Question Number 227670 by mr W last updated on 12/Feb/26
Commented by mr W last updated on 13/Feb/26
$${an}\:{interesting}\:{old}\:{question}\:{reposted}: \\ $$$${what}\:{is}\:{the}\:{largest}\:{length}\:{of}\:{a}\:{thin} \\ $$$${stick}\:{which}\:{can}\:{rest}\:{with}\:{one}\:{end} \\ $$$${inside}\:{the}\:{cup}\:{as}\:{shown}? \\ $$$${assume}\:{the}\:{interior}\:{of}\:{the}\:{cup}\:{has} \\ $$$${the}\:{parabolic}\:{form}\:{and}\:{there}\:{is} \\ $$$${no}\:{friction}\:{between}\:{stick}\:{and}\:{cup}. \\ $$
Commented by ajfour last updated on 13/Feb/26
$${I}\:{remember},\:{sir},\:{i}\:{shall}\:{reattempt} \\ $$$${i}\:{am}\:{good},\:{though}\:{been}\:{absent} \\ $$$${sometime}. \\ $$
Commented by mr W last updated on 14/Feb/26
$${good}\:{to}\:{know}\:{that}\:{you}\:{are}\:{good}\:{sir}! \\ $$
Commented by vnm last updated on 15/Feb/26
$$ \\ $$who knows how to make this link clickable? https://www.desmos.com/calculator/m9xh6pswei
Commented by vnm last updated on 16/Feb/26
$$\mathrm{6}.\mathrm{1772932766} \\ $$
Commented by mr W last updated on 16/Feb/26
$${that}'{s}\:{right}! \\ $$$${exact}\:{value}:\:{L}_{{max}} =\frac{\mathrm{15}\sqrt{\mathrm{30}+\mathrm{6}\sqrt{\mathrm{5}}}}{\mathrm{16}} \\ $$
Commented by mr W last updated on 16/Feb/26
$${what}\:{do}\:{you}\:{get}\:{for}\:{example}: \\ $$$${R}=\mathrm{1},\:{H}=\mathrm{2},\:{L}_{{max}} =? \\ $$
Commented by mr W last updated on 15/Feb/26
https://www.desmos.com/calculator/m9xh6pswei
Answered by mr W last updated on 16/Feb/26
Commented by mr W last updated on 18/Feb/26
Commented by mr W last updated on 16/Feb/26
$${when}\:{the}\:{stick}\:{rests}\:{with}\:{one}\:{end} \\ $$$${inside}\:{the}\:{cup},\:{its}\:{gravity}\:{force} \\ $$$${and}\:{the}\:{two}\:{contact}\:{forces}\:{between} \\ $$$${the}\:{stick}\:{and}\:{the}\:{cup}\:{must}\:{be}\:{in} \\ $$$${equilibrium},\:{i}.{e}.\:{their}\:{lines}\:{of}\:{action} \\ $$$${must}\:{intersect}\:{at}\:{a}\:{point},\:{say}\:{S}. \\ $$
Commented by mr W last updated on 16/Feb/26
Commented by mr W last updated on 16/Feb/26
$${we}\:{can}\:{see},\:{when}\:{the}\:{lower}\:{end}\:{of} \\ $$$${the}\:{stick}\:{lies}\:{near}\:{the}\:{bottum}\:{of} \\ $$$${the}\:{cup},\:{this}\:{point}\:{S}\:{lies}\:{above}\:{the} \\ $$$${cup}.\:{the}\:{lowest}\:{position}\:{of}\:{the}\:{lower} \\ $$$${end}\:{of}\:{the}\:{stick}\:{is}\:{there},\:{when}\: \\ $$$${AM}={MB},\:{i}.{e}.\:{C}\:{coincides}\:{with}\:{B}. \\ $$$${we}\:{can}\:{also}\:{see},\:{when}\:{the}\:{lower}\:{end} \\ $$$${of}\:{the}\:{stick}\:{lies}\:{near}\:{the}\:{rim}\:{of}\:{the} \\ $$$${cup},\:{this}\:{point}\:{S}\:{lies}\:{also}\:{above}\:{the} \\ $$$${cup}.\:{when}\:{the}\:{lower}\:{end}\:{reaches}\:{the} \\ $$$${rim}\:{of}\:{the}\:{cup},\:{S},\:{M}\:{and}\:{B}\:{become} \\ $$$${the}\:{same}\:{point}.\:{in}\:{this}\:{case}\: \\ $$$${L}=\mathrm{2}{AB}=\mathrm{2}×\mathrm{2}{R}=\mathrm{4}{R}. \\ $$$${when}\:{the}\:{lower}\:{end}\:{lies}\:{at}\:{some} \\ $$$${position}\:{between},\:{the}\:{point}\:{S}\:{may} \\ $$$${lie}\:{under}\:{the}\:{cup}\:{and}\:{the}\:{stick}\:{will} \\ $$$${get}\:{longer}. \\ $$
Commented by mr W last updated on 17/Feb/26
Commented by mr W last updated on 17/Feb/26
![let η=(H/R), ξ=(x/R) in the given coordinate system, the equation of the parabolic cup is y=H((x/R))^2 =Rηξ^2 A(−ξR, Hξ^2 ) B(R, H) tan θ=−(dy/dx)=−((2H)/R)(−(x/R))=2ηξ tan φ=((H−Hξ^2 )/(R+ξR))=η(1−ξ) AB=((R(1+ξ))/(cos φ))=R(1+ξ)(√(1+η^2 (1−ξ)^2 )) ϕ=θ+φ−(π/2) BS=AB tan ϕ=((AB)/(−tan (θ+φ))) BM=BS tan φ=((AB tan φ)/(−tan (θ+φ))) L=2AM=2(AB+BM) =2AB[1−((tan φ)/(tan (θ+φ)))] =2AB[((tan θ(1+tan^2 φ))/(tan θ+tan φ))] =2R(1+ξ)(√(1+η^2 (1−ξ)^2 ))×((2ηξ[1+η^2 (1−ξ)^2 ])/(2ηξ+η−ηξ)) =4Rξ[1+η^2 (1−ξ)^2 ]^(3/2) L≥AB=R(1+ξ)(√(1+η^2 (1−ξ)^2 )) 4ξ[1+η^2 (1−ξ)^2 ]≥1+ξ ⇒ξ^3 −2ξ^2 +(1+(3/(4η^2 )))ξ−(1/(4η^2 ))≥0 this gives us the closest possible position of the lower end of the stick to the bottum of the cup and thus the smallest length of stick. let ξ_(min) =λ+(2/3) (λ+(2/3))^3 −2(λ+(2/3))^2 +(1+(3/(4η^2 )))(λ+(2/3))−(1/(4η^2 ))=0 λ^3 −3((1/9)−(1/(4η^2 )))λ+2((1/(27))+(1/(8η^2 )))=0 λ=(1/2)[((((√((1/η^2 )−(1/3)+((32η^2 )/(27))))−1)(1/η^2 )−(8/(27))))^(1/3) −((((√((1/η^2 )−(1/3)+((32η^2 )/(27))))+1)(1/η^2 )+(8/(27))))^(1/3) ] ⇒ξ_(min) =(2/3)+(1/2)[((((√((1/η^2 )−(1/3)+((32η^2 )/(27))))−1)(1/η^2 )−(8/(27))))^(1/3) −((((√((1/η^2 )−(1/3)+((32η^2 )/(27))))+1)(1/η^2 )+(8/(27))))^(1/3) ] L_(min) =4Rξ_(min) [1+η^2 (1−ξ_(min) )^2 ]^(3/2) for ξ=1 we have L=4R. let =(L/(4R))=ξ[1+η^2 (1−ξ)^2 ]^(3/2) for maximum of L: (d /dξ)=[1+η^2 (1−ξ)^2 ]^(3/2) −3ξ[1+η^2 (1−ξ)^2 ]^(1/2) η^2 (1−ξ)=0 4ξ^2 −5ξ+1+(1/η^2 )=0 ⇒ξ=(1/8)(5±(√(9−((16)/η^2 )))) (+ for local L_(min) and − for local L_(max) ) such that L_(max) exists, η≥(4/3). L_(max) =(R/2)(5−(√(9−((16)/η^2 ))))[1+(η^2 /(64))(3+(√(9−((16)/η^2 ))))^2 ]^(3/2) such that L_(max) ≥4R (1/2)(5−(√(9−((16)/η^2 ))))[1+(η^2 /(64))(3+(√(9−((16)/η^2 ))))^2 ]^(3/2) ≥4 (5−(√(9−((16)/η^2 ))))[1+(η^2 /(64))(3+(√(9−((16)/η^2 ))))^2 ]^(3/2) ≥8 ⇒η≥1.5242375972 summary: if η≥1.5242375972, the largest length is L_(max) =(R/2)(5−(√(9−((16)/η^2 ))))[1+(η^2 /(64))(3+(√(9−((16)/η^2 ))))^2 ]^(3/2) otherwise L_(max) =4R example: R=1, H=2 η=2 L_(max) =((15(√(30+6(√5))))/(16))≈6.1773 ξ_(min) ≈0.0581, L_(min) ≈2.2567](Q227755.png)
$${let}\:\eta=\frac{{H}}{{R}},\:\xi=\frac{{x}}{{R}} \\ $$$${in}\:{the}\:{given}\:{coordinate}\:{system}, \\ $$$${the}\:{equation}\:{of}\:{the}\:{parabolic}\:{cup}\:{is} \\ $$$${y}={H}\left(\frac{{x}}{{R}}\right)^{\mathrm{2}} ={R}\eta\xi^{\mathrm{2}} \\ $$$${A}\left(−\xi{R},\:{H}\xi^{\mathrm{2}} \right) \\ $$$${B}\left({R},\:{H}\right) \\ $$$$\mathrm{tan}\:\theta=−\frac{{dy}}{{dx}}=−\frac{\mathrm{2}{H}}{{R}}\left(−\frac{{x}}{{R}}\right)=\mathrm{2}\eta\xi \\ $$$$\mathrm{tan}\:\phi=\frac{{H}−{H}\xi^{\mathrm{2}} }{{R}+\xi{R}}=\eta\left(\mathrm{1}−\xi\right) \\ $$$${AB}=\frac{{R}\left(\mathrm{1}+\xi\right)}{\mathrm{cos}\:\phi}={R}\left(\mathrm{1}+\xi\right)\sqrt{\mathrm{1}+\eta^{\mathrm{2}} \left(\mathrm{1}−\xi\right)^{\mathrm{2}} } \\ $$$$\varphi=\theta+\phi−\frac{\pi}{\mathrm{2}} \\ $$$${BS}={AB}\:\mathrm{tan}\:\varphi=\frac{{AB}}{−\mathrm{tan}\:\left(\theta+\phi\right)} \\ $$$${BM}={BS}\:\mathrm{tan}\:\phi=\frac{{AB}\:\mathrm{tan}\:\phi}{−\mathrm{tan}\:\left(\theta+\phi\right)} \\ $$$${L}=\mathrm{2}{AM}=\mathrm{2}\left({AB}+{BM}\right) \\ $$$$\:\:\:=\mathrm{2}{AB}\left[\mathrm{1}−\frac{\mathrm{tan}\:\phi}{\mathrm{tan}\:\left(\theta+\phi\right)}\right] \\ $$$$\:\:\:=\mathrm{2}{AB}\left[\frac{\mathrm{tan}\:\theta\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi\right)}{\mathrm{tan}\:\theta+\mathrm{tan}\:\phi}\right] \\ $$$$\:\:\:=\mathrm{2}{R}\left(\mathrm{1}+\xi\right)\sqrt{\mathrm{1}+\eta^{\mathrm{2}} \left(\mathrm{1}−\xi\right)^{\mathrm{2}} }×\frac{\mathrm{2}\eta\xi\left[\mathrm{1}+\eta^{\mathrm{2}} \left(\mathrm{1}−\xi\right)^{\mathrm{2}} \right]}{\mathrm{2}\eta\xi+\eta−\eta\xi} \\ $$$$\:\:\:=\mathrm{4}{R}\xi\left[\mathrm{1}+\eta^{\mathrm{2}} \left(\mathrm{1}−\xi\right)^{\mathrm{2}} \right]^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$ \\ $$$${L}\geqslant{AB}={R}\left(\mathrm{1}+\xi\right)\sqrt{\mathrm{1}+\eta^{\mathrm{2}} \left(\mathrm{1}−\xi\right)^{\mathrm{2}} } \\ $$$$\mathrm{4}\xi\left[\mathrm{1}+\eta^{\mathrm{2}} \left(\mathrm{1}−\xi\right)^{\mathrm{2}} \right]\geqslant\mathrm{1}+\xi \\ $$$$\Rightarrow\xi^{\mathrm{3}} −\mathrm{2}\xi^{\mathrm{2}} +\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{4}\eta^{\mathrm{2}} }\right)\xi−\frac{\mathrm{1}}{\mathrm{4}\eta^{\mathrm{2}} }\geqslant\mathrm{0} \\ $$$${this}\:{gives}\:{us}\:{the}\:{closest}\:{possible} \\ $$$${position}\:{of}\:{the}\:{lower}\:{end}\:{of}\:{the}\:{stick} \\ $$$${to}\:{the}\:{bottum}\:{of}\:{the}\:{cup}\:{and}\:{thus}\:{the} \\ $$$${smallest}\:{length}\:{of}\:{stick}. \\ $$$${let}\:\xi_{{min}} =\lambda+\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\left(\lambda+\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} −\mathrm{2}\left(\lambda+\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} +\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{4}\eta^{\mathrm{2}} }\right)\left(\lambda+\frac{\mathrm{2}}{\mathrm{3}}\right)−\frac{\mathrm{1}}{\mathrm{4}\eta^{\mathrm{2}} }=\mathrm{0} \\ $$$$\lambda^{\mathrm{3}} −\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{9}}−\frac{\mathrm{1}}{\mathrm{4}\eta^{\mathrm{2}} }\right)\lambda+\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{27}}+\frac{\mathrm{1}}{\mathrm{8}\eta^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$\lambda=\frac{\mathrm{1}}{\mathrm{2}}\left[\sqrt[{\mathrm{3}}]{\left(\sqrt{\frac{\mathrm{1}}{\eta^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{32}\eta^{\mathrm{2}} }{\mathrm{27}}}−\mathrm{1}\right)\frac{\mathrm{1}}{\eta^{\mathrm{2}} }−\frac{\mathrm{8}}{\mathrm{27}}}−\sqrt[{\mathrm{3}}]{\left(\sqrt{\frac{\mathrm{1}}{\eta^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{32}\eta^{\mathrm{2}} }{\mathrm{27}}}+\mathrm{1}\right)\frac{\mathrm{1}}{\eta^{\mathrm{2}} }+\frac{\mathrm{8}}{\mathrm{27}}}\right] \\ $$$$\Rightarrow\xi_{{min}} =\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}\left[\sqrt[{\mathrm{3}}]{\left(\sqrt{\frac{\mathrm{1}}{\eta^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{32}\eta^{\mathrm{2}} }{\mathrm{27}}}−\mathrm{1}\right)\frac{\mathrm{1}}{\eta^{\mathrm{2}} }−\frac{\mathrm{8}}{\mathrm{27}}}−\sqrt[{\mathrm{3}}]{\left(\sqrt{\frac{\mathrm{1}}{\eta^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{32}\eta^{\mathrm{2}} }{\mathrm{27}}}+\mathrm{1}\right)\frac{\mathrm{1}}{\eta^{\mathrm{2}} }+\frac{\mathrm{8}}{\mathrm{27}}}\right] \\ $$$${L}_{{min}} =\mathrm{4}{R}\xi_{{min}} \left[\mathrm{1}+\eta^{\mathrm{2}} \left(\mathrm{1}−\xi_{{min}} \right)^{\mathrm{2}} \right]^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$ \\ $$$${for}\:\xi=\mathrm{1}\:{we}\:{have}\:{L}=\mathrm{4}{R}. \\ $$$$ \\ $$$${let}\:\:=\frac{{L}}{\mathrm{4}{R}}=\xi\left[\mathrm{1}+\eta^{\mathrm{2}} \left(\mathrm{1}−\xi\right)^{\mathrm{2}} \right]^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$${for}\:{maximum}\:{of}\:{L}: \\ $$$$\frac{{d}\:}{{d}\xi}=\left[\mathrm{1}+\eta^{\mathrm{2}} \left(\mathrm{1}−\xi\right)^{\mathrm{2}} \right]^{\frac{\mathrm{3}}{\mathrm{2}}} −\mathrm{3}\xi\left[\mathrm{1}+\eta^{\mathrm{2}} \left(\mathrm{1}−\xi\right)^{\mathrm{2}} \right]^{\frac{\mathrm{1}}{\mathrm{2}}} \eta^{\mathrm{2}} \left(\mathrm{1}−\xi\right)=\mathrm{0} \\ $$$$\mathrm{4}\xi^{\mathrm{2}} −\mathrm{5}\xi+\mathrm{1}+\frac{\mathrm{1}}{\eta^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow\xi=\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{5}\pm\sqrt{\mathrm{9}−\frac{\mathrm{16}}{\eta^{\mathrm{2}} }}\right) \\ $$$$\left(+\:{for}\:{local}\:{L}_{{min}} \:{and}\:−\:{for}\:{local}\:{L}_{{max}} \right) \\ $$$${such}\:{that}\:{L}_{{max}} \:{exists},\:\eta\geqslant\frac{\mathrm{4}}{\mathrm{3}}. \\ $$$${L}_{{max}} =\frac{{R}}{\mathrm{2}}\left(\mathrm{5}−\sqrt{\mathrm{9}−\frac{\mathrm{16}}{\eta^{\mathrm{2}} }}\right)\left[\mathrm{1}+\frac{\eta^{\mathrm{2}} }{\mathrm{64}}\left(\mathrm{3}+\sqrt{\mathrm{9}−\frac{\mathrm{16}}{\eta^{\mathrm{2}} }}\right)^{\mathrm{2}} \right]^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$ \\ $$$${such}\:{that}\:{L}_{{max}} \geqslant\mathrm{4}{R} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{5}−\sqrt{\mathrm{9}−\frac{\mathrm{16}}{\eta^{\mathrm{2}} }}\right)\left[\mathrm{1}+\frac{\eta^{\mathrm{2}} }{\mathrm{64}}\left(\mathrm{3}+\sqrt{\mathrm{9}−\frac{\mathrm{16}}{\eta^{\mathrm{2}} }}\right)^{\mathrm{2}} \right]^{\frac{\mathrm{3}}{\mathrm{2}}} \geqslant\mathrm{4} \\ $$$$\left(\mathrm{5}−\sqrt{\mathrm{9}−\frac{\mathrm{16}}{\eta^{\mathrm{2}} }}\right)\left[\mathrm{1}+\frac{\eta^{\mathrm{2}} }{\mathrm{64}}\left(\mathrm{3}+\sqrt{\mathrm{9}−\frac{\mathrm{16}}{\eta^{\mathrm{2}} }}\right)^{\mathrm{2}} \right]^{\frac{\mathrm{3}}{\mathrm{2}}} \geqslant\mathrm{8} \\ $$$$\Rightarrow\eta\geqslant\mathrm{1}.\mathrm{5242375972} \\ $$$$ \\ $$$$\underline{{summary}:} \\ $$$${if}\:\eta\geqslant\mathrm{1}.\mathrm{5242375972},\:{the}\:{largest}\:{length}\:{is} \\ $$$${L}_{{max}} =\frac{{R}}{\mathrm{2}}\left(\mathrm{5}−\sqrt{\mathrm{9}−\frac{\mathrm{16}}{\eta^{\mathrm{2}} }}\right)\left[\mathrm{1}+\frac{\eta^{\mathrm{2}} }{\mathrm{64}}\left(\mathrm{3}+\sqrt{\mathrm{9}−\frac{\mathrm{16}}{\eta^{\mathrm{2}} }}\right)^{\mathrm{2}} \right]^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$${otherwise} \\ $$$${L}_{{max}} =\mathrm{4}{R} \\ $$$$ \\ $$$${example}:\:{R}=\mathrm{1},\:{H}=\mathrm{2} \\ $$$$\eta=\mathrm{2} \\ $$$${L}_{{max}} =\frac{\mathrm{15}\sqrt{\mathrm{30}+\mathrm{6}\sqrt{\mathrm{5}}}}{\mathrm{16}}\approx\mathrm{6}.\mathrm{1773} \\ $$$$\xi_{{min}} \approx\mathrm{0}.\mathrm{0581},\:{L}_{{min}} \approx\mathrm{2}.\mathrm{2567} \\ $$