Question

A spinning cylinder with angular velocity ${\omega }_o$ of mass $M$ and radius $R$ is lowered on a rough inclined plane of angle ${30}^o$ with the horizontal and $\mu =1/\sqrt{3}$. The cylinder is released at a height of $3R$ from horizontal. The total time taken by the cylinder to reach the bottom of the incline is $(\frac{{\omega }_{0}R}{g}+x\sqrt{\frac{R}{g}})$. Find the value of $x$.

Answer 1

$Beforethepurerollingstartsf=\mu mgcos\theta =\frac{mg}{2}Forceequationalongtheinclineismgsin\theta -f=ma\Rightarrow a=0.thus,thecyclinderstaysinitsinitialpositionuntilpurerollingstarts.timetakentillthepurerollingstartsist=\frac{{\omega }_0}{\alpha }=\frac{{R\omega }_0}{g}Afterpurerollingstartstheequationofmotionsaremgsin{30}^0-f=mafR=\frac{1}{2}M{R}^2\alpha anda=\alpha R\Rightarrow f=\frac{Ma}{2}anda=\frac{g}{3}\frac{3R}{sin{30}^0}=\frac{1}{2}\frac{g}{3}{t}^2\Rightarrow t=6\sqrt{\frac{R}{g}}totaltime=\frac{{R\omega }_0}{g}+6\sqrt{\frac{R}{g}}\Rightarrow x=6$