Question Number 61994 by ajfour last updated on 13/Jun/19
Commented by ajfour last updated on 13/Jun/19
$${Find}\:{the}\:\boldsymbol{{r}}\:{of}\:{largest}\:{size}\:{sphere}\: \\ $$$${that}\:{can}\:{rest}\:{against}\:{the}\:{rim}\:{of} \\ $$$${cylinder}\:{on}\:{one}\:{side}\:{and}\:{the} \\ $$$${stick}\:{on}\:{the}\:{other}\:{side}. \\ $$
Answered by mr W last updated on 13/Jun/19
$$\frac{{r}+{h}}{{r}}=\frac{\sqrt{\left(\mathrm{2}{R}\right)^{\mathrm{2}} +{h}^{\mathrm{2}} }}{\mathrm{2}{R}} \\ $$$$\Rightarrow{r}_{{max}} =\frac{{h}}{\:\sqrt{\mathrm{1}+\left(\frac{{h}}{\mathrm{2}{R}}\right)^{\mathrm{2}} }−\mathrm{1}} \\ $$
Commented by ajfour last updated on 13/Jun/19
$${not}\:{surprising},\:{seems}\:{natural}\:{Sir}. \\ $$$${I}\:{just}\:{viewed}\:{your}\:{conclusion}\:{sir}. \\ $$
Commented by mr W last updated on 13/Jun/19
Commented by mr W last updated on 13/Jun/19
$${this}\:{is}\:{maybe}\:{not}\:{the}\:{largest}\:{possible} \\ $$$${sphere},\:{since}\:{the}\:{stick}\:\left({blue}\:{line}\right)\:{can} \\ $$$${also}\:{be}\:{in}\:{other}\:{positions}.\:{it}\:{only}\: \\ $$$${needs}\:{to}\:{touch}\:{the}\:{sphere}. \\ $$
Commented by mr W last updated on 13/Jun/19
Commented by ajfour last updated on 13/Jun/19
$${O}'\:{yes},\:{understood}\:{sir},\:{sphere} \\ $$$${can}\:{be}\:{balanced}\:{at}\:{one}\:{point}\:{even}. \\ $$$${Thanks}. \\ $$
Commented by mr W last updated on 13/Jun/19
![position of stick is θ. it touches the sphere at point P. C(−R, 0, h+r) A(−R cos θ, −R sin θ, 0) B(R cos θ, R sin θ, h) eqn. of line AB (stick): ((x+R cos θ)/(2R cos θ))=((y+R sin θ)/(2R sin θ))=(z/h)=k ⇒x=R cos θ(2k−1) ⇒y=R sin θ(2k−1) ⇒z=hk distance from point C to line AB=d: D=d^2 =[R cos θ (2k−1)+R]^2 +[R sin θ (2k−1))]^2 +[hk−h−r]^2 =2R^2 [2k^2 −(2k−1)(1−cos θ)]+(hk−h−r)^2 =4R^2 [k^2 −(2k−1) sin^2 (θ/2)]+(hk−h−r)^2 (dD/dk)=0 4R^2 (k−sin^2 (θ/2))+(hk−h−r)h=0 ⇒k=(((h+r)h+4R^2 sin^2 (θ/2))/(4R^2 +h^2 ))=((1+4((R/h))^2 sin^2 (θ/2)+(r/h))/(1+4((R/h))^2 )) ((r/h))^2 =4((R/h))^2 [k^2 −(2k−1) sin^2 (θ/2)]+(k−1−(r/h))^2 let ρ=(R/h), λ=(r/h) ⇒k=1+((λ−4ρ^2 cos^2 (θ/2))/(1+4ρ^2 )) 0=4ρ^2 [((λ/(1+4ρ^2 ))−((4ρ^2 cos^2 (θ/2))/(1+4ρ^2 )))^2 −2((λ/(1+4ρ^2 ))−((4ρ^2 cos^2 (θ/2))/(1+4ρ^2 ))) cos^2 (θ/2)−cos^2 (θ/2)]+((λ/(1+4ρ^2 ))−((4ρ^2 cos^2 (θ/2))/(1+4ρ^2 ))−2λ)((λ/(1+4ρ^2 ))−((4ρ^2 cos^2 (θ/2))/(1+4ρ^2 ))) ⇒max. λ is when θ=0](https://www.tinkutara.com/question/Q62009.png)
$${position}\:{of}\:{stick}\:{is}\:\theta.\:{it}\:{touches}\:{the} \\ $$$${sphere}\:{at}\:{point}\:{P}. \\ $$$${C}\left(−{R},\:\mathrm{0},\:{h}+{r}\right) \\ $$$${A}\left(−{R}\:\mathrm{cos}\:\theta,\:−{R}\:\mathrm{sin}\:\theta,\:\mathrm{0}\right) \\ $$$${B}\left({R}\:\mathrm{cos}\:\theta,\:{R}\:\mathrm{sin}\:\theta,\:{h}\right) \\ $$$${eqn}.\:{of}\:{line}\:{AB}\:\left({stick}\right): \\ $$$$\frac{{x}+{R}\:\mathrm{cos}\:\theta}{\mathrm{2}{R}\:\mathrm{cos}\:\theta}=\frac{{y}+{R}\:\mathrm{sin}\:\theta}{\mathrm{2}{R}\:\mathrm{sin}\:\theta}=\frac{{z}}{{h}}={k} \\ $$$$\Rightarrow{x}={R}\:\mathrm{cos}\:\theta\left(\mathrm{2}{k}−\mathrm{1}\right) \\ $$$$\Rightarrow{y}={R}\:\mathrm{sin}\:\theta\left(\mathrm{2}{k}−\mathrm{1}\right) \\ $$$$\Rightarrow{z}={hk} \\ $$$${distance}\:{from}\:{point}\:{C}\:{to}\:{line}\:{AB}={d}: \\ $$$$\left.{D}={d}^{\mathrm{2}} =\left[{R}\:\mathrm{cos}\:\theta\:\left(\mathrm{2}{k}−\mathrm{1}\right)+{R}\right]^{\mathrm{2}} +\left[{R}\:\mathrm{sin}\:\theta\:\left(\mathrm{2}{k}−\mathrm{1}\right)\right)\right]^{\mathrm{2}} +\left[{hk}−{h}−{r}\right]^{\mathrm{2}} \\ $$$$=\mathrm{2}{R}^{\mathrm{2}} \left[\mathrm{2}{k}^{\mathrm{2}} −\left(\mathrm{2}{k}−\mathrm{1}\right)\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\right]+\left({hk}−{h}−{r}\right)^{\mathrm{2}} \\ $$$$=\mathrm{4}{R}^{\mathrm{2}} \left[{k}^{\mathrm{2}} −\left(\mathrm{2}{k}−\mathrm{1}\right)\:\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}\right]+\left({hk}−{h}−{r}\right)^{\mathrm{2}} \\ $$$$\frac{{dD}}{{dk}}=\mathrm{0} \\ $$$$\mathrm{4}{R}^{\mathrm{2}} \left({k}−\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}\right)+\left({hk}−{h}−{r}\right){h}=\mathrm{0} \\ $$$$\Rightarrow{k}=\frac{\left({h}+{r}\right){h}+\mathrm{4}{R}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}{\mathrm{4}{R}^{\mathrm{2}} +{h}^{\mathrm{2}} }=\frac{\mathrm{1}+\mathrm{4}\left(\frac{{R}}{{h}}\right)^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}+\frac{{r}}{{h}}}{\mathrm{1}+\mathrm{4}\left(\frac{{R}}{{h}}\right)^{\mathrm{2}} } \\ $$$$\left(\frac{{r}}{{h}}\right)^{\mathrm{2}} =\mathrm{4}\left(\frac{{R}}{{h}}\right)^{\mathrm{2}} \left[{k}^{\mathrm{2}} −\left(\mathrm{2}{k}−\mathrm{1}\right)\:\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}\right]+\left({k}−\mathrm{1}−\frac{{r}}{{h}}\right)^{\mathrm{2}} \\ $$$${let}\:\rho=\frac{{R}}{{h}},\:\lambda=\frac{{r}}{{h}} \\ $$$$\Rightarrow{k}=\mathrm{1}+\frac{\lambda−\mathrm{4}\rho^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}{\mathrm{1}+\mathrm{4}\rho^{\mathrm{2}} } \\ $$$$\mathrm{0}=\mathrm{4}\rho^{\mathrm{2}} \left[\left(\frac{\lambda}{\mathrm{1}+\mathrm{4}\rho^{\mathrm{2}} }−\frac{\mathrm{4}\rho^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}{\mathrm{1}+\mathrm{4}\rho^{\mathrm{2}} }\right)^{\mathrm{2}} −\mathrm{2}\left(\frac{\lambda}{\mathrm{1}+\mathrm{4}\rho^{\mathrm{2}} }−\frac{\mathrm{4}\rho^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}{\mathrm{1}+\mathrm{4}\rho^{\mathrm{2}} }\right)\:\mathrm{cos}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}−\mathrm{cos}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}\right]+\left(\frac{\lambda}{\mathrm{1}+\mathrm{4}\rho^{\mathrm{2}} }−\frac{\mathrm{4}\rho^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}{\mathrm{1}+\mathrm{4}\rho^{\mathrm{2}} }−\mathrm{2}\lambda\right)\left(\frac{\lambda}{\mathrm{1}+\mathrm{4}\rho^{\mathrm{2}} }−\frac{\mathrm{4}\rho^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}{\mathrm{1}+\mathrm{4}\rho^{\mathrm{2}} }\right) \\ $$$$\Rightarrow{max}.\:\lambda\:{is}\:{when}\:\theta=\mathrm{0} \\ $$
Commented by mr W last updated on 13/Jun/19
Commented by mr W last updated on 13/Jun/19
$${yes}\:{sir}! \\ $$$${optimum}\:{in}\:{the}\:{nature}\:{is}\:{always}\:{in} \\ $$$${symmetry}. \\ $$