Question and Answers Forum

All Questions      Topic List

Mechanics Questions

Previous in All Question      Next in All Question      

Previous in Mechanics      Next in Mechanics      

Question Number 71421 by ajfour last updated on 15/Oct/19

Commented by ajfour last updated on 15/Oct/19

$$Findspeedofsphereasitrolls$$$$downthewedge,andisaboutto$$$$touchtheground.Alsofindthe$$$$leastcoefficientoffriction,$$$$sothatspherecomesrollingall$$$$theway;wedge{isn}^{\prime }tfixed,$$$$assumethatgroundissmooth.$$

Answered by mr W last updated on 16/Oct/19

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

$$v=\frac{ds}{dt}=\omega r$$$$A=r\alpha$$$$\alpha =\frac{d\omega }{dt}=\frac{d\omega }{ds}\times \frac{ds}{dt}=\frac{v}{r}\times \frac{dv}{ds}$$$$I=\frac{2{mr}^2}{5}$$$$I\alpha =rf$$$$\alpha =\frac{rf}{I}=\frac{5f}{2mr}$$$$$$$$Ma=-f\cos \theta +N\sin \theta$$$$a=\frac{-f\cos \theta +N\sin \theta }{M}$$$$$$$$f=mg\sin \theta +ma\cos \theta -mr\alpha$$$$\frac{7}{2}f=mg\sin \theta +ma\cos \theta$$$$N=mg\cos \theta -ma\sin \theta$$$$\frac{7}{2}f=mg\sin \theta +\frac{m}{M}(-f\cos \theta +N\sin \theta )\cos \theta$$$$\Rightarrow (\frac{7}{2}+\frac{m}{M}{\cos }^2\theta )f-\frac{m}{M}\sin \theta \cos \theta N=mg\sin \theta$$$$N=mg\cos \theta -\frac{m}{M}(-f\cos \theta +N\sin \theta )\sin \theta$$$$\Rightarrow -\frac{m}{M}\sin \theta \cos \theta f+(1+\frac{m}{M}{\sin }^2\theta )N=mg\cos \theta$$$$\Rightarrow f=\frac{mg\sin \theta (1+\frac{m}{M}{\sin }^2\theta )-mg\cos \theta (-\frac{m}{M}\sin \theta \cos \theta )}{(\frac{7}{2}+\frac{m}{M}{\cos }^2\theta )(1+\frac{m}{M}{\sin }^2\theta )-{(-\frac{m}{M}\sin \theta \cos \theta )}^2}$$$$\Rightarrow f=\frac{mg\sin \theta (1+\frac{m}{M})}{\frac{7}{2}+\frac{m}{M}(\frac{5}{2}{\sin }^2\theta +1)}$$$$$$$$\Rightarrow N=\frac{mg\cos \theta (\frac{7}{2}-\frac{m}{M}{\cos }^2\theta )-mg\sin \theta (-\frac{m}{M}\sin \theta \cos \theta )}{(\frac{7}{2}+\frac{m}{M}{\cos }^2\theta )(1+\frac{m}{M}{\sin }^2\theta )-{(-\frac{m}{M}\sin \theta \cos \theta )}^2}$$$$\Rightarrow N=\frac{mg\cos \theta (\frac{7}{2}+\frac{m}{M})}{\frac{7}{2}+\frac{m}{M}(\frac{5}{2}{\sin }^2\theta +1)}$$$$f⩽\mu N$$$$\frac{mg\sin \theta (1+\frac{m}{M})}{\frac{7}{2}+\frac{m}{M}(\frac{5}{2}{\sin }^2\theta +1)}⩽\mu \frac{mg\cos \theta (\frac{7}{2}+\frac{m}{M})}{\frac{7}{2}+\frac{m}{M}(\frac{5}{2}{\sin }^2\theta +1)}$$$$\Rightarrow \mu ⩾\frac{1+\frac{m}{M}}{\frac{7}{2}+\frac{m}{M}}\times \tan \theta$$$$\Rightarrow \alpha =\frac{5f}{2mr}=\frac{5g}{2r}\times \frac{\sin \theta (1+\frac{m}{M})}{\frac{7}{2}+\frac{m}{M}(\frac{5}{2}{\sin }^2\theta +1)}$$$$\Rightarrow \frac{v}{r}\times \frac{dv}{ds}=\frac{5g}{2r}\times \frac{\sin \theta (1+\frac{m}{M})}{\frac{7}{2}+\frac{m}{M}(\frac{5}{2}{\sin }^2\theta +1)}$$$$\Rightarrow vdv=\frac{\frac{5}{2}g\sin \theta (1+\frac{m}{M})}{\frac{7}{2}+\frac{m}{M}(\frac{5}{2}{\sin }^2\theta +1)}ds=\lambda ds$$$$with\lambda =\frac{\frac{5}{2}g\sin \theta (1+\frac{m}{M})}{\frac{7}{2}+\frac{m}{M}(\frac{5}{2}{\sin }^2\theta +1)}$$$$=\frac{5g\sin \theta }{\frac{5M+2M+2m}{M+m}+\frac{5m(1-{\cos }^2\theta )}{M+m}}$$$$=\frac{g\sin \theta }{\frac{7}{5}-\frac{m{\cos }^2\theta }{M+m}}$$$$\Rightarrow \frac{v^2}{2}=\lambda s$$$$\Rightarrow v=\sqrt{2\lambda s}=\sqrt{\frac{2gs\sin \theta }{\frac{7}{5}-\frac{m{\cos }^2\theta }{M+m}}}$$$$whentheballisabouttotouchthe$$$$ground,s=\frac{h}{\sin \theta }-\frac{r}{\tan \frac{\theta }{2}}$$$$\Rightarrow v=\sqrt{\frac{2g(h-2r{\cos }^2\frac{\theta }{2})}{\frac{7}{5}-\frac{m{\cos }^2\theta }{M+m}}}$$

Commented by ajfour last updated on 18/Oct/19

Commented by ajfour last updated on 18/Oct/19

$$s=\frac{h}{\sin \theta }-r\tan \frac{\theta }{2}$$

Commented by ajfour last updated on 18/Oct/19

$$mgs\sin \theta =\frac{{MV}^2}{2}+\frac{1}{2}\left(\frac{2{mr}^2}{5}\right)\frac{v^2}{r^2}$$$$+\frac{m}{2}[{(v\cos \theta -V)}^2+v^2\sin {}^2\theta ]$$$$\mathrm{\&}mv\cos \theta =(M+m)V$$$$\Rightarrow V=\frac{mv\cos \theta }{M+m}$$$$\mathrm{\&}v\cos \theta -V=\frac{Mv\cos \theta }{M+m}$$$$substitutinginfirsteq.$$$$mgs\sin \theta ={mv}^2(\frac{1}{5}+\frac{\sin {}^2\theta }{2})$$$$+\frac{M}{2}{\left(\frac{mv\cos \theta }{M+m}\right)}^2+\frac{m}{2}{\left(\frac{Mv\cos \theta }{M+m}\right)}^2$$$$\Rightarrow mgs\sin \theta ={mv}^2(\frac{1}{5}+\frac{\sin {}^2\theta }{2})$$$$+\frac{{Mmv}^2\cos {}^2\theta }{2(M+m)}$$$$\Rightarrow v^2=\frac{gs\sin \theta }{\frac{1}{5}+\frac{\sin {}^2\theta }{2}+\frac{M\cos {}^2\theta }{2(M+m)}}$$$$v=\sqrt{\frac{gs\sin \theta }{\frac{1}{5}+\frac{\sin {}^2\theta }{2}+\frac{M\cos {}^2\theta }{2(M+m)}}}$$$$v=\sqrt{\frac{2gs\sin \theta }{\frac{7}{5}-\frac{m\cos {}^2\theta }{M+m}}}$$$$Andwiths=\frac{h}{\sin \theta }-r\tan \frac{\theta }{2}$$$$v=\sqrt{\frac{g(h-r\tan \frac{\theta }{2}\sin \theta )}{\frac{1}{5}+\frac{\sin {}^2\theta }{2}+\frac{M\cos {}^2\theta }{2(M+m)}}}.$$$$v\cos \theta -V=\frac{Mv\cos \theta }{M+m}$$$$\Rightarrow v_{actual}^2={\left(\frac{Mv\cos \theta }{M+m}\right)}^2+{\left(v\sin \theta \right)}^2$$$$v_{actual}=v\sqrt{{\left(\frac{M}{M+m}\right)}^2\cos {}^2\theta +\sin {}^2\theta }$$$$thisisspeedofsphere$$$$relativetowedge..$$$$$$

Commented by mr W last updated on 17/Oct/19

$$yourmethodismoreeffective,very$$$$nice!buti{didn}^{\prime }tfindwheremyerror$$$$is.$$

Commented by ajfour last updated on 18/Oct/19

$$N\sin \theta -f\cos \theta =Ma...\left(i\right)$$$$N+ma\sin \theta =mg\cos \theta ...\left(ii\right)$$$$mg\sin \theta +ma\cos \theta =f+m\alpha r...\left(iii\right)$$$$rf=\frac{2}{5}{mr}^2\alpha ....\left(iv\right)$$$$\Rightarrow ma\sin {}^2\theta +Ma+f\cos \theta$$$$=mg\sin \theta \cos \theta$$$$\Rightarrow (M+m\sin {}^2\theta )\left(\frac{f-mg\sin \theta +m\alpha r}{m\cos \theta }\right)$$$$+f\cos \theta =mg\sin \theta \cos \theta$$$$f=\frac{mg\sin \theta \cos \theta +\frac{(M+m\sin {}^2\theta )(g\sin \theta -\alpha r)}{\cos \theta }}{\cos \theta +\frac{M+m\sin {}^2\theta }{m\cos \theta }}$$$$=\frac{m[(M+m)g\sin \theta -\alpha r(M+m\sin {}^2\theta )]}{M+m}$$$$=\frac{2}{5}mr\alpha \left[from\left(iv\right)\right]\Rightarrow$$$$\alpha r[M+m\sin {}^2\theta +\frac{2}{5}(M+m)]$$$$=(M+m)g\sin \theta$$$$\Rightarrow \alpha r=\frac{g\sin \theta }{\frac{7}{5}-\frac{m\cos {}^2\theta }{M+m}}$$$$\frac{vdv}{ds}=\alpha r\Rightarrow v=\sqrt{2\alpha rs}$$$$\Rightarrow v=\sqrt{\frac{2gs\sin \theta }{\frac{7}{5}-\frac{m\cos {}^2\theta }{M+m}}}.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com