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Homework Help: 2 hanging masses on Pully with friction time to reach floor?

  1. Apr 23, 2010 #1
    1. The problem statement, all variables and given/known data

    http://session.masteringphysics.com/problemAsset/1070551/3/12.P70.jpg

    The two blocks in the figure are connected by a massless rope that passes over a pulley. The pulley is 15 cm in diameter and has a mass of 2.6 kg. As the pulley turns, friction at the axle exerts a torque of magnitude 0.46 N-m.

    If the blocks are released from rest, how long does it take the 4.0 kg block to reach the floor?

    2. Relevant equations

    Tnet=I[tex]\alpha[/tex]

    T=FD

    F=ma

    [tex]\alpha[/tex]=a/r

    Moment of inertia of a disc:
    I=.5mr2

    g is gravity!
    3. The attempt at a solution

    T1 is tension of the left mass, T2 is tension of the right mass
    m1 is mass of left mass, m2 is mass of right mass
    T1=-m1a+m1g
    T2=m2a+m2g

    [tex]\Sigma[/tex]T= TT1 - TPully friction - TT2

    I[tex]\alpha[/tex]=TT1 - TPully friction - TT2

    Ia/r=(-m1a+m1g)r - .46 - (m2a+m2g)r

    Ia/r= -m1ar + m1gr - .46 -m2ar -m2gr

    Ia = -m1ar2 + m1gr2 - .46r -m2ar2 -m2gr2

    Ia + m1ar2 + m2ar2 = m1gr2 - .46r -m2gr2

    a(I + m1r2 + m2r2) = m1gr2 - .46r -m2gr2

    a = (m1gr2 - .46r -m2gr2) / (I + m1r2 + m2r2)

    I=.5mr2

    I = (.5)(2.6)(.075)2
    I = .0073125

    m1 = 4kg
    m2 = 2kg
    r = .075m
    g=9.8

    a = ((4)(9.8)(.0752) - .46(.075) -(2)(9.8)(.075)2) / (.0073125) + (4)(.075)2 + (2)(.075)2

    a = (.2205 - .0345 - .11025) / (.0073125 + .0225 + .01125)

    a = .07575/.0410625

    a = 1.8447

    yf=yi+vit+.5at^2

    0=1 + 0 + .92235t^2

    t=1.04124 seconds

    is this correct?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 23, 2010 #2
    Just double checked with a slightly different method. Looks right to me.
     
  4. Apr 23, 2010 #3
    Second opinion please anyone!

    Just want to make sure! Dont have many tries left on the problem
     
  5. Apr 23, 2010 #4
    If you wanted to be pedantic, to be more correct you would round to two significant figures (1.0 rather than 1.04124). Also, at this step, probably a typo but there should be a negative sign or have the 1 moved to the other side:

    0=1 + 0 + .92235t^2

    change to

    0=-1 + 0 + .92235t^2

    The answer should be right though.
     
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