Time?? It's quite different !!

I post this Same diagram 2nd Time :)

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Let mass density of ring is " $\sigma$ " Consider on the disc an ring element of radius " x " and width " dx " and on this ring consider another circular arc element of width " $Rd\theta$ " which subtends angle of " $d\theta$ " at the centre of ring element. it's mass is " dm " and clearly friction on this element acts tangentially . So we calculate torque on this due to friction and then integrate it !

$\displaystyle{dm\quad =\quad \sigma xd\theta (dx)\\ \\ df\quad =\quad \mu (dm)g\\ \quad \quad \quad =\quad \sigma \mu g(xdx)d\theta}$ .

Now torque due to this is:

$\displaystyle{d\tau \quad =\quad (df)x\quad \\ \\ d\tau \quad =\quad \sigma \mu g({ x }^{ 2 }dx)d\theta \\ \\ \int { d\tau \quad } =\quad \sigma \mu g\int _{ \theta =0 }^{ \theta =2\pi }{ d\theta } \int _{ 0 }^{ R }{ { x }^{ 2 }dx } \\ \\ \tau \quad \quad =\quad \cfrac { 2\pi \sigma \mu g{ R }^{ 3 } }{ 3 } \quad \quad \quad ........\quad (1)\\ \\ \tau \quad =\quad I\alpha \\ \\ \quad \quad =\quad \cfrac { m{ R }^{ 2 } }{ 2 } \alpha \quad \quad \quad \\ \quad \quad =\quad \cfrac { { \sigma \pi R }^{ 2 } }{ 2 } \alpha \quad \quad \quad \quad \quad .......(2)\\ \\ \Rightarrow \quad \alpha \quad =\quad \cfrac { 4\mu g }{ 3R } \quad \quad .\quad .\quad .\quad .\quad .\quad (3)}$ .

Using angular momentum conservation of both system about their common centre :

$\displaystyle{I\omega \quad =\quad (I\quad +\quad I){ \omega }^{ " }\quad \\ \\ { \omega }^{ " }\quad =\quad \cfrac { \omega }{ 2 } \quad \quad .\quad .\quad .\quad .\quad \quad (4)}$ .

Now using equation of motion ( Since angular accl. is constant )

$\displaystyle{\cfrac { \omega }{ 2 } \quad =\quad \omega \quad -\quad \alpha T\\ \\ \Rightarrow \quad \quad \boxed { T\quad =\quad \cfrac { 3\omega R }{ 8\mu g } } }$ .

Let the final angular velocity be $\displaystyle{\omega'}$ . Hence by the conservation of angular momentum we get $I \omega = (I + I) \omega'$ or simply $\Rightarrow \omega'= ^{\omega}/_{2}$ Now consider a ring of radius $\displaystyle{x}$ and thickness $\displaystyle{dx}$ Frictional force on this ring is $d \mathcal{F} = \mu g \times \Bigg(2 \pi x dx \dfrac{\mathcal{M}}{\pi \mathcal{R}^{2}}\Bigg)$ Where $\displaystyle{\mathcal{M}}$ is the mass of one ring $d \tau = d \mathcal{F} x$ $= \dfrac{2 \mu \mathcal{M} g}{\mathcal{R}^{2}} x^2 dx$ . Total torque, $\tau = \int d \tau = \dfrac{2 \mu \mathcal{M} g}{\mathcal{R}^{2}} \int_{0}^{\mathcal{R}}x^2 dx$ or $I \Bigg( \dfrac{-d \omega}{dt}\Bigg) = \dfrac{2 \mu \mathcal{M} g \mathcal{R} }{3}$ Negative sign indicates that the angular velocity of the disc decreases with time. Hence $dt = \dfrac{-3 I }{2 \mu g \mathcal{M} \mathcal{R}} \times d \omega$ Solving the integral we get $\boxed{ t = \dfrac{3}{8} \times \dfrac{\mathcal{R} \omega}{\mu g} }$