Problem Statement: We need to find the initial deceleration of a rotating disc (angular velocity $\omega_0$) which is given a small translational velocity $v_0$, where $v_0 \ll R\omega_0$.

The velocity of any point on the disc is $\vec{v} = \vec{v}_0 + \vec{v}_{rot}$. Since $v_0 \ll \omega r$, the magnitude of the velocity is dominated by rotation: $|\vec{v}| \approx \omega r$.
The friction force opposes the net velocity vector.

We need the component of friction opposing the translational motion (along the x-axis).
The x-component of velocity at a point defined by $(r, \theta)$ is: $$ v_x = v_0 – \omega r \sin \theta $$
The x-component of the friction force $dF_x$ is proportional to the fraction of velocity in the x-direction: $$ dF_x = – (\mu g dm) \frac{v_x}{|\vec{v}|} \approx – \mu g dm \frac{v_0 – \omega r \sin \theta}{\omega r} $$ $$ dF_x = – \mu g dm \left( \frac{v_0}{\omega r} – \sin \theta \right) $$

Substitute $dm = \sigma r dr d\theta$ and integrate over the whole disc ($0 \to R$, $0 \to 2\pi$).
The term containing $\sin \theta$ integrates to zero over a full cycle.

$$ F_{net} = \int – \mu g \left( \frac{v_0}{\omega r} \right) \sigma r dr d\theta $$ $$ F_{net} = – \frac{\mu g v_0 \sigma}{\omega} \int_0^R dr \int_0^{2\pi} d\theta $$ $$ F_{net} = – \frac{\mu g v_0 \sigma}{\omega} (R) (2\pi) $$

Substitute mass density $\sigma = \frac{M}{\pi R^2}$:

$$ F_{net} = – \frac{\mu g v_0}{\omega} \left( \frac{M}{\pi R^2} \right) (2\pi R) = – \frac{2 \mu M g v_0}{\omega R} $$

Using Newton’s Second Law $F = Ma$: