In a continuous printing process suppose the total length of paper is $l$.

If $r_1$ is the inner radius of roll then, $\pi r^2-\pi r_1^2=l\times b$             ...(i)
And at this moment, $v=r\omega$     ...(ii)    


Differentiating equation (i) w.r.t. time,
$\Rightarrow \pi \times 2r\frac{dr}{dt}=\frac{dl}{dt}·b$
$\Rightarrow \frac{dr}{dt}=\frac{v.b}{2\pi r}$          ...(iii)

Differentiating equation (ii) w.r.t time
$\Rightarrow \frac{dv}{dt}=r\left(\frac{d\omega }{dt}\right)+\omega \left(\frac{dr}{dt}\right)$
$\text{⇒}0=r\alpha +\omega \frac{dr}{dt}$           $\left(\because \frac{\text{dr}}{\text{dt}}\text{is}-\text{ve}\right)$
$\text{⇒}\frac{dr}{dt}=\frac{r\alpha }{\omega }$
$\text{⇒}\alpha =\frac{\omega }{r}\left(\frac{dr}{dt}\right)$           ...(iv)

Using (iii) and (iv), $\frac{vb.\omega }{2\pi r^2}=\frac{v^{2}b}{2\pi r^3}=\alpha$