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Acceleration of a pulley. Torque, and moment of inertia.

  1. Dec 3, 2009 #1
    1. The problem statement, all variables and given/known data
    There are two blocks, one of mass m1 and the other of m2. They are connected by a rope that runs over a pully of radius R and interia I. m2 slides down a frictionless incline at an angle theta, and m1 hangs virtically while being pulled up by the tension in the rope. Find the expression for acceleration.


    2. Relevant equations
    Fnet=m2gsin(theta)-m1g
    F/M=a


    3. The attempt at a solution

    With the previous two equations I came to this conclusion...

    (m2gsin(theta)-m1g)/(m1+m2+I)

    The right answer has I/R^2 instead of just I.

    I figured it would be just I because the mass of an object is its inertia.

    Howwoud I derive the I/R^2?
     
  2. jcsd
  3. Dec 3, 2009 #2

    ideasrule

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    Homework Helper

    How did you get an equation with an "I" when your initial equations:

    Fnet=m2gsin(theta)-m1g
    F/M=a

    didn't have an "I" anywhere? Anyhow, for mechanics problems, you should always draw free-body diagrams for every object (including the pulley!), write down Newton's second law for each object, and then solve the resulting equations. That way, you'll see very clearly where the I/R^2 came from.
     
  4. Dec 3, 2009 #3
    I got the I from thinking that I of the pulley would be the same thing as mass for the blocks because isn't mass just the inertia of an object? and I put my net force over the total mass.

    I think I may have figured it out but I also may be making a mistake here...

    T1 and T2 are equal to the tensions on each side, and alpha=a/R

    T2R-T1R=Ialpha

    R(T2-T1)=Ialpha

    Tnet=I(a/R)/R

    ma=Ia/(R^2)

    m=I/R^2
     
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