A disk rotating about a point on its circumference

This is the third time I am posting the same figure (:P)

Let P be the fixed point. Consider a small element (an arc of radius x) in red on the disk. The force of friction is tangential at every point on the arc, as the slipping tendency is tangential.

Clearly, $dm = 2 \sigma \theta x dx$

Using cosine rule in $\triangle OPQ,$

$\cos \theta = \dfrac{x}{2R} \Rightarrow \theta = \cos^{-1} \bigg(\dfrac{x}{2R}\bigg)$

$df = \mu_{k} dm g = 2\mu_{k} \sigma g x \cos^{-1}\bigg(\dfrac{x}{2R}\bigg) dx$

Clearly, $|d \tau_{P}| = x df = 2\mu_{k} \sigma g x^2 \cos^{-1}\bigg(\dfrac{x}{2R}\bigg) dx$

Hence, $|\tau_{P}| = 2\mu_{k} \sigma g \displaystyle \int_{0}^{2R}x^2 \cos^{-1}\bigg(\dfrac{x}{2R}\bigg) dx$

= $16 \mu_{k} \sigma g R^3 \displaystyle \int_{0}^{1} z^2 \cos^{-1} z dz$

Apply By parts on the integral taking $\cos^{-1} z$ as first part and $z^2$ as second part to transform it to:

$\displaystyle \int_{0}^{1} \dfrac{z^3}{3 \sqrt{1 - z^2}} dz$

Put $1 - z^2 = t^2$ to get it as :

$\displaystyle \dfrac{1}{3}\int_{0}^{1} (1 - t^2) dt = \dfrac{2}{9}$

Putting this value, we get:

$|\tau_{P}| = \dfrac{32}{9} \mu_{k} \sigma g R^3$

$I_{P} = \dfrac{3}{2} mR^2 = \dfrac{3}{2} \sigma \pi R^4$

Now, $\alpha = \dfrac{\tau_{P}}{I_{P}} = \dfrac{64}{27} \dfrac{\mu_{k}g}{\pi R}$

Now, $t = \dfrac{\omega}{\alpha} = \boxed{\dfrac{27}{64} \dfrac{\pi R \omega_{0}}{\mu_{k} g}}$

Hence, $a+b = 27+64 = 91$