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Question Number 70423 by ajfour last updated on 04/Oct/19

Commented by ajfour last updated on 04/Oct/19

$$Withinahollowsphere,mass$$$$M_0,radiusR,reststwosmaller$$$$spheresmasses,Mandm,radii$$$$b,a,respectively.Findequilibrium$$$$angles\alpha ,\beta .$$

Answered by ajfour last updated on 04/Oct/19

Commented by ajfour last updated on 04/Oct/19

$$Letthenormalreactionbetween$$$$thetwosmallerspheresbeF.$$$$R\sin \gamma =(R-b)\cos \alpha -(R-a)\cos \beta$$$$....\left(i\right)$$$$R\cos \gamma =(R-b)\sin \alpha +(R-a)\sin \beta$$$$....\left(ii\right)$$$$Fromconcurrencyofforces$$$$F=\frac{Mg\sin \alpha }{\cos (\gamma +\alpha )}=\frac{mg\sin \beta }{\cos (\beta -\gamma )}$$$$\Rightarrow M(\frac{1}{\tan \beta }+\tan \gamma )$$$$=m(\frac{1}{\tan \alpha }-\tan \gamma )$$$$\Rightarrow \tan \gamma =\frac{m\cot \alpha -M\cot \beta }{m+M}..\left(I\right)$$$$Alsofrom\left(i\right)\mathrm{\&}\left(ii\right)$$$$\tan \gamma =\frac{(R-b)\cos \alpha -(R-a)\cos \beta }{(R-b)\sin \alpha +(R-a)\sin \beta }$$$$......\left(II\right)$$$$letR-a=p,R-b=q,M=\mu m,$$$$r=a+b$$$$fromcosinerulein△ACB$$$$\cos (\alpha +\beta )=\frac{p^2+q^2-r^2}{2pq}=\mathit{k}..\left(1\right)$$$$Andfrom\left(I\right)\mathrm{\&}\left(II\right)$$$$\frac{\cot \alpha -\mu \cot \beta }{1+\mu }=\frac{q\cos \alpha -p\cos \beta }{q\sin \alpha +p\sin \beta }$$$$\Rightarrow (q\sin \alpha +p\sin \beta )(\sin \beta \cos \alpha -\mu \cos \beta \sin \alpha )$$$$=(1+\mu )(q\cos \alpha -p\cos \beta )\left(\sin \alpha \sin \beta \right)$$$$\Rightarrow$$$$q\sin \alpha \cos \alpha \sin \beta -\mu q\sin {}^2\alpha \cos \beta$$$$+p\cos \alpha \sin {}^2\beta -\mu p\sin \alpha \sin \beta \cos \beta$$$$=(1+\mu )q\sin \alpha \cos \alpha \sin \beta$$$$-(1+\mu )p\sin \alpha \sin \beta \cos \beta$$$$\Rightarrow$$$$\mu q\sin \alpha \cos \alpha \sin \beta -p\sin \alpha \sin \beta \cos \beta$$$$=p\cos \alpha \sin {}^2\beta -\mu q\sin {}^2\alpha \cos \beta$$$$\Rightarrow$$$$\left(\mu q\sin \alpha \right)\sin (\alpha +\beta )$$$$=p\sin \beta \sin (\alpha +\beta )$$$$\Rightarrow \frac{\sin \alpha }{\sin \beta }=\frac{p}{\mu q}(hastobeashort$$$$waytothis)$$$$Nowusing\left(1\right)$$$$\cos \alpha \sqrt{1-\sin {}^2\beta }-\sin \alpha \sin \beta =k$$$$\Rightarrow$$$$\cos \alpha \sqrt{1-\frac{{\mu }^2{q}^2}{p^2}\sin {}^2\alpha }$$$$-\frac{\mu q}{p}\sin {}^2\alpha =k$$$$(1-\sin {}^2\alpha )(1-\frac{{\mu }^2{q}^2}{p^2}\sin {}^2\alpha )$$$$={(k+\frac{\mu q}{p}\sin {}^2\alpha )}^2$$$$letscall\sin \alpha =x,\frac{\mu q}{p}=c$$$$\Rightarrow 1-(1+c^2)x^2=k^2+2{ckx}^2$$$$\Rightarrow x^2=\frac{1-k^2}{(c^2+2ck+1)}$$$$y^2=c^2{x}^2=\frac{c^2(1-k^2)}{(c^2+2ck+1)}$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$\Rightarrow \mathit{\alpha }={\sin }^{-1}\sqrt{\frac{1-k^2}{c^2+2ck+1}}$$$$\mathit{\beta }={\sin }^{-1}\sqrt{\frac{c^2(1-k^2)}{c^2+2ck+1}}$$$$c=\frac{M(R-b)}{m(R-a)}$$$$k=\frac{{(R-a)}^2+{(R-b)}^2-{(a+b)}^2}{2(R-a)(R-b)}$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}◼$$$$Ifasanexample$$$$\mu =8,b=2a=\frac{R}{2}\left(samedensities\right)$$$$c=\frac{16}{3},k=\frac{\frac{9}{16}+\frac{4}{16}-\frac{9}{16}}{\frac{12}{16}}=\frac{1}{3}$$$$\sin \alpha =\sqrt{\frac{1-\frac{1}{9}}{\frac{256}{9}+2\left(\frac{1}{3}\right)\left(\frac{16}{3}\right)+1}}$$$$\alpha ={\sin }^{-1}\left(\frac{2\sqrt{2}}{\sqrt{297}}\right)\approx 9.446°$$$$\beta ={\sin }^{-1}\left(\sqrt{\frac{2048}{2673}}\right)\approx 61.082°.$$

Answered by mr W last updated on 04/Oct/19

Commented by mr W last updated on 04/Oct/19

$$p=R-a$$$$q=R-b$$$$r=a+b$$$$M=\mu m$$$$\cos \theta =\frac{q^2+r^2-p^2}{2qr}=\frac{q^2+\frac{r^2}{{(1+\mu )}^2}-h^2}{2q\frac{r}{1+\mu }}$$$$\frac{q^2+r^2-p^2}{1+\mu }=q^2+\frac{r^2}{{(1+\mu )}^2}-h^2$$$$\Rightarrow h^2=\frac{\mu q^2}{1+\mu }+\frac{p^2}{1+\mu }-\frac{\mu r^2}{{(1+\mu )}^2}$$$$\Rightarrow h^2=\frac{1}{{(1+\mu )}^2}[\mu (1+\mu )q^2+(1+\mu )p^2-\mu r^2]$$$$\cos \alpha =\frac{q^2+h^2-\frac{r^2}{{(1+\mu )}^2}}{2qh}$$$$=\frac{(1+2\mu )q^2+p^2-r^2}{2q\sqrt{\mu (1+\mu )q^2+(1+\mu )p^2-\mu r^2}}$$$$\Rightarrow \alpha ={\cos }^{-1}\frac{(1+2\mu )q^2+p^2-r^2}{2q\sqrt{\mu (1+\mu )q^2+(1+\mu )p^2-\mu r^2}}$$$$\cos \beta =\frac{p^2+h^2-\frac{{\mu }^2{r}^2}{{(1+\mu )}^2}}{2ph}$$$$=\frac{\mu q^2+(2+\mu )p^2-\mu r^2}{2p\sqrt{\mu (1+\mu )q^2+(1+\mu )p^2-\mu r^2}}$$$$\Rightarrow \beta ={\cos }^{-1}\frac{\mu q^2+(2+\mu )p^2-\mu r^2}{2p\sqrt{\mu (1+\mu )q^2+(1+\mu )p^2-\mu r^2}}$$$$$$$$example:$$$$\mu =8,a=\frac{R}{4},b=\frac{R}{2}$$$$p=R-a=\frac{3R}{4},q=R-b=\frac{2R}{4},r=\frac{3R}{4}$$$$\cos \alpha =\frac{17}{3\sqrt{33}}\Rightarrow \sin \alpha =\frac{2\sqrt{2}}{3\sqrt{33}}\Rightarrow 9.446°$$$$\cos \beta =\frac{25}{9\sqrt{33}}\Rightarrow \sin \beta =\frac{32\sqrt{2}}{9\sqrt{33}}\Rightarrow 61.083°$$

Commented by ajfour last updated on 04/Oct/19

$$sirpleaseconsidertheexample$$$$ifhavechosenandbringout$$$$valuesfor\alpha ,\beta .$$

Commented by mr W last updated on 04/Oct/19

$$theresultsarethesamesir.$$

Commented by ajfour last updated on 04/Oct/19

$$Great,ihaddoneblunders,$$$$edited,corrected.Yoursolution$$$$isshorter.ThankyouSir.$$

Commented by mr W last updated on 04/Oct/19

$$itreatedthetwoballsasasingle$$$$objectwhosecenterofmassmust$$$$beverticallybeneaththecenterof$$$$hollowspherewheninequilibrium.$$

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