1. Not finding help here? Sign up for a free 30min 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!

Inertia of a beam with added masses

  1. Apr 16, 2014 #1
    1. The problem statement, all variables and given/known data

    A beam of mass m and length L with a moment of inertia (mL^2)/12 carries additional masses of m/2 at one end and at its centre. The moment of inertia about its centre is now...

    (a) - (5/24)mL^2
    (b) - 0
    (c) - (1/12)mL^2
    (d) - (1/3)mL^2
    (e) - (17/96)L^2

    2. Relevant equations

    I'm not sure what equations to use.

    3. The attempt at a solution

    And I have no solution.
     
  2. jcsd
  3. Apr 16, 2014 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Well, what formulas have you studied? Most people will list something.

    Can you at least figure out what the center of mass will be?
     
  4. Apr 16, 2014 #3
    Thanks SteamKing for the reply.
    I'm revising for exams in two weeks or so and I can't remember doing any questions like this, although we did some theory on centres of gravity and inertia. They were really proofs though. No I don't understand how to get the centre of mass. Although I could give it a go;

    Is it 1/8* L from the centre of the beam to the end of the beam with the additional mass?

    → m * g * (1/2) * y = m * g * (3/2) * x

    where x is the distance of the centre of gravity from the centre of the beam and y is the distance of the centre of gravity from end of the beam with the additional mass.

    → y = 3x; y + x = L * (1/2);

    → L * (1/2) - x = 3x

    → x = L * (1/8)
     
  5. Apr 16, 2014 #4

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Well, the proofs should tell you something. I mean, you were trying to prove that the c.o.m. of
    a given shape was related to its dimensions, right? Ditto for the moment of inertia.
    Was one of the proofs called the 'parallel axis theorem' by any chance?

    In any event, your calculations for the c.o.m. are wrong. You want to calculate the
    first moment of all the masses about a common origin, and then divide this moment by the total mass.
    You do know what a moment is, don't you?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Inertia of a beam with added masses
  1. Inertia of I beam (Replies: 2)

Loading...