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Question Number 131071 by mr W last updated on 01/Feb/21

Commented by mr W last updated on 01/Feb/21

$$asmallballofmassmisfixedon$$$$theinnerwallofahollowcylinder$$$$ofradiusRandmassM.$$$$thecylinderisreleasedfromrest$$$$atthepositionasshownandrolls$$$$ontheground.$$$$findthecontactforceNintermsof$$$$x.$$

Answered by ajfour last updated on 01/Feb/21

Commented by ajfour last updated on 01/Feb/21

$$mgR(1-\cos \theta )={Mu}^2+K$$$$K=\frac{1}{2}m[{(\omega R\cos \theta +u)}^2+{\left(\omega R\sin \theta \right)}^2]$$$$N-(M+m)g=m\frac{d^{2}y}{{dt}^2}$$$$y=R+R\cos \theta$$$$\frac{dy}{dt}=-\omega R\sin \theta$$$$\frac{d^{2}y}{{dt}^2}=-{\omega }^{2}R\cos \theta -\alpha R\sin \theta$$$$\theta =\frac{x}{R}$$$$\omega R=u$$$$mgR(1-\cos \theta )={Mu}^2+K$$$$K={mu}^2(1+\cos \theta )$$$$u^2=\frac{mgR(1-\cos \theta )}{M+m(1+\cos \theta )}={\omega }^2{R}^2$$$$N=(M+m)g+m\frac{d^{2}y}{{dt}^2}$$$$\frac{d^{2}y}{{dt}^2}=-{\omega }^{2}R\cos \theta -\alpha R\sin \theta$$$$u^2=\frac{mgR(1-\cos \theta )}{M+m(1+\cos \theta )}={\omega }^2{R}^2$$$$\alpha =\frac{\omega d\omega }{d\theta };\theta =\frac{x}{R}$$$$$$

Commented by ajfour last updated on 01/Feb/21

$$\frac{1}{2}{Mu}^2+\frac{1}{2}\left({MR}^2\right){\omega }^2={Mu}^2$$

Commented by mr W last updated on 01/Feb/21

$$yes.isee.$$

Answered by mr W last updated on 01/Feb/21

Commented by mr W last updated on 02/Feb/21

$$x=R\theta$$$$\omega =\frac{d\theta }{dt}$$$$\alpha =\frac{d\omega }{dt}$$$$OS=f=\frac{m}{M+m}\times R=\frac{R}{\frac{M}{m}+1}$$$$let\lambda =\frac{M}{m}+1$$$$\Rightarrow f=\frac{R}{\lambda }$$$$I_B={MR}^2+{MR}^2+m{\left(2R\cos \frac{\theta }{2}\right)}^2$$$$=2{MR}^2+2{mR}^2(\cos \theta +1)$$$$=2R^2[M+m(1+\cos \theta )]$$$$\frac{1}{2}{I}_B{\omega }^2=mgR(1-\cos \theta )$$$$R^2[M+m(1+\cos \theta )]{\omega }^2=mgR(1-\cos \theta )$$$${\omega }^2=\frac{g(1-\cos \theta )}{R(\frac{M}{m}+1+\cos \theta )}=\frac{g(1-\cos \theta )}{R(\lambda +\cos \theta )}$$$$\Rightarrow \omega =\sqrt{\frac{g}{R}}\sqrt{\frac{1-\cos \theta }{\lambda +\cos \theta }}$$$$\frac{d\theta }{dt}=\sqrt{\frac{g}{R}}\sqrt{\frac{1-\cos \theta }{\lambda +\cos \theta }}$$$$\int \sqrt{\frac{\lambda +\cos \theta }{1-\cos \theta }}d\theta =\sqrt{\frac{g}{R}}\int dt$$$$\Rightarrow t=\sqrt{\frac{R}{g}}{\int }_0^{\theta }\sqrt{\frac{\lambda +\cos \theta }{1-\cos \theta }}d\theta$$$$periodT$$$$T=2\sqrt{\frac{R}{g}}{\int }_0^{\pi }\sqrt{\frac{\lambda +\cos \theta }{1-\cos \theta }}d\theta$$$$$$$$\frac{d}{dt}\left(I_B\omega \right)=(M+m)gf\sin \theta$$$$I_B\alpha +{\omega }^2\frac{{dI}_B}{d\theta }=\frac{gR}{\lambda }(M+m)\sin \theta$$$$2R^2[M+m(1+\cos \theta )]\alpha -2{\omega }^2{R}^{2}m\sin \theta =\frac{gR}{\lambda }(M+m)\sin \theta$$$$(\lambda +1+\cos \theta )\alpha =(\frac{g}{2R}+{\omega }^2)\sin \theta$$$$\alpha =\frac{g}{2R}\times \frac{(\lambda +2-\cos \theta )\sin \theta }{(\lambda +\cos \theta )(\lambda +1+\cos \theta )}$$$$$$$$v_{y,S}=\omega f\sin \theta =\frac{R}{\lambda }\omega \sin \theta$$$$a_{y,S}=\frac{{dv}_y}{dt}=\frac{R}{\lambda }(\alpha \sin \theta +{\omega }^2\cos \theta )$$$$a_{y,S}=\frac{g(1-\cos \theta )[{\cos }^2\theta +3(\lambda +1)\cos \theta +\lambda +2]}{2\lambda (\lambda +\cos \theta )(\lambda +1+\cos \theta )}$$$$(M+m)g-N=(M+m)a_{y,S}$$$$N=(M+m)(g-a_{y,S})$$$$\Rightarrow \frac{N}{(M+m)g}=1-\frac{(1-\cos \theta )[{\cos }^2\theta +3(\lambda +1)\cos \theta +\lambda +2]}{2\lambda (\lambda +\cos \theta )(\lambda +1+\cos \theta )}$$$$\frac{N_{max}}{(M+m)g}=1+\frac{2}{\lambda (\lambda -1)}$$

Commented by mr W last updated on 02/Feb/21

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