a
\(\tau= I \alpha \Rightarrow\left(12 t -3 t ^{2}\right) 1.5=4.5 \alpha\)

\(\alpha=4 t - t ^{2}\)

\(\frac{ d \omega}{ dt }=4 t - t ^{2} \Rightarrow \omega=\int_{0}^{ t }\left(4 t - t ^{2}\right) dt\)

\(\omega=2 t ^{2}- t ^{3} / 3\)

For \(\omega=0=2 t^{2}-\frac{t^{3}}{3} \Rightarrow t^{2}\left(2-\frac{t}{3}\right)=0\)

\(t=0,6\)

\(\frac{d \theta}{d t}=2 t^{2}-\frac{t^{3}}{3} \Rightarrow \theta=\int_{0}^{6}\left(2 t^{2}-\frac{t^{3}}{3}\right) d t\)

\(=\left[\frac{2 t^{3}}{3}-\frac{t^{4}}{12}\right]_{0}^{6}\)

\(=6^{3}\left(\frac{2}{3}-\frac{6}{12}\right)=6^{3}\left(\frac{8-6}{12}\right)\)

\(=\frac{6^{3}}{6}=36\)

\(\therefore K =18\)