Question

A spinning cylinder of mass $M$ and radius $R$ is lowered on 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. If 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

From t=0 till the moment when pure rolling starts friction acting is the maximum possible friction-$f=\mu mg$

Angular acceleration-

$\mu mg\cos \theta R=\frac{m{R}^2}{2}\alpha$

At the moment at which pure rolling starts,

$\frac{(mg\sin \theta -\mu mg\cos \theta )t}{m}=-{\omega }_{0}R+\alpha tt=\frac{{\omega }_{0}R}{g}mg\sin \theta -\mu mg\cos \theta =0$

So distance moved along the plane during this time is 0.

$\frac{3}{2}M{R}^2\alpha =Mg\sin \theta ta=R\alpha =\frac{2g\sin \theta }{3}s=\frac{3R}{\sin \theta }t=\sqrt{\frac{2s}{a}}=6\sqrt{\frac{R}{g}}$