Some data from the pulley.

The gate trip time of a 35 mm Atwood’s pulley with 60 g on each side equal to the trip times for 105 mm Atwood’s with 20 g on each side; and 20 g on the 105 side and 60 g on 35 mm side?

I placed 80 g (77.366 g) on the left side and 60 g (59.185 g) on the right side at the 35 mm diameter of the 4 radii pulley. The trip time of the photo gate after a quarter rotation were 0571, .0565, .0561 sec. Average .0566 sec

I placed 20.04 g on the left side at the 35 mm diameter of the 4 radii pulley. And I placed 19.8 g on the left side at the 105 mm diameter and 19.8 g on the right side at the 105 mm diameter of the 4 radii pulley. The trip times of the photo gate after a quarter rotation were .0568, .0560, .0571, .0569, .0564 sec. Average .0566 sec

I placed 78.84 g on the left side at the 35 mm diameter of the 4 radii pulley. And I placed 20.23 g on the right side at the 105 mm diameter of the 4 radii pulley.

The trip times of the photo gate after a quarter rotation were .0568, .0563, .0568 sec. Average .0566 sec

All the 20 g drive masses are at the 35 mm diameter. So the 20g at the 105 mm diameter is equal to the 60 g at the 35 mm diameter.

I took the position of proving that it was (105 mm / 35 mm = 3 3² equal 9) 9 times harder to rotate the balanced masses at the 105 mm diameter. That would mean that 1/9 the mass of 60 g each side at 35; would be equal to 6.66 g each side at 105 mm. So I placed 6.960 grams on the right and 6.960 grams on the left side at the 105 mm diameter. I accelerated this with 20.04 g at the 35 mm diameter.

The same arrangement tripped at .0468, .0468, .0467 sec instead of the .0566 sec for the 60 g on each side at the 35 mm diameter.

So mr is off by a few 10,000th (.0566 to .0566) and mr² is off by 100 / 10,000th (.0468 to .0566). This means that the likelihood of mr being correct is about 99.9% and the probability of mr² being correct is somewhere around zero.

And this proves that energy is not a conserved quantity.