Finding mass of unbalanced meter stick from torque

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SUMMARY

The discussion focuses on calculating the mass of an unbalanced meter stick using torque principles. A 100g mass is hung at the 0m point, and the fulcrum is moved to the 0.33m point for balance. The calculated torque on the left side is 0.3234 Nm, while the moment arm on the right side is incorrectly assumed to be 0.34m. The resulting mass calculation for the right side yields 0.0971 kg, which does not align with the total measured mass of the meter stick at 0.19 kg, indicating a misunderstanding of the moment arm and torque balance.

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Homework Statement


Unbalanced meter stick has hanging mass of 100g at 0m point. I then moved fulcrum to balance it and it was balanced at the 0.33m point. (The hanging mass and 0m point is on left side)
I want to get the mass of the portion of the meter stick on right side, but I am confused as to what the moment arm on right side would be.
.

Homework Equations


ccw torque = cw torque (balanced stick)

The Attempt at a Solution


The torque on left side is 0.1kg(9.8m/s/s)(0.33m) = .3234NmI think the moment arm for right side is (1 - 0.33)/2 = 0.34m, because center of mass will be at 0.67m point on meter stick.
But when I calculate mass from this, it does not agree with my already weighed mass of stick
mass of meter stick on right side of fulcrum = 0.3234Nm / (9.8m/s/s * 0.34m) = 0.0971
mass of entire meter stick = 0.19kg (weighing it on scale)

I know that the mass of only part of the meter stick will be less than the mass of the whole meter stick, but my values are different by almost 50%. Thanks for any help
 
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What about the torque from the left side of the stick?
 
A diagram would help.
 
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