Converting Rotational Inertia into Linear Equivalent Mass I am working on a dynamic model of a very simple "lever" system, consisting of a solid body pivoting around a point, and a linear actuator acting on the body making it rotate around such pivot. (Sketched it up in GeoGebra here. A slider at the top controls movement.)
I have determined the dynamics of the system follow an equation of the form:
$$J\cdot \ddot\theta = M(x)$$
Where $J$ is the moment of inertia of the body around the pivot, $\theta$ is the rotation angle around the pivot and $M(x)$ is the moment applied on the body as a function of $x$, the extension of the actuator. All of these variables are known as functions of $x$.
My question is, how do I turn this equation into an equation of the form:
$$m_{eq}\cdot\ddot x = F(x)$$
Where $m_{eq}$ is an equivalent mass (possibly function of $x$) and $F(x)$ is the force exerted by the actuator (known).
Can I simply do a transformation of the form $J=m_{eq}\cdot r^2$ where $r$ is the distance from the pivot to the point where the actuator applies force (always constant)?
 A: 
Can I simply do a transformation of the form $J=meq⋅r2$ where $r$ is the distance from the pivot to the point where the actuator applies force (always constant)?

Unfortunately it's far from that simple.

Look at the geometry with the actuator extended (above). Note my definition of $\theta$ and that $|P'P|=r$. The actuator force $F$ does not act perpendicularly to $OP$ and the 'useful' moment about $O$ is only:
$$\tau=F\times |OP| \times \cos\theta$$
Now look at the geometry without the actuator extended (below), where $|P'P|=0$, that is $|O'P|=r_0$:

Because $OPO'$ is a right angled triangle:
$$|OO'|^2=r_0^2+|OP|^2$$
Using the trigonometry of non right angled triangles:
$$|OO'|^2=|O'P|^2+|OP|^2-2|O'P||OP|\cos\theta$$
$$\cos\theta=\frac{|O'P|^2+|OP|^2-|OO'|^2}{2|O'P||OP|}$$
$$\cos\theta=\frac{(r+r_0)^2+|OP|^2-r_0^2-|OP|^2}{2r_0|OP|}$$
$$\cos\theta=\frac{r^2+2rr_0}{2r_0|OP|}$$
So that:
$$\tau=F\frac{r^2+2rr_0}{2r_0}$$
The angular acceleration $\alpha=\dot{\omega}$ of the solid body is then given by:
$$\tau=I\alpha$$
Where $I$ is the moment of inertia of the solid body about the point $O$.
A: There is a simple geometric relationship for planar mechanics for the effective mass a force sees. Below I show the relationship between an applied force $F$ and the acceleration of the body along the line of action of the force $\ddot{x}$.
Consider the two cases below:



*

*A pinned object about point A, with distance $c$ from the pivot to the center of mass, and $d$ from the pivot to the line of action of the force $F_P$.
$$ F_P = \left( \frac{I_C + m c^2}{d^2}  \right)  \ddot{x}_P =\left( \frac{I_A }{d^2}  \right)  \ddot{x}_P $$
$I_C$ is the mass moment of inertia about C, and $I_A$ about A.

*A sliding object along point A, with angle $\psi$ to the line of action of the force $F_P$.
$$ F_P = \left( \frac{m}{\cos(\psi)^2}  \right) \ddot{x}_P  $$
The stuff inside the parenthesis is the effective linear mass along the force line of action. As a check, you can see that if $d=0$ the mass is infinite since the force is acting against the pivot. Similarly when the body is sliding perpendicularly to the force and $\psi = \tfrac{\pi}{2}$.
