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Question Number 80057 by ajfour last updated on 30/Jan/20

Commented by ajfour last updated on 30/Jan/20

$$Ifreleasedasshowninupper$$$$diagram,afterwhattimedoes$$$$theuppersphereloosecontact,$$$$withthelowerones?$$

Commented by ajfour last updated on 30/Jan/20

$$\left(assumefrictionlesssurfaces\right).$$

Answered by mr W last updated on 30/Jan/20

Commented by mr W last updated on 04/Feb/20

$$R=radiusofcircles$$$$y=2R\sin \theta$$$$x=2R\cos \theta$$$$let\omega =\frac{d\theta }{dt},\alpha =\frac{d\omega }{dt}=\omega \frac{d\omega }{d\theta }$$$$V,A=velocityandacceleration\downarrow$$$$v,a=velocityandacceleration\leftarrow \to$$$$$$$$V=-\frac{dy}{dt}=-2R\cos \theta \frac{d\theta }{dt}=-2R\cos \theta \omega$$$$v=\frac{dx}{dt}=-2R\sin \theta \frac{d\theta }{dt}=-2R\sin \theta \omega$$$$\frac{1}{2}{mV}^2+2\times \frac{1}{2}{mv}^2=mg(\sqrt{3}R-2R\sin \theta )$$$$2R(1+{\sin }^2\theta ){\omega }^2=g(\sqrt{3}-2\sin \theta )$$$$\Rightarrow {\omega }^2=\frac{g}{2R}\times \frac{\sqrt{3}-2\sin \theta }{1+{\sin }^2\theta }$$$$\Rightarrow \omega =-\sqrt{\frac{g}{2R}}\sqrt{\frac{\sqrt{3}-2\sin \theta }{1+{\sin }^2\theta }}$$$$$$$$A=\frac{dV}{dt}=2R(\sin \theta {\omega }^2-\cos \theta \alpha )$$$$a=\frac{dv}{dt}=2R(-\cos \theta {\omega }^2-\sin \theta \alpha )$$$$$$$$mg-2N\sin \theta =mA$$$$mg-2N\sin \theta =2mR(\sin \theta {\omega }^2-\cos \theta \alpha )...\left(i\right)$$$$$$$$N\cos \theta =ma$$$$N\cos \theta =2mR(-\cos \theta {\omega }^2-\sin \theta \alpha )...\left(ii\right)$$$$$$$$\left(i\right)\cos \theta +2\times \left(ii\right)\sin \theta :$$$$\frac{g}{2R}\cos \theta =-{\omega }^2\sin \theta \cos \theta -(1+{\sin }^2\theta )\alpha$$$$\alpha =-\frac{(\frac{g}{2R}+{\omega }^2\sin \theta )\cos \theta }{(1+{\sin }^2\theta )}$$$$\Rightarrow \alpha =-\frac{g}{2R}\times \frac{(1+\sqrt{3}\sin \theta -{\sin }^2\theta )\cos \theta }{{(1+{\sin }^2\theta )}^2}$$$$put{\omega }^{2}and\alpha into\left(ii\right):$$$$\Rightarrow \frac{N}{mg}=\frac{{\sin }^3\theta +3\sin \theta -\sqrt{3}}{{(1+{\sin }^2\theta )}^2}$$$$$$$$forN=0\left(lossofcontact\right):$$$$\Rightarrow {\sin }^3\theta +3\sin \theta -\sqrt{3}=0$$$$\Rightarrow \sin \theta =\sqrt[3]{\frac{\sqrt{7}+\sqrt{3}}{2}}-\sqrt[3]{\frac{\sqrt{7-\sqrt{3}}}{2}}$$$$\Rightarrow {\theta }_1={\sin }^{-1}(\sqrt[3]{\frac{\sqrt{7}+\sqrt{3}}{2}}-\sqrt[3]{\frac{\sqrt{7-\sqrt{3}}}{2}})$$$$\Rightarrow {\theta }_1\approx 0.556506rad(\approx 31.885421°)$$$$$$$$\omega =\frac{d\theta }{dt}=-\sqrt{\frac{g}{2R}}\sqrt{\frac{\sqrt{3}-2\sin \theta }{1+{\sin }^2\theta }}$$$${\int }_0^{t}dt=-\sqrt{\frac{2R}{g}}{\int }_{\frac{\pi }{3}}^{\theta }\sqrt{\frac{1+{\sin }^2\theta }{\sqrt{3}-2\sin \theta }}d\theta$$$$\Rightarrow t\left(\theta \right)=\sqrt{\frac{2R}{g}}{\int }_{\theta }^{\frac{\pi }{3}}\sqrt{\frac{1+{\sin }^2\theta }{\sqrt{3}-2\sin \theta }}d\theta$$$$t_1=t\left({\theta }_1\right)=\sqrt{\frac{2R}{g}}{\int }_{{\theta }_1}^{\frac{\pi }{3}}\sqrt{\frac{1+{\sin }^2\theta }{\sqrt{3}-2\sin \theta }}d\theta$$$$\approx 1.670696\sqrt{\frac{2R}{g}}$$$$$$$$thatmeansattime{t}_1=1.670696\sqrt{\frac{2R}{g}}$$$$andat{\theta }_1=31.885421°thecontactgetslost.$$$$$$$$V=\sqrt{2Rg}\cos \theta \sqrt{\frac{\sqrt{3}-2\sin \theta }{1+{\sin }^2\theta }}$$$$at\theta ={\theta }_1:$$$$\frac{V_1}{\sqrt{2Rg}}=\cos {\theta }_1\sqrt{\frac{\sqrt{3}-2\sin {\theta }_1}{1+{\sin }^2{\theta }_1}}\approx 0.617121$$$$thatmeansatthemomentoflossof$$$$contactthevelocitydownwardsis$$$$V_1=0.617121\sqrt{2Rg}.$$

Commented by mr W last updated on 31/Jan/20

Commented by ajfour last updated on 31/Jan/20

$$yourmethodisstraightand$$$$clear,Sir,mustbetrue,unless$$$$thereissomething,thatwe$$$$needtoamendinourconcepts..$$

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