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Calculating moment of inertia about a door hinge.à

  1. Dec 4, 2011 #1
    The problem statement, all variables and given/known data
    A solid door of mass 39.30 kg is 2.34 m high, 1.68 m wide, and 3.23 cm thick.

    What is the moment of inertia of the door about the axis through its hinges?

    If the edge of the door has a tangential speed of 76.5 cm/s, what is the rotational kinetic energy of the door?

    The attempt at a solution

    I don't really know where to start. Should I find the center of mass of the door? Because I = Ʃmr^2?

    I know how to solve for the second half (Ek = (1/2)Iω^2), but I'm not sure how to calculate I.
  2. jcsd
  3. Dec 4, 2011 #2
    The moment of inertia of a thin rod of mass m and lenght L about an axis at one of its ends and perpendicular to the rod is given by
    I = [itex]\frac{mL^{2}}{3}[/itex]

    Now let us imagine the door to be a set of thin rods parallel to the axis passing through the hinges.
  4. Dec 4, 2011 #3
    I have the answer to the first question, but I'm not nearly as well off with the second question as I had assumed.

    I got 37 kgm^2 for the first question.

    I assumed you would take that value and take 76.5 cm/s (.765 m/s) as ω and plug it into
    E = (1/2)Iω^2

    I get 10.8 when I do that (I used m/s and I'm assuming the units are joules), but that's incorrect.

    Any advice?
  5. Dec 4, 2011 #4
    [itex]\omega[/itex] is the ANGULAR speed i.e. measured in ANGLE per second i.e. rad/s and so you have the convert the TANGENTIAL speed given to ANGULAR speed.
  6. Dec 4, 2011 #5
    Oh! Okay. How do I calculate angular speed from tangential speed?
  7. Dec 4, 2011 #6
    angle (in radians) rotated = distance along the arc / radius
    therefore distance along arc = radius x angle
    i.e. distance along arc/time = radius x (angle /time)
    v = r x w
    v in m/s because linear or tangential
    w in rad/s because angular
  8. Dec 4, 2011 #7
    So to get velocity, I take the linear velocity and multiply it by the width of the door to get ω, and then plug that value into the energy equation?

    Edit: when I do that I get 30.6 J and that is incorrect.
  9. Dec 4, 2011 #8
    v = R[itex]\omega[/itex]
    therefore .... = [itex]\omega[/itex]
  10. Dec 4, 2011 #9
    Got it! Thanks!
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