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Finding Moment of Inertia Using Integration

  1. Aug 6, 2013 #1
    1. The problem statement, all variables and given/known data

    A cylinder with radius R and mass M has density that increases linearly with distance r from the cylinder axis, ρ = αr, where α is a positive constant. Calculate the moment of inertia of the cylinder about a longitudinal axis through its center in terms of M and R.

    This is from my textbook, but my textbook has only explained how to find inertias from uniform masses leading up to this question. What's even worse is that this question is given the easiest rating.

    2. Relevant equations

    I = Ʃ(mi*r^2i)

    Normally I would find the integral of r^2*dm, but my book says that it is only for a uniform distribution of mass.

    3. The attempt at a solution

    I = Ʃ(mi*r^2i) = m(1)*r^2(1) + m(2)*r^2(2) + ...

    I just cannot figure out how to incorporate an increasing density into this.
     
    Last edited: Aug 6, 2013
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  3. Aug 6, 2013 #2

    SteamKing

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  4. Aug 7, 2013 #3
    I couldn't find non-uniform increasing density in the link you provided. I feel fairly comfortable with simple objects that have a uniform density. But I just have no idea what to with a changing density. Here's my work.

    I = ∫r^2*dm

    ρ = αr (according to the question)

    ρ = αr = dm/dV then dm = α*r*dV

    I = ∫r^2*α*r*dV

    dV = 2*∏*L*r*dr

    I = ∫r^2*α*r*2*∏*L*r*dr
    = ∫r^4*α*2*∏*L*dr
    = α*2*∏*L*∫r^4*dr
    = (ρ/r)*2*∏*L*∫r^4*dr
    = (ρ/r)*2*∏*L*(r^5)/5
    = ρ*2*∏*L*(r^4)/5
    = (M/V)*2*∏*L*(r^4)/5
    = (M*2*∏*L*r^4)/(5*∏*L*R^2)
    = (2*M*R^2)/5

    My textbook has I = (3*M*R^2)/5. I do not feel confident in my answer.
     
  5. Aug 7, 2013 #4

    TSny

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    I think everything's fine up to here. Of course, you should have definite limits of integration in mind for the integral.

    But your next step shown below is odd.

    You can't replace the constant α by ρ/r because r is your variable of integration. Leave the integral in terms of α. But then you will need to do a separate integration to relate α to M and R.
     
  6. Aug 7, 2013 #5
    The reason why I did this is because the question said that α is a constant. So then I just made sure that α = p/r stayed as a constant. I feel quite lost.

    I don't why I would do that, and I don't know how I would do that. I feel like this question is beyond anything that we learnt up until now in the textbook. But it is a question among many relatively easy ones that the chapter has a direct explanation for.
     
  7. Aug 7, 2013 #6

    TSny

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    OK, let's just take it step by step. You had a very good start where you got to the integral

    I = ∫(r2)(αr)(L2∏rdr) = α2∏L∫r4dr.

    (Note that we have formatting buttons for superscripts and subscripts that you can use to make the expressions easier to read.)

    Now, α is just a constant factor that you can leave alone for now. Go ahead and evaluate the integral with the correct lower and upper limits for r.
     
  8. Aug 7, 2013 #7
    I = α*2*∏*l*∫r4dr = (α*2*∏*L*r5)/5

    I can't figure out how to show the proper scripts for the lower and upper limits for the integral. The lower is 0 and the upper is R.
     
    Last edited: Aug 7, 2013
  9. Aug 7, 2013 #8

    TSny

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    [EDIT] Shouldn't your result for the integral be expressed in terms of the limits of integration?
     
  10. Aug 7, 2013 #9
    I am just trying to figure out how to show that.
     
  11. Aug 7, 2013 #10

    TSny

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    Don't worry about showing the limits of integration on the integral. Just worry about using those limits when evaluating the integral.
     
  12. Aug 7, 2013 #11
    [itex]\int[/itex][itex]^{R}_{0}[/itex] How did you get your subscripts on top and on the bottom like that.
     
  13. Aug 7, 2013 #12

    TSny

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    Let's not worry about how to type in the limits on the integral. What is the answer to ##\int_0^R r^4dr##?
     
  14. Aug 7, 2013 #13
    Ok, it is r = R, and r = 0, right?
     
  15. Aug 7, 2013 #14
  16. Aug 7, 2013 #15

    TSny

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    Those are the upper and lower limits of the integral. Do you know how to evaluate an integral at its limits? So far, you have gotten to the point of evaluating

    ##I = a2\pi L \int_0^Rr^4dr##.

    Can you carry out the integration and evaluate at the limits?
     
  17. Aug 7, 2013 #16

    TSny

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    Good. So, what is your expression for the moment of inertia at this point?
     
  18. Aug 7, 2013 #17
    I = a*2*∏*L*R5/5
     
  19. Aug 7, 2013 #18

    TSny

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    OK. Now you just need to figure out how to express the constant ##a## in terms of ##M## and ##R##. Can you use the density function to set up another integral that would represent the total mass of the cylinder?
     
  20. Aug 7, 2013 #19
    I = a*2*∏*L*R5/5 and α = ρ/r = m/(∏*L*r3)

    = (m*2*∏*L*R5)/(∏*L*r3*5)

    = (m*2*R2)/5

    Hmmm, I'm stuck.
     
  21. Aug 7, 2013 #20

    TSny

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    You don't want to substitute α = ρ/r. That would put the variable r into the expression. But I should be a fixed constant. Note that r and R have distinct meanings, the first is a variable while the second is a constant.

    The total mass M must be the sum (integral) of all the elements of mass dm: M = ∫dm. You know how to substitute for dm in terms of the density function and dV. Try it here and see if you can evaluate the integral.
     
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