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Question Number 214928 by ajfour last updated on 24/Dec/24

Commented by ajfour last updated on 24/Dec/24

https://youtu.be/_ME_QSE6MIo?si=JxKs4xa21FSBKJSp Do you think the analysis is all correct here?

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

$$yes,ithink{it}^{\prime }scorrect.$$

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

$$centerofrod(x,y)$$$$x=\frac{L\cos \theta }{2}$$$$y=\frac{L\sin \theta }{2}$$$$\omega =-\frac{d\theta }{dt}$$$$u=\frac{dx}{dt}=\frac{\omega L\sin \theta }{2}(\to )$$$$v=-\frac{dy}{dt}=\frac{\omega L\cos \theta }{2}(\downarrow )$$$$\frac{mgL(\sin {\theta }_0-\sin \theta )}{2}=\frac{m(u^2+v^2)}{2}+\frac{{\omega }^2}{2}\times \frac{{mL}^2}{12}$$$$gL(\sin {\theta }_0-\sin \theta )=\frac{{\omega }^2{L}^2}{4}+\frac{{\omega }^2{L}^2}{12}$$$$\Rightarrow {\omega }^2=\frac{3g(\sin {\theta }_0-\sin \theta )}{L}$$$$\Rightarrow \omega \frac{d\omega }{d\theta }=-\frac{3g\cos \theta }{2L}$$$$a_x=\frac{du}{dt}=-\omega \frac{du}{d\theta }=-\frac{\omega L}{2}(\omega \cos \theta +\sin \theta \frac{d\omega }{d\theta })$$$$a_x=\frac{3g\cos \theta (3\sin \theta -2\sin {\theta }_0)}{4}$$$$N={ma}_x$$$$\Rightarrow \frac{N}{mg}=\frac{3\cos \theta (3\sin \theta -2\sin {\theta }_0)}{4}$$$$N=0:$$$$3\sin \theta -2\sin {\theta }_0=0$$$$\Rightarrow \sin \theta =\frac{2\sin {\theta }_0}{3}$$

Commented by ajfour last updated on 24/Dec/24

$$Awesome!Ihadgivenupalmost.$$$$Thankyousir.$$

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