IIT JEE , AIEEE Forum - Physics , Mechanics

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(b)

Assuming the initial spin to be clockwise.

The inital slip of the cylinder is down the plane. So friction acts up at its full magnitude $\mu mg cos \theta$

Now, as $\mu = tan\theta \Rightarrow \mu mgcos\theta=mgsin\theta$

Hence the cylinder had no acceleration. So the CM doesn't develop any velocity and so the cylinder doesn't move.

$\frac{1}{2}mR^2\alpha=\mu mgcos\theta R$

So,

$\omega={\omega}_{0}-\frac{2\mu g cos\theta}{R}t$

The spinning stops at $T=\frac{{\omega}_{0} R}{2\mu g cos\theta}$

After this the cylinder moves down