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Question Number 143666 by ajfour last updated on 16/Jun/21

Commented by ajfour last updated on 19/Jun/21

$$Iftheupperdiscisreleased$$$$atangle\alpha ,find\alpha suchthat$$$$disclosescontactwith$$$$thefixedhemispherejust$$$$whenithitstheground.$$$$(assumefrictioncoeff.\mu )$$

Answered by mr W last updated on 17/Jun/21

Commented by mr W last updated on 19/Jun/21

$$\phi =\frac{a+b}{b}\theta$$$$\frac{d\phi }{dt}=\frac{a+b}{b}\times \frac{d\theta }{dt}=\frac{(a+b)\omega }{b}$$$$mg\sin \theta -N=m(a+b){\omega }^2$$$$\Rightarrow \frac{N}{mg}=\sin \theta -\frac{(a+b){\omega }^2}{g}$$$$\frac{1}{2}I{\left(\frac{d\phi }{dt}\right)}^2+\frac{1}{2}m{(a+b)}^2{\omega }^2=mg(a+b)(\sin \alpha -\sin \theta )$$$$\frac{1}{2}\times \frac{{mb}^2}{2}\frac{{(a+b)}^2{\omega }^2}{b^2}+\frac{m{(a+b)}^2{\omega }^2}{2}=mg(a+b)(\sin \alpha -\sin \theta )$$$$I=\frac{{mb}^2}{2}(asdisc,notsphere)$$$$\Rightarrow {\omega }^2=\frac{4g}{3(a+b)}(\sin \alpha -\sin \theta )$$$$\frac{N}{mg}=\sin \theta -\frac{(a+b){\omega }^2}{g}=0\left(lossofcontact\right)$$$$\sin \theta =\frac{4}{3}(\sin \alpha -\sin \theta )$$$$\Rightarrow \sin \alpha =\frac{7}{4}\sin \theta$$$$\sin \theta =\frac{b}{a+b}whenballhitstheground$$$$\Rightarrow \sin \alpha =\frac{7b}{4(a+b)}$$

Answered by ajfour last updated on 18/Jun/21

$$Energyconservation$$$$mg\{(a+b)\sin \theta -b\}$$$$=\frac{{mb}^2{\omega }^2}{4}+\frac{{mv}^2}{2}$$$$v=\omega b=-(a+b)\frac{d\theta }{dt}$$$$\Rightarrow g\{(a+b)\sin \theta -b\}=\frac{3b^2{\omega }^2}{4}$$$$N=0whenatbottom\Rightarrow$$$$mg\sin \varphi =\frac{m{\omega }^2{b}^2}{(a+b)}$$$$\sin \varphi =\frac{b}{a+b}$$$$\Rightarrow {\omega }^2=\frac{g}{b}$$$$\sin \theta =\frac{\frac{3b}{4}+b}{a+b}=\frac{7b}{4(a+b)}$$

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