1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Applying the parallel axis theorem to find inertia

  1. Dec 2, 2015 #1
    1. The problem statement, all variables and given/known data
    Calculate the moment of inertia of a uniform rigid rod of length L and mass M, about an axis perpendicular to the rod through one end.

    2. Relevant equations
    Parallel axis theorem: I = Icm + MD2
    Long thin rod with rotation axis through centre: Icm = 1/12 ML2
    Long thin rod with rotation axis through end: I = 1/3 ML2

    3. The attempt at a solution

    I know this is a straightforward substitution problem into the parallel axis theorem EQN.
    However, for this question, I'm not sure why the answer key uses Icm of the rotation axis through the CENTRE. The question specifically states that the rotation axis is at the end....

    The answer given is I = 1/3ML2;
    I have calculated I = 7/12ML2, using the Icm of a long thin rod with the rotation axis through the end.
    D = 1/2L, since the centre of mass of a rod is right down the middle
    I = Icm + MD2
    I = 1/3 ML2 + M(1/2L)2
    I = 7/12ML2

    To reiterate, I'm just confused about why Icm was used for am axis through the centre, rather than through the end.
    Last edited: Dec 2, 2015
  2. jcsd
  3. Dec 2, 2015 #2
    Can you show your calculation?
  4. Dec 2, 2015 #3
    Just edited my post with the full calculations.
  5. Dec 2, 2015 #4
    Why Icm[/SUB=1/3ML2. Different from you stated in relevant equations.
  6. Dec 3, 2015 #5
    I'm not sure which part of the solution you're referring to.
    I = Icm + MD2 is the parallel axis theory, and I plugged in I = 1/3 ML2 the Icm component of that equation.
  7. Dec 4, 2015 #6


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You seem to be applying the parallel axis theorem twice over.
    The MoI about the rod's centre, axis perpendicular to the rod, is ML2/12. By the parallel axis theorem (or otherwise) the MoI about one end is ML2/12+M(L/2)2=ML2/3. You quoted this as one of your known equations, and it is directly the answer to the question. There is no need to go adding another ML2/4.
    Note that the Icm in the parallel axis theorem is the MoI for an axis through the object's mass centre. It is not valid to substitute in there the MoI about any other axis.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted