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Homework Help: Moment of inertia of a rod about an axis

  1. Jun 2, 2006 #1
    Find the moment of inertia of a rod about an axis through its center if the mass per unit length is lambda = lambda (sub zero) time X.
    Answer (I = (1/8) M L^2)

    This problem is totally throwing me off. Normally lambda is equal to (M/L) so im not sure what this new formula is doing to the problem. I tried substituting and came up with (1/32)ML^3 which is horribly off from the answer. My idea was to add the x term from the new lambda giving me x^3. I then brought lambda (sub zero) out of the integral and integrated from (-L/2) to positive (L/2). Which brings us back to the wrong answer. Any ideas as to where I’m going wrong would be much appreciated!


  2. jcsd
  3. Jun 2, 2006 #2
    I found this thread on the internet, but I guess it still doesnt clear things up.

    "Okay, I'll do the first one for you.

    The line density is given as [tex]\lambda=\lambda_{0}x[/tex]

    It's no wonder that you are stuck with this exercise, because this expression is by itself meaningless.
    The correct expression must be:

    [tex]\lambda=\lambda_{0}|x|[/tex], -L/2<=x<=L/2
    i.e the absolute value of x, rather than x itself.
    (I assume you gave us ALL the information present in the exercise?)
    We gain:

    [tex]I=\int_{-\frac{L}{2}}^{\frac{L}{2}}\lambda_{0}|x|{x}^{2}dx= \lambda_{0}\frac{2}{4}(\frac{L}{2})^{4}[/tex]

    The mass M of the rod is readily calculated:

    [tex]M=\int_{-\frac{L}{2}}^{\frac{L}{2}}\lambda_{0}|x|dx=\lambda _{0}(\frac{L}{2})^{2}[/tex]

    Combining the expressions yiels the desired result"

    I guess at this point I just need a little help understanding the integration that took place. The absolute value of X is a little confusing. Thanks again in advance!

    Here is the problem again in more readable text:

    Find the moment of inertia of a rod about an axis through its center if the mass per unit length is [tex]\lambda=\lambda_o x[/tex].

    Answer: [tex]I=\frac 1 8 ML^2[/tex]

  4. Jun 2, 2006 #3


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    Gold Member

    It just means that the density is given as 0 at the axis and as [tex]\lambda_0 x[/tex] at a distance x away from the axis.

    You have to use the absolute value because you are integrating x from -L/2 to L/2, but you are concerned about the distance. The distance to the point -L/2 on the rod is L/2 so it follows that if x is -L/2 and you want the distance to x you want the absolute value.
    Last edited: Jun 2, 2006
  5. Jun 2, 2006 #4
    Oh, so the density of the rod is not constant. That part makes sense, but why was an equation for the mass given? The other thread explains that combining the equations gives the result, but it just isnt making sense yet. Thanks so much for your help,

  6. Jun 2, 2006 #5


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    Gold Member

    It depends on how you interpret the problem. It says the mass per unit length is [tex]\lambda=\lambda_0 x[/tex]. I assumed (as did the site where you got the solution from) that x is the distance from the axis to any point on the rod. So if you are looking for the density a distance L/4 away then it would be [tex]\lambda_0\frac{L}{4}[/tex]

    You need to combine your equations because the calculated moment of inertia [tex]I=\lambda_{0}\frac{2}{4}(\frac{L}{2})^{4}[/tex] is still in terms of [tex]\lambda_0[/tex]. You can solve for [tex]\lambda_0[/tex] in terms of M using the second integration that gave [tex]M=\lambda _{0}(\frac{L}{2})^{2}[/tex]
    Last edited: Jun 2, 2006
  7. Jun 2, 2006 #6

    Doc Al

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    Staff: Mentor

    Since the integrand is symmetric with respect to the origin, just integrate the positive side and double it.

    I find that statement odd. If [itex]\lambda_o[/itex] has units of [itex]\lambda[/itex], then [itex]\lambda_o x[/itex] cannot. I presume it should read (I haven't checked it) something like: [tex]\lambda=\lambda_o |x|/(L/2)[/tex]
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