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MI moment of inertia on pulley

  1. Nov 12, 2005 #1
    A pulley has a moment of inertia of 5kg metre squared and a radius of 0.5m. The cord suporting the masses m1 and m2 does not slip and is hung on each side of the pulley with a cord thru the pulley. Assume that the axle is frictionless. How to find the acceleration of each mass when m1=2 and m2=5?

    Why cant the method be as follows Total torque=Moment of inertia * angular accel

    hence, 7g(0.5)=5alpha and alpha =0.7g. since a=r(alpha) then we know r=0.5 so we get a=3.43 but why is this not the correct ans?
     
  2. jcsd
  3. Nov 12, 2005 #2

    Doc Al

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    Staff: Mentor

    Realize that the total torque is not 7g(0.5): the two masses pull in opposite directions.
     
  4. Nov 12, 2005 #3
    Oh thanks alot! How could I have made this mistake???!!! ARGH!
     
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