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How to linearise this equation for experiment?

  1. May 2, 2017 #1
    1. The problem statement, all variables and given/known data
    I have a rotating platform that spins as a mass attached to a wheel rotates the larger platform. The mass accelerates to the ground which spins the platform essentially. I am trying to calculate the moment of inertia of another mass which will be attached to the rotating platform. To do this I have a formula which I calculated:

    I = (r^2 * m * (g - a))/a
    Where I is the moment of inertia. r is the radius at which the string acts on the smaller "gear" wheel. m is the mass of the hanging object, g is the acceleration due to gravity and a is the actual acceleration of the object.

    I have two questions:
    1) Is this formula correct?
    2) I need to linearise the equation so that I can perform linear regression by collecting data on the acceleration of the mass as it falls. How can I do this? All my attempts at making it linear have failed. I am allowed to use the approximation that I/mr^2 is much greater than 1 but I do not have to, it is optional

    2. Relevant equations
    I used:
    α=a/r
    τ=Iα=rT
    F = ma
    Fg - T = ma
    where T is the tension in the string and Fg is the force on the mass due to gravity

    To get:
    I = (r2 * m * (g - a))/a

    3. The attempt at a solution
    The closest I can get is basically this:
    I = (r2*m*g)/a - r2*m
    But that is a hyperbola and I do not know how I can linearly regress using that
     
  2. jcsd
  3. May 2, 2017 #2

    kuruman

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    Let me understand the experiment first. You suspended different masses m, you measured the acceleration of the hanging mass a and you need to extract the moment of inertia I for a certain number of such measurements. Right?
     
  4. May 2, 2017 #3
    Yes, so the mass falls and applies a torque to a gear which spins a larger platform. So we suspend different masses and measure the acceleration of them as they fall in order to calculate the moment of inertia
     
  5. May 2, 2017 #4

    kuruman

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    Right. So your dependent variable is the acceleration and the independent variable is the hanging mass. Your equation can be solved for the acceleration to yield
    $$a=\frac{mR^2g}{I+mR^2}~\rightarrow~\frac{1}{a}=\frac{I+mR^2}{mR^2 g}$$
    Hence
    $$\frac{1}{a}=\frac{I}{mR^2g}+\frac{1}{g}$$
    Can you see how to linearize this?

    Edited to fix typos.
     
  6. May 2, 2017 #5
    Sorry, not quite. Can I also ask what happens to the g on the I/mR2 part at the end? But I do notice that that is the thing I can assume to be >> 1
     
  7. May 2, 2017 #6

    kuruman

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    Please note that I edited the equation I posted to add a "g" in the first denominator on the right side. The correct equation is
    $$\frac{1}{a}=\frac{I}{mR^2g}+\frac{1}{g}$$
    Suppose you let ##y=1/a## and ##x = 1/m## and you substitute in the above equation. What do you get?
     
  8. May 2, 2017 #7
    Sorry, must have read before you fixed it.

    Ohh, ok so I would end up with:
    y = Ix/R2g + 1/g

    Which is a linear equation with I/(R^2g) being the gradient of the line right?!
     
  9. May 2, 2017 #8
    I also have a follow up question. This gets me the moment of inertia of the platform itself right? So how can I calculate the moment of Inertia of a block placed on the platform?
     
  10. May 2, 2017 #9

    kuruman

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    Calculate or measure? If "calculate", use the definition of moment of inertia; if "measure", repeat your measurements for different hanging masses with the block sitting on top of the platform. This will give you the combined platform + block moment of inertia from which you can subtract the platform value that you have already found. Ideally, you do both and compare your "calculated" value with your "measured" value.
     
  11. May 2, 2017 #10
    Yes, I meant measure. So it is just as simple as subtraction? Excellent, thank you. So just do the experiment twice once with the block and once without and then subtract the platform only from the platform and box result. Thank you for all of your help. You have saved me a lot of stress tomorrow
     
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