Force required for Angular Accleration Suppose I have 4' x 4' piece of wood.  Its mass is 150kg evenly distributed.  I drill a hole right at a corner and place rod through the hole to create a rotational axis.  I apply a tangental force at the opposite corner of the wood. This other corner is diagonally across from the hole that contains the axis  How much torque would I need to accelerate the rotation at 30 rotations per second per second?  Assume the wood is on a frictionless surface and im applying the torque at the opposite corner from the corner which is the axis.
 A: Consider the general case of a rectangular block, and the three possible rotations, a) about the center G, b) about an edge E, or c) about a corner N.

What you are asking is the mass moment of inertia for each of these scenarios, which relates torque about each of the axes, to angular acceleration.
$$ \begin{aligned}  \tau_G & = I_G \dot{\omega} & \tau_E & = I_E \dot{\omega} & \tau_N & = I_N \dot{\omega} \end{aligned} $$
Using a table for the mass moment of inertia and the parallel axis theorem you have
$$ \begin{aligned}  I_G & = \boxed{ \frac{m}{12} \left( a^2 + b^2 \right)} \\ I_E & = I_G + m \left( \tfrac{a}{2} \right)^2 =\boxed{ \frac{m}{12} \left( 4 a^2 + b^2 \right)} \\ I_N &= I_G + m \left(\tfrac{a}{2}\right)^2 + m\left(\tfrac{b}{2}\right)^2  = \boxed{\frac{m}{3} \left(a^2+b^2\right)}  \end{aligned} $$
So now you have to measure the piece you have (remember a 4×4 is not exactly 4 inches by 4 inches) and weight it. Get everything in consistent units ${\rm (N·m) = (kg \,m^2)·(rad/sec^2)}$ and find your answer.
