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Question Number 214449 by ajfour last updated on 08/Dec/24

Commented by ajfour last updated on 09/Dec/24

https://youtu.be/mVW8FiNpKjk?si=q68kdETxCuqwyZ3-

Answered by mr W last updated on 14/Dec/24

Commented by mr W last updated on 15/Dec/24

$$(X,Y)=centeroflowercylinder$$$$(x,y)=centerofuppercylinder$$$$X=-r\sin \theta$$$$Y=r$$$$x=r\sin \theta$$$$y=r+2r\cos \theta$$$$let\omega =\frac{d\theta }{dt}$$$$V_x=-\frac{dX}{dt}=\omega r\cos \theta$$$$V_y=0$$$$v_x=\frac{dx}{dt}=\omega r\cos \theta$$$$v_y=-\frac{dy}{dt}=2\omega r\sin \theta$$$$bothcylindersobtainnorotation.$$$$2mgr(1-\cos \theta )=\frac{m{\omega }^2{r}^2{\cos }^2\theta }{2}+\frac{m{\omega }^2{r}^2({\cos }^2\theta +4{\sin }^2\theta )}{2}$$$$2g(1-\cos \theta )={\omega }^{2}r(1+{\sin }^2\theta )$$$$\Rightarrow {\omega }^2=\frac{2g(1-\cos \theta )}{r(1+{\sin }^2\theta )}$$$$2\omega \frac{d\omega }{d\theta }=\frac{2g}{r}[\frac{\sin \theta }{1+{\sin }^2\theta }-\frac{2(1-\cos \theta )\sin \theta \cos \theta }{{(1+{\sin }^2\theta )}^2}]$$$$\omega \frac{d\omega }{d\theta }=\frac{g\sin \theta ({\cos }^2\theta -2\cos \theta +2)}{r{(1+{\sin }^2\theta )}^2}$$$$A_x=\frac{{dV}_x}{dt}=\omega \frac{{dV}_x}{d\theta }=\omega r(-\omega \sin \theta +\cos \theta \frac{d\omega }{d\theta })$$$$A_x=g[-\frac{2\sin \theta (1-\cos \theta )}{(1+{\sin }^2\theta )}+\frac{\sin \theta \cos \theta ({\cos }^2\theta -2\cos \theta +2)}{{(1+{\sin }^2\theta )}^2}]$$$$A_x=\frac{g\sin \theta (-{\cos }^3\theta +6\cos \theta -4)}{{(1+{\sin }^2\theta )}^2}$$$$N\sin \theta ={mA}_x$$$$\Rightarrow \frac{N}{mg}=\frac{-{\cos }^3\theta +6\cos \theta -4}{{(1+{\sin }^2\theta )}^2}$$$$whenN=0,$$$${\cos }^3\theta -6\cos \theta +4=0$$$$(\cos \theta -2)({\cos }^2\theta +2\cos \theta -2)=0$$$$0<\cos \theta <1\Rightarrow onlyonerootissuitable$$$$\cos \theta =\sqrt{3}-1$$$$\Rightarrow \theta ={\cos }^{-1}(\sqrt{3}-1)\approx 42.941°$$$$==================$$$$V_x=\omega r\cos \theta =\cos \theta \sqrt{\frac{2gr(1-\cos \theta )}{1+{\sin }^2\theta }}$$$$V_{x,final}=\sqrt{(3\sqrt{3}-5)gr}\approx 0.4429\sqrt{gr}$$$$thisisthefinalspeedofthelower$$$$cylinderafterseparationfromthe$$$$uppercylinder.$$

Commented by ajfour last updated on 15/Dec/24

Really glad for the same conclusion, sir.

Answered by ajfour last updated on 09/Dec/24

Commented by ajfour last updated on 04/Jan/25

$$mg\left(2r\right)(1-\cos \theta )={mV}^2+\frac{1}{2}{mv}^2$$$$V\sin \theta =v\cos \theta -V\sin \theta \mathrm{\&}$$$$mg\cos \theta -(N=0)-(mA\sin \theta =0)$$$$=m\frac{{(2V\cos \theta +v\sin \theta )}^2}{2r}$$$$\Rightarrow 2V\cos \theta +v\sin \theta =\sqrt{2gr\cos \theta }$$$$2V\sin \theta =v\cos \theta$$$$\Rightarrow v(\frac{\cos {}^2\theta }{\sin \theta }+\sin \theta )=\sqrt{2gr\cos \theta }$$$$v^2=2gr\cos \theta \sin {}^2\theta$$$$V^2=\frac{gr\cos {}^3\theta }{2}hence$$$$2(1-\cos \theta )=\frac{\cos {}^3\theta }{2}+\cos \theta (1-\cos {}^2\theta )$$$$If\cos \theta =c$$$$\Rightarrow 2-2c=-\frac{c^3}{2}+c$$$$c^3-6c+4=0$$$$hasjustonerootc=\sqrt{3}-1$$$$\theta ={\cos }^{-1}(\sqrt{3}-1)$$$$V^2=\frac{gr\cos {}^3\theta }{2}$$$$V=\sqrt{gr(3\sqrt{3}-5)}$$$$$$

Commented by ajfour last updated on 09/Dec/24

https://youtu.be/pHHTfPtRV98?si=JtR34MdARBkxuPIV

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