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Calculating Spring Constant From Plot of T^2 versus r^2

  1. May 1, 2017 #1
    1. The problem statement, all variables and given/known data
    Hi,

    I recently performed a lab experiment for calculating the moment of inertia and spring constant of a couple of equal masses on a steel rod. I had to do this experiment at the exact same time as performing another experiment so I was unable to perform any calculations in the lab, just simply gather data.

    The first part of the experiment requires the calculation of the torsional spring constant (k) using a plot of T2 v r2 and as I was unable to perform calculations in the lab, I was hoping someone could provide a little guidance here.

    2. Relevant equations
    T2=4π2(I/k)
    r2=I/2m
    y=Δx+c (where Δis the gradient (wrong symbol, I know))

    (T is the period of one single oscillation, r is the radius of the masses from the centre of mass)

    3. The attempt at a solution
    I know that T2 = 4π2(I/k) and that r2 = I/2m (because there were two equally spaced masses on a thin rod of negligible mass). If I then plot T2 v r2 in JLineFit and equate this to the the equation of a straight line, I should find that T2 = Δr2+c.

    Subbing in my values for T2 and r2 and solving for k, I get the equation to be:
    2(I/k) = (ΔI/2m)+c

    so if I start by multiplying through by k I get

    2I=k((ΔI/2m)+c)

    so

    2I/((ΔI/2m)+c)=k

    so

    k = 4π2/((Δ/2m)+c)

    All I really want to know is if my rearranging is correct (do the I's cancel?). I get very confused here!

    Thanks
     
    Last edited: May 1, 2017
  2. jcsd
  3. May 1, 2017 #2

    kuruman

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    What is the relevance of this equation to your experiment? What do the symbols mean in terms of something that you measured?
    How do you use the symbol Δ? Usually it means a change.
     
  4. May 1, 2017 #3
    it is the equation of a straight line (y=mx+c) I should have used ∇.

    T is the time period of one single oscillation, r is the radius of the mass.

    Thanks
     
  5. May 1, 2017 #4

    scottdave

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    OK, so since c is an arbitrary constant (to be determined from the data), then k*c can also be just an arbitrary constant. When you divided through by I to cancel, you would have an c/I, which is also an arbitrary constant, as well. This could simplify some things for you when you had 4π²I=k((ΔI/2m)+c) = =k*(ΔI/2m) + (k*c), where k*c, or even k*c/I can be just an arbitrary constant (to be determined).
     
  6. May 1, 2017 #5

    kuruman

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    To make things simple: You have ##T^2=4 \pi^2 I/k## and ##I = 2mr^2##. If you replace the ##I## in the first equation, you get
    ##T^2=8 \pi^2 mr^2/k##. Now let ##y=T^2##, ##slope=8 \pi^2 m/k## and ##r^2 = x##. Then you have the standard straight line equation ##y=slope*x##. I used "slope" instead of m in order not to confuse the "slope" m with the "mass" m. You can add an intercept ##c##, if you wish, but you need to explain what it means to have some non-zero period when r = 0. Once you get a number for the slope, you can solve ##slope=8 \pi^2 m/k## to find ##k##. This process is known as linearization.
     
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