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Question Number 139013 by ajfour last updated on 21/Apr/21
Answered by mr W last updated on 21/Apr/21
Commented by mr W last updated on 21/Apr/21
$${assume}: \\ $$$${air}\:{bubble}\:{is}\:{very}\:{small}\:{compared}\:{with}\: \\ $$$${the}\:{sphere}. \\ $$$${there}\:{is}\:{no}\:{resistance}\:{with}\:{water}. \\ $$$${there}\:{is}\:{no}\:{resistance}\:{with}\:{surface} \\ $$$${of}\:{sphere}. \\ $$$${r}_{\mathrm{0}} ={radius}\:{of}\:{bubble}\:{at}\:{O}. \\ $$$${r}={radius}\:{of}\:{bubble}\:{at}\:{T}. \\ $$$${m}={mass}\:{of}\:{air}\:{bubble}. \\ $$$$\rho_{{W}} ={density}\:{of}\:{water}. \\ $$$${h}={equivalent}\:{depth}\:{of}\:{center}\:{of}\:{sphere}. \\ $$$${let}\:\lambda=\frac{{h}}{{R}} \\ $$$$\frac{{V}}{{V}_{\mathrm{0}} }=\frac{{r}^{\mathrm{3}} }{{r}_{\mathrm{0}} ^{\mathrm{3}} }=\frac{{h}+{R}}{{h}+{R}\:\mathrm{cos}\:\theta}=\frac{\lambda+\mathrm{1}}{\lambda+\mathrm{cos}\:\theta} \\ $$$${F}=\frac{\mathrm{4}\pi\rho_{{W}} {gr}^{\mathrm{3}} }{\mathrm{3}} \\ $$$${F}=\frac{\mathrm{4}\pi\rho_{{W}} {gr}_{\mathrm{0}} ^{\mathrm{3}} }{\mathrm{3}}\left(\frac{\lambda+\mathrm{1}}{\lambda+\mathrm{cos}\:\theta}\right) \\ $$$${F}={m}_{\mathrm{0}} {g}\left(\frac{\lambda+\mathrm{1}}{\lambda+\mathrm{cos}\:\theta}\right)\:{with}\:{m}_{\mathrm{0}} =\frac{\mathrm{4}\pi\rho_{{W}} {r}_{\mathrm{0}} ^{\mathrm{3}} }{\mathrm{3}} \\ $$$${F}_{{R}} =\left({F}−{mg}\right)\:\mathrm{cos}\:\theta \\ $$$${F}_{\theta} =\left({F}−{mg}\right)\:\mathrm{sin}\:\theta \\ $$$$\omega=\frac{{d}\theta}{{dt}} \\ $$$$\alpha=\frac{{d}\omega}{{dt}}=\omega\frac{{d}\omega}{{d}\theta} \\ $$$${mR}\alpha=\left({F}−{mg}\right)\:\mathrm{sin}\:\theta \\ $$$${R}\alpha=\left(\frac{{m}_{\mathrm{0}} }{{m}}×\frac{\lambda+\mathrm{1}}{\lambda+\mathrm{cos}\:\theta}−\mathrm{1}\right){g}\:\mathrm{sin}\:\theta \\ $$$$\alpha=\omega\frac{{d}\omega}{{d}\theta}=\left(\frac{{m}_{\mathrm{0}} }{{m}}×\frac{\lambda+\mathrm{1}}{\lambda+\mathrm{cos}\:\theta}−\mathrm{1}\right)\frac{{g}}{{R}}\:\mathrm{sin}\:\theta \\ $$$$\int_{\mathrm{0}} ^{\omega} \omega{d}\omega=\int_{\mathrm{0}} ^{\theta} \left(\frac{{m}_{\mathrm{0}} }{{m}}×\frac{\lambda+\mathrm{1}}{\lambda+\mathrm{cos}\:\theta}−\mathrm{1}\right)\frac{{g}}{{R}}\:\mathrm{sin}\:\theta{d}\theta \\ $$$$\Rightarrow\omega^{\mathrm{2}} =\frac{\mathrm{2}{g}}{{R}}\int_{\mathrm{0}} ^{\theta} \left(\frac{{m}_{\mathrm{0}} }{{m}}×\frac{\lambda+\mathrm{1}}{\lambda+\mathrm{cos}\:\theta}−\mathrm{1}\right)\mathrm{sin}\:\theta{d}\theta \\ $$$${mR}\omega^{\mathrm{2}} =\left({F}−{mg}\right)\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow\omega^{\mathrm{2}} =\frac{{g}}{{R}}\left(\frac{{m}_{\mathrm{0}} }{{m}}×\frac{\lambda+\mathrm{1}}{\lambda+\mathrm{cos}\:\theta}−\mathrm{1}\right)\mathrm{cos}\:\theta \\ $$$$\left(\frac{{m}_{\mathrm{0}} }{{m}}×\frac{\lambda+\mathrm{1}}{\lambda+\mathrm{cos}\:\theta}−\mathrm{1}\right)\mathrm{cos}\:\theta=\mathrm{2}\int_{\mathrm{0}} ^{\theta} \left(\frac{{m}_{\mathrm{0}} }{{m}}×\frac{\lambda+\mathrm{1}}{\lambda+\mathrm{cos}\:\theta}−\mathrm{1}\right)\:\mathrm{sin}\:\theta{d}\theta \\ $$$$\left(\frac{\lambda+\mathrm{1}}{\lambda+\mathrm{cos}\:\theta}−\frac{{m}}{{m}_{\mathrm{0}} }\right)\mathrm{cos}\:\theta=\mathrm{2}\int_{\mathrm{0}} ^{\theta} \left(\frac{\lambda+\mathrm{1}}{\lambda+\mathrm{cos}\:\theta}−\frac{{m}}{{m}_{\mathrm{0}} }\right)\:\mathrm{sin}\:\theta{d}\theta \\ $$$${since}\:{density}\:{of}\:{air}\:{is}\:{much}\:{smaller} \\ $$$${than}\:{that}\:{of}\:{water},\:\frac{{m}}{{m}_{\mathrm{0}} }=\frac{\rho_{{Air}} }{\rho_{{W}} }\approx\mathrm{0}. \\ $$$$\frac{\lambda+\mathrm{1}}{\lambda+\mathrm{cos}\:\theta}\mathrm{cos}\:\theta=\mathrm{2}\int_{\mathrm{0}} ^{\theta} \frac{\lambda+\mathrm{1}}{\lambda+\mathrm{cos}\:\theta}\:\mathrm{sin}\:\theta{d}\theta \\ $$$$\frac{\mathrm{cos}\:\theta}{\lambda+\mathrm{cos}\:\theta}=−\mathrm{2}\int_{\mathrm{0}} ^{\theta} \frac{{d}\left(\lambda+\mathrm{cos}\:\theta\right)}{\lambda+\mathrm{cos}\:\theta} \\ $$$$\Rightarrow\frac{\mathrm{cos}\:\theta}{\lambda+\mathrm{cos}\:\theta}=\mathrm{2ln}\:\frac{\lambda+\mathrm{1}}{\lambda+\mathrm{cos}\:\theta} \\ $$$${we}\:{get}\:\theta\:{from}\:{it}. \\ $$$$\frac{{b}}{{a}}=\frac{{r}}{{r}_{\mathrm{0}} }=\sqrt[{\mathrm{3}}]{\frac{\lambda+\mathrm{1}}{\lambda+\mathrm{cos}\:\theta}} \\ $$$$ \\ $$$${examples}: \\ $$$$\lambda=\frac{{h}}{{R}}=\mathrm{1}:\:\theta\approx\mathrm{49}.\mathrm{90}°,\:\frac{{b}}{{a}}\approx\mathrm{1}.\mathrm{07} \\ $$$$\lambda=\frac{{h}}{{R}}=\mathrm{5}:\:\theta\approx\mathrm{48}.\mathrm{69}°,\:\frac{{b}}{{a}}\approx\mathrm{1}.\mathrm{53} \\ $$$$\lambda=\frac{{h}}{{R}}=\mathrm{10}:\:\theta\approx\mathrm{48}.\mathrm{46}°,\:\frac{{b}}{{a}}\approx\mathrm{1}.\mathrm{88} \\ $$
Commented by ajfour last updated on 22/Apr/21
$${Thanks}\:{sir},\:{beautiful}\:{answer}; \\ $$$${I}\:{haven}'{t}\:{had}\:{enough}\:{leisure}\:{time}.. \\ $$