Given, $M=20 \mathrm{~kg}$
$$
\begin{gathered}
\omega=100 \mathrm{rad} / \mathrm{s} \\
R=0.25 \mathrm{~m}
\end{gathered}
$$
Moment of inertia of the solid cylinder about its axis of symmetry.
$$
\begin{gathered}
\quad I=\frac{1}{2} M R^{2} \\
=\frac{1}{2} \times 20 \times(0.25)^{2} \\
=10 \times 0.0625 \\
=0.625 \mathrm{~kg}-\mathrm{m}^{2}
\end{gathered}
$$
Kinetic energy associated with the rotation of the cylinder is given by
$$
K=\frac{1}{2} / \omega^{2}
$$
$$
\begin{aligned}
&=\frac{1}{2} \times 0.625 \times(100)^{2} \\
&=0.3125 \times 10000 \\
&=3125 \mathrm{~J}
\end{aligned}
$$
Angular momentum, $L=/ \omega$
$$
\begin{aligned}
&=0.625 \times 100 \\
&=62.5 \mathrm{~J}-\mathrm{s}
\end{aligned}
$$