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Angular and linear acceleration

  1. Apr 1, 2009 #1
    1. The problem statement, all variables and given/known data

    An Atwood machine is constructed using a
    hoop with spokes of negligible mass. The
    2.4 kg mass of the pulley is concentrated on
    its rim, which is a distance 20.9 cm from the
    axle. The mass on the right is 1.36 kg and on
    the left is 1.79 kg.
    What is the magnitude of the linear accel-
    eration of the hanging masses? The accelera-
    tion of gravity is 9.8 m/s^2

    Answer in units of m/s^2


    2. Relevant equations
    F=ma
    [tex]\tau[/tex]=I [tex]\alpha[/tex]
    [tex]\alpha[/tex] R=v


    3. The attempt at a solution

    Ok I found the net force action on the system as the difference of the two weight forces so:
    F1=1.79*9.8=17.542N F2=1.36*9.8=13.328N
    Fnet=4.214N

    Now I said that F=ma, but since the wheel has a mass, and thus a moment of inertia I said:
    F=(m1+m2)a+I[tex]\alpha[/tex] where m1 and m2 are the weights and I is the moment of inertia and alpha is the angular acceleration of the wheel
    I think this is probably where I am mistaken...but continuing from here

    I=mr^2=2.4*(.209^2)=.104834

    and I say a=[tex]\alpha[/tex]r

    4.214=(3.15 kg)a+I*a/r
    4.214=3.15a+.501598a => 4.214=(3.6516)a
    a=1.15402 m/s^2

    and needless to say this is incorrect :(

    Any help would be wonderful! Thanks a lot
     

    Attached Files:

  2. jcsd
  3. Apr 1, 2009 #2
    You need to draw a FBD for each of the three bodies (two falling masses and the wheel) and let that guide you to writing your equations.
     
  4. Apr 1, 2009 #3
    Ok, I have figured it out by drawing a FBD and treating the weights as torques on the wheel. But I am still curious and would like to know at what point what I attempted becomes invalid?
     
  5. Apr 1, 2009 #4
    I'll let you figure that out.
     
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