second condition of equilibrium

second condition of equilibrium - mass, which was 0.70...

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Akshay Patel Period 10 Honors Physics Lab- Second Condition of Equilibrium Objective: How can the second condition of equilibrium be applied to determine the mass of a meter stick? Data: Meter Stick Mass 1 with slider Mass 2 with slider Distance from axis 0.35m 0.15m 0.70m Mass .078kg 0.1365kg 0.0665kg Conclusion: The second condition of equilibrium can be applied to use to find the mass of a meter stick. The second condition of equilibrium states that the sum of all torques must be zero, so there is no angular acceleration. During this lab, we knew that there was no angular acceleration when we thought the meter stick was balanced. The sum of the torques of this lab is zero and the axis starts at 35 cm. There was a negative torque from the 0.1365kg mass that was 0.20 away from the axis and also from the meter stick, which was 0.35 meters away from the axis and the torque came from the ruler since it is the center of gravity of the meter stick. There was a positive torque from the .0665-kilogram
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Unformatted text preview: mass, which was 0.70 meters away from the axis. We solved for the mass, which is shown, in my calculations. From the calculations you can see the second condition of equilibrium applied. We used this to find the actual mass of the meter stick. We found the mass of the meter stick, as 0.078kg while the actual is 0.2kg, which gives us a percent error of 39.0%. There were quite a few sources of error from looking at the percent error. One possible source of error was the meter stick never being truly balanced. During the lab, it seemed like it would be balanced for a few seconds, then fall to one of the sides. Also, there could have been error in the placement of the sliders and the masses that we knew. The placement couldve been a little off, and that couldve affected other results in the lab....
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