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Moment of inertia car problem

  1. Aug 19, 2008 #1
    1. The problem statement, all variables and given/known data

    A driver starts his car with the door on the passenger’s side
    wide open ([tex]\omega[/tex] = 0) the 36-kg door has a centroidal radius of gyration
    k= 250 mm, and its mass center is located at a distance r = 440 mm from
    its vertical axis of rotation. Knowing that the driver maintains a constant
    acceleration of 2 m/s2, determine the angular velocity of the door as it
    slams shut ([tex]\omega[/tex] = 90°).

    R = 440mm = .44m
    k = 250mm = .25m
    a = 2 m/s
    m = 36kg


    2. Relevant equations

    [tex]\Sigma[/tex]M = Ig[tex]\alpha[/tex] + [tex]\Sigma[/tex] r x mag ... (1)

    [tex]\omega[/tex] = [tex]\omega[/tex]0 + 2[tex]\alpha[/tex][tex]\theta[/tex]

    a = R[tex]\alpha[/tex]

    3. The attempt at a solution

    [tex]\Sigma[/tex]M = Ig[tex]\alpha[/tex] + [tex]\Sigma[/tex] r x mag

    ::: [tex]\Sigma[/tex]M = Wg R cos([tex]\theta[/tex])
    * even though i did this...what do i use for theta? (90? degrees)
    * also, i think the only thing that contributes to the moment is the weight of the door?

    ::: Ig[tex]\alpha[/tex] + [tex]\Sigma[/tex] r x mag

    = I[tex]\alpha[/tex] + maR
    = mk2 [tex]\alpha[/tex] + mR2[tex]\alpha[/tex]

    after i complete (1) i get this when i solve for alpha:

    [tex]\alpha[/tex] = Rg/ (k2 + R2cos([tex]\theta[/tex])

    and then i get stuck here. i didn't know what to use for the theta, but i think im suppose to find [tex]\alpha[/tex] and then use the kinematic equation:

    [tex]\omega[/tex] = [tex]\omega[/tex]0 + 2[tex]\alpha[/tex][tex]\theta[/tex]
    a = R[tex]\alpha[/tex]

    to find omega, and i know omega initial = 0 ??

    i feel like im missing an important part of the equation because im not getting the right answer...and i don't quite understand it. Im a little confused on the cosine part, am i suppose to use that for the acceleration, and not the moment of the weight? if someone can please help?!? i would appreciate it!! Thanks!! :D :D
  2. jcsd
  3. Aug 19, 2008 #2


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    Homework Helper

    Have you thought about approaching this problem from the frame of reference of the accelerating car?

    In that case wouldn't the change in potential energy - the center of mass moving in the accelerating frame from 0 to the radius of the center of mass below the door post (viewed from the top) - translate into the kinetic energy of the door moving at 1/2*Iw^2 ?

    Looks like you can calculate I from the radius of gyration and its mass. Rg^2 = I/m.

    Not sure however if this is the way they might want you to solve it for the course you are taking.
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