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!

Rotational inertia Problem

  1. May 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Three identical balls, with masses of M, 2M, and 3M, are fastened to a massless rod of length
    L as shown. The rotational inertia about the left end of the rod is:

    http://img13.imageshack.us/img13/1424/1234sjy.jpg [Broken]

    2. Relevant equations

    I = MR^2

    3. The attempt at a solution

    I total = (3M)(0)^2 + (2M)(L/2)^2 + (M)(L)^2

    I total = 3ML^2/2

    It says the answer is 3ML^2/4 though. Thanks for the help
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 3, 2009 #2
    Hmm...

    [tex]
    I = mr^{2}
    [/tex]

    [tex]
    I = (3M)0^{2} + 2M(\frac{L}{2})^{2}) + ML^{2}
    [/tex]

    Therefore;

    [tex]
    I = \frac{1}{2}ML^{2} + ML^{2}
    [/tex]

    [tex]
    I = \frac{3}{2} ML^{2}
    [/tex]

    I get the same as you.

    The definition of the moment of inertia is;

    SUM( miri2 ) So I'm preaty certian that we are following the correct method. So perhaps it is some arithmetical mistake we are both making.

    2/4 is 1/2 yes and 1/2mr2 + mr2 is mr2(1/2 + 1)
    (1/2 + 1) = 3/2. Nope. I think your answer book might be flawed.

    UNLESS you are meant to use the distribution of mass in a sphere for the first mass (3M). Mindyou, because they havn't given you a radius for the sphere I would assume not.

    Haths
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Rotational inertia Problem
Loading...