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.

  • $\begingroup$ reason i am using the wood as an example is I have a similar situation I need to calculate this for as I am working on a project . $\endgroup$ – Mike Nov 23 '18 at 23:39

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.


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