How does moment of inertia transfer through a series of levers? (Specifically, a Piano Action) Ultimate Goal: Calculate the mass moment of inertia that a finger experiences as it depresses a piano key. 
Background on the question is at the end if you need, but I'll keep this high-level so that you don't need to be a Registered Piano Technician to answer this.

Imagine you have two interconnected levers (such that the output of one is mechanically linked to the input of the other). Each lever, which carries arbitrary weights, has its own moment of inertia about its pivot point. These individual moments of inertia can easily be calculated using thin rod approximations, parallel axis theorem, etc.
The tricky part to me comes in when you try to calculate the total moment of inertia of the system about the pivot of the first lever. Recall that, since the levers are connected, the first lever's I will be subject to the I of the second lever in some way. What I'm looking for is, as the title suggests, how this moment of inertia transfers.
At first I thought I could just use the the basic definition of moment of inertia:
$$I=\int_V{r^2dm}$$
with V as the solid lever system (though difficult by hand, this is easily done with a SolidWorks model), but then I read this article about reflection of moment of inertia through a system with a motor, two gears, and a load at the other end. It suggested that the total I at the motor is:
$$I_{Total}=\frac{I_2}{N^2}+I_1$$
With I1 being the moment of inertia of the motor about its CM, I2 being the moment of inertia of the load about its CM, and N being the gear ratio. Since levers are analogous to gears, I predicted that the transfer of moment of inertia through our imagined 2-lever system would be the same as the above equation, only instead of a gear ratio, we'd have:
$$N=\frac{B_1}{A_1}*\frac{B_2}{A_2}$$
where B1 is the length of the rear segment of lever 1, A1 is the length of the front segment of lever 1, and so on.
However, when I compare this to calculations done by some guy on this website (85% of the way down the page) for the same purpose, he got (adapted to fit this example):
$$I_{Total}=I_2(\frac{B_1}{A_2})^2+I_1$$
which makes his N
$$N=\frac{A_2}{B_1}$$
So it seems like, to calculate his gear ratio (or rather, mechanical advantage) he only involved the input segment of the second lever, and the output segment of the first lever, whereas I involved all segments.
So, finally, the questions:


*

*Is it indeed incorrect to use the simple integration form of moment of inertia (shown above) for a lever system?

*Is it incorrect to extend this gear-train moment of inertia formula to levers?

*If none of the above formulae are correct, what is the correct way to calculate the moment of inertia about the first lever's pivot in a system of levers?


I think I might be misunderstanding the conversion from gear ratio to the analogous gear ratio for levers, but I'm surprised I haven't been able to find any good sources about this online.

BACKGROUND:
The grand piano action consists of a system of 3 interconnected levers that looks like this:

Where L1 is input segment of lever 1, L2 is output segment of lever 1, L3 is input segment of lever 2, and so on. Levers 1, 2, and 3 are blue, red, and green, respectively. Finger presses down on L1, this transfers force through levers that eventually throws the hammer (end of L6) towards the string.
 A: What is the effective mass that each force $F_x$ and $F_y$ feels in the mechanism below?

We have a single pinned rigid body at point P, with normal force $F_x$ and perpendicular force $F_y$ acting through point B, and a lumped mass $m_a$ attached to another point A. Additionally, the body has mass $m_c$ and mass moment of inertia $I_c$ located at the center of mass point C.
The effective mass that the normal force $F_x$ sees is infinite as it is using against the pivot. But the perpendicular force $F_y$ sees an effective mass of $$m_y = \frac{I_c + m_c r_c^2 + m_a r_a^2}{r_b^2} \tag{1}$$
The is you can write an expression like $F_y = m_y \ddot{y}$  where the direction y is along the axis of $F_y$.
This is derived from the kinematics and the equations of motion of the body, relating the torque applied on the pivot $\tau = F_y r_b$ with the total MMOI about the pivot times angular acceleration.
But what about a force at B along some other direction. Then you redefine the distance $r_b$ to be the moment arm (perpendicular distance) of the force


Now comes the more interesting part. You have two connected bodies, with one degree of freedom. Like a slider-crank mechanism. Let's try to figure out the effective mass of the piston force $F$ as a function of the crank angle $\varphi$. 

Each linkage has individual mass properties of $m_1$, $I_1$, $m_2$ and $I_2$ as well as the location of the center of mass relative to a pin of $c_1$ and $c_2$. The overall lengths of the links are $\ell_1$ and $\ell_2$. NOTE: actual cranks have $c_2$ as a negative value, or zero.
Here we need to consider the kinematics first in order to find the acceleration of each center of mass, but disregard any velocity related terms since we are only interested in acceleration effects. It is like the mechanism is at rest when we apply the force.
We find that $\sin \beta = \frac{\ell_2}{\ell_1} \sin \varphi$ and how the angular accelerations depend on the acceleration of point C.
$$ \begin{aligned} \ddot \varphi & = - \frac{\cos \beta}{\ell_2 \sin(\beta+\varphi)} \ddot y & \ddot \beta & = - \frac{\cos \varphi}{\ell_1 \sin(\beta+\varphi)} \ddot y \end{aligned} \tag{2}$$
Then we write the equations of motion at point A and at point B for the two links. We combine the two force directions with the out of plane torque balance. Also, we can apply a torque $\tau_A$ to see what effect that has on the motion.
$$ \begin{pmatrix}A_{x}\\
A_{y}\\
\tau_{A}
\end{pmatrix}-\begin{bmatrix}1\\
 & 1\\
-\ell_{2}\cos\varphi & \ell_{2}\sin\varphi & 1
\end{bmatrix}\begin{pmatrix}B_{x}\\
B_{y}\\
0
\end{pmatrix}=\begin{pmatrix}c_{2}m_{2}\cos\varphi\\
-c_{2}m_{2}\sin\varphi\\
I_{2}+m_{2}c_{2}^{2}
\end{pmatrix}\ddot{\varphi} \tag{3}$$
and 
$$ \begin{pmatrix}B_{x}\\
B_{y}\\
0
\end{pmatrix}-\begin{bmatrix}1\\
 & 1\\
-\ell_{1}\cos\beta & -\ell_{1}\sin\beta & 1
\end{bmatrix}\begin{pmatrix}C_{x}\\
-F\\
0
\end{pmatrix}=\begin{pmatrix}-c_{1}m_{1}\cos\beta\\
-c_{1}m_{1}\sin\beta\\
I_{1}+m_{1}c_{1}^{2}
\end{pmatrix}\ddot{\beta} \tag{4}$$
Which is 6 equations to be solved for 6 unknowns, four pin forces $A_x$, $A_y$, $B_x$, $B_y$, the piston side load $C_x$ and the acceleration $\ddot y$.
Unfortunately, we cannot simply state that $F = m_{\rm eff} \ddot y$ because the slider is going to move ever without a force applied due to the fact that the connecting rod is always under rotational acceleration, from equation (2).
Then there are other issues, like the fact that $\ddot y$ flips signs for the same direction of $F$ for different crank angles. Also when the linkages are all in-line then the effective mass is infinite.
My point is that doing the mathematical modeling of an articulated system of bodies becomes quickly a daunting task. In planar scenarios, you have $3n$ force balance equations, in addition to $n$ constraint conditions (such as zero torque at pins) and $3n$ kinematic relationships needed to connect everything together.
So a planar system with at least $n=4$ bodies such as the piano action mechanism requires consideration for 28 simultaneous linear equations and an equal number of unknown variables. Whilst in 3D that would be 52 equations.
Lastly, a little hope here. There is a systematic method of answering what is the effective mass matrix of a body connected to several other bodies using the Featherstone algorithm. This has been implemented in several game engines and other computer software as it reduced such problems to only solving $n$ equations, that are vastly more complex to put together.
A: 
let say you have 2 levers. for each lever you know the center of mass (CM) the inertia tensor of the CM. to obtain the total inertia you must define one inertial frame (red one).
**Lever 1 **
$$I_1=\left[ \begin {array}{ccc} J_{{{\it x1}}}&0&0\\ 0&J
_{{{\it y1}}}&0\\ 0&0&J_{{{\it z1}}}\end {array}
 \right]
$$
the rotation matrix between lever 1 and I-system is:
$$S_1=\left[ \begin {array}{ccc} \cos \left( \varphi _{{1}} \right) &-\sin
 \left( \varphi _{{1}} \right) &0\\ \sin \left( 
\varphi _{{1}} \right) &\cos \left( \varphi _{{1}} \right) &0
\\ 0&0&1\end {array} \right]
$$
thus $I_1$ transfer to I-system is (parallel axis transformation): 
$$I_{1I}=S_1\,I_1\,S_1^T-m_1\,\tilde{{R}}_1\,\tilde{{R}}_1$$
where :
$$\tilde{{R}}_1=\left[ \begin {array}{ccc} 0&0&-\sin \left( \varphi _{{1}} \right) L_
{{1}}\\ 0&0&\cos \left( \varphi _{{1}} \right) L_{{1
}}\\ \sin \left( \varphi _{{1}} \right) L_{{1}}&-
\cos \left( \varphi _{{1}} \right) L_{{1}}&0\end {array} \right] 
$$
**Lever 2 **
$$I_2=\left[ \begin {array}{ccc} J_{{{\it x2}}}&0&0\\ 0&J
_{{{\it y2}}}&0\\ 0&0&J_{{{\it z2}}}\end {array}
 \right] 
$$
$$S_2=\left[ \begin {array}{ccc} \cos \left( \varphi _{{2}} \right) &\sin
 \left( \varphi _{{2}} \right) &0\\ -\sin \left( 
\varphi _{{2}} \right) &\cos \left( \varphi _{{2}} \right) &0
\\ 0&0&1\end {array} \right] 
$$
$$I_{2I}=S_2\,I_2\,S_2^T-m_2\,\tilde{{R}}_2\,\tilde{{R}}_2$$
where:
$$\tilde{{R}}_2=\left[ \begin {array}{ccc} 0&0&-\sin \left( \varphi _{{2}} \right) L_
{{2}}\\0&0&-\cos \left( \varphi _{{2}} \right) L_{{
2}}\\ \sin \left( \varphi _{{2}} \right) L_{{2}}&
\cos \left( \varphi _{{2}} \right) L_{{2}}&0\end {array} \right] 
$$
the total inertia is:
$$I_T=I_{1I}+I_{2I}$$
your case is 2D , thus $J_{xi}=0\,,J_{y_i}=0$ and you get
$$I_{Tz}=J_{{{\it z1}}}+{\it m1}\,{L_{{1}}}^{2}+J_{{{\it z2}}}+{\it m2}\,{L_{{2
}}}^{2}
$$
edit

parallel axis transformation 3D case 


*

*$I (3\times 3)$ center of mass inertia

*$S (3\times 3)$ transformation matrix between CM and O-system

*$\vec{R} (3\times 1 )$ the distance between CM and O-system

*m rigid body mass


the equation to obtain the inertial tensor in o-system is:
$$\boxed{I_O=S\,I_{\text{CM}}\,S^T-m\,\widetilde{\left(S\,\vec{R}\right)}\,\widetilde{\left(S\,\vec{R}\right)}}$$
where :
$$S\,\vec{R}=\begin{bmatrix}
  x \\
  y \\
  z \\
\end{bmatrix}$$
and 
$$\widetilde{\left(S\,\vec{R}\right)}=\left[ \begin {array}{ccc} 0&-z&y\\ z&0&-x
\\ -y&x&0\end {array} \right] 
$$
Effective mass of the piston

we want to obtain the equation $M_y\,\ddot{y}=F$, thus $M_y$ is  the effective mass:
I) Position Vectors:
$$\vec{R}_C=\begin{bmatrix}
  0 \\
  y \\
\end{bmatrix}$$
$$\vec{R}_{\text{Crank}}=c2\,\begin{bmatrix}
  \cos(\varphi) \\
  \sin(\varphi) \\
\end{bmatrix}$$
$$\vec{R}_{\text{Conrod}}=(l2-c2)\,\begin{bmatrix}
  \sin(\beta) \\
  -\cos(\beta)+y \\
\end{bmatrix}$$
with:
$$y={\it l2}\,\cos \left( \varphi  \right) +\sqrt {{{\it l1}}^{2}-{{\it l2
}}^{2}\sin \left( \varphi  \right) }
\tag 1$$
$$\beta=\arcsin \left( {\frac {{\it l2}\,\sin \left( \varphi  \right) }{{\it 
l1}}} \right) 
$$
II) Kinetic Energy 
$$T=\frac{1}{2}\,m_{\text{Piston}}\dot{y}^2+
\frac{1}{2}\,m_{\text{Crank}}\dot{v}_{\text{Crank}}^2+
\frac{1}{2}\,J_{\text{Crank}}\dot{\varphi}^2+
\,m_{\text{Conrod}}\dot{v}_{\text{Conrod}}^2+
\frac{1}{2}\,J_{\text{Conrod}}\dot{\beta}^2$$
where $v^2=\vec{\dot{R}}^T\,\vec{\dot{R}}$
thus:
$$T=T(\varphi\,,\dot{\varphi}^2)$$
the equation of motion:
$$M_\varphi\,\ddot{\varphi}=\frac{\partial y}{\partial \varphi}\,F+\mathbb{0}(\dot{\varphi})\tag 2$$
with equation (1) you get
$$\ddot{y}=\frac{\partial y}{\partial \varphi}\,\ddot{\varphi}+\mathbb{0}(\dot{\varphi})\tag 3$$
thus ( equation (2) and (3) ) you obtain:
$$\underbrace{M_\varphi\,\left[\frac{\partial y}{\partial \varphi}\right]^{-2}}_{M_y(\varphi)}\,\ddot{y}=F$$
where:
$$M\varphi=\frac{\partial}{\partial\dot{\varphi}}\left(\frac{\partial T}{\partial\dot{\varphi}}\right)$$
