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Angular Acceleration of a Pulley with Mass

  1. Aug 12, 2017 #1
    1. The problem statement, all variables and given/known data
    A pulley hangs of mass, m, and radius, R, hangs from the ceiling. Two blocks of masses, m1 and m2 are connected by a massless, non-stretchable rope on the pulley (assume no slipping). What is the angular acceleration of the pulley and what is the ratio of the tension forces on the vertical portions of the rope while the blocks are moving?

    2. Relevant equations
    τ=Iα
    F=ma
    Icyl=0.5mR2
    atan=αR

    3. The attempt at a solution
    Because the pulley has a mass, the tension forces on each side of the rope are different (if the pulley were massless then the tensions on each side would be the same, to my knowledge). The torque determines the angular acceleration:

    τ=Iα

    Since the tension work in opposite directions, the tensions, T1 and T2, act in opposite directions (where T1 is on the side of block 1 and T2 is on the side of block 2).

    τ=Iα=T1-T2

    I can use Newton's Second Law to try and find the two unknown tension forces:

    F1=m1a=m1g-T1
    F2=m2a=T2-m2g

    Solving for T:

    T1=m1g-m1a
    T2=m2g+m2a

    Plugging this into Iα=T1-T2:

    m1g-m1a-m2g+m2a=Iα

    Here is where I start to get more unsure. I still have the unknown linear acceleration, a. I think I can relate it to α using a=αr since there is no slipping. Then my equation would be:

    m1g-m1αR-m2g+m2αR=0.5mR2α

    Then, I can isolate and solve for α. I like to try and see if my answers make intuitive sense, and I'm not sure this one does:

    α= g(m1-m2) / 0.5mR2+m1R-m2R

    It seems angular acceleration increases as the mass of block 1 increases. This makes sense because if the weight of block 1 is greater, the pulley will rotate faster. However, the denominator also increases as m1 increases and decreases as m2 increases, which seems counterintuitive to me.

    The ratio of tensions (T1/T2) is as simple as m1(g-a)/m2(g-a). I'm not sure how to think about this, but I'm confident in my work on this part.

    Thanks for checking over my work!
     
  2. jcsd
  3. Aug 12, 2017 #2

    Doc Al

    User Avatar

    Staff: Mentor

    You are on the right track, but fix that equation. (You left out a factor of R on the right hand side. You need torques on both sides of the equation.)
     
  4. Aug 12, 2017 #3
    Oops right I forgot that τ=FtanR.

    When fixing this, my equation becomes:

    R(T1-T2)=Iα

    Plugging in my tension values from Newton's Second Law and assuming a=Rα:

    R(m1(g-a)-m2(g+a))=0.5mR2α

    I can cancel out an R. I also seem to have forgotten to distribute the negative through T2 in my original post.

    m1g-m1Rα-m2g-m2Rα=0.5mRα

    Isolating and solving for angular acceleration now gets:

    α= g(m1-m2) / R(0.5m+m1+m2)

    My ratio of tensions should be unchanged.
     
  5. Aug 12, 2017 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Looks good to me, assuming you can treat the pulley as a solid disk.

    One of those minus signs should be a + sign. You can also express this ratio in terms of the known masses, but the expression might be messy. (I didn't work it out.)
     
  6. Aug 12, 2017 #5
    Oh oops, I just made a typo. It should be:

    T1/T2 = m1(g-a) / m2(g+a)

    We always treat pulleys as solid cylinders. Thanks so much for your help!
     
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