= Solution
With clockwise rotation taken positive and $v_0=b\omega_0$, the total <kinetic energy> is
$$
\boxed{K_0=\frac M2v_0^2+\frac I2\omega_0^2=\frac M2(a^2+b^2)\omega_0^2.}
$$
<Angular momentum> needs an origin. Its spin about the centre is $I\omega_0$, in the clockwise direction. About a fixed point in the horizontal rail plane, spin and orbital contributions add to
$$
\boxed{L_{\rm rail}=I\omega_0+Mbv_0=M(a^2+b^2)\omega_0.}
$$
Here the drum centre is height $b$ above that plane. The signed clockwise convention fixes the displayed signs; the associated vector points perpendicular to the vertical plane of motion.
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