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Question Number 145483 by ajfour last updated on 05/Jul/21

Commented by ajfour last updated on 05/Jul/21

$$Ifreleasedinthisorientation,$$$$howlongshallittaketheother$$$$wheelofthecratetohitthe$$$$ground?\left(nofrictionanywhere\right)$$

Answered by mr W last updated on 05/Jul/21

Commented by mr W last updated on 06/Jul/21

$$AC=p=\frac{\sqrt{a^2+b^2}}{2}$$$$v_x=0$$$$\phi ={\tan }^{-1}\frac{a}{b}={\tan }^{-1}\frac{1}{\lambda }$$$$y_C=p\sin (\theta +\phi )$$$$v=\frac{{dy}_C}{dt}=p\omega \cos (\theta +\phi )$$$$\frac{1}{2}{mv}^2+\frac{1}{2}I{\omega }^2=mg(\frac{ab}{\sqrt{a^2+b^2}}-p\sin (\theta +\phi ))$$$$[1+3{\cos }^2(\theta +\phi )]{\omega }^2=\frac{12g}{\sqrt{a^2+b^2}}(\frac{2ab}{a^2+b^2}-\sin (\theta +\phi ))$$$$let\lambda =\frac{b}{a}$$$$[1+3{\cos }^2(\theta +\phi )]{\omega }^2=\frac{12g}{a\sqrt{1+{\lambda }^2}}(\frac{2\lambda }{1+{\lambda }^2}-\sin (\theta +\phi ))$$$$$$$$\omega =\frac{d\theta }{dt}=\sqrt{\frac{12g}{a\sqrt{1+{\lambda }^2}}}\times \sqrt{\frac{\frac{2\lambda }{1+{\lambda }^2}-\sin (\theta +\phi )}{1+3{\cos }^2(\theta +\phi )}}$$$$\Rightarrow t=\sqrt{\frac{a\sqrt{1+{\lambda }^2}}{12g}}{\int }_0^{\phi }\sqrt{\frac{1+3{\cos }^2(\theta +\phi )}{\frac{2\lambda }{1+{\lambda }^2}-\sin (\theta +\phi )}}d\theta$$$$\Rightarrow t=\sqrt{\frac{a\sqrt{1+{\lambda }^2}}{12g}}{\int }_{\phi }^{2\phi }\sqrt{\frac{1+3{\cos }^2\varphi }{\frac{2\lambda }{1+{\lambda }^2}-\sin \varphi }}d\varphi$$$$withb=2a,\lambda =2:$$$$\Rightarrow t=\sqrt{\frac{\sqrt{5}a}{12g}}{\int }_{{\tan }^{-1}\frac{1}{2}}^{{2\tan }^{-1}\frac{1}{2}}\sqrt{\frac{1+3{\cos }^2\varphi }{\frac{4}{5}-\sin \varphi }}d\varphi$$$$\approx 1.149799\sqrt{\frac{a}{g}}$$

Commented by ajfour last updated on 06/Jul/21

$$yes,sir,iwasmistaken,thanks.$$

Commented by mr W last updated on 06/Jul/21

$$icansolvetheintegralonly$$$$numerically.$$

Commented by mr W last updated on 06/Jul/21

Commented by ajfour last updated on 06/Jul/21

Thanks for the solution, too; impeccable!

Commented by mr W last updated on 06/Jul/21

$$y_c=\omega p\cos (\theta +\phi )?$$

Answered by ajfour last updated on 06/Jul/21

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