Landing gear dynamics: analyzing moments and gyroscopic effects
In this post, I explore the dynamics of an aircraft landing gear under an applied moment. I revisit a previously analyzed landing gear kinematic model and shift my focus to the moments acting on the system. I consider the wheel’s angular momentum and use the moment equation to relate applied moments to the wheel’s rotational behavior and the arm’s motion. I then calculate the reaction moments at the connection point between the wheel and the arm, a critical location for bearing design. Finally, I examine the resulting equations and discuss the counterintuitive presence of a gyroscopic moment, even though the input velocities are confined to two axes.
I consider a landing gear with a wheel of radius r and thickness h. The wheel spins with angular velocity \omega_s about its axis, and the arm rotates with angular velocity \omega_y about a perpendicular axis. I analyze the system using an intermediate frame fixed to the arm.
The equation used is Euler second law equation with an intermediate frame about the center of mass:
\sum \mathbf{M}_C = \frac{\mathrm{d}\mathbf{L}_C}{\mathrm{d}t}\bigg|_1 = \frac{\mathrm{d}\mathbf{L}_C}{\mathrm{d}t}\bigg|_4 + \boldsymbol{\omega}_{4/1} \times \mathbf L_C
where \mathbf{M}_C is the sum of moments about point C, \mathbf{L}_C is the angular momentum about C, and \boldsymbol{\omega}_{4/1} is the angular velocity of the intermediate frame 4 with respect to the inertial frame 1.
I determine the angular momentum of the wheel:
\mathbf L_C = \mathbf I_C \boldsymbol \omega_{3/1}
where \mathbf{I}_C is the inertia tensor and \boldsymbol{\omega}_{3/1} is the absolute angular velocity of the wheel. For this symmetric wheel, the principal moments of inertia are I_1 = \frac{1}{2}mr^2 and I_2 = \frac{1}{12}m(3r^2 + h^2). I calculate the time derivative of the angular momentum and consider the applied moment \mathbf{M}_0 along the y-axis.
After applying the moment equation and resolving the components, I arrive at these equations:
\begin{aligned} 0 &= I_1 \dot \omega_s \\ M_{C_y} &= I_2 \dot \omega_y \\ M_{C_z} &= -I_2 \omega_y \omega_s \end{aligned}
From these equations, I observe that the wheel’s spin rate \omega_s is constant. The moment M_{C_y} directly influences the arm’s angular acceleration \dot{\omega}_y. The third equation reveals a gyroscopic moment M_{C_z} about the axis perpendicular to both the wheel’s spin axis and the arm’s rotation axis. This moment arises even though the input velocities are only along the other two axes.
For more insights into this topic, you can find the details here.