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Question Number 214722 by mr W last updated on 18/Dec/24

Commented by mr W last updated on 18/Dec/24

$$threeidenticalcylindersare$$$$releasedfromthestateasshown.$$$$thereisnofrictionanywhere.$$$$find$$$$1)thefinalspeedofthelower$$$$cylinders$$$$2)afterwhattimeandwithwhat$$$$speedthemiddlecylinderhitsthe$$$$ground$$

Answered by mr W last updated on 19/Dec/24

Commented by mr W last updated on 20/Dec/24

$$m=massofeachcylinder$$$$\omega =\frac{d\theta }{dt}$$$$X=2r\sin \theta$$$$Y=2r\cos \theta$$$$U=\frac{dX}{dt}=2r\omega \cos \theta$$$$V=-\frac{dY}{dt}=2r\omega \sin \theta$$$$2mgr(\cos \frac{\pi }{6}-\cos \theta )=2\times \frac{{mU}^2}{2}+\frac{{mV}^2}{2}$$$$4gr(\frac{\sqrt{3}}{2}-\cos \theta )=2\times 4r^2{\omega }^2{\cos }^2\theta +4r^2{\omega }^2{\sin }^2\theta$$$$g(\frac{\sqrt{3}}{2}-\cos \theta )=r{\omega }^2(1+{\cos }^2\theta )$$$${\omega }^2=\frac{g(\frac{\sqrt{3}}{2}-\cos \theta )}{r(1+{\cos }^2\theta )}$$$$2\omega \frac{d\omega }{d\theta }=\frac{g}{r}[\frac{\sin \theta }{1+{\cos }^2\theta }+\frac{2\cos \theta \sin \theta (\frac{\sqrt{3}}{2}-\cos \theta )}{{(1+{\cos }^2\theta )}^2}]$$$$\omega \frac{d\omega }{d\theta }=\frac{g\sin \theta (1+\sqrt{3}\cos \theta -{\cos }^2\theta )}{2r{(1+{\cos }^2\theta )}^2}$$$$A=\frac{dU}{dt}=2r\omega (-\omega \sin \theta +\cos \theta \frac{d\omega }{d\theta })$$$$A=\frac{g\sin \theta ({\cos }^3\theta +3\cos \theta -\sqrt{3})}{{(1+{\cos }^2\theta )}^2}$$$$N\sin \theta =mA$$$$\Rightarrow N=\frac{mg({\cos }^3\theta +3\cos \theta -\sqrt{3})}{{(1+{\cos }^2\theta )}^2}$$$$N=0:$$$${\cos }^3\theta +3\cos \theta -\sqrt{3}=0$$$$\cos \theta =\sqrt[3]{\frac{\sqrt{7}+\sqrt{3}}{2}}-\sqrt[3]{\frac{\sqrt{7}-\sqrt{3}}{2}}=\lambda \approx 0.5282$$$$U=\cos \theta \sqrt{\frac{2(\sqrt{3}-2\cos \theta )}{1+{\cos }^2\theta }}\sqrt{gr}$$$$V=\sin \theta \sqrt{\frac{2(\sqrt{3}-2\cos \theta )}{1+{\cos }^2\theta }}\sqrt{gr}$$$$dt=\frac{d\theta }{\omega }=\sqrt{\frac{r(1+{\cos }^2\theta )}{g(\frac{\sqrt{3}}{2}-\cos \theta )}}d\theta$$$$T_1=\sqrt{\frac{r}{g}}{\int }_{\frac{\pi }{6}}^{\theta }\sqrt{\frac{1+{\cos }^2\theta }{\frac{\sqrt{3}}{2}-\cos \theta }}d\theta$$$$T_1=\sqrt{\frac{r}{g}}{\int }_{\theta }^{\frac{\pi }{6}}\sqrt{\frac{1+{\cos }^2\theta }{(\frac{\sqrt{3}}{2}-\cos \theta )(1-{\cos }^2\theta )}}d\left(\cos \theta \right)$$$$T_1=\sqrt{\frac{r}{g}}{\int }_{\lambda }^{\frac{\sqrt{3}}{2}}\sqrt{\frac{1+{\xi }^2}{(1-{\xi }^2)(\frac{\sqrt{3}}{2}-\xi )}}d\xi$$$$\approx 2.36272\sqrt{\frac{r}{g}}$$$$h=r(1+2\cos \theta )$$$$V_2=\sqrt{V^2+2gh}$$$$V_2=\sqrt{gr}\sqrt{\frac{2(1-{\cos }^2\theta )(\sqrt{3}-2\cos \theta )}{1+{\cos }^2\theta }+2(1+2\cos \theta )}$$$$\approx 2.20741\sqrt{gr}$$$$T_2=\frac{V_2-V}{g}$$$$T_2=[\sqrt{\frac{2(1-{\cos }^2\theta )(\sqrt{3}-2\cos \theta )}{1+{\cos }^2\theta }+2(1+2\cos \theta )}-\sin \theta \sqrt{\frac{2(\sqrt{3}-2\cos \theta )}{1+{\cos }^2\theta }}]\sqrt{\frac{r}{g}}$$$$\approx 1.33510\sqrt{\frac{r}{g}}$$$$T=T_1+T_2\approx 3.69782\sqrt{\frac{r}{g}}$$$$\underset{―}{camparation:}$$$$freefallofthemiddlecylinder$$$$fromsamehight(1+\sqrt{3})r:$$$$V_2=\sqrt{2g(1+\sqrt{3})r}\approx 2.33754\sqrt{gr}$$$$T=\sqrt{\frac{2(1+\sqrt{3})r}{g}}\approx 2.33754\sqrt{\frac{r}{g}}$$

Commented by ajfour last updated on 20/Dec/24

https://youtube.com/shorts/iyQMnvfGLVI?si=T6BLqABWNk_2EqJT

Commented by mr W last updated on 20/Dec/24

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