How do I calculate the force necessary to accelerate a point on a rigid body by a certain amount? Given a rigid body of uniform density with moment of inertia tensor $I$ and mass $m$, I'm trying to figure out how much force $F_p$ I need to apply at point $\vec p$ so that that point accelerates by a known $\vec a_p$. The direction of the force is given, so I really just need the magnitude.
I know that the center of mass $\vec c$ of the rigid body will accelerate by $\vec a_c = \vec F_p / m$ and that $\vec F_p$ will produce a torque around $\vec c$ given by $\vec \tau = \vec F_p \times \vec r = I \vec \alpha$ (where $\vec r = \vec p - \vec c$). So I think I could express $\vec a_p$ as $$\vec a_p = \vec a_c + \vec \alpha \times \vec r = \vec F_p/m+(\vec F_p \times \vec r)I^{-1} \times \vec r$$
However, I don't know how to untangle $F_p$ from here or if there's another way I should be calculating this. I'm a programmer, not really a physics expert, so I'd appreciate any help.
 A: Rev 0
What you are looking for is the dynamic effective mass of a body along a direction and through a point. The is the same calculation used to find out contact impulses for 3D bodies.
This dynamic mass is found as follows
$$ m^\star_p = \frac{1}{ \frac{1}{m}+\left(\hat{n}\times\overline{c}\right)\cdot\mathrm{I}^{-1}_c\left(\hat{n}\times\overline{c}\right) } $$
where $m$ is the mass of the body, $\mathrm{I}_c$ is the 3×3 mass moment of inertia tensor about the center of mass and along the inertial frame directions, $\hat{n}$ is the direction of the force or impulse action, and $\overline{c}$ is the relative position of the center of mass to the contact point.
In 2D the above becomes
$$ m^\star_p = \frac{1}{ \tfrac{1}{m} + \tfrac{d^2}{I_c} } $$
where $d$ is the perpendicular distance between the line of action and the center of mass.
Once this is known then $\overline{F}_p = m^\star_p \, \overline{a}_p$.
Rev 1
I realized that the above is applicable only if the direction of force is specified (like in a contact). But in your case, $\overline{F}_p$ could be in any direction, which makes the problem a bit simpler.
Consider the 3×3 mass matrix
$$ \mathrm{M}_p = \left[\tfrac{1}{m} \bf{1}-[\overline{c}\times]\mathrm{I}_{c}^{-1}[\overline{c}\times]\right]^{-1} $$
Here $[\overline{c}\times]$ is the 3×3 cross product matrix defined as
$$ \overline{c} = \pmatrix{x\\y\\z} $$
$$ [\overline{c}\times] = \begin{bmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{bmatrix}$$ 
then I can show that
$$ \overline{F}_{p} =\mathrm{M}_p\,\overline{a}_{p} $$
In 2D when $\overline{c} = \pmatrix{c_x \\ c_y}$ then
$$ \mathrm{M}_p = \begin{bmatrix} \tfrac{1}{m} + \tfrac{c_y^2}{I_c} & -\tfrac{c_x c_y}{I_c} \\ - \tfrac{c_x c_y}{I_c} & \tfrac{1}{m} + \tfrac{c_x^2}{I_c} \end{bmatrix}^{-1} $$

Proof
Consider the equations of motion of the offset force $\overline{F}_p$
$$\begin{aligned}\overline{F}_{p} & =m\,\left(\overline{a}_{p}+\overline{\alpha}\times\overline{c}\right)\\
\overline{F}_{p}\times\overline{c} & =\mathrm{I}_{c}\overline{\alpha}
\end{aligned}$$
Solving the 2nd equation and plugging into the first one yields
$$\begin{aligned}\overline{F}_{p}&=m\,\overline{a}_{p}+\overline{c}\times m\left(\mathrm{I}_{c}^{-1}\left(\overline{c}\times\overline{F}_{p}\right)\right)\\\overline{F}_{p}-\overline{c}\times m\left(\mathrm{I}_{c}^{-1}\left(\overline{c}\times\overline{F}_{p}\right)\right)&=m\,\overline{a}_{p}\\ m\left[\tfrac{1}{m}-\overline{c}\times\mathrm{I}_{c}^{-1}\overline{c}\times\right]\overline{F}_{p}&=m\, \overline{a}_{p}\\ \overline{F}_{p}&=\left[\tfrac{1}{m}-[\overline{c}\times]\mathrm{I}_{c}^{-1}[\overline{c}\times]\right]^{-1}\overline{a}_{p}\end{aligned}$$
Addendum
The two scenarios above are related with the following expression
$$ m^\star_p = \frac{ 1} {\hat{n}^\top \mathrm{M}_p^{-1} \hat{n} } $$
