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Question Number 88967 by ajfour last updated on 14/Apr/20

Commented by ajfour last updated on 14/Apr/20

$$Astickoflength2rslides$$$$againstafixedsmoothcylindrical$$$$surface.Itisreleasedwithone$$$$atA\left(intangentialmanner\right),$$$$andlosescontactthereafterat$$$$B.Find\theta .coefficintoffriction$$$$betweenstickandgroundis\mu .$$

Answered by mr W last updated on 14/Apr/20

Commented by mr W last updated on 14/Apr/20

$$radius=r$$$$lengthl=2r$$$$\sin \beta =\frac{r(1+\sin \theta )}{l}=\frac{1+\sin \theta }{2}$$$$\Rightarrow \beta ={\sin }^{-1}\left(\frac{1+\sin \theta }{2}\right)$$$$att=0:\beta =\frac{\pi }{2}-{\theta }_0$$$$\sin (\frac{\pi }{2}-{\theta }_0)=\frac{1+\sin {\theta }_0}{2}$$$$2\cos {\theta }_0-\sin {\theta }_0=1$$$$\Rightarrow {\theta }_0={\cos }^{-1}\frac{1}{\sqrt{5}}-{\cos }^{-1}\frac{2}{\sqrt{5}}\approx 36.87°$$$$\phi =\frac{\pi }{2}-\theta$$$$\varphi ={\tan }^{-1}\mu$$$$$$$$\frac{d\beta }{d\theta }=\frac{\cos \theta }{2\cos \beta }=\frac{\cos \theta }{\sqrt{(1-\sin \theta )(3+\sin \theta )}}$$$$\omega =-\frac{d\beta }{dt}=-\frac{d\beta }{d\theta }\times \frac{d\theta }{dt}=-\frac{\cos \theta }{\sqrt{(1-\sin \theta )(3+\sin \theta )}}\times \frac{d\theta }{dt}$$$$\alpha =\frac{d\omega }{dt}=\frac{d\omega }{d\theta }\times \frac{d\theta }{dt}=-\frac{\sqrt{(1-\sin \theta )(3+\sin \theta )}}{\cos \theta }\times \frac{\omega d\omega }{d\theta }$$$$EB=r\cos \theta +l\cos \beta =r[\cos \theta +2\sqrt{1-{\left(\frac{1+\sin \theta }{2}\right)}^2}]$$$$BC=r+r[\cos \theta +\sqrt{(1-\sin \theta )(3+\sin \theta )}]\tan \theta$$$$BC=r[1+\sin \theta +\tan \theta \sqrt{(1-\sin \theta )(3+\sin \theta )}]$$$$\frac{DB}{\sin \phi }=\frac{BC}{\sin (\phi +\varphi )}$$$$DB=\frac{r\cos \theta [1+\sin \theta +\tan \theta \sqrt{(1-\sin \theta )(3+\sin \theta )}]}{\cos (\theta -\varphi )}$$$$DB=\frac{r[1+\sin \theta +\tan \theta \sqrt{(1-\sin \theta )(3+\sin \theta )}]}{\cos \varphi +\tan \theta \sin \varphi }$$$$I=\frac{m{\left(2r\right)}^2}{12}=\frac{{mr}^2}{3}$$$$I\alpha =mg(\frac{l\cos \beta }{2}-DB\sin \varphi )$$$$-\frac{r}{3}\times \frac{\sqrt{(1-\sin \theta )(3+\sin \theta )}}{\cos \theta }\times \frac{\omega d\omega }{d\theta }=g(\frac{\sqrt{(1+\sin \theta )(3+\sin \theta )}}{2}-\frac{1+\sin \theta +\tan \theta \sqrt{(1-\sin \theta )(3+\sin \theta )}}{\cot \varphi +\tan \theta })$$$$\frac{\omega d\omega }{d\theta }=-\frac{3g}{r}(\frac{\sqrt{(1+\sin \theta )(3+\sin \theta )}}{2}-\frac{1+\sin \theta +\tan \theta \sqrt{(1-\sin \theta )(3+\sin \theta )}}{\cot \varphi +\tan \theta })\frac{\cos \theta }{\sqrt{(1-\sin \theta )(3+\sin \theta )}}$$$$\frac{\omega d\omega }{d\theta }=-\frac{3g}{r}[\frac{\cos \theta }{2}-\frac{\sin \theta }{\cot \varphi +\tan \theta }-\frac{(1+\sin \theta )\cos \theta }{(\cot \varphi +\tan \theta )\sqrt{(1-\sin \theta )(3+\sin \theta )}}]$$$$\frac{{\omega }^2}{2}=-\frac{3g}{r}{\int }_{{\theta }_0}^{\theta }[\frac{\cos \theta }{2}-\frac{\sin \theta }{\cot \varphi +\tan \theta }-\frac{(1+\sin \theta )\cos \theta }{(\cot \varphi +\tan \theta )\sqrt{(1-\sin \theta )(3+\sin \theta )}}]d\theta$$$${\omega }^2=-\frac{6g}{r}{\int }_{{\theta }_0}^{\theta }[\frac{\cos \theta }{2}-\frac{\sin \theta }{\cot \varphi +\tan \theta }-\frac{(1+\sin \theta )\cos \theta }{(\cot \varphi +\tan \theta )\sqrt{(1-\sin \theta )(3+\sin \theta )}}]d\theta$$$$......$$

Commented by ajfour last updated on 14/Apr/20

$$ThanksSir,ishallsoonbe$$$$attempting..$$

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