Answer
$$ H_G=12.5 \mathrm{Mg} \cdot \mathrm{m}^2 / \mathrm{s}
$$
Work Step by Step
$$
I=1800(1.2)^2=2592 \mathrm{~kg} \cdot \mathrm{m}^2 \quad I_z=1800(0.8)^2=1152 \mathrm{~kg} \cdot \mathrm{m}^2
$$
Applying the third of Eqs. $21-36$ with $\theta=5^{\circ} \quad \psi=6 \mathrm{rad} / \mathrm{s}$
$$
\begin{aligned}
& \psi=\frac{I-I_z}{H_z} H_G \cos \theta \\
& 0=\frac{2592-1152}{2592(1152)} H_G \cos 5^{\circ} \\
& H_G=12.5 \mathrm{Mg} \cdot \mathrm{m}^2 / \mathrm{s}
\end{aligned}
$$
