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Atwood's pulley-moment of inertia

  1. Mar 31, 2005 #1
    Ok I've done an experiment in which both a standard Atwood's pulley was used, two different masses suspended on either side of the pulley. And an accelerating cart, where the cart was on a horizontal surface, and a weight was attached on the other side of the pulley hanging down. The mass was kept constant by transferring weight from one mass to the other.

    From the experimental data I made graphs of force vs acceleration, and was able to calculate the moment of inertia.

    Here is my question, I found that the moment of inertia found for the accelerating car is considerably smaller than for the Atwood’s pulley. But I really don't know how to explain the significance of this.

    Anyone know???
     
  2. jcsd
  3. Mar 31, 2005 #2
    if you used the same pulley, shouldnt the moment of inertia be the same??

    For a circular disc the Moment of inertia is [tex] I = \frac{1}{2} ML^2 [/tex] where L is the radial vector about which the torque is exerted. But in btoh your described cases the moment of inertia is the same since the string is wrapped around the pulley's end (am i right in assuming this?)

    Perhasp you have made an error in calculations?
     
  4. Apr 1, 2005 #3
    One is suspended over and therefore the force is based on the mass difference between the weights. And in the other case, because the horizontal surface is perpendicular to the movement of m2 which is suspended on the other side, the force is only a result of m2 and gravity. This somehow changes the acceleration, and therefore the inertia. I think?

    Anyone know if I'm on the right track here?
     
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