# Torque on a fixed reference of falling masses

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1. Sep 15, 2015

### FG_313

I had a little thought experiment, in which there are two objects with the same masses near each other (same height) on freefall. If I set up a point that is on an instant besides the two masses and call it the center of torque, I get that the torque produced by the nearest one's weight is T=F.d, where d is the distance to the imaginary point. The other one is T'=Fd', where d'>d and F=mg. I find that the torque T'>T and so I expect a'>a, where a is angular acceleration. But for a small instant of time, the geometry of the problem will tell you that the angle of the first mass made with the point (call it o) is larger than the angle the other one is making, because they are falling on the same rate. And so o'<o and a'>a. That for me is very difficult to understand why... Am I missing something? Thanks in advance!

2. Sep 15, 2015

### Staff: Mentor

Hi FG_313, welcome to PF!

Why do you expect that? What is the formula relating torque and angular acceleration? Does it involve any other quantity?

3. Sep 16, 2015

### FG_313

Thanks for the quick answer, don't know why I missed something like that. If the moment of inertia is considered the result makes sense.

4. Sep 16, 2015

Exactly!