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!

Moments of inertia

  1. May 1, 2017 #1
    Hi,

    I need some assistance understanding moments of inertia. I am doing some review for an upcoming exam, however I am slightly stumped by this question. I have already tried googling "moments of inertia" to try and understand the concept better, but I am having issues knowing when to apply which formulas.

    I have attached my specific question and the "answer" to the question, hopefully someone can help me understand how to get the answer.

    ----

    Two identical slender rods of length l and mass m are linked together at 90deg, as shown in figure Q4a (image attached), to form a link in a mechanism. Point B is midway between A and C, and points B, D, E, F and G are equally spaced along the lower link.

    i) Determine the position of the center of gravity of the link

    ii) Find the moment of inertia of the link about the point B

    iii) Find the moment of inertia of the link about the point D

    iv) Find the radius of gyration about the point G

    fxfGv.png



    ---

    ANSWERS:

    i) Center of gravity is at D

    ii) IB=(5/12)ml2 kg m2

    iii) ID=(7/24)ml2 kg m2

    iv) k= sqrt(17/24) l m


    ----

    *My attempts*


    i)


    I know inherently that the centre of gravity is at D. How do i prove it mathematically?

    ----

    ii)


    I get the moment of inertial about B like so:

    I[subBb[/sub] = (ml2)/3 + ((ml2/3) - (ml2)/4)

    But i don't understand why I am supposed to subtract. Is it because the centre of gravity is l/2 below B?

    ----
    iii)


    For ID I get the answer by doing:

    ID = ((ml2)/3 - (ml2)/4) + ((ml2)/(3*16)) + (9ml2)/(16*3)

    Why do I subtract (ml2)/4) if that rod is above D?


    ----

    iv)


    I'm trying to use IG to find the radius of gyration:

    So for IG

    IG = (7/24)ml2 - ml2


    Here I am using the parallel axis theorem, but why do I subtract the (md2) of the AC rod from the Moment of inertia about D?
     
  2. jcsd
  3. May 1, 2017 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    It is obviously on the line BG, but where? You seem to have forgotten how to find a mass centre. It is the point about which the net first moment is zero. E.g. in the y direction, if D is the mass centre then measuring the yi offset of each element mi from D you want Σyimi=0.
    By what logic? Normally you would just add up the MoIs of the simple parts. What is the MoI of AC about B? What is the MoI of BG about B?
    Again, how do you explain the logic behind that?
    (Can you see how to use the results of (i) and (ii) and the parallel axis theorem to answer (iii) quickly?)
    I have no idea why you would do such a subtraction, especially since it gives the wrong answer.
     
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: Moments of inertia
  1. Moment of inertia (Replies: 3)

  2. Moment of inertia (Replies: 2)

  3. Moment of Inertia (Replies: 2)

  4. Moment of Inertia (Replies: 1)

Loading...