1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

diagram

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.