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Motor Raising a Mass

  1. Jul 24, 2008 #1
    1. The problem statement, all variables and given/known data

    When the motor shown in the figure raises the m = 1050 kg mass, it produces a tension of 1.42×104 N in the cable on the right side of the pulley. The pulley has a moment of inertia of 79.8 kgm2 and a radius of 0.440 m. The cable rides over the pulley without slipping. Determine the acceleration of the mass. (Draw free-body diagrams of the mass and the pulley. Do not assume that the tension in the cable is the same on both sides of the pulley.)


    2. Relevant equations

    I=(1/2)M*R^2 (for disks rotating about their center)

    a_obj=|a_angular|*R

    3. The attempt at a solution

    Many people in my class are having problems completing this problem. I really didn't know where to start so I used the pulley's moment of inertia to calculate it's mass, but from there I don't know where to go. The different tensions on either side of the pulley are also really racking my brain. Any help would be appreciated!
     

    Attached Files:

  2. jcsd
  3. Jul 24, 2008 #2
    I don't see the figure yet, but I guess that you should use Newton's 2nd law for the mass and the pulley. It looks like this:

    [tex] F_{\text{tension}} = ma + I \frac{a}{R}[/tex]

    Then just express [tex]a[/tex] in terms of known quantities.
     
  4. Jul 25, 2008 #3

    Doc Al

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    Staff: Mentor

    Analyze the forces acting on (1) the mass, and (2) the pulley. (Just call the unknown tension T.) Apply Newton's 2nd law to each and you'll get two equations which you can then solve together to find the acceleration.
     
  5. Oct 22, 2010 #4
    same question.
    moment of inertia 72.8 kg·m2
    radius 0.788 m
    m = 1140 kg
    1.19×10^4 N in the cable on the right side of the pulley (T2)
    acceleration of m=?
    i got it as 0.701m/s^2, but got it wrong can someone check my work.
    T1-mg=ma, T1=ma+mg
    r(T1-T2)=Ia/r^2, sub T1=ma+mg, T2=11900, m=1140, g=9.81, I=72.8, solve for a
     
  6. Oct 23, 2010 #5

    Doc Al

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    Staff: Mentor

    This is correct.
    But this is not. You forgot to cancel the 'r' on the left side.
     
  7. Oct 23, 2010 #6
    i actually just copied that down wrong, my bad.
    i have..
    r(T1-T2)=Ia/r
    r(ma+mg-T2)=Ia/r

    a=(T2-mg)/(m-(I/r^2))
    my answer is 0.701 m/s^2, which is wrong.. cant seem to find where i went wrong.
    thanks in advance
     
  8. Oct 23, 2010 #7

    Doc Al

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    Staff: Mentor

    You have a sign error. Since T2 > T1, your value for 'a' will be negative here (but positive elsewhere). Instead, be consistent about 'a' being positive. (Just swap T1 and T2 in this equation.)
     
  9. Oct 24, 2010 #8
    If you have two 'a' values in r(T2-(ma+mg)) = Ia/r, how do you solve for a?
     
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