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Moments of Inertia, Centres of Mass help !

  1. May 14, 2008 #1
    Moments of Inertia, Centres of Mass... help plz!

    1. The problem statement, all variables and given/known data

    A rod of length l rotates about an axis that runs perpendicularly to the rod through one end. The linear density p(x) of the rod in terms of the distance x from the axis is given by:

    p(x) = p0(1 + (x^2/l^2))

    where p0 is a constant. In terms of l and p0, find the total mass M of the rod, the distance R of the centre of mass of the rod from x=0, and the moment of inertia I and radius of gyration k of the rod about the axis.

    3. The attempt at a solution

    For the mass M, I guessed what you needed to do was integrate the p(x) expression between the limits of 0 and l with respect to x. I did this and got (4/3)lp0. Is this the right method?

    For the centre of mass, I really don't know! I don't know how to do it for this type of problem, only simple (uniform density) problems.

    For the moment of inertia, based on my limited understanding, I decided to take the expression for p(x) and perform the same integration as above but with an expression x^2 in there, so I did this:

    p0 * [integral between limits 0 and l] (x^2 + (x^4/l^2) dx

    Please don't ask me to explain why I did this, it's based on guesswork and trying to copy the method of other problems I've seen, which were a lot more simple! Anyway, the answer I got was (8/15)(l^3)p0. Am I anywhere close?

    For the radius of gyration I know the formula I = mk^2 (where k = radius of gyration) so that should be easy if I have the correct mass and inertia. If I find it based on the answers I've got for M and I here (which I expect are wrong), the answer is sqrt((2l^2)/5).

    Sorry, I would try to figure out how to type all the symbols and notation etc properly, but I'm under extreme exam pressure and don't have time! Your help would be greatly welcomed!
  2. jcsd
  3. May 14, 2008 #2


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    You have mass right. You have moment of inertia right. You have radius of gyration right. At this point, given your lack of confidence, I'm pretty impressed. The center of mass is the integral of p(x)*x divided by the mass. I think you'll get that right too. Now you just have to figure out why you are right all the time. You are not doing well at portraying yourself as clueless.
    Last edited: May 14, 2008
  4. May 15, 2008 #3
    Sorry, I didn't mean to portray myself as clueless, I just haven't done a problem like this before and was convinced I was going to get it all wrong and look like an idiot. I still don't understand it very well - for instance, I really don't understand why I had to put that extra x^2 in the integral to work out the moment of inertia, but I'm glad it gave me the right answer! Thanks for the formula for the centre of mass. I guess that means it's (9/16)l then?

    Believe me, when it comes to Mechanics I definitely am not "right all the time", but this has given me a bit more confidence so thanks!
  5. May 15, 2008 #4


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    Yes, (9/16)*l. It really isn't that hard to see where these formulas are coming from. If you know how to find these quantities for a set of point masses (review this), the extension to a continuous body is pretty straight forward.
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