Variable friction on the Horizontal disc

Image 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 " $xd\theta$ " 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:

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