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Question Number 136984 by mr W last updated on 28/Mar/21

Commented by mr W last updated on 28/Mar/21

$$seeQ136382$$

Commented by mr W last updated on 30/Mar/21

$$r=(b-y)\tan \gamma$$$$dx=rd\theta$$$$dy=\tan \varphi dx$$$$dz=dr=-\tan \gamma dy=-\tan \gamma \tan \varphi dx$$$$dz=-\tan \gamma \tan \varphi rd\theta$$$$\omega =\frac{d\theta }{dt}$$$$I=\frac{3{Ma}^2}{10}$$$$E=\frac{1}{2}I{\omega }^2$$$$dE=-Mgdz=Mg\tan \gamma \tan \varphi rd\theta$$$$\frac{1}{2}\times \frac{3{Ma}^2}{10}\times 2\omega \frac{d\omega }{dt}=Mg\tan \gamma \tan \varphi r\frac{d\theta }{dt}$$$$\frac{d\omega }{dt}=\alpha =\frac{10g\tan \varphi }{3ab}\times r=\lambda r$$$$with\lambda =\frac{10g\tan \varphi }{3ab}$$$$v_x=r\omega$$$$a_x=\frac{{dv}_x}{dt}=\omega \frac{dr}{dt}+r\frac{d\omega }{dt}$$$$=-r{\omega }^2\tan \varphi \tan \gamma +\lambda r^2$$$$att=0:\omega =0$$$$\Rightarrow a_x=\lambda r^2=\frac{10g\tan \varphi r^2}{3ab}$$$$withr=(b-x\tan \varphi )\frac{a}{b}$$$$===================$$$$\omega \frac{d\omega }{d\theta }=\lambda r$$$$\omega d\omega =\lambda dx$$$$\frac{{\omega }^2}{2}=\lambda (x-x_0)$$$$\omega =\frac{dx}{rdt}=\sqrt{2\lambda (x-x_0)}$$$$\frac{dx}{(b-x\tan \varphi )\tan \gamma dt}=\sqrt{2\lambda (x-x_0)}$$$$\frac{dx}{\sqrt{2\lambda }\tan \varphi \tan \gamma (\frac{b}{\tan \varphi }-x)\sqrt{x-x_0}}=dt$$$$\Rightarrow t=\frac{1}{\frac{2\tan \varphi }{b}\sqrt{\frac{5ga\tan \varphi }{3b}}}{\int }_{x_0}^x\frac{dx}{(\frac{b}{\tan \varphi }-x)\sqrt{x-x_0}}$$$$\Rightarrow t=\frac{1}{\frac{2\tan \varphi }{b}\sqrt{\frac{5ag}{3}(1-\frac{x_0\tan \varphi }{b})}}\ln \frac{\sqrt{\frac{b}{\tan \varphi }-x_0}+\sqrt{x-x_0}}{\sqrt{\frac{b}{\tan \varphi }-x_0}-\sqrt{x-x_0}}$$$$\Rightarrow t=\frac{1}{2\sqrt{\frac{5g\tan \gamma \tan \varphi }{3x_m}}}\times \frac{1}{\sqrt{1-{\xi }_0}}\ln \frac{\sqrt{1-{\xi }_0}+\sqrt{\xi -{\xi }_0}}{\sqrt{1-{\xi }_0}-\sqrt{\xi -{\xi }_0}}$$$$\Rightarrow t=\sqrt{\frac{3x_m}{20g\tan \varphi \tan \gamma }}\mathrm{\Phi }$$$$\Rightarrow \mathrm{\Phi }=\frac{1}{\sqrt{1-{\xi }_0}}\ln \frac{\sqrt{1-{\xi }_0}+\sqrt{\xi -{\xi }_0}}{\sqrt{1-{\xi }_0}-\sqrt{\xi -{\xi }_0}}$$$$x_m=\frac{b}{\tan \varphi }$$$$\xi =\frac{x}{x_m},{\xi }_0=\frac{x_0}{x_m}$$

Commented by ajfour last updated on 29/Mar/21

$$Superbsolutionsir.Thanks.$$

Commented by mr W last updated on 30/Mar/21

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