A good rotational motion question!

The solution assumes that the angular velocity "linearly" increases with time and "instantly" stop when its rim velocity reaches the belt velocity. That "discontinuous" slope seems aphysical. Perhaps there's more to it than what has been consdiered.

The disc will reach constant angular velocity when $v=\omega \times r$ where $r= \mathrm{\ radius\ of\ disc} \$ , $\mathrm{\omega= angular\ velocity \ of \ disc}$ , $v = \mathrm{speed\ of\ belt}$ . Here kinetic friction will provide the torque. As we all know the equation $\tau=I \times \alpha$ and $\tau=r \times \ F$ .

Now, $F=\mu_kN$

$N=mg$

$I=\dfrac {Mr^2}{2}$ $(\mathrm{For\ disc, \ Moment\ of\ Inertia}=\dfrac {Mr^2}{2})$

Now equating $r \times F = I \times \alpha$ and substituting the values we get $\alpha = 66.67\ \mathrm{rad/sec^2}.$

Using the equation of rotational motion $\omega_f^2-\omega_i^2=2\alpha\theta$ where $\omega_i=0$ and $\omega_f=\dfrac{v}{r}$

$v=18m/s (given)$ , substitute the values of $v,\alpha, r$ in the rotational equation of motion to get $\theta=431.97 \ radians$

Therefore No of revolutions = $\dfrac {\theta}{2\pi}\ = \dfrac {431.97}{2\pi}\ \approx \boxed{\dfrac {216}{\pi}}$