Alvarez+lab+19

In this lab we used the properties of rotational equilibrium and use a known mass in order to find an unknown mass by knowing how far away the masses are from the pivot point.

Cases III:

Pivot Point: 0.502 m Moving Mass: 0.201 kg Weight of moving mass: 1.970 N Mass of Hanger: 0.017 kg

The stationary masses (B,D,E,F,R) were all at .1 m and were .4 m away from the pivot point.

B- the moving mass was placed at 0.969 m ( had .04 kg added to be able to fit it onto the meter stick D- the moving mass was placed at 0.983 m E- the moving mass was placed at 0.946 m F- the moving mass was placed at 0.929 m R- the moving mass was placed at 0.901 m

calculated stationary mass = actual mass + mass of hanger = ma + mh = ((lever arm of moving mass * weight of moving mass)/(lever arm of stationary mass*9.8)) - mass of hanger = ((rm * wm)/(rs * g)) - mh = (-(rm * 1.97)/(0.402 * .98)) – 0.017 <span style="font-family: 'Times New Roman',Times,serif;">i.e.) (-(-0.467 * 1.97)/(0.402 * .98)) – 0.017 = .216

<span style="font-family: 'Times New Roman',Times,serif;">Calculated Values

<span style="font-family: 'Times New Roman',Times,serif;">B- .257 m <span style="font-family: 'Times New Roman',Times,serif;">D- .224 m <span style="font-family: 'Times New Roman',Times,serif;">E- .205 m <span style="font-family: 'Times New Roman',Times,serif;">F- .197 m <span style="font-family: 'Times New Roman',Times,serif;">R- .133 m

<span style="font-family: 'Times New Roman',Times,serif;">Actual Values

<span style="font-family: 'Times New Roman',Times,serif;">B- .257 m <span style="font-family: 'Times New Roman',Times,serif;">D- .271 m <span style="font-family: 'Times New Roman',Times,serif;">E- .206 m <span style="font-family: 'Times New Roman',Times,serif;">F- .197 m <span style="font-family: 'Times New Roman',Times,serif;">R- .132 m

<span style="font-family: 'Times New Roman',Times,serif;">percent error = |(theoretical value – actual value)/actual value|*100 <span style="font-family: 'Times New Roman',Times,serif;">i.e.) |(0.694 – 0.690)/.690|*100 = 0.58%

<span style="font-family: 'Times New Roman',Times,serif;">Percent error <span style="font-family: 'Times New Roman',Times,serif;">B- .19% <span style="font-family: 'Times New Roman',Times,serif;">D- 17.52% <span style="font-family: 'Times New Roman',Times,serif;">E- .47% <span style="font-family: 'Times New Roman',Times,serif;">F- .24% <span style="font-family: 'Times New Roman',Times,serif;">R- .39%

Case IV:

Pivot point; 0.293 m Actual mass of meter stick: 0.135 kg

<span style="font-family: 'Trebuchet MS',sans-serif;">m1 <span style="font-family: 'Trebuchet MS',sans-serif;">Mass #: m1 <span style="font-family: 'Trebuchet MS',sans-serif;">Mass (kg): .097 <span style="font-family: 'Trebuchet MS',sans-serif;">Mass Position (m): 0 <span style="font-family: 'Trebuchet MS',sans-serif;">Lever Arm (m): .293 <span style="font-family: 'Trebuchet MS',sans-serif;">Weight Force (N): .951 <span style="font-family: 'Trebuchet MS',sans-serif;">Torque (m*N): .279

<span style="font-family: 'Trebuchet MS',sans-serif;">m2 <span style="font-family: 'Trebuchet MS',sans-serif;">Mass #: m2 <span style="font-family: 'Trebuchet MS',sans-serif;">Mass (kg): .135 <span style="font-family: 'Trebuchet MS',sans-serif;">Mass Position (m): 0.5 <span style="font-family: 'Trebuchet MS',sans-serif;">Lever Arm (m): .207 <span style="font-family: 'Trebuchet MS',sans-serif;">Weight Force (N): 1.323 <span style="font-family: 'Trebuchet MS',sans-serif;">Torque (m*N): -.274

<span style="font-family: 'Trebuchet MS',sans-serif;">percent difference

<span style="font-family: 'Times New Roman',Times,serif;">Counter Clockwise Torque (m*N): .279

<span style="font-family: 'Times New Roman',Times,serif;">Clockwise Torque (m*N): -.274

<span style="font-family: 'Times New Roman',Times,serif;">Difference of Magnitudes (m*N): .005

<span style="font-family: 'Times New Roman',Times,serif;">Average (m*N): .28

<span style="font-family: 'Times New Roman',Times,serif;">Percent Difference (%) 1.69

<span style="font-family: 'Times New Roman',Times,serif;">Questions:

<span style="font-family: 'Times New Roman',Times,serif;">1) As shown through this lab the closer a mass gets to the pivot point, the less torque that it will have. That goes the same way if you go the other way, the further a mass gets to the pivot point, the more torque that it will have. So to make a larger force produce little or no torque either get it closer to the pivot point or at the pivot point itself. An to make a smaller force produce a larger torque just move it away from the pivot point.

<span style="font-family: 'Times New Roman',Times,serif;">2) There is a support force that acts on the meter stick in order to keep the masses from falling to the table, this is not included in the calculations because this force and the force of gravity cancel which makes it have a net force of zero.

<span style="font-family: 'Times New Roman',Times,serif;">3) If an uniform meter stick is at rotational equilibrium when 220 g is suspended at 5.0 cm. 120 g is suspended at 90, and the support stand is placed at the 40 cm mark then in order to solve for the mass of the meter stick you must... <span style="font-family: 'Times New Roman',Times,serif;">((.4m–.05m)*.2kg*9.8m/s2)=((.9m–.4m)*.12kg*9.8m/s2) + ((.5m-.4m)*MASS*9.8m/s2) once you solve for mass then you have found the mass of the meter stick.

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