1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Negative Moment of Inertia from Atwood Machine Experiment?

  1. Mar 29, 2016 #1
    I'm currently completing an Atwood Machine Experiment with two 100g weights on either side of the pulley, with a variance in weight created by attaching dimes and pennies to either side of the weights. The point of the experiment is to validate:

    (m1 – m2)g = (m1 + m2 + I/R2)

    1. The problem statement, all variables and given/known data
    The experiment involved varying the weights between the sides and timing how long M1 takes to hit the ground from being dropped from a height where m2 is just hovering above the ground. This was my data.
    http://puu.sh/nYt7k/cef6642fe9.png [Broken]
    n represents the number of dimes attached to the bottom of m2. As the number decreases, the dimes removed are added to the m1 side.

    height from pulley to ground: 0.31m
    pulley radius: 0.025m

    2. Relevant equations
    m1g - T1 = m1 a
    T2 - m2 g = m2 a
    (T1-T2) R = Iα
    From this system:

    (m1 - m2)*g = (m1 + m2 + I/R2 )*a

    To graph the results, we can use:

    y= (m1-m2)g
    mx = (m1 + m2 + I/R2 )*a

    3. The attempt at a solution
    http://puu.sh/nYtGJ/5ebdb6e1a8.png [Broken]

    Trendline equation for (m1-m2)g vs a: y= 194.179672x + 47.74182125
    Moment of Inertia was calculated using: slope = (m1 + m2 + I/R2 )

    when solving for inertia by using the slope of the graph created by the a as the independent axis and (m1-m2)g as the dependent axis, I get a negative number. I'm quite lost at this point and I'm not sure how to proceed. It doesn't make sense to me that inertia is negative. Could this be because i'm not accounting for the fact that the y intercept isn't 0 in our combined inertia equation?

    Thanks in advance for the help,
    Firefox
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Mar 29, 2016 #2
    What are you trying to express with with equation?
     
  4. Mar 29, 2016 #3
    Sorry about that, I'm not used to the formatting here yet!

    It should read:
    (T1-T2) R = Iα
     
  5. Mar 29, 2016 #4
    What do you think your dependent and independent variables are for you data analysis?
     
  6. Mar 29, 2016 #5
    I think (m1-m2)g is my dependent variable and my independent is a (acceleration).
     
  7. Mar 29, 2016 #6
    But don't you vary the masses and measure the acceleration? that implies to me that the variation in the masses is the independent variable. An that is a problem since the equation not only depends on the difference of the masses but also on their sum which are independent of one another.
     
  8. Mar 29, 2016 #7
    You make a good point, I'll admit, I didn't critically think about this point; I was following instructions from my lab manual. I've switched the variables around to see if it yields a different result, and alas, it seems I am still getting a negative inertia.

    http://puu.sh/nYDQv/d35faf1af2.png [Broken]
     
    Last edited by a moderator: May 7, 2017
  9. Mar 29, 2016 #8
    The equation that relates a to the difference is not linear. It should be of the form (I think)

    a = g⋅Δ/(2M +I/R2 - Δ) Where Δ is the mass difference and M is the sum of the two masses which is kept constant so we have only one independent variable.

    Oops Just noticed that a should = g⋅Δ/(2M +I/R2 + Δ)
     
  10. Mar 29, 2016 #9
    I'm not quire sure how you arrived at that equation. Do you mind explaining further?
     
  11. Mar 29, 2016 #10
    a the measured acceleration is the y- axis and Δ the difference in the masses of the weights is the x-axis. M is the sum of the two masses and is kept constant. You will be fitting a function of the form
    y = gx/( k + x) where k= 2M + I/R2 : k clearly should be a positive number.
     
  12. Mar 29, 2016 #11
    of course you just need to put in a value for a, Δ , and M to find I.
     
  13. Mar 29, 2016 #12

    gneill

    User Avatar

    Staff: Mentor

    A quick question; you wrote:
    Is the pulley height that of the center of the pulley? The bottom? Something else?
    You don't specify the starting height of the descending mass specifically. Can you clarify?

    It seems to me that if your total distance for the mass movement was less than about 26 cm then your calculated moment of inertia value would be positive.
     
  14. Mar 29, 2016 #13
    Wow ! Thank you! I've been staring at my data, and you have made realized this blunder. my h value is from the center of the pulley to the ground, which now I realize, is not the height in which the weight begins to fall from. Cheers! Mystery solved.

    Also, thanks to gleem for the help as well!

    I'm still quite new, is there a rep system where I can click for you guys?
     
  15. Mar 29, 2016 #14

    gneill

    User Avatar

    Staff: Mentor

    I'm glad it worked out. I couldn't see any obvious issues with the mechanics of your analysis so it occurred to me to look for what might be missing in the data.

    As for a "rep system", you can "like" individual posts, and once per year there's a member appreciation vote for various categories. There's also the Feedback and Announcements forum in the PF Lounge area where members can express their opinions on their experiences here at Physics Forums.

    Good luck with your lab report!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Negative Moment of Inertia from Atwood Machine Experiment?
  1. Atwood Machine Lab (Replies: 5)

Loading...