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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

$$assume:$$$$airbubbleisverysmallcomparedwith$$$$thesphere.$$$$thereisnoresistancewithwater.$$$$thereisnoresistancewithsurface$$$$ofsphere.$$$$r_0=radiusofbubbleatO.$$$$r=radiusofbubbleatT.$$$$m=massofairbubble.$$$${\rho }_W=densityofwater.$$$$h=equivalentdepthofcenterofsphere.$$$$let\lambda =\frac{h}{R}$$$$\frac{V}{V_0}=\frac{r^3}{r_0^3}=\frac{h+R}{h+R\cos \theta }=\frac{\lambda +1}{\lambda +\cos \theta }$$$$F=\frac{4\pi {\rho }_W{gr}^3}{3}$$$$F=\frac{4\pi {\rho }_W{gr}_0^3}{3}\left(\frac{\lambda +1}{\lambda +\cos \theta }\right)$$$$F=m_{0}g\left(\frac{\lambda +1}{\lambda +\cos \theta }\right)with{m}_0=\frac{4\pi {\rho }_W{r}_0^3}{3}$$$$F_R=(F-mg)\cos \theta$$$$F_{\theta }=(F-mg)\sin \theta$$$$\omega =\frac{d\theta }{dt}$$$$\alpha =\frac{d\omega }{dt}=\omega \frac{d\omega }{d\theta }$$$$mR\alpha =(F-mg)\sin \theta$$$$R\alpha =(\frac{m_0}{m}\times \frac{\lambda +1}{\lambda +\cos \theta }-1)g\sin \theta$$$$\alpha =\omega \frac{d\omega }{d\theta }=(\frac{m_0}{m}\times \frac{\lambda +1}{\lambda +\cos \theta }-1)\frac{g}{R}\sin \theta$$$${\int }_0^{\omega }\omega d\omega ={\int }_0^{\theta }(\frac{m_0}{m}\times \frac{\lambda +1}{\lambda +\cos \theta }-1)\frac{g}{R}\sin \theta d\theta$$$$\Rightarrow {\omega }^2=\frac{2g}{R}{\int }_0^{\theta }(\frac{m_0}{m}\times \frac{\lambda +1}{\lambda +\cos \theta }-1)\sin \theta d\theta$$$$mR{\omega }^2=(F-mg)\cos \theta$$$$\Rightarrow {\omega }^2=\frac{g}{R}(\frac{m_0}{m}\times \frac{\lambda +1}{\lambda +\cos \theta }-1)\cos \theta$$$$(\frac{m_0}{m}\times \frac{\lambda +1}{\lambda +\cos \theta }-1)\cos \theta =2{\int }_0^{\theta }(\frac{m_0}{m}\times \frac{\lambda +1}{\lambda +\cos \theta }-1)\sin \theta d\theta$$$$(\frac{\lambda +1}{\lambda +\cos \theta }-\frac{m}{m_0})\cos \theta =2{\int }_0^{\theta }(\frac{\lambda +1}{\lambda +\cos \theta }-\frac{m}{m_0})\sin \theta d\theta$$$$sincedensityofairismuchsmaller$$$$thanthatofwater,\frac{m}{m_0}=\frac{{\rho }_{Air}}{{\rho }_W}\approx 0.$$$$\frac{\lambda +1}{\lambda +\cos \theta }\cos \theta =2{\int }_0^{\theta }\frac{\lambda +1}{\lambda +\cos \theta }\sin \theta d\theta$$$$\frac{\cos \theta }{\lambda +\cos \theta }=-2{\int }_0^{\theta }\frac{d(\lambda +\cos \theta )}{\lambda +\cos \theta }$$$$\Rightarrow \frac{\cos \theta }{\lambda +\cos \theta }=2\ln \frac{\lambda +1}{\lambda +\cos \theta }$$$$weget\theta fromit.$$$$\frac{b}{a}=\frac{r}{r_0}=\sqrt[3]{\frac{\lambda +1}{\lambda +\cos \theta }}$$$$$$$$examples:$$$$\lambda =\frac{h}{R}=1:\theta \approx 49.90°,\frac{b}{a}\approx 1.07$$$$\lambda =\frac{h}{R}=5:\theta \approx 48.69°,\frac{b}{a}\approx 1.53$$$$\lambda =\frac{h}{R}=10:\theta \approx 48.46°,\frac{b}{a}\approx 1.88$$

Commented by ajfour last updated on 22/Apr/21

$$Thankssir,beautifulanswer;$$$$I{haven}^{\prime }thadenoughleisuretime..$$

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