Question Number 135818 by mr W last updated on 16/Mar/21
Commented by mr W last updated on 16/Mar/21
$${unsolved}\:{old}\:{question} \\ $$$${see}\:{Q}\mathrm{74557} \\ $$
Answered by mr W last updated on 16/Mar/21
Commented by mr W last updated on 17/Mar/21
Commented by mr W last updated on 21/Mar/21
$${length}\:{of}\:{string}\:={L} \\ $$$${radius}\:{of}\:{hemisphere}\:={R} \\ $$$${let}\:\lambda=\frac{{L}}{{R}} \\ $$$$\mathrm{cos}\:\phi=\frac{{L}}{\mathrm{2}{R}}=\frac{\lambda}{\mathrm{2}} \\ $$$$\Rightarrow\phi=\mathrm{cos}^{−\mathrm{1}} \frac{\lambda}{\mathrm{2}} \\ $$$${when}\:{the}\:{mass}\:{is}\:{released}\:{at}\:{point}\:{A}, \\ $$$${it}\:{tries}\:{to}\:{move}\:{directly}\:{to}\:{point}\:{C}. \\ $$$${since}\:{OC}=\sqrt{\mathrm{2}}{R}>{L},\:{tension}\:{is}\:{the} \\ $$$${string}. \\ $$$${the}\:{position}\:{P}\:{of}\:{the}\:{mass}\:{is}\:{described} \\ $$$${by}\:\theta. \\ $$$${let}\:{OP}\:'=\rho \\ $$$${x}_{{P}} =\rho\:\mathrm{cos}\:\theta \\ $$$${y}_{{P}} =\rho\:\mathrm{sin}\:\theta \\ $$$${we}\:{have} \\ $$$${OP}={L}: \\ $$$${x}_{{P}} ^{\mathrm{2}} +{y}_{{P}} ^{\mathrm{2}} +{z}_{{P}} ^{\mathrm{2}} ={L}^{\mathrm{2}} \\ $$$$\Rightarrow\rho^{\mathrm{2}} +{z}_{{P}} ^{\mathrm{2}} ={L}^{\mathrm{2}} \\ $$$${P}\:\:{lies}\:{on}\:{the}\:{hemisphere}: \\ $$$$\left({x}_{{P}} −{R}\right)^{\mathrm{2}} +{y}_{{P}} ^{\mathrm{2}} +{z}_{{P}} ^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$${x}_{{P}} ^{\mathrm{2}} −\mathrm{2}{Rx}_{{P}} +{y}_{{P}} ^{\mathrm{2}} +{z}_{{P}} ^{\mathrm{2}} =\mathrm{0} \\ $$$${L}^{\mathrm{2}} −\mathrm{2}{Rx}_{{P}} =\mathrm{0} \\ $$$$\Rightarrow{x}_{{P}} =\frac{{L}^{\mathrm{2}} }{\mathrm{2}{R}}=\frac{\lambda{L}}{\mathrm{2}}={constant} \\ $$$${i}.{e}.\:{the}\:{locus}\:{of}\:{the}\:{mass}\:{is}\:{a}\:{vertical} \\ $$$${semicircle}\:{with}\:{radius}\:{r},\: \\ $$$${with}\:{r}={L}\sqrt{\mathrm{1}−\frac{\lambda^{\mathrm{2}} }{\mathrm{4}}}=\frac{{L}\sqrt{\mathrm{4}−\lambda^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\Rightarrow\rho=\frac{{x}_{{P}} }{\mathrm{cos}\:\theta}=\frac{\lambda{L}}{\mathrm{2}\:\mathrm{cos}\:\theta} \\ $$$${z}_{{P}} ^{\mathrm{2}} ={L}^{\mathrm{2}} −\rho^{\mathrm{2}} ={L}^{\mathrm{2}} −\frac{\lambda^{\mathrm{2}} {L}^{\mathrm{2}} }{\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$$\Rightarrow{z}_{{P}} =−{L}\sqrt{\mathrm{1}−\frac{\lambda^{\mathrm{2}} }{\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta}}=−\frac{{L}\sqrt{\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\lambda^{\mathrm{2}} }}{\mathrm{2}\:\mathrm{cos}\:\theta} \\ $$$$\Rightarrow{y}_{{P}} =\frac{\lambda{L}}{\mathrm{2}\:\mathrm{cos}\:\theta}×\mathrm{sin}\:\theta=\frac{\lambda{L}\:\mathrm{tan}\:\theta}{\mathrm{2}} \\ $$$${let}\:\omega=\frac{{d}\theta}{{dt}} \\ $$$${v}_{{x}} =\frac{{dx}_{{P}} }{{dt}}=\frac{{d}\theta}{{dt}}×\frac{{dx}_{{P}} }{{d}\theta}=\mathrm{0} \\ $$$${v}_{{y}} =\frac{{dy}_{{P}} }{{dt}}=\frac{{d}\theta}{{dt}}×\frac{{dy}_{{P}} }{{d}\theta}=\frac{\lambda{L}\omega}{\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$${v}_{{z}} =\frac{{dz}_{{P}} }{{dt}}=\frac{{d}\theta}{{dt}}×\frac{{dz}_{{P}} }{{d}\theta}=\frac{\lambda^{\mathrm{2}} {L}\omega\:\mathrm{sin}\:\theta}{\:\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\theta\sqrt{\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\lambda^{\mathrm{2}} }} \\ $$$${v}_{{P}} ^{\mathrm{2}} ={v}_{{x}} ^{\mathrm{2}} +{v}_{{y}} ^{\mathrm{2}} +{v}_{{z}} ^{\mathrm{2}} =\frac{\lambda^{\mathrm{2}} \left(\mathrm{4}−\lambda^{\mathrm{2}} \right){L}^{\mathrm{2}} \omega^{\mathrm{2}} }{\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta\:\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\lambda^{\mathrm{2}} \right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{mv}_{{P}} ^{\mathrm{2}} =−{mgz}_{{P}} \\ $$$${v}_{{P}} ^{\mathrm{2}} =\frac{{gL}\sqrt{\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\lambda^{\mathrm{2}} }}{\mathrm{cos}\:\theta} \\ $$$$\frac{\lambda^{\mathrm{2}} \left(\mathrm{4}−\lambda^{\mathrm{2}} \right){L}^{\mathrm{2}} \omega^{\mathrm{2}} }{\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta\:\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\lambda^{\mathrm{2}} \right)}=\frac{{gL}}{\mathrm{cos}\:\theta}\sqrt{\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\lambda^{\mathrm{2}} } \\ $$$$\omega^{\mathrm{2}} =\frac{\mathrm{4}{g}\:\mathrm{cos}\:\theta\:\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\lambda^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\lambda^{\mathrm{2}} \left(\mathrm{4}−\lambda^{\mathrm{2}} \right){L}} \\ $$$$\Rightarrow\omega=\sqrt{\frac{\mathrm{4}{g}}{\lambda^{\mathrm{2}} \left(\mathrm{4}−\lambda^{\mathrm{2}} \right){L}}}×\:\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\lambda^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{4}}} \sqrt{\mathrm{cos}\:\theta} \\ $$$$\Rightarrow\frac{{d}\theta}{{dt}}=\sqrt{\frac{\mathrm{4}{g}}{\lambda^{\mathrm{2}} \left(\mathrm{4}−\lambda^{\mathrm{2}} \right){L}}}×\:\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\lambda^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{4}}} \sqrt{\mathrm{cos}\:\theta} \\ $$$$\Rightarrow\sqrt{\frac{\lambda^{\mathrm{2}} \left(\mathrm{4}−\lambda^{\mathrm{2}} \right){L}}{\mathrm{4}{g}}}×\frac{{d}\theta}{\:\:\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\lambda^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{4}}} \sqrt{\mathrm{cos}\:\theta}}={dt} \\ $$$$\Rightarrow\sqrt{\frac{\lambda^{\mathrm{2}} \left(\mathrm{4}−\lambda^{\mathrm{2}} \right){L}}{\mathrm{4}{g}}}×\int_{\mathrm{0}} ^{\phi} \frac{{d}\theta}{\:\:\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\lambda^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{4}}} \sqrt{\mathrm{cos}\:\theta}}=\frac{{T}}{\mathrm{2}} \\ $$$$\Rightarrow{T}=\sqrt{\frac{\lambda^{\mathrm{2}} \left(\mathrm{4}−\lambda^{\mathrm{2}} \right){L}}{{g}}}\:\int_{\mathrm{0}} ^{\mathrm{cos}^{−\mathrm{1}} \frac{\lambda}{\mathrm{2}}} \frac{{d}\theta}{\:\:\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\lambda^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{4}}} \sqrt{\mathrm{cos}\:\theta}} \\ $$$${or} \\ $$$${T}=\xi\sqrt{\frac{{L}}{{g}}} \\ $$$${with}\:\xi=\lambda\sqrt{\mathrm{4}−\lambda^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{\mathrm{cos}^{−\mathrm{1}} \frac{\lambda}{\mathrm{2}}} \frac{{d}\theta}{\:\:\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\lambda^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{4}}} \sqrt{\mathrm{cos}\:\theta}} \\ $$$$\left({this}\:{integral}\:{is}\:{divergent}\:{for}\:\lambda<\sqrt{\mathrm{2}}\:!\right) \\ $$$$ \\ $$$$\boldsymbol{{example}}:\:{L}=\sqrt{\mathrm{2}}{R},\:\lambda=\sqrt{\mathrm{2}} \\ $$$$\xi=\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{d}\theta}{\:\:\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{2}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \sqrt{\mathrm{cos}\:\theta}}\approx\mathrm{3}.\mathrm{118169} \\ $$$${T}\approx\mathrm{3}.\mathrm{118169}\sqrt{\frac{{L}}{{g}}} \\ $$$$==================== \\ $$$${T}=\sqrt{\frac{\lambda^{\mathrm{2}} \left(\mathrm{4}−\lambda^{\mathrm{2}} \right){L}}{{g}}}\:\int_{\mathrm{0}} ^{\mathrm{cos}^{−\mathrm{1}} \frac{\lambda}{\mathrm{2}}} \frac{{d}\theta}{\:\:\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\lambda^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{4}}} \sqrt{\mathrm{cos}\:\theta}} \\ $$$${L}=\frac{\mathrm{2}{r}}{\:\sqrt{\mathrm{4}−\lambda^{\mathrm{2}} }} \\ $$$${T}=\lambda\sqrt[{\mathrm{4}}]{\mathrm{4}−\lambda^{\mathrm{2}} }\:\sqrt{\frac{\mathrm{2}{r}}{{g}}}\int_{\mathrm{0}} ^{\mathrm{cos}^{−\mathrm{1}} \frac{\lambda}{\mathrm{2}}} \frac{{d}\theta}{\:\:\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\lambda^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{4}}} \sqrt{\mathrm{cos}\:\theta}} \\ $$$${x}_{{P}} \mathrm{tan}\:\theta={r}\:\mathrm{sin}\:\varphi \\ $$$$\mathrm{tan}\:\theta=\frac{\sqrt{\mathrm{4}−\lambda^{\mathrm{2}} }}{\lambda}\:\:\mathrm{sin}\:\varphi \\ $$$$\Rightarrow\mathrm{4}−\frac{\lambda^{\mathrm{2}} }{\mathrm{cos}^{\mathrm{2}} \:\theta}=\left(\mathrm{4}−\lambda^{\mathrm{2}} \right)\mathrm{cos}^{\mathrm{2}} \:\varphi \\ $$$$\Rightarrow\frac{{d}\theta}{\mathrm{cos}^{\mathrm{2}} \:\theta}=\frac{\sqrt{\mathrm{4}−\lambda^{\mathrm{2}} }\:\mathrm{cos}\:\varphi\:{d}\varphi}{\lambda} \\ $$$$ \\ $$$${T}=\eta\sqrt{\frac{\mathrm{2}{r}}{\:{g}}} \\ $$$$\eta=\lambda\sqrt[{\mathrm{4}}]{\mathrm{4}−\lambda^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{\mathrm{cos}^{−\mathrm{1}} \frac{\lambda}{\mathrm{2}}} \frac{{d}\theta}{\:\:\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\lambda^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{4}}} \sqrt{\mathrm{cos}\:\theta}} \\ $$$$\eta=\left(\mathrm{4}−\lambda^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{4}}} \:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{cos}\:\varphi\:{d}\varphi}{\:\:\left(\mathrm{4}−\frac{\lambda^{\mathrm{2}} }{\mathrm{cos}^{\mathrm{2}} \:\theta}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} } \\ $$$$\eta=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\varphi}{\:\:\sqrt{\mathrm{cos}\:\varphi}}\:\checkmark \\ $$
Commented by mr W last updated on 17/Mar/21
Commented by mr W last updated on 17/Mar/21
$$\omega=\frac{{d}\varphi}{{dt}} \\ $$$${v}={r}\omega \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} ={mgr}\:\mathrm{cos}\:\varphi \\ $$$${v}^{\mathrm{2}} =\mathrm{2}{gr}\:\mathrm{cos}\:\varphi \\ $$$$\omega^{\mathrm{2}} =\frac{\mathrm{2}{g}}{{r}}\:\mathrm{cos}\:\varphi \\ $$$$\omega=\frac{{d}\varphi}{{dt}}=\sqrt{\frac{\mathrm{2}{g}}{{r}}}×\sqrt{\mathrm{cos}\:\varphi} \\ $$$$\sqrt{\frac{{r}}{\mathrm{2}{g}}}×\frac{{d}\varphi}{\:\sqrt{\mathrm{cos}\:\varphi}}={dt} \\ $$$$\sqrt{\frac{{r}}{\mathrm{2}{g}}}×\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\varphi}{\:\sqrt{\mathrm{cos}\:\varphi}}=\frac{{T}}{\mathrm{2}} \\ $$$$\Rightarrow{T}=\sqrt{\frac{\mathrm{2}{r}}{{g}}}×\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\varphi}{\:\sqrt{\mathrm{cos}\:\varphi}} \\ $$$$\:\:\:\:\:\:\:\:=\frac{{B}\left(\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{1}}{\mathrm{2}}\right)}{\:\sqrt{\mathrm{2}}}\sqrt{\frac{{r}}{{g}}} \\ $$$$\:\:\:\:\:\:\:\:\approx\mathrm{3}.\mathrm{708149}\sqrt{\frac{{r}}{{g}}} \\ $$
Commented by mr W last updated on 17/Mar/21
$${applying}\:{this}\:{for}\:{r}={L}\sqrt{\mathrm{1}−\frac{\lambda^{\mathrm{2}} }{\mathrm{4}}}\:{we}\:{get} \\ $$$${T}\approx\mathrm{3}.\mathrm{708149}\left(\mathrm{1}−\frac{\lambda^{\mathrm{2}} }{\mathrm{4}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \sqrt{\frac{{L}}{{g}}} \\ $$$${with}\:\lambda=\frac{{L}}{{R}} \\ $$$${examples}: \\ $$$$\lambda=\sqrt{\mathrm{2}}:\:\:\:{T}\approx\mathrm{3}.\mathrm{118169}\:\checkmark \\ $$$$\lambda=\mathrm{1}:\:\:\:\:\:\:{T}\approx\mathrm{3}.\mathrm{450821} \\ $$$$\lambda=\frac{\mathrm{1}}{\mathrm{2}}:\:\:\:\:{T}\approx\mathrm{3}.\mathrm{648799} \\ $$
Commented by ajfour last updated on 17/Mar/21
$${Thank}\:{you},\:{Sir},\:{looks}\:{great}, \\ $$$${shall}\:{try}\:{to}\:{follow}\:{as}\:{soon}\:{as}\:{i} \\ $$$${get}\:{well},\:{since}\:{yesterday}\:{i}\:{m} \\ $$$${down}\:{in}\:{fever}.. \\ $$
Commented by mr W last updated on 17/Mar/21
$${sorry}\:{to}\:{hear}\:{that}\:{sir},\:{wish}\:{you}'{ll} \\ $$$${get}\:{well}\:{soon}! \\ $$