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I Rotational Inertia

  1. Apr 26, 2017 #1
    A 24 kg solid door is 220 cm tall, 95 cm wide. What is the door's moment of inertia for rotation about a vertical axis inside the door, 17 cm from one edge?

    I've looked at examples of how to do this problem. They connect it to center mass but I get confused when determining what the center mass is. They also talk about parallel axis theorem, but my professor has not introduced this approach.

    The formula that I have been using is I= (1/3)mr^2
  2. jcsd
  3. Apr 26, 2017 #2
    If you're like me and look at other examples for help:
    I was using the wrong eq. use I=(1/12)mL^2 + m (L/2 -d)^2
    L is the length of the door and d is the distance from the edge
  4. Apr 28, 2017 #3


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    That looks right.

    From here..

    The moment of inertia about the center of mass (Icm) of the door is..
    Icm = (1/12)mL2

    The centre of mass will be the middle of the door (L/2 from the edge) for a door of uniform thickness.

    Then the parallel axis theorem says..
    I = Icm + mx2

    where x is the distance between the centre of mass and the required axis. In your case x = (L/2)-d so the total moment of inertial is

    I = (1/12)mL2 + m((L/2)-d)2.......................................(1)

    Which is your equation.

    You should also be able to get the same answer without using parallel axis theorem...

    You could also approach this problem by pretending you had TWO doors, a narrow one (width d) and a wide one (width=L-d) connected together at a common hinge and add together the moments of inertia of each.

    The moment of inertia of a door about it's edge is..
    Ih = (1/3)mL2

    In this case you have two doors of different width L..

    The narrow one..
    In = (1/3)md^2
    The wide one..
    Iw = (1/3)m(L-d)2

    So the total moment of inertial would be..
    I = In + Iw

    I = (1/3)md^2 + (1/3)m(L-d)2......................................(2)

    I'll let you prove (1) and (2) are the same.
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