The rotor of a turbojet engine of an aircraft has a mass 180 kg and polar moment of inertia 10 kg m² about the rotor axis. The rotor rotates at a constant speed of 1100 rad/s in the clockwise direction when viewed from the front of the aircraft. The aircraft while flying at a speed of 800 km per hour takes a turn with a radius of 1.5 km to the left. The gyroscopic moment exerted by the rotor on the aircraft structure and the direction of motion of the nose when the aircraft turns, are

The correct answer is: 1629.6 N-m and the nose goes down

1. Identifications of Given Parameters:
Mass of the rotor, $m=180\mathrm{kg}$
Polar moment of inertia of the rotor, $I=10\mathrm{kg}·m^2$
Angular velocity of the rotor, $\omega =1100\mathrm{rad/s}$ (Clockwise when viewed from the front)
Linear speed of the aircraft, $v=800\mathrm{km/h}=800\times \frac{5}{18}\mathrm{m/s}\approx 222.22\mathrm{m/s}$
Radius of the turn, $R=1.5\mathrm{km}=1500m$

2. Calculation of Angular Velocity of Precession (${\omega }_p$):
The angular velocity of precession of the aircraft during the turn is given by:
$${\omega }_p=\frac{v}{R}$$
Substituting the values:
$${\omega }_p=\frac{222.22}{1500}\approx 0.14815\mathrm{rad/s}$$

3. Calculation of Gyroscopic Moment (Couple):
The gyroscopic couple $C$ exerted by the rotor on the aircraft structure is calculated as:
$$C=I\times \omega \times {\omega }_p$$
Substituting the values:
$$C=10\times 1100\times 0.14815=1629.6\mathrm{N-m}$$

4. Determination of the Direction of Motion of the Nose:
Let us establish a coordinate system where:
• The axis of flight (forward direction) is along the positive $x$-axis.
• The left side of the aircraft is along the positive $y$-axis.
• The vertically upward direction is along the positive $z$-axis.

Since the rotor rotates clockwise when viewed from the front (looking towards the rear), the angular momentum vector $\overrightarrow{H}$ points along the negative $x$-direction:
$$\overrightarrow{H}=-I\omega \hat{i}$$

As the aircraft turns left, it rotates counter-clockwise about the vertical $z$-axis. Therefore, the precession angular velocity vector points vertically upward:
$${\overrightarrow{\omega }}_p={\omega }_p\hat{k}$$

The active gyroscopic couple vector is given by:
$${\overrightarrow{C}}_{\mathrm{active}}={\overrightarrow{\omega }}_p\times \overrightarrow{H}=\left({\omega }_p\hat{k}\right)\times (-I\omega \hat{i})=-I\omega {\omega }_p\hat{j}$$

The reactive gyroscopic couple exerted by the rotor on the aircraft structure is equal and opposite to the active couple:
$${\overrightarrow{C}}_{\mathrm{reactive}}=-{\overrightarrow{C}}_{\mathrm{active}}=I\omega {\omega }_p\hat{j}$$

Since $\hat{j}$ points to the left, a reactive couple vector pointing to the left (by the right-hand rule) generates a torque that acts to push the nose (front of the aircraft) downwards and lift the tail upwards.
Thus, the nose goes down.