Lab Report: Newtons Second Law Atwoods Machine

18Atwoods Machine EX-9973Page 6 of 7

Atwoods Machine EX-9973Page 1 of 7

Lab Report: Newtons Second Law Atwoods MachineName: Sample Data

DATA TABLE for Procedure A: Total Mass Constant

Run

Descending

Ascending

Experimental

(Slope of the v vs. t plot.)

Trial 1Trial 2Trial 3Average

129.927.10.3350.3410.3340.337

231.925.10.9640.9670.9450.959

335.921.12.122.221.952.097

437.020.12.572.322.512.467

[g][g][m/s2]

RunDescending

Ascending

Total Mass

Net Force

Experimental

(Average)

Theoretical

Percent difference

129.927.157.00.0270.3370.48135.4%

231.925.10.0670.9591.16919.8%

335.921.10.1452.0972.54519.3%

437.020.10.1662.4672.90116.2%

[g][g][g][N][m/s2][m/s2]

DATA TABLE for PROCEDURE B: Net Force Constant

Run

Descending

Ascending

Experimental

(Slope of the v vs. t plot.)

Trial 1Trial 2Trial 3Average

514.910.01.441.441.331.403

616.912.01.241.231.221.230

717.913.01.171.171.181.173

837.832.90.5090.5020.5100.507

[g][g][m/s2]

RunDescending

Ascending

Total Mass

Net Force

Experimental

(Average)

Theoretical

Percent difference

514.910.024.90.0481.4031.92931.5%

616.912.028.91.2301.66229.9%

717.913.030.91.1731.55427.9%

837.832.970.70.5070.67929.0%

[g][g][g][N][m/s2][m/s2]

DATA for PROCEDURE C: The Effect of the Pulley

Mass of string used in Procedures A & B: 0.4 gDiameter of the Super Pulley: 5.1 cm

Radius of the Super Pulley, 2.54 cm

RunFrom Previous ActivitiesCalculate

Experimental Acceleration

Net Force

Total Mass

10.3370.0270.0570.0082

20.9590.0670.0570.0120

32.0970.1450.0570.0255

42.4670.1660.0570.0248

51.4030.0480.0250.0131

61.2300.0480.0290.0125

71.1730.0480.0310.0118

80.5070.0480.0710.0122

[m/s2][N][kg][N]

From the plot:

Slope, = 0.0081 kg

y-intercept, = 0.0041 N

Calculation:

Rotational Inertia of the pulley, = 52.3 x 10-7 kg m2

QUESTIONS Procedure A: Total Mass Constant1. Look at the data: as the net force increased, what happened to the acceleration? Did it increase, decrease or stay constant? As the net force increased, the acceleration increased as well.

Did a change in the net force produce a change in acceleration by the same factor? Do your results agree with Newtons 2nd Law?The factor was not always exactly the same. For example, Trial 1 has Fnet = 0.027 N, and Trial 4 has Fnet = 0.166 N. That is an increase in the net force by a factor of about 6. (0.166/0.027 = 6.2). In comparison, the acceleration increased from 0.337 m/s2 to 2.467 m/2, an increase by a factor of approximately 7.Use this grid to make a plot of Net Force vs. Experimental Acceleration and draw the best fitting line.

Calculate the slope of the best-fitting line. What does the slope of the best-fit line represent? (Hint: what are the units of the slope?)Slope:0.14 N 0.04 N_________________________ = 0.067 kg2.0 m/s2 0.5 m/s2The slope represents the total mass of the system. The units are N/m/s2, which is equivalent to kg.

QUESTIONS Procedure B: Net Force Constant1. Look at the data: as the total mass increased, what happened to the acceleration? Did it increase, decrease or stay constant? As the total mass increased, the acceleration decreased.

Did a change in the total mass produce a change in acceleration by the same factor? Do your results agree with Newtons 2nd Law?The factor was almost exactly the same. For example, Trial 1 has Mass = 24.9 g, and Trial 4 has Mass = 70.7 g. That is an increase in the total mass by a factor of about 2.8. (70.7/24.8 = 2.8). In comparison, the acceleration decreased from 1.403 m/s2 to 0.507 m/2, a decrease by a factor of approximately 1/2.8 (1.403/0.507 = 2.8).

QUESTIONS Procedure C: The Rotational Inertia of the Pulleys

1. The motion and mass of the string that moves the system was never considered in any part of the theoretical analysis. Looking at your results, is it reasonable to ignore the mass of the string as part of the total mass of the system? Discuss.The mass of the string is extremely small compared to the calculated excess mass of the system. It is reasonable to assume that the mass of the string does not considerably affect the results and it is OK to ignore it. The excess mass seems to be the effect of the rotational inertia of the pulley and it is also very small, as expected.

1. Looking back at the ideal case: was it safe to assume that the system is essentially frictionless? Yes, it was safe to assume this. There is friction in the system, but it is minimal.

1. What has a larger impact on the percent differences found in procedures A & B, the small excess mass or the small amount of friction? Discuss.The small amount of friction varies from being 3% to 15% of the net force across the trials. The small excess mass varies from being 11% to 32% of the total mass. It seems that the effects of the rotational inertia of the pulley are more significant than any friction in the system in each of the trials.

1.

Assuming frictional forces only act to oppose the linear motion of the masses, and that each mass receives the same amount of friction , prove that .

The equation of motion of the descending mass, , is:

The equation of motion for the ascending mass, , is:

Adding the two equations together:

That is,

PASCO 200418 - 1 of 16Written by Cecilia A. Hernndez2010