- #36

jbriggs444

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Reading from the attachments...Viii said:Hello. I assumed the z axis. I chose it randomly, since it's not stated anywhere which one shall I take. By the way, love your answer. It's very detailed.

The first attachment shows a set of coordinate axes (x, y and z) drawn isometrically beside an isometric drawing of a portion of the skeleton of a cube. Unfortunately, the axes are drawn slanted differently from the cube. However we can see that the "z" direction is normal to the "bottom" face of the cube.

So we know the direction that the axis of rotation points -- vertical. But that does not answer the question I was trying to ask.

Are we considering rotation around a vertical axis through the centers of the top and bottom faces?

Are we considering rotation around a vertical axis along the center line of the right hand face?

Are we considering rotation around a vertical axis along one vertical edge?

**It matters because the moment of inertia of a cube is different depending on what axis you are using**.

I had provided a clickable link with my question. If you had clicked on that link, you would have been taken to a page that gives two different formulas for the moment of inertia of a cube. Let me extract the information from that link and provide it directly here.

https://www.vedantu.com/iit-jee/moment-of-inertia-of-a-cube said:I = 1/6 ma^{2}= ma^{2}/6when the axis of rotation passes through the center

I = 2 mb^{2}/ 3when the axis of rotation passes through its edge

Way back in the original post in this thread, you wrote that the moment of inertia of a cube is:

"Iz=(2*m*(b^2))/3". That looks very much like the second of the formulas above.

**That means that it is the wrong formula. Because it assumes the wrong axis of rotation.**