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Question Number 131823 by mr W last updated on 09/Feb/21

Commented by mr W last updated on 10/Feb/21

$$asmallsphereofmassmisreleased$$$$fromrestatpositionasshown.$$$$boththesphereandthesemicylinder$$$$movewithoutfriction.$$$$findthetimethesphereneedsto$$$$hittheground.$$

Answered by mr W last updated on 09/Feb/21

Commented by mr W last updated on 10/Feb/21

Commented by mr W last updated on 10/Feb/21

Commented by mr W last updated on 09/Feb/21

$${\theta }_0={\cos }^{-1}\frac{R-r}{R+r}$$$${\theta }_1={\sin }^{-1}\frac{r}{R+r}$$

Commented by mr W last updated on 11/Feb/21

$$let\omega =\frac{d\theta }{dt},\alpha =\frac{d\omega }{dt},\mu =\frac{M}{m}$$$$x_Q=r+(R+r)\cos \theta$$$$V=\frac{{dx}_Q}{dt}=\omega \frac{{dx}_Q}{d\theta }=-(R+r)\omega \sin \theta$$$$A=\frac{dV}{dt}=-(R+r)(\alpha \sin \theta +{\omega }^2\cos \theta )$$$$$$$$y_P=(R+r)\sin \theta$$$$v=-\frac{{dy}_P}{dt}=-\omega \frac{{dy}_P}{d\theta }=-(R+r)\omega \cos \theta$$$$a=\frac{dv}{dt}=-(R+r)(\alpha \cos \theta -{\omega }^2\sin \theta )$$$$$$$$\frac{{MV}^2+{mv}^2}{2}=mg(R+r)(\sin {\theta }_0-\sin \theta )$$$$\frac{M}{m}{V}^2+v^2=2g(R+r)(\sin {\theta }_0-\sin \theta )$$$$$$$$[\frac{M}{m}{\sin }^2\theta +{\cos }^2\theta ]{\omega }^2{(R+r)}^2=2g(R+r)(\sin {\theta }_0-\sin \theta )$$$$\Rightarrow {\omega }^2=\frac{2g}{R+r}\times \frac{\sin {\theta }_0-\sin \theta }{\mu {\sin }^2\theta +{\cos }^2\theta }$$$$$$$$N\cos \theta =MA$$$$\Rightarrow N=-\frac{M(R+r)}{\cos \theta }(\alpha \sin \theta +{\omega }^2\cos \theta )$$$$mg-N\sin \theta =ma$$$$N=\frac{m(g-a)}{\sin \theta }$$$$\Rightarrow N=\frac{m(R+r)}{\sin \theta }(\frac{g}{R+r}+\alpha \cos \theta -{\omega }^2\sin \theta )$$$$$$$$\frac{m(R+r)}{\sin \theta }(\frac{g}{R+r}+\alpha \cos \theta -{\omega }^2\sin \theta )=-\frac{M(R+r)}{\cos \theta }(\alpha \sin \theta +{\omega }^2\cos \theta )$$$$\frac{g}{R+r}+\alpha \cos \theta -{\omega }^2\sin \theta =-\frac{\mu \sin \theta }{\cos \theta }(\alpha \sin \theta +{\omega }^2\cos \theta )$$$$\alpha =-\frac{\cos \theta }{\mu {\sin }^2\theta +{\cos }^2\theta }[\frac{g}{R+r}+(\mu -1)\sin \theta {\omega }^2]$$$$\Rightarrow \alpha =-\frac{g}{R+r}\times \frac{\cos \theta }{\mu {\sin }^2\theta +{\cos }^2\theta }[1+\frac{2(\mu -1)\sin \theta (\sin {\theta }_0-\sin \theta )}{\mu {\sin }^2\theta +{\cos }^2\theta }]$$$$$$$$\Rightarrow N=-M(R+r)(\frac{\sin \theta }{\cos \theta }\alpha +{\omega }^2)$$$$\Rightarrow N=Mg\{\frac{\sin \theta }{\mu {\sin }^2\theta +{\cos }^2\theta }[1+\frac{2(\mu -1)\sin \theta (\sin {\theta }_0-\sin \theta )}{\mu {\sin }^2\theta +{\cos }^2\theta }]-\frac{2(\sin {\theta }_0-\sin \theta )}{\mu {\sin }^2\theta +{\cos }^2\theta }\}$$$$\Rightarrow \frac{N}{mg}=\frac{\mu }{\mu {\sin }^2\theta +{\cos }^2\theta }[\sin \theta -\frac{2(\sin {\theta }_0-\sin \theta )}{\mu {\sin }^2\theta +{\cos }^2\theta }]$$$$$$$$sayN=0at\theta ={\theta }_2.$$$$N=0\Rightarrow \sin \theta -\frac{2(\sin {\theta }_0-\sin \theta )}{\mu {\sin }^2\theta +{\cos }^2\theta }=0$$$$\sin \theta =\frac{2(\sin {\theta }_0-\sin \theta )}{(\mu -1){\sin }^2\theta +1}$$$${\sin }^3\theta +\frac{3}{\mu -1}\sin \theta -\frac{2}{\mu -1}\sin {\theta }_0=0$$$$if\mu =1:$$$$3\sin \theta -2\sin {\theta }_0=0$$$$\Rightarrow \sin \theta =\frac{2}{3}\sin {\theta }_0$$$$\Rightarrow {\theta }_2={\sin }^{-1}\left(\frac{2\sin {\theta }_0}{3}\right)$$$$for\mu \ne 1:$$$$let\lambda =\frac{1}{\mu -1}=\frac{m}{M-m}$$$${\sin }^3\theta +3\lambda \sin \theta -2\lambda \sin {\theta }_0=0$$$$\mathrm{\Delta }={\lambda }^3+{\lambda }^2{\sin }^2{\theta }_0={\lambda }^2(\lambda +{\sin }^2{\theta }_0)$$$$for\lambda +{\sin }^2{\theta }_0=\frac{1}{\mu -1}+{\sin }^2{\theta }_0>0,i.e.$$$$for\mu >1:$$$$\sin \theta =\sqrt[3]{\lambda (\sqrt{\lambda +{\sin }^2{\theta }_0}+\sin {\theta }_0)}-\sqrt[3]{\lambda (\sqrt{\lambda +{\sin }^2{\theta }_0}-\sin {\theta }_0)}$$$$\Rightarrow {\theta }_2={\sin }^{-1}[\sqrt[3]{\frac{1}{\mu -1}(\sqrt{\frac{1}{\mu -1}+{\sin }^2{\theta }_0}+\sin {\theta }_0)}-\sqrt[3]{\frac{1}{\mu -1}(\sqrt{\frac{1}{\mu -1}+{\sin }^2{\theta }_0}-\sin {\theta }_0)}]$$$$for\mu <1:$$$$\sin \theta =\frac{2}{\sqrt{1-\mu }}\sin [\frac{1}{3}{\sin }^{-1}\left(\sqrt{1-\mu }\sin {\theta }_0\right)+\frac{2k\pi }{3}]$$$$\Rightarrow {\theta }_2={\sin }^{-1}\left\{\frac{2}{\sqrt{1-\mu }}\sin \left[\frac{1}{3}{\sin }^{-1}\left(\sqrt{1-\mu }\sin {\theta }_0\right)\right]\right\}$$$$weget{\theta }_2>{\theta }_1,seediagram.$$$$thatmeansthespherelossescontact$$$$beforeithitstheground.afterthis$$$$pointthespherehasfreefall.$$$$$$$$T_1=timefor\theta from{\theta }_{0}to{\theta }_2$$$$T_2=timeforfreefallafter{\theta }_2$$$$\Rightarrow \omega =\frac{d\theta }{dt}=-\sqrt{\frac{2g}{R+r}}\times \sqrt{\frac{\sin {\theta }_0-\sin \theta }{\mu {\sin }^2\theta +{\cos }^2\theta }}$$$$\Rightarrow T_1=\sqrt{\frac{R+r}{2g}}{\int }_{{\theta }_2}^{{\theta }_0}\sqrt{\frac{1+(\mu -1){\sin }^2\theta }{\sin {\theta }_0-\sin \theta }}d\theta$$$$$$$$v=\sqrt{2g(R+r)}\times \cos \theta \sqrt{\frac{\sin {\theta }_0-\sin \theta }{\mu {\sin }^2\theta +{\cos }^2\theta }}$$$$v_2=\sqrt{2g(R+r)}\times \cos {\theta }_2\sqrt{\frac{\sin {\theta }_0-\sin {\theta }_2}{\mu {\sin }^2{\theta }_2+{\cos }^2{\theta }_2}}$$$$\mathrm{\Delta }h=(R+r)\sin {\theta }_2-r$$$$\mathrm{\Delta }h=v_2{T}_2+\frac{1}{2}{gT}_2^2$$$$T_2=\frac{1}{g}(\sqrt{v_2^2+2g\mathrm{\Delta }h}-v_2)$$$$$$$$totaltime:$$$$T=T_1+T_2$$

Commented by mr W last updated on 10/Feb/21

Commented by mr W last updated on 10/Feb/21

$$Godblessyoutoo!$$

Commented by otchereabdullai@gmail.com last updated on 10/Feb/21

$$\mathrm{Amen}$$

Commented by otchereabdullai@gmail.com last updated on 10/Feb/21

$$\mathrm{You}\mathrm{are}\mathrm{too}\mathrm{much}\mathrm{prof}\mathrm{W}\mathrm{God}\mathrm{bless}\mathrm{you}$$

Commented by mr W last updated on 10/Feb/21

Commented by ajfour last updated on 10/Feb/21

$$N\cos \theta =MA$$$$mg-N\sin \theta =ma$$$$a\sin \theta =A\cos \theta$$$$\Rightarrow mg=ma+\frac{MA\sin \theta }{\cos \theta }$$$$\Rightarrow mg=ma+\frac{Ma\sin {}^2\theta }{\cos {}^2\theta }$$$$a=\frac{g}{1+\lambda \tan {}^2\theta }where\lambda =\frac{M}{m}$$$$A=\frac{g\tan \theta }{1+\lambda \tan {}^2\theta }$$$$\Rightarrow NormalreactionNisnot$$$$zerobeforeballtouchesground;$$$$soitappearsfromthe$$$$accelerationexpressions...$$$$What{d}^{\prime }yathinkSir?$$$$\frac{dv}{dt}=\frac{g}{1+\lambda \tan {}^2\theta }$$$$y=(R+r)\sin \theta$$$$v=(-\frac{d\theta }{dt})(R+r)\cos \theta$$$$\Rightarrow \int vdv=-g(R+r)\int \frac{\cos \theta d\theta }{1+\lambda \tan {}^2\theta }$$$$\frac{v^2}{2}=g(R+r){\int }_{{\theta }_0}^{\theta }\frac{(1-\sin {}^2\theta )d\left(\sin \theta \right)}{1+(\lambda -1)\sin {}^2\theta }$$$$\frac{v^2}{2}=\frac{g(R+r)}{\lambda -1}\int \frac{(1-s^2)ds}{s^2+\frac{1}{\lambda -1}}$$$$\frac{v^2(\lambda -1)}{2g(R+r)}=-\int ds+\frac{1}{\lambda -1}\int \frac{ds}{s^2+\frac{1}{\lambda -1}}$$$$\frac{v^2(\lambda -1)}{2g(R+r)}=-s+\frac{1}{\sqrt{\lambda -1}}{\tan }^{-1}\left(s\sqrt{\lambda -1}\right))$$$$$$

Commented by mr W last updated on 10/Feb/21

$$thankssofarsir!$$$$iagreewithv\sin \theta =V\cos \theta ,butit$$$${doesn}^{\prime }tfollowthata\sin \theta =A\cos \theta .$$$$pleaserecheck!$$

Commented by mr W last updated on 11/Feb/21

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