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Question Number 214815 by mr W last updated on 20/Dec/24

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

$$asolidballwithmass\mathit{m}andradius\mathit{r}$$$$isreleasedfromthetopofa$$$$quarter-cylindricalwedgewith$$$$mass\mathit{M}andradius\mathit{R}asshown.$$$$thereisenoughfrictionbetween$$$$ballandwedge,butthereisno$$$$frictionbetweenwedgeandground.$$$$findthefinalspeedofthewedge.$$

Answered by aleks041103 last updated on 21/Dec/24

Commented by aleks041103 last updated on 21/Dec/24

Commented by mr W last updated on 21/Dec/24

$$thankssir!$$$$butigotacubicfinalequationfor$$$$\cos \theta .$$$$example:\frac{M}{m}=2$$$$\Rightarrow \cos \theta \approx 0.449192\Rightarrow \theta \approx 63.308°$$

Answered by mr W last updated on 21/Dec/24

Commented by mr W last updated on 21/Dec/24

$$centerofquartercircle(X,0)$$$$centerofball(x,y)$$$$letk=\frac{M}{m}+1$$$$x=X+(R+r)\sin \theta$$$$y=(R+r)\cos \theta$$$$rotationofball:$$$$\phi =(1+\frac{R}{r})\theta$$$$\frac{d\phi }{dt}=(1+\frac{R}{r})\frac{d\theta }{dt}=(1+\frac{R}{r})\omega$$$$\alpha =\frac{d^2\phi }{{dt}^2}=(1+\frac{R}{r})\omega \frac{d\omega }{d\theta }$$$$U=-\frac{dX}{dt}(\leftarrow )$$$$u=\frac{dx}{dt}=-U+(R+r)\omega \cos \theta (\to )$$$$v=-\frac{dy}{dt}=(R+r)\omega \sin \theta (\downarrow )$$$$MU=mu=-mU+m(R+r)\omega \cos \theta$$$$\Rightarrow U=\frac{(R+r)\omega \cos \theta }{k}$$$$mg(R+r)(1-\cos \theta )=\frac{{MU}^2}{2}+\frac{m(u^2+v^2)}{2}+\frac{I}{2}{\left(\frac{d\phi }{dt}\right)}^2$$$$2g(R+r)(1-\cos \theta )=(\frac{M}{m}+1)U^2+{(R+r)}^2{\omega }^2-2U(R+r)\omega \cos \theta +\frac{2{(R+r)}^2{\omega }^2}{5}$$$$2g(R+r)(1-\cos \theta )=(\frac{M}{m}+1)U^2+\frac{7{(R+r)}^2{\omega }^2}{5}-2U(R+r)\omega \cos \theta$$$$2g(1-\cos \theta )=(R+r){\omega }^2(\frac{7}{5}-\frac{{\cos }^2\theta }{\frac{M}{m}+1})$$$$\Rightarrow {\omega }^2=\frac{2kg(1-\cos \theta )}{(R+r)(\frac{7k}{5}-{\cos }^2\theta )}$$$$\omega \frac{d\omega }{d\theta }=\frac{kg\sin \theta ({\cos }^2\theta -2\cos \theta +\frac{7k}{5})}{(R+r){(\frac{7k}{5}-{\cos }^2\theta )}^2}$$$$-fr=I\alpha$$$$-fr=\frac{2{mr}^2}{5}\times (1+\frac{R}{r})\omega \frac{d\omega }{d\theta }$$$$f=-\frac{2m(R+r)}{5}\times \frac{kg\sin \theta ({\cos }^2\theta -2\cos \theta +\frac{7k}{5})}{(R+r){(\frac{7k}{5}-{\cos }^2\theta )}^2}$$$$f=-\frac{2kmg\sin \theta ({\cos }^2\theta -2\cos \theta +\frac{7k}{5})}{5{(\frac{7k}{5}-{\cos }^2\theta )}^2}$$$$A=\frac{dU}{dt}=\frac{(R+r)\omega }{k}[-\omega \sin \theta +\cos \theta \frac{d\omega }{d\theta }]$$$$A=\frac{g\sin \theta (-{\cos }^3\theta +\frac{21k}{5}\cos \theta -\frac{14k}{5})}{{(\frac{7k}{5}-{\cos }^2\theta )}^2}$$$$N\sin \theta +f\cos \theta =MA$$$$N\sin \theta -\frac{2kmg\sin \theta ({\cos }^2\theta -2\cos \theta +\frac{7k}{5})\cos \theta }{5{(\frac{7k}{5}-{\cos }^2\theta )}^2}=\frac{Mg\sin \theta (-{\cos }^3\theta +\frac{21k}{5}\cos \theta -\frac{14k}{5})}{{(\frac{7k}{5}-{\cos }^2\theta )}^2}$$$$N=0:$$$$2k({\cos }^2\theta -2\cos \theta +\frac{7k}{5})\cos \theta +5(k-1)(-{\cos }^3\theta +\frac{21k}{5}\cos \theta -\frac{14k}{5})$$$$5(-3k+5){\cos }^3\theta -20k{\cos }^2\theta +(119k^2-105)\cos \theta -70k(k-1)=0$$$$\frac{r}{R+r}⩽\cos \theta <1$$$$\frac{U}{\sqrt{g(R+r)}}=\cos \theta \sqrt{\frac{2(1-\cos \theta )}{k(\frac{7k}{5}-{\cos }^2\theta )}}$$

Commented by aleks041103 last updated on 21/Dec/24

$$Inyourenergyconservationeqnyoumissed$$$$toaccountfortherotationalkineticenergy.$$

Commented by mr W last updated on 22/Dec/24

$${you}^{\prime }reright.now{it}^{\prime }sfixed.$$

Commented by ajfour last updated on 21/Dec/24

Awesome Watch this another geometry video i made. https://youtu.be/JOiE3jnIDvs?si=ADUmXZ0DRrw7MxD_

Commented by mr W last updated on 21/Dec/24

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