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Linear acceleration using atwood machine

  1. May 31, 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


    2. Relevant equations
    F=ma
    =I
    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 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=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
     

    Attached Files:

  2. jcsd
  3. May 31, 2009 #2

    nrqed

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    The forces producing a torque on the pulley are the tensions in the rope. You are assuming that the tensions are equal to the weights, which is clearly not the case (otherwise the masses would move at constant speed).

    This equation can't be right since the units don't match.

    You need to set up three equations : one for each mass and one for the pulley.
     
  4. May 31, 2009 #3
    Would the equation be something ike this

    m1g-T1=m1a
    T2-m2g=m2a

    where T1=T2

    Torque=I(alpha)
    Torque=Tr
    alpha=a/r
    therefore
    (a/r)=(Tr/I)
     
  5. May 31, 2009 #4

    nrqed

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    yes, that's a good start
    No, the tensions are different
    Yes, except that T is T1-T2
    Yes, that last one is good.

    So you end up with three equations for three unknowns: T1, T2 and a.
     
  6. May 31, 2009 #5
    ok i got accerlation=...

    a=(g(m1-m2)+(T2-T1))/(m1+m2)=((T1-T2)*R^2)/I

    um exactly how am i going to do this because I cannot get rid of either T1 or T2
     
  7. May 31, 2009 #6

    nrqed

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    You can isolate T1-T2.
    Once you have T1-T2, you plug that back in one of these expressions and you have the acceleration
     
  8. May 31, 2009 #7
    Ok so now i have
    T1-T2=(g(m1-m2)M)/((m1+m2)+M)

    should i move the T2 to the other side and plug T1 somewhere in the 3 equation if so which one
    or plug (T1-T2) in the equation that you are refering to which I do not see where to plug in
     
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