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Question Number 29322 by ajfour last updated on 07/Feb/18

Commented by ajfour last updated on 07/Feb/18

$$Findaccelerationofsphereas$$$$afunctionofangle\theta .$$$$Systemisreleasedat\theta ={\theta }_0(<90°);$$$$\left(Allsurfacesarefrictionless\right).$$

Commented by ajfour last updated on 07/Feb/18

$$Pleasesolveorcheckmysolution$$$$mrWSir.$$

Commented by ajfour last updated on 07/Feb/18

Commented by ajfour last updated on 07/Feb/18

$$x=R\cot \frac{\theta }{2}$$$$v=-\frac{\omega R}{2}\mathrm{cosec}{}^2\frac{\theta }{2}$$$$\Rightarrow {\mathit{v}}^2=\frac{{\mathit{\omega }}^2{\mathit{R}}^2}{4}\mathrm{cosec}{}^4\frac{\theta }{2}....\left(i\right)$$$$a=\frac{dv}{dt}=-\frac{\alpha R}{2}\mathrm{cosec}{}^2\frac{\theta }{2}$$$$+\frac{{\omega }^{2}R}{2}\mathrm{cosec}{}^2\frac{\theta }{2}\cot \frac{\theta }{2}...\left(ii\right)$$$$N\sin \theta =ma....\left(iii\right)$$$$NR\cot \frac{\theta }{2}-\frac{MgL\cos \theta }{2}=\left(\frac{{ML}^2}{3}\right)\alpha$$$$$$$$\Rightarrow \alpha =\frac{3}{{ML}^2}(\frac{maR\cot \frac{\theta }{2}}{\sin \theta }-\frac{MgL\cos \theta }{2})$$$$.....\left(iv\right)$$$$\frac{MgL}{2}(\sin {\theta }_0-\sin \theta )=\frac{1}{2}{mv}^2+\frac{1}{2}\left(\frac{{ML}^2}{3}\right){\omega }^2$$$$using\left(i\right):$$$$\frac{MgL}{2}(\sin {\theta }_0-\sin \theta )=\frac{m{\omega }^2{R}^2}{8}\mathrm{cosec}{}^4\frac{\theta }{2}$$$$+\frac{{ML}^2{\omega }^2}{6}$$$${\omega }^2=\frac{\frac{MgL}{2}(\sin {\theta }_0-\sin \theta )}{\frac{{ML}^2}{6}+\frac{{mR}^2}{8}\mathrm{cosec}{}^4\frac{\theta }{2}}....\left(A\right)$$$$Usingeq.\left(A\right),\left(iii\right),\left(iv\right)ineq.\left(\mathit{i}\mathit{i}\right)$$$$a=\frac{R}{2}\mathrm{cosec}{}^2\frac{\theta }{2}\left(\frac{3}{{ML}^2}\right)(\frac{MgL\cos \theta }{2}-\frac{maR\cot \frac{\theta }{2}}{\sin \theta })$$$$+\frac{R}{2}\mathrm{cosec}{}^2\frac{\theta }{2}\cot \frac{\theta }{2}\left[\frac{\frac{MgL}{2}(\sin {\theta }_0-\sin \theta )}{\frac{{ML}^2}{6}+\frac{{mR}^2}{8}\mathrm{cosec}{}^4\frac{\theta }{2}}\right]$$$$$$$$a[1+\frac{3{mR}^2\mathrm{cosec}{}^2\frac{\theta }{2}\cot \frac{\theta }{2}}{2{ML}^2\sin \theta }]=$$$$\frac{R}{2}\mathrm{cosec}{}^2\frac{\theta }{2}[\frac{3g\cos \theta }{2L}+\frac{\frac{MgL}{2}(\sin {\theta }_0-\sin \theta )\cot \frac{\theta }{2}}{\frac{{ML}^2}{6}+\frac{{mR}^2}{8}\mathrm{cosec}{}^4\frac{\theta }{2}}].$$

Commented by mrW2 last updated on 07/Feb/18

$$solutioniscorrectsir!$$

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