Rotating cylinder against rough walls

Relevant wiki: Torque - Dynamical Behavior

Let the mass of cylinder be $M$ and let the normal reaction with wall be $N$ and with the floor be $N'$ .

For horizontal and vertical equilibrium, we can write, $N'+kN = Mg$ and $N = kN'$

Solving these two equations we get $N' = \dfrac{Mg}{(k^2)+1}.$

Now, using the torque equations -

$k(N+N')R = \dfrac{M(R^2)}{2} \alpha$ Here, $\alpha$ is the angular acceleration.

Substituting the values of $N$ and $N'$ , we get, $\alpha = \dfrac{2k(k+1)g}{R((k^2)+1)}.$

Now, using $\omega^2 = 2 \alpha \theta$ , where $\theta$ is the angular displacement. If the number of rotations are $n$ then $\theta = n \times (2(\pi)).$

On solving, we get, $n = \dfrac{(\omega^2)R(k^2+1)}{8(\pi)k(k+1)g}.$

Hence, the answer is $\boxed{14}$ .