What is the inertia caused by angular momentum when twisted on it's rotating axis? I would like to provide a more thorough answer to this question here
https://aviation.stackexchange.com/q/3709
but I realized I don't know enough about angular momentum. If an airplane wheel is rotating at 100 rpm, and the wheel weighs 10kg, with a diameter of 50cm and a uniform mass (approximations applicable to a standard small aircraft), what is the difference in force necessary to bring the plane to 20 degrees of bank as opposed to when the wheels are stopped?
I know that this involves calculating angular momentum, which I have at 5kgm/s per wheel, so 10kgm/s total, I'm just not sure how I would quantify the affect of this angular momentum when trying to bank the aircraft 20 degrees over a course of 5 seconds (replicating first turn in the airport pattern).
I bet the following terms are involved: $\sin(20), 5s, 10kgm/s.$
Not sure if it's relevant, but we can assume the aircraft wheels are suspended 1 meter below the aircraft.
 A: You asked for it.
Here is how you find the forces and moments acting on the wheel from the plane to gauge the effect of spin. Reverse the sign to find the forces acting to the plane by the wheel.


*

*Consider a coordinate system with $z$ axis along the plane axis, and $y$ up (opposite gravity). The plane precession angle is $\varphi$ (as it turns about $y$), the plane bank angle $\psi$ and the wheel spin angle $\theta$. The turn radius is $R$, the wheel is hanging a distance $\ell$ from the center of mass of the plane and it has a radius $r$.

*The 3×3 orientation matrix of the wheel spindle is $E = {\rm Ry}(\psi){\rm Rz}(\varphi)$, but for evaluation purposes we will always use $\psi=0$, $\dot{\psi} = \Omega$ and $\ddot{\psi}=0$ for a steady normal turn. $E=\begin{bmatrix}\cos\varphi & -\sin\varphi & 0 \\ \sin\varphi & \cos\varphi & 0 \\ 0 & 0 & 1 \end{bmatrix}$

*The mass moment of inertia tensor along the spindle and wheel is $I_{body} = \begin{bmatrix} I_{disk} & 0 & 0 \\ 0 & \frac{1}{2} I_{disk} & 0 \\ 0 & 0 & \frac{1}{2} I_{disk} \end{bmatrix}$ where $I_{disk}=\frac{m}{2} r^2$ for a uniform disk. The mass moment of inertia on the world coordinates is $I = E I_{body} E^\top$ yielding $I= I_{disk} \begin{bmatrix} \frac{\cos^2(\varphi)+1}{2}  & \frac{\sin\varphi\cos\varphi}{2} & 0 \\ \frac{\sin\varphi\cos\varphi}{2} & \frac{\sin^2(\varphi)+1}{2} & 0 \\ 0 & 0 & \frac{1}{2} \end{bmatrix}$

*The position of the wheel center of mass is $\vec{r} = {\rm Ry}(\psi) \left( \hat{i} R - {\rm Rz}(\varphi) \hat{j} \ell \right) = (R+\ell \sin\varphi, -\ell \cos\varphi, 0)$. In general a varying bank has $\dot\varphi \ne 0$.

*The velocity of the wheel center of mass is $\vec{v} = \dot{\vec{r}} = (\ell \dot\varphi \cos\varphi,\ell \dot\varphi \sin\varphi, -(\ell \sin\varphi+R) \Omega )$ and the rotational velocity $\vec\omega = \hat{j} \Omega +  {\rm Ry}(\psi) ( \hat{k} \dot\varphi + {\rm Rz}(\varphi) \hat{i}\dot\theta) = (\dot\theta \cos\varphi, \Omega+\dot\theta \sin\varphi,\dot\varphi)$. Remember that wheel spin speed is $\dot\theta$ and turn rate is $\Omega$.

*Similarly the acceleration of the wheel center of mass is $\vec{a} = (-\ell (\Omega^2 +\dot\varphi^2)\sin\varphi - \Omega^2 R, \ell\dot\varphi^2\cos\varphi,-2\ell\Omega\dot\varphi\cos\varphi)$ and the rotational acceleration $\vec\alpha = (\dot\varphi (\Omega-\dot\theta \sin\varphi), \dot\varphi\dot\theta \cos\varphi,-\Omega \dot\theta \cos\varphi)$

*The force needed to keep the center of mass in motion is $\vec{F} = m \vec{a} = (-m \ell (\Omega^2 +\dot\varphi^2)\sin\varphi - m \Omega^2 R, m \ell\dot\varphi^2\cos\varphi,-2 m \ell\Omega\dot\varphi\cos\varphi)$ and not related to wheel spin (of course).

*The moment applied to the spindle to keep the wheel in motion is $\vec{M} = I \vec{\alpha} + \vec{\omega} \times I \vec{\omega} = $ $$\boxed{ \vec{M} = \begin{pmatrix} I_{disk} \dot\varphi (\Omega \cos^2\varphi - \dot\theta\sin\varphi) \\ I_{disk} \dot\varphi \cos\varphi ( \Omega \sin\varphi + \dot\theta) \\ - I_{disk} \Omega \left(\dot\theta + \frac{\sin\varphi}{2} \Omega \right) \cos\varphi)\end{pmatrix}}$$


So if the spin is $\dot\theta \ge \frac{\sin\varphi}{2} \Omega$ then the effects along the axis of the plane of the wheel spinning are more that the effect of the wheel orbiting around the turn. Otherwise it is less. There is component of the reaction moment along $\ell$ which is equal to $I_{disk} \dot\varphi \dot\theta$ and a reaction moment along the spin axis of the wheel equal to $I_{disk} \Omega \dot\varphi  \cos\varphi$. The former will cause the plane to yaw when a bank is introduced $\dot\varphi \ne 0$ and the wheel is spinning $\dot\theta$.
