}−\mu^{\mathrm{2}} \right)\lambda^{\mathrm{2}} } \\ $$
Commented by mr W last updated on 16/Nov/25
Commented by mr W last updated on 17/Nov/25
when the ball is dropped from a point on the red curve, the ball will always return to this point.
$${when}\:{the}\:{ball}\:{is}\:{dropped}\:{from}\:{a} \\ $$$${point}\:{on}\:{the}\:{red}\:{curve},\:{the}\:{ball}\:{will} \\ $$$${always}\:{return}\:{to}\:{this}\:{point}. \\ $$
Commented by mr W last updated on 16/Nov/25
Answered by mr W last updated on 17/Nov/25
Commented by mr W last updated on 17/Nov/25
Q(−d_1 , k_1 ) k_1 =b((√(1+(d_1 ^2 /a^2 )))−1)=μa((√(1+λ_1 ^2 ))−1) tan θ_1 =((b^2 d_1 )/(a^2 b(√(1+(d_1 ^2 /a^2 )))))=((μλ_1 )/( (√(1+λ_1 ^2 )))) say the velocity at Q is u. t=((d_1 +d)/(u sin θ_1 )) k−k_1 =u cos θ_1 ((d_1 +d)/(u sin θ_1 ))−(g/2)(((d_1 +d)/(u sin θ_1 )))^2 ((k−k_1 )/(d_1 +d))=(1/(tan θ_1 ))−((g(d_1 +d))/(2u^2 sin^2 θ_1 )) ((g(d_1 +d))/(2u^2 sin^2 θ_1 ))=(1/(tan θ_1 ))−((k−k_1 )/(d_1 +d)) ⇒(u^2 /(ga))=(((λ_1 +λ)[1+(1+μ^2 )λ_1 ^2 ])/(2μ^2 λ_1 ^2 [((√(1+λ_1 ^2 ))/(μλ_1 ))−((μ((√(1+λ^2 ))−(√(1+λ_1 ^2 ))))/(λ_1 +λ))])) v_(Px) =u sin θ_1 v_(Py) =u cos θ_1 −g(((d_1 +d)/(u sin θ_1 ))) ((−v_(Py) )/v_(Px) )=(1/(tan 2θ)) ((−u cos θ_1 +g(((d_1 +d)/(u sin θ_1 ))))/(u sin θ_1 ))=(1/(tan 2θ)) −(1/(tan θ_1 ))+((g(d_1 +d))/(u^2 sin^2 θ_1 ))=(1/(tan 2θ)) −(1/(2 tan θ_1 ))−((k−k_1 )/(d_1 +d))+(1/(tan θ_1 ))=(1/(2 tan 2θ)) (1/(tan θ_1 ))−(1/(tan 2θ))=((2(k−k_1 ))/(d_1 +d)) ((√(1+λ_1 ^2 ))/(μλ_1 ))−((1+(1−μ^2 )λ^2 )/(2μλ(√(1+λ^2 ))))=((2[μ((√(1+λ^2 ))−1)−μ((√(1+λ_1 ^2 ))−1)])/((λ_1 +λ))) ((√(1+λ_1 ^2 ))/λ_1 )−((2μ^2 ((√(1+λ^2 ))−(√(1+λ_1 ^2 ))))/(λ_1 +λ))=((1+(1−μ^2 )λ^2 )/(2λ(√(1+λ^2 )))) for a given d (λ), we can find a corresponding d_1 (λ_1 ) through this equation. u=(1/(μλ_1 ))(√((ga(λ_1 +λ)[1+(1+μ^2 )λ_1 ^2 ])/(2[((√(1+λ_1 ^2 ))/( μλ_1 ))−((μ((√(1+λ^2 ))−(√(1+λ_1 ^2 ))))/(λ_1 +λ))]))) mgk_1 +((mu^2 )/2)=mgh (h/a)=(k_1 /a)+(u^2 /(2ga)) ⇒(h/a)=μ((√(1+λ_1 ^2 ))−1)+(((λ_1 +λ)[1+(1+μ^2 )λ_1 ^2 ])/(4μ^2 λ_1 ^2 [((√(1+λ_1 ^2 ))/(μλ_1 ))−((μ((√(1+λ^2 ))−(√(1+λ_1 ^2 ))))/(λ_1 +λ))]))
$${Q}\left(−{d}_{\mathrm{1}} ,\:{k}_{\mathrm{1}} \right) \\ $$$${k}_{\mathrm{1}} ={b}\left(\sqrt{\mathrm{1}+\frac{{d}_{\mathrm{1}} ^{\mathrm{2}} }{{a}^{\mathrm{2}} }}−\mathrm{1}\right)=\mu{a}\left(\sqrt{\mathrm{1}+\lambda_{\mathrm{1}} ^{\mathrm{2}} }−\mathrm{1}\right) \\ $$$$\mathrm{tan}\:\theta_{\mathrm{1}} =\frac{{b}^{\mathrm{2}} {d}_{\mathrm{1}} }{{a}^{\mathrm{2}} {b}\sqrt{\mathrm{1}+\frac{{d}_{\mathrm{1}} ^{\mathrm{2}} }{{a}^{\mathrm{2}} }}}=\frac{\mu\lambda_{\mathrm{1}} }{\:\sqrt{\mathrm{1}+\lambda_{\mathrm{1}} ^{\mathrm{2}} }} \\ $$$${say}\:{the}\:{velocity}\:{at}\:{Q}\:{is}\:{u}. \\ $$$${t}=\frac{{d}_{\mathrm{1}} +{d}}{{u}\:\mathrm{sin}\:\theta_{\mathrm{1}} } \\ $$$${k}−{k}_{\mathrm{1}} ={u}\:\mathrm{cos}\:\theta_{\mathrm{1}} \:\frac{{d}_{\mathrm{1}} +{d}}{{u}\:\mathrm{sin}\:\theta_{\mathrm{1}} }−\frac{{g}}{\mathrm{2}}\left(\frac{{d}_{\mathrm{1}} +{d}}{{u}\:\mathrm{sin}\:\theta_{\mathrm{1}} }\right)^{\mathrm{2}} \\ $$$$\frac{{k}−{k}_{\mathrm{1}} }{{d}_{\mathrm{1}} +{d}}=\frac{\mathrm{1}}{\mathrm{tan}\:\theta_{\mathrm{1}} }−\frac{{g}\left({d}_{\mathrm{1}} +{d}\right)}{\mathrm{2}{u}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta_{\mathrm{1}} } \\ $$$$\frac{{g}\left({d}_{\mathrm{1}} +{d}\right)}{\mathrm{2}{u}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta_{\mathrm{1}} }=\frac{\mathrm{1}}{\mathrm{tan}\:\theta_{\mathrm{1}} }−\frac{{k}−{k}_{\mathrm{1}} }{{d}_{\mathrm{1}} +{d}} \\ $$$$\Rightarrow\frac{{u}^{\mathrm{2}} }{{ga}}=\frac{\left(\lambda_{\mathrm{1}} +\lambda\right)\left[\mathrm{1}+\left(\mathrm{1}+\mu^{\mathrm{2}} \right)\lambda_{\mathrm{1}} ^{\mathrm{2}} \right]}{\mathrm{2}\mu^{\mathrm{2}} \lambda_{\mathrm{1}} ^{\mathrm{2}} \left[\frac{\sqrt{\mathrm{1}+\lambda_{\mathrm{1}} ^{\mathrm{2}} }}{\mu\lambda_{\mathrm{1}} }−\frac{\mu\left(\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }−\sqrt{\mathrm{1}+\lambda_{\mathrm{1}} ^{\mathrm{2}} }\right)}{\lambda_{\mathrm{1}} +\lambda}\right]} \\ $$$$ \\ $$$${v}_{{Px}} ={u}\:\mathrm{sin}\:\theta_{\mathrm{1}} \\ $$$${v}_{{Py}} ={u}\:\mathrm{cos}\:\theta_{\mathrm{1}} −{g}\left(\frac{{d}_{\mathrm{1}} +{d}}{{u}\:\mathrm{sin}\:\theta_{\mathrm{1}} }\right) \\ $$$$\frac{−{v}_{{Py}} }{{v}_{{Px}} }=\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{2}\theta} \\ $$$$\frac{−{u}\:\mathrm{cos}\:\theta_{\mathrm{1}} +{g}\left(\frac{{d}_{\mathrm{1}} +{d}}{{u}\:\mathrm{sin}\:\theta_{\mathrm{1}} }\right)}{{u}\:\mathrm{sin}\:\theta_{\mathrm{1}} }=\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{2}\theta} \\ $$$$−\frac{\mathrm{1}}{\mathrm{tan}\:\theta_{\mathrm{1}} }+\frac{{g}\left({d}_{\mathrm{1}} +{d}\right)}{{u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta_{\mathrm{1}} }=\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{2}\theta} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{tan}\:\theta_{\mathrm{1}} }−\frac{{k}−{k}_{\mathrm{1}} }{{d}_{\mathrm{1}} +{d}}+\frac{\mathrm{1}}{\mathrm{tan}\:\theta_{\mathrm{1}} }=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{tan}\:\mathrm{2}\theta} \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\theta_{\mathrm{1}} }−\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{2}\theta}=\frac{\mathrm{2}\left({k}−{k}_{\mathrm{1}} \right)}{{d}_{\mathrm{1}} +{d}} \\ $$$$\frac{\sqrt{\mathrm{1}+\lambda_{\mathrm{1}} ^{\mathrm{2}} }}{\mu\lambda_{\mathrm{1}} }−\frac{\mathrm{1}+\left(\mathrm{1}−\mu^{\mathrm{2}} \right)\lambda^{\mathrm{2}} }{\mathrm{2}\mu\lambda\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }}=\frac{\mathrm{2}\left[\mu\left(\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }−\mathrm{1}\right)−\mu\left(\sqrt{\mathrm{1}+\lambda_{\mathrm{1}} ^{\mathrm{2}} }−\mathrm{1}\right)\right]}{\left(\lambda_{\mathrm{1}} +\lambda\right)} \\ $$$$\frac{\sqrt{\mathrm{1}+\lambda_{\mathrm{1}} ^{\mathrm{2}} }}{\lambda_{\mathrm{1}} }−\frac{\mathrm{2}\mu^{\mathrm{2}} \left(\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }−\sqrt{\mathrm{1}+\lambda_{\mathrm{1}} ^{\mathrm{2}} }\right)}{\lambda_{\mathrm{1}} +\lambda}=\frac{\mathrm{1}+\left(\mathrm{1}−\mu^{\mathrm{2}} \right)\lambda^{\mathrm{2}} }{\mathrm{2}\lambda\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }} \\ $$$${for}\:{a}\:{given}\:{d}\:\left(\lambda\right),\:{we}\:{can}\:{find}\:{a} \\ $$$${corresponding}\:{d}_{\mathrm{1}} \:\left(\lambda_{\mathrm{1}} \right)\:{through}\:{this} \\ $$$${equation}. \\ $$$$ \\ $$$${u}=\frac{\mathrm{1}}{\mu\lambda_{\mathrm{1}} }\sqrt{\frac{{ga}\left(\lambda_{\mathrm{1}} +\lambda\right)\left[\mathrm{1}+\left(\mathrm{1}+\mu^{\mathrm{2}} \right)\lambda_{\mathrm{1}} ^{\mathrm{2}} \right]}{\mathrm{2}\left[\frac{\sqrt{\mathrm{1}+\lambda_{\mathrm{1}} ^{\mathrm{2}} }}{\:\mu\lambda_{\mathrm{1}} }−\frac{\mu\left(\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }−\sqrt{\mathrm{1}+\lambda_{\mathrm{1}} ^{\mathrm{2}} }\right)}{\lambda_{\mathrm{1}} +\lambda}\right]}} \\ $$$${mgk}_{\mathrm{1}} +\frac{{mu}^{\mathrm{2}} }{\mathrm{2}}={mgh} \\ $$$$\frac{{h}}{{a}}=\frac{{k}_{\mathrm{1}} }{{a}}+\frac{{u}^{\mathrm{2}} }{\mathrm{2}{ga}} \\ $$$$\Rightarrow\frac{{h}}{{a}}=\mu\left(\sqrt{\mathrm{1}+\lambda_{\mathrm{1}} ^{\mathrm{2}} }−\mathrm{1}\right)+\frac{\left(\lambda_{\mathrm{1}} +\lambda\right)\left[\mathrm{1}+\left(\mathrm{1}+\mu^{\mathrm{2}} \right)\lambda_{\mathrm{1}} ^{\mathrm{2}} \right]}{\mathrm{4}\mu^{\mathrm{2}} \lambda_{\mathrm{1}} ^{\mathrm{2}} \left[\frac{\sqrt{\mathrm{1}+\lambda_{\mathrm{1}} ^{\mathrm{2}} }}{\mu\lambda_{\mathrm{1}} }−\frac{\mu\left(\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }−\sqrt{\mathrm{1}+\lambda_{\mathrm{1}} ^{\mathrm{2}} }\right)}{\lambda_{\mathrm{1}} +\lambda}\right]} \\ $$
Commented by fantastic2 last updated on 17/Nov/25
great sir(though i understand nothing)
$${great}\:{sir}\left({though}\:{i}\:{understand}\:{nothing}\right) \\ $$
Commented by mr W last updated on 17/Nov/25
the question is if a ball can return back to its starting point, when it is dropped from a height onto to a hyperbola ground.
$${the}\:{question}\:{is}\:{if}\:{a}\:{ball}\:{can}\:{return} \\ $$$${back}\:{to}\:{its}\:{starting}\:{point},\:{when}\:{it}\:{is} \\ $$$${dropped}\:{from}\:{a}\:{height}\:{onto}\:{to}\:{a} \\ $$$${hyperbola}\:{ground}. \\ $$
Commented by fantastic2 last updated on 17/Nov/25
complex...
$${complex}… \\ $$
Commented by mr W last updated on 17/Nov/25
i have showed that at least from following two heights the ball will return to its starting position.
$${i}\:{have}\:{showed}\:{that}\:{at}\:{least}\:{from} \\ $$$${following}\:{two}\:{heights}\:{the}\:{ball}\:{will} \\ $$$${return}\:{to}\:{its}\:{starting}\:{position}. \\ $$
Commented by mr W last updated on 17/Nov/25
Commented by mr W last updated on 17/Nov/25

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