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Moment of Inertia of the disk

  1. Mar 24, 2008 #1
    1. The problem statement, all variables and given/known data

    A uniform disk of radius 0.12 m is mounted on a frictionless, horizontal axis. A light cord wrapped around the disk supports a 1 kg object, as shown in the figure from rest the object falls with a downward acceleration of 2.3 m/s^2.The acceleration of gravity is 9.8 m/s^2 When released What is the moment of inertia of the disk? in units of kg m^2

    2. Relevant equations

    Moment of Inertia of a disk by itself= 1/2MR^2
    I= Torque/Angular Acceleration

    3. The attempt at a solution

    1/2*(M)*(.12^2) = I

    If there is downward acceleration of 2.3 m/s^2 is that added to the 9.8 m/s^2 from gravity? 12.1 m/s^2

    Im having a hard time knowing where to start with how this 1kg object affects the moment of inertia of the disk, and how I find M in the disk.

    Attached Files:

  2. jcsd
  3. Mar 24, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    You'll need to separately analyze the forces on the hanging mass and on the disk.

    Start with the hanging mass. Apply Newton's 2nd law to find the tension in the cord.
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