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Moment of Inertia for a Square and a pipe of variable density

  1. Jul 7, 2012 #1
    1. The problem statement, all variables and given/known data

    First, there's a slender rod with length L that has a mass per unit length that varies with distance from the left end, where x=0, according to dm/dx = yx where y has units of kg/m^2. (a) Calculate the total mass of the rod in terms of y and L (Which I've already done and is .5yL^2) (b) Use I = ∫r^2*dm to calculate the moment of inertia of the rod for an axis at the left end, perpendicular to the rod. Express your answer in terms of total mass M and length L. (c) Repeat b for an axis at the right end, perpendicular to the rod. Express your answer in terms of total mass M and length L.

    Second, a thin, uniform rod is bent into a square of side length a. If the total mass is M, find the moment of inertia about an axis through the center and perpendicular to the plane of the square.

    2. Relevant equations

    I= ∫r^2 dm.

    Ip = Icm + Md^2 (parallel-axis theorem)

    3. The attempt at a solution

    For the square I tried to use the Parallel-Axis Theorem by calculating the moment of inertia of a corner with 2 sides coming out of it. Using the Parallel-axis theorem to move the center to the middle and then doubling it to account for the two other sides. That got me 4Ma^3/3+Ma^2/2 which is wrong.

    For the rod of variable mass, I tried to use the integrals ∫r^2*y*L and ∫r^2*y*(L-x) but those didn't work.

    What should I have done?
     
  2. jcsd
  3. Jul 7, 2012 #2

    TSny

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    Welcome to Physics Forums!

    I would not use a corner with 2 sides because then it get's a little involved using the parallel axis theorem. For instance, you would have to first find Icm for this corner which requires locating the cm of the 2 sides (which is not at the corner) and finding the moment of inertia about that cm point. Then you would add (M/2)d^2 where d is distance from center of the square to the cm of the corner with 2 sides and M is the total mass of the square. Then double to take care of 2 corners.

    Instead, I would recommend using the parallel axis theorem to find the moment of inertia of one side of the square about the center of the square and then multiplying by 4.

    For the rod, it would help if you showed us more of your work in getting the integrands of the integrals in terms of one variable of integration.
     
  4. Jul 7, 2012 #3
    Thanks for your suggestion for the square, I found the correct answer as Ma^2 /3. Thanks for helping. As for the rod problem, I start with ∫r^2*dm replace dm with rho*dv to get ∫r^2*ρ*dv then since I know the density equals y*x I replace p with y*x and integrate. I'm pretty sure I'm doing something wrong there but fortunately, I managed to luck into the answer for when the rod's being held at the left end (ML^2/2). Do you think I could use the parallel-axis theorem to find the moment of inertia for moving it to the other side?
     
  5. Jul 7, 2012 #4
    I tried my suggestion and couldn't get it to work. How should I have set up the integral in my original attempt?
     
  6. Jul 7, 2012 #5

    TSny

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    You want to evaluate ∫r2dm.

    How would you express r in terms of the variable x? Remember, r is the distance from an element of mass dm to the axis (at x = L).
     
  7. Jul 7, 2012 #6
    I'm still thinking it should be L-x, since the density gets lower as it gets farther from the end.
     
  8. Jul 7, 2012 #7

    TSny

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    Yes, r = L-x. But the reason does not have to do with the density. It's because if x denotes the location of an element of mass dm, then the distance from the element to the right end of the rod is r = L-x.

    So, now you know what to use for r. All you need to do now is substitute an expression for dm in terms of x and dx and integrate.
     
  9. Jul 7, 2012 #8
    According to the problem statement, dm = yxdx.

    Edit:Substituting yxdx for dm I get ∫(L-x)^2*yxdx. Integrating from 0 to L I get L^3*yx/3. Multiplying by M/(2yL^2) I get ML^2 /24. How's that?
     
  10. Jul 7, 2012 #9

    TSny

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    So, what do you get when you evaluate the integral?
     
  11. Jul 7, 2012 #10
    I tried ML^2 /24, but that's incorrect.
     
  12. Jul 7, 2012 #11

    TSny

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    Let's see.

    ∫r2dm = ∫(L-x)2(yxdx) where limits are from x = 0 to x = L.

    Is this what you evaluated?

    I get (y/12)L4 which may be expressed in terms of M as (1/6)ML2
     
  13. Jul 7, 2012 #12
    Yes, turns out I multiplied wrong. The correct answer is ML^2/6.
     
  14. Jul 7, 2012 #13

    TSny

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    x should not appear in your answer.
     
  15. Jul 7, 2012 #14
    That was just a typo. Thanks for your help.
     
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