Torque of pulley about its axis is τ=Iα, where, I is moment of inertia and α is angular acceleration.
Torque of a force about its axis is τ=FR, here, R is radius of pulley.
Then, (12t−3t2)1.5=4.5α
⇒α=4t−t2
Angular acceleration in terms of angular velocity is
⇒α=dtdω=4t−t2
⇒ω=∫0t(4t−t2)dt
⇒ω=2t2−3t3
For ω=0,2t2−3t3=0⇒t2(2−3t)=0
⇒t=0or6s
Now, using ω=dtdθ, we have
dtdθ=2t2−3t3⇒θ=∫06(2t2−3t3)dt
=[32t3−12t4]06
=63(32−126)=63(128−6)=663=36
Number of revolutions =2π36=π18
∴K=18