Discussion
A rotor of moment of inertia \( 0.002 \ \mathrm{kg\,m^2} \) spins at \( 1000 \ \mathrm{rad/s} \). Its angular momentum is
Answer: C
Correct answer: \( 2.0 \ \mathrm{kg\,m^2/s} \)
Angular momentum is the product of moment of inertia and angular velocity:
\[ H = I\omega = 0.002 \times 1000 = 2.0\ \mathrm{kg\,m^2/s} \]
It is the quantity that determines both rigidity and precession rate: a larger \( H \) means the same disturbing torque produces proportionately less precession. Designers therefore spin the rotor as fast as the bearings allow and put the mass at the rim, where it contributes most to \( I \).
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