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Question Number 138336 by mr W last updated on 12/Apr/21

Commented by mr W last updated on 12/Apr/21

$$uniformrodwithlengthL(>s)and$$$$massm,cubewithsidelengthsand$$$$massM.oneendoftherodishinged.$$$$nofriction.$$$$motionbeginnsat\theta =90°fromrest.$$$$findthemaximalvelocityofthe$$$$cube.$$

Answered by mr W last updated on 14/Apr/21

Commented by mr W last updated on 15/Apr/21

$$\omega =-\frac{d\theta }{dt}$$$$I=\frac{{mL}^2}{3}$$$$v=\omega L\sin \theta$$$$\frac{1}{2}\left(\frac{{mL}^2}{3}\right){\omega }^2+\frac{1}{2}M{\left(\omega L\sin \theta \right)}^2=mg(1-\sin \theta )\frac{L}{2}$$$${\omega }^2(1+\frac{3M{\sin }^2\theta }{m})=\frac{3g}{L}(1-\sin \theta )$$$$\lambda =\frac{3M}{m}$$$$\omega =\sqrt{\frac{3g}{L}}\sqrt{\frac{1-\sin \theta }{1+\lambda {\sin }^2\theta }}=\sqrt{\frac{3g}{L}}\mathrm{\Phi }\left(\theta \right)$$$$\mathrm{\Phi }\left(\theta \right)=\sqrt{\frac{1-\sin \theta }{1+\lambda {\sin }^2\theta }}$$$$\alpha =\frac{d\omega }{dt}=\frac{d\theta }{dt}\times \frac{d\omega }{d\theta }=-\omega \frac{d\omega }{d\theta }=-\frac{3g}{L}\mathrm{\Phi }\left(\theta \right){\mathrm{\Phi }}^{\prime }\left(\theta \right)$$$$\left(\frac{{mL}^2}{3}\right)\alpha =mg\frac{L}{2}\cos \theta -NL\sin \theta$$$$N=\frac{mg}{\sin \theta }(\frac{\cos \theta }{2}+\mathrm{\Phi }\left(\theta \right){\mathrm{\Phi }}^{\prime }\left(\theta \right))$$$$N=0$$$$\Rightarrow \mathrm{\Phi }\left(\theta \right){\mathrm{\Phi }}^{\prime }\left(\theta \right)+\frac{\cos \theta }{2}=0\ast )$$$$v=\sqrt{3gL}\mathrm{\Phi }\left(\theta \right)\sin \theta$$$$$$$$\ast )canbeformedto:$$$$\frac{2}{3}-\sin \theta -\frac{M}{m}{\sin }^3\theta =0$$$$orwith\mu =\frac{m}{M}$$$${\sin }^3\theta +\mu \sin \theta -\frac{2\mu }{3}=0$$$$\Rightarrow \sin \theta =\sqrt[3]{\frac{\mu }{3}}(\sqrt[3]{\sqrt{1+\frac{\mu }{3}}+1}-\sqrt[3]{\sqrt{1+\frac{\mu }{3}}-1})$$$$$$$$example:$$$$\frac{M}{m}=1$$$$N=0at\theta \approx 0.5508(31.558°)$$$$v_{max}\approx 0.4637\sqrt{gL}$$

Commented by ajfour last updated on 15/Apr/21

$${\omega }^{2}L\cos \theta =\alpha L\sin \theta$$$$\frac{mgL\cos \theta }{2}=\frac{{mL}^2\alpha }{3}$$$$\frac{mgL}{2}(1-\sin \theta )=\frac{{mL}^2{\omega }^2}{6}+\frac{{Mv}^2}{2}$$$$\Rightarrow \frac{3g}{2}=\frac{{\omega }^{2}L}{\sin \theta }$$$$\mathrm{\&}\omega L\sin \theta =v$$$$\Rightarrow gL(1-\sin \theta )=\frac{gL\sin \theta }{2}$$$$+\frac{3MgL\sin {}^3\theta }{2m}$$$$2(1-\sin \theta )=\sin \theta +3\left(\frac{M}{m}\right)\sin {}^3\theta$$$$\theta isobtainedfromthiseq.$$$$Nowv=\sqrt{\frac{3gL\sin {}^3\theta }{2}}$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$

Commented by ajfour last updated on 15/Apr/21

$$yessir,thanks;corrected.$$

Commented by mr W last updated on 15/Apr/21

$$nicesolutionsir!$$

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