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Moment of Inertia and two uniform disks

  1. Apr 12, 2008 #1
    1. The problem statement, all variables and given/known data

    Two uniform disks are connected by a light inextensible string supported by a massless pulley on a frictionless axis. (Very ideal LOL) The string is attached to a point on th circumference of disk A and wound around the disk B like a yo-yo.

    Their moment of inertia does not equal 1/2mr^2 and the disks do not have uniform radial density.

    For each disk A and B, the following are true:

    I = .0254024 kgm^2
    m = 1 kg
    r = .2m
    height = 1.6 m
    g = 9.8 m/s^2

    Which disk will reach the floor fit and what is the time interval for this disk to reach the floor?

    2. Relevant equations

    t = sqrt(2x/a) (Maybe?)

    3. The attempt at a solution

    As the spinning object would have a slower decent, object A would hit the ground first. (I think)

    Past this, I'm totally lost. Anyone mind handing me a flashlight?
  2. jcsd
  3. Apr 13, 2008 #2

    Doc Al

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    Staff: Mentor

    But what's the acceleration?

    What determines the acceleration of each disk? What forces act?
  4. Apr 13, 2008 #3
    That would be gravity, correct?

    Disk A (the one tethered to the string) would have g pulling down and B pulling it up.
    Disk B (the yo-yo) would have g pulling down, A pulling it up, and centrifugal force (though it isn't really a force, so maybe I'm wrong)

    Am I stumbling the right direction?
  5. Apr 13, 2008 #4

    Doc Al

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    Staff: Mentor

    Do you mean that a = g? No. (The acceleration would equal g if the objects were in free fall, but they're not.)
    When analyzing the forces on A, forget about B. It's the string that pulls up on A.

    I'll forget you ever said the phrase "centrifugal force". :yuck:

    Again: What forces act on A? On B?
  6. Apr 13, 2008 #5
    Oh no.

    *tries not to be obviously deficient*

    Ok, A has g pulling down and the string pulling up.
    B has g pulling down, the string pulling up, and um...something else having to do with the spin? o.o Torque maybe? (Please don't make fun of me :D)
  7. Apr 13, 2008 #6

    Doc Al

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    Staff: Mentor

    Good. The weight acts down; the string tension acts up.
    The same two forces act on B.

    Applying Newton's 2nd law will give you one equation.
    So far, we've just analyzed the translational motion. The same forces will create a torque on B. Analyze the torque on B and apply Newton's 2nd law for rotation.
  8. Apr 13, 2008 #7
    I have the same problem. I understand Tr=Ia/r (a/r=alpha) so you get mg-T= ma and T=Ia/r^2 so you get a=mg/(m+I/r^2), but when i put that a into sqrt(2h/a) i didn't get the right answer, perhaps i'm doing something wrong
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