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Question Number 213838 by ajfour last updated on 18/Nov/24

Commented by ajfour last updated on 18/Nov/24

$$Asolidballisreleasedoverafixed$$$$cylindricalwedgeasshown.Friction$$$$issufficient.Ifjustaftertheball$$$$leavesthecurvedsurfacedueto$$$$Normalreactionvanishing,it$$$$thereupontouchesthegroundbelow$$$$thenfindtheradiusratior/R.$$

Answered by mr W last updated on 18/Nov/24

Commented by mr W last updated on 20/Nov/24

$$rotationofball$$$$\varphi =(1+\frac{R}{r})\theta$$$$\frac{d\varphi }{dt}=(1+\frac{R}{r})\omega with\omega =\frac{d\theta }{dt}$$$$\alpha =\frac{d^2\varphi }{{dt}^2}=(1+\frac{R}{r})\omega \times \frac{d\omega }{d\theta }$$$$(\frac{2{mr}^2}{5}+{mr}^2)\times (1+\frac{R}{r})\omega \times \frac{d\omega }{d\theta }=mgr\sin \theta$$$$\omega d\omega =\frac{5g}{7(r+R)}\sin \theta d\theta$$$${\int }_0^{\omega }\omega d\omega =\frac{5g}{7(r+R)}{\int }_0^{\theta }\sin \theta d\theta$$$$\frac{{\omega }^2}{2}=\frac{5g(1-\cos \theta )}{7(r+R)}$$$$\Rightarrow {\omega }^2=\frac{10g(1-\cos \theta )}{7(r+R)}$$$$-+-+-+-+-+-+-$$$$alternatively:$$$$mg(r+R)(1-\cos \theta )=\frac{1}{2}\left(\frac{2{mr}^2}{5}\right){(1+\frac{R}{r})}^2{\omega }^2+\frac{m}{2}\times r^2\times {(1+\frac{R}{r})}^2{\omega }^2$$$$2g(1-\cos \theta )=\frac{7}{5}(r+R){\omega }^2$$$$\Rightarrow {\omega }^2=\frac{10g(1-\cos \theta )}{7(r+R)}$$$$-+-+-+-+-+-+-$$$$mg\cos \theta -N=\frac{m}{r+R}\times {(1+\frac{R}{r})}^2{r}^2{\omega }^2$$$$N=mg\cos \theta -m(r+R)\times \frac{10g(1-\cos \theta )}{7(r+R)}$$$$N=\frac{(17\cos \theta -10)mg}{7}$$$$N=0\Rightarrow 17\cos \theta -10=0\Rightarrow \cos \theta =\frac{10}{17}$$$$\Rightarrow \theta ={\cos }^{-1}\frac{10}{17}\approx 53.97°$$$$\theta =\phi$$$$\Rightarrow \cos \phi =\cos \theta =\frac{10}{17}$$$$\frac{r}{r+R}=\frac{10}{17}$$$$\Rightarrow \frac{r}{R}=\frac{1}{\frac{17}{10}-1}=\frac{10}{7}✓$$

Commented by mr W last updated on 18/Nov/24

$$angle\theta atwhichtheballislosing$$$$thecontactwiththewedgeis$$$$independentfromtheirradii!$$

Commented by ajfour last updated on 18/Nov/24

$$lossinG.P.E.=gainintotalK.E.$$$$mgR=\frac{1}{2}\left(\frac{7}{5}{mr}^2\right){\omega }^2$$$$mg\cos \theta =m{\omega }^2{r}^2/(R+r)$$$$nowdividing$$$$\frac{R}{\cos \theta }=\frac{7}{10}(R+r)$$$$now\cos \theta =\frac{r}{R+r}$$$$\Rightarrow \frac{R}{r}=\frac{7}{10}★$$

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