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Question Number 23472 by ajfour last updated on 31/Oct/17

Commented by ajfour last updated on 31/Oct/17

$$Q.23390(continued..)$$

Answered by ajfour last updated on 31/Oct/17

$$Applying\mathrm{\Sigma }{F}_x=\mathrm{\Sigma }{ma}_{x}fromthe$$$$groundframe:$$$$f=(M+m)\left(\alpha R\right)+\frac{m\left(\alpha R\right)\cos \theta }{2}$$$$-m{\omega }^2\left(\frac{R}{2}\right)\sin \theta ...\left(i\right)$$$$Applying\mathrm{\Sigma }\tau =I_{net}\alpha fromthe$$$$frameofcentreofring:$$$$\frac{mgR\sin \theta }{2}-fR-\left(m\alpha R\right)\left(\frac{R}{2}\cos \theta \right)$$$$=({MR}^2+\frac{{mR}^2}{3})\alpha ...\left(ii\right)$$$$eliminating\mathit{f}from\left(i\right)and\left(ii\right):$$$$\frac{mg\sin \theta }{2}-(M+m)\left(\alpha R\right)$$$$-\frac{m\left(\alpha R\right)\cos \theta }{2}+\frac{m{\omega }^{2}R\sin \theta }{2}$$$$-\frac{m\left(\alpha R\right)\cos \theta }{2}=(M+\frac{m}{3})\left(\alpha R\right)$$$$\Rightarrow \left(\alpha R\right)[M+\frac{m}{3}+M+m+m\cos \theta ]$$$$=\frac{m\sin \theta }{2}({\omega }^{2}R+g)$$$$\Rightarrow \alpha =\frac{m({\omega }^2+g/R)\sin \theta }{2[2M+\frac{4m}{3}+m\cos \theta ]}$$$$\Rightarrow \frac{\omega d\omega }{d\theta }=\frac{m({\omega }^2+\frac{g}{R})\sin \theta }{2[2M+\frac{4m}{3}+m\cos \theta ]}$$$${\int }_0^{\omega }\frac{2\omega d\omega }{{\omega }^2+\frac{g}{R}}={\int }_0^{\theta }\frac{m\sin \theta }{2M+\frac{4m}{3}+m\cos \theta }$$$$\ln \left(\frac{{\omega }^2+\frac{g}{R}}{\frac{g}{R}}\right)=\ln \left(\frac{2M+\frac{4m}{3}+m}{2M+\frac{4m}{3}+m\cos \theta }\right)$$$$\frac{{\omega }^{2}R}{g}+1=1-\frac{m(1-\cos \theta )}{2M+\frac{4m}{3}+m\cos \theta }$$$${\mathit{\omega }}^2=\frac{\mathit{g}}{\mathit{R}}\left(\frac{1-\cos \mathit{\theta }}{\frac{2\mathit{M}}{\mathit{m}}+\frac{4}{3}+\cos \mathit{\theta }}\right)$$$$And\mathit{\alpha }=\frac{1}{2}\frac{\mathit{d}\left({\mathit{\omega }}^2\right)}{\mathit{d}\mathit{\theta }}.$$$$Toobtainthetimeforthereturn,$$$$\Rightarrow \mathit{\omega }=\frac{\mathit{d}\mathit{\theta }}{\mathit{d}\mathit{t}}=\sqrt{\frac{g}{R}}{\left[\frac{1-\cos \theta }{b^2+\cos \theta }\right]}^{1/2}$$$$\Rightarrow T=4\sqrt{\frac{R}{g}}{\int }_0^{\pi }{\left[\frac{b^2+\cos \theta }{1-\cos \theta }\right]}^{1/2}d\theta$$$$where{b}^2=\frac{2M}{m}+\frac{4}{3}.$$$$....mightcontinue...$$

Commented by mrW1 last updated on 31/Oct/17

$$\mathrm{how}\mathrm{is}\mathrm{it}\mathrm{to}\mathrm{explain},\mathrm{that}\mathrm{the}\mathrm{time}$$$$\mathrm{from}\theta =0\mathrm{is}\mathrm{not}\mathrm{convergent}?$$

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