A pulley of radius 2m is rotated about its axis by a force =(20t -...

We have,

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

or $2\left(20t - 5t^{2}\right) = 10\alpha$

$\alpha = 4t - t^{2}$

But $\alpha = \dfrac{d\omega}{dt},$ so

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

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

Integrating $\omega = 2t^{2} - \dfrac{t3}{3}$

$\omega$ will be& zero& at $t = 6s$

So, $\omega = \dfrac{d\theta}{dt} = 2t^{2} - \dfrac{t3}{3}$

or $d\theta = \left(2t^{2} - \dfrac{t3}{3}\right)dt$

Again, integrating both sides
$\theta = \dfrac{2t^{3}}{3} - \dfrac{t4}{12}$

since. $t = 6s$

so, $\theta = \dfrac{2 \times {6}^{3}}{3} - \dfrac{64}{12} = 36$

Number of turns, $n = \dfrac{θ}{2\pi} = \dfrac{36}{2\pi} = 5. 73$