A pulley of radius 2   m is rotated about its axis by a force = 20 t - 5 t 2 newton (where t is…

We have,

Torque $=r\times F=l\alpha$

or $2\left(20t-5t^2\right)=10\alpha$

$\alpha =4t-t^2$

But $\alpha =\frac{d\omega }{dt},$ so

$\frac{d\omega }{dt}=4t-t^2$

$d\omega =\left(4t-t^2\right)dt$

Integrating $\omega =2t^2-\frac{t^3}{3}$

$\omega$ will be zero at $t=6 \mathrm{s}$

So, $\omega =\frac{d\theta }{dt}=2t^2-\frac{t^3}{3}$

or $d\theta =\left(2t^2-\frac{t^3}{3}\right)dt$

Again, integrating both sides
$\theta =\frac{2t^3}{3}-\frac{t^4}{12}$

since. $t=6 \mathrm{s}$

so, $\theta =\frac{2\times 6^3}{3}-\frac{6^4}{12}=36$

Number of turns, $n=\frac{\theta }{2\pi }=\frac{36}{2\pi }=5.73$