Disc Rotated Flatly?

Lets take an elemental part at distance $x$ from center whose width is $dx$ .

So Mass of elemental part $dm = \dfrac{M 2\pi x dx}{\pi R^2}$

So $dN = dm g$

$\implies dN = \dfrac{Mg 2\pi x dx}{\pi R^2}$ [Normal reaction of the element]

$dF = \dfrac{Mg \mu 2\pi x dx}{\pi R^2}$ [Friction on the element]

Then Torque due to friction which stops it is

$dF.x = \dfrac{Mg \mu 2\pi x^2 dx}{\pi R^2}$

$\implies d\tau = \dfrac{Mg \mu 2\pi x^2 dx}{\pi R^2}$

Integrating from $0 \rightarrow R$ we get

$\implies \displaystyle{ \int_{0}^{\tau}d\tau = \int_{0}^{R}\dfrac{Mg \mu 2\pi x^2 dx}{\pi R^2} }$

$\tau = \dfrac{2Mg \mu R^3}{3R^2}$

$\tau = \dfrac{2 Mg \mu R}{3}$

$I \alpha = \dfrac{2 Mg \mu R}{3}$

$\alpha = \dfrac{2 Mg \mu R}{3I}$

Now by equation of rotation we know

$0=\omega_{o}-\alpha t$

$\dfrac{\omega_{o}}{\alpha} = t$

$\implies \dfrac{\omega_{o}}{\dfrac{2 Mg \mu R}{3I}}= t$

Now for disc, $I=\dfrac{1 MR^2}{2}$

Putting it we get

$\implies \dfrac{3\omega_{o}R}{4 g \mu} = t$

$\implies \boxed{t=\dfrac{3\omega_{o}R}{4 g \mu}}$

Putting values we get

$\boxed{t=10s}$

Mass element(dm)

Assuming uniform mass distribution,

$\frac {M}{\pi R^{2}} = \frac {dm}{rd\theta dr} \Rightarrow dm = \frac {Mrd\theta dr}{\pi R^{2}}$

Now, since $d\tau = \mu dmgr = \frac {\mu Mgr^{2}d\theta dr}{\pi R^{2}}$ ,

$\therefore \tau = \int d\tau = \int_{0}^{2\pi} d\theta \int_{0}^{R} \frac{\mu Mgr^{2} dr}{\pi R^{2}} = \frac {2\mu Mg}{R^{2}} \left(\frac {R^{3}}{3}\right) = \frac {2}{3} \mu MgR$

Using $\tau = I\alpha$

$\alpha = \frac {\tau}{I_{disc}} = \frac {4\mu g}{3R}$

as $I_{disc} = \frac {MR^{2}}{2}.$ Finally,

$\omega_{0} = \frac {4\mu gt}{3R} \Rightarrow \boxed{t = \frac {3\omega_{0}R}{4\mu g}}$

Evaluating at the given values, we get $\boxed {t = 10 \text {s.}}$ As you can see, stoppage time is independent of mass.