Centripetal Force Lab of string and mass

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SUMMARY

The discussion centers on a centripetal force lab where participants measured the relationship between mass and frequency using a string with a radius of 50.0 cm. The first mass used was 29.0 g, and after conducting 20 trials over 15.75 seconds, the slope of the line of best fit calculated via Excel was 3.83. A key takeaway is that the mass of the rubber stopper will not approximate the slope of the line of best fit, as it is influenced by the factor of 4π²r. Additionally, participants were advised to plot against frequency squared (f²) rather than frequency (f) for accurate results.

PREREQUISITES
  • Understanding of centripetal force and its formula: F = m4π²r/T²
  • Familiarity with data plotting tools, specifically Excel for graphing
  • Knowledge of frequency and period relationships in physics
  • Basic skills in algebra for solving equations
NEXT STEPS
  • Learn how to accurately plot data in Excel, focusing on creating lines of best fit
  • Study the relationship between frequency and period in circular motion
  • Explore the implications of centripetal force in practical experiments
  • Investigate the effects of varying mass on centripetal acceleration
USEFUL FOR

Students conducting physics experiments, educators teaching centripetal force concepts, and anyone interested in understanding the dynamics of circular motion and data analysis in lab settings.

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Homework Statement


I had to do a lab, the diagram of which is shown.
ucm_app.gif


The radius (r) of the string was 50.0 cm, and our first mass was 29.0 g. The time for 20 trials was 15.75 s. We repeated the process with five other masses and recorded all the same information. Then, we had to plot the weight (F) of the masses versus the frequency and find the slope of the line of best fit. Our slope came out to be 3.83 (according to Excel).
We also have to use this information to find the mass of the rubber stopper. My attempt at the solution is below but I'm not sure if I did it right. Also, someone else in my class told me that the mass of the stopper should be close to the slope of the line of best fit, but mine was way off. Is that true?

Homework Equations


F = m4π2r/T2


The Attempt at a Solution


F = m4π2r/T2
0.284N = m(4)(π2)(.5m)/(.62s)
m = 0.0089 kg
 
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Interesting experiment (how did you manage to keep the mass on the bottom from oscillating up and down??). Looking at your calculations, there was a tiny error regarding Period and Frequency. Frequency is 1/T where T = 15.75 s for your trial. But you divided by this number (1/15.75) instead of multiplying it! Perhaps you should reevaluate your calculations a bit and then try it out again. I can tell you now that the mass of the stopper will not be close to the line of the best fit. It will differ by a factor of 4pi^2r (As you have unknowingly indicated in your equation).

Also, you should try to plot against f^2 and not f (frequency)

Best of luck!
PL
 
Last edited:

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