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Question Number 58257 by ajfour last updated on 20/Apr/19

Commented by ajfour last updated on 20/Apr/19

$$\mathrm{Find}equilibriumangles\alpha \mathrm{and}\beta \mathrm{if}$$$$\mathrm{the}\mathrm{hemisphere}\mathrm{is}\mathrm{fixed}\mathrm{and}$$$$\mathrm{frictionless}\mathrm{and}\mathrm{string}\mathrm{has}\mathrm{length}l.$$

Answered by mr W last updated on 20/Apr/19

Commented by mr W last updated on 20/Apr/19

$$AC=\sqrt{R^2+h^2-2Rh\cos \alpha }$$$$BC=\sqrt{R^2+h^2-2Rh\cos \beta }$$$$\sqrt{R^2+h^2-2Rh\cos \alpha }+\sqrt{R^2+h^2-2Rh\cos \beta }=l$$$$with\kappa =\frac{R}{l},\lambda =\frac{h}{l}$$$$\Rightarrow \sqrt{{\kappa }^2+{\lambda }^2-2\kappa \lambda \cos \alpha }+\sqrt{{\kappa }^2+{\lambda }^2-2\kappa \lambda \cos \beta }=1...\left(i\right)$$$$Mg\sin \alpha =T\cos (\mathrm{\angle }OAC-\frac{\pi }{2})=T\sin \mathrm{\angle }OAC$$$$\sin \mathrm{\angle }OAC=\frac{h\sin \alpha }{\sqrt{R^2+h^2-2Rh\cos \alpha }}$$$$\Rightarrow Mg=\frac{Th}{\sqrt{R^2+h^2-2Rh\cos \alpha }}$$$$\Rightarrow Th=Mg\sqrt{R^2+h^2-2Rh\cos \alpha }$$$$similarly$$$$\Rightarrow Th=mg\sqrt{R^2+h^2-2Rh\cos \beta }$$$$with\mu =\frac{M}{m}$$$$\Rightarrow \mu \sqrt{{\kappa }^2+{\lambda }^2-2\kappa \lambda \cos \alpha }=\sqrt{{\kappa }^2+{\lambda }^2-2\kappa \lambda \cos \beta }..\left(ii\right)$$$$$$$$witha=\sqrt{{\kappa }^2+{\lambda }^2-2\kappa \lambda \cos \alpha },b=\sqrt{{\kappa }^2+{\lambda }^2-2\kappa \lambda \cos \beta }$$$$\Rightarrow a+b=1$$$$\Rightarrow \mu a=b$$$$\Rightarrow a=\sqrt{{\kappa }^2+{\lambda }^2-2\kappa \lambda \cos \alpha }=\frac{1}{1+\mu }=\delta =\frac{m}{M+m}$$$$\Rightarrow \cos \alpha =\frac{{\kappa }^2+{\lambda }^2-{\delta }^2}{2\kappa \lambda }$$$$\Rightarrow \alpha ={\cos }^{-1}\frac{{\kappa }^2+{\lambda }^2-{\delta }^2}{2\kappa \lambda }={\cos }^{-1}\frac{R^2+h^2-{\left(\frac{ml}{M+m}\right)}^2}{2Rh}$$$$similarly$$$$\Rightarrow \beta ={\cos }^{-1}\frac{R^2+h^2-{\left(\frac{Ml}{M+m}\right)}^2}{2Rh}$$

Commented by ajfour last updated on 21/Apr/19

$$\mathcal{SUPER}\mathcal{AWESOME}!Sir.$$

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