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Question Number 41233 by ajfour last updated on 03/Aug/18

Commented by ajfour last updated on 03/Aug/18

$$Find\mathbf{v}asafunctionof\theta .$$$$Initially\theta =0°.$$

Answered by MrW3 last updated on 04/Aug/18

$$let\omega =\frac{d\theta }{dt}$$$$COMofrightrod(x_2,y_2):$$$$x_2=(2l-\frac{l}{2})\cos \theta =\frac{3l}{2}\cos \theta$$$$y_2=\frac{l}{2}\sin \theta$$$$v_{2x}=\frac{{dx}_2}{dt}=-\frac{3l}{2}\sin \theta \omega$$$$v_{2y}=\frac{{dy}_2}{dt}=\frac{l}{2}\cos \theta \omega$$$$\Rightarrow v_2^2=\frac{l^2}{4}(9{\sin }^2\theta +{\cos }^2\theta ){\omega }^2=\frac{l^2{\omega }^2}{4}(8{\sin }^2\theta +1)=\frac{l^2{\omega }^2}{4}(5-4\cos 2\theta )$$$$$$$$\frac{1}{2}\left(\frac{{ml}^2}{3}\right){\omega }^2+\frac{1}{2}\left(\frac{{ml}^2}{12}\right){\omega }^2+\frac{1}{2}m\frac{l^2{\omega }^2}{4}(5-4\cos 2\theta )=2mg\frac{l}{2}\sin \theta$$$$(5-3\cos 2\theta ){\omega }^{2}l=6g\sin \theta$$$$\Rightarrow \omega =\sqrt{\frac{6g\sin \theta }{l(5-3\cos 2\theta )}}$$$$$$$$rightendpointofrightrod(x_3,0):$$$$x_3=2l\cos \theta$$$$v_{3x}=\frac{{dx}_3}{dt}=-2l\sin \theta \omega$$$$\mathit{v}=-v_{3x}$$$$\Rightarrow \mathit{v}=2\sin \theta \sqrt{\frac{6gl\sin \theta }{5-3\cos 2\theta }}$$

Commented by ajfour last updated on 04/Aug/18

$$ExcellentSir,thanksalot.$$

Commented by tanmay.chaudhury50@gmail.com last updated on 04/Aug/18

$$excellent...$$

Commented by MrW3 last updated on 04/Aug/18

$$thankyouboth!$$

Answered by MrW3 last updated on 05/Aug/18

Commented by MrW3 last updated on 06/Aug/18

$$\mathrm{A}notherway:$$$$$$$$BC=2l\sin \theta$$$${CM}^2={BC}^2+{BM}^2-2BC\times BM\cos (90°-\theta )$$$$=4l^2{\sin }^2\theta +\frac{l^2}{4}-2\times 2l\sin \theta \times \frac{l}{2}\times \sin \theta$$$$=\frac{l^2}{4}(8{\sin }^2\theta +1)$$$$I_c=\frac{{ml}^2}{12}+m\times \frac{l^2}{4}(8{\sin }^2\theta +1)$$$$=\frac{{ml}^2}{3}(1+{6\sin }^2\theta )$$$$$$$$\frac{1}{2}\left(\frac{{ml}^2}{3}\right){\omega }^2+\frac{1}{2}\times \frac{{ml}^2}{3}(1+6{\sin }^2\theta ){\omega }^2=2mg\times \frac{l}{2}\sin \theta$$$$(1+3{\sin }^2\theta ){\omega }^{2}l=3g\sin \theta$$$$\Rightarrow \omega =\sqrt{\frac{3g\sin \theta }{(1+3{\sin }^2\theta )l}}$$$$$$$$\Rightarrow \mathit{v}=BC\times \omega =2\sin \theta \sqrt{\frac{3gl\sin \theta }{1+3{\sin }^2\theta }}$$$$=2\sin \theta \sqrt{\frac{6gl\sin \theta }{5-3\cos 2\theta }}$$

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