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Torque and Tangential Acceleration

  1. Feb 26, 2007 #1
    1. The problem statement, all variables and given/known data

    My ENTIRE AP physics class is stumped on the way to correctly write the formula to find acceleration for a falling block (mass of "m") attached by a string (tension of "t") to a fixed, rotating pulley (mass of "M", radius of "r"). Our teacher told us that for a block attached on the left side of a pulley (rotating counterclockwise) the tangential acceleration ("a") of the pulley should be positive (+) and for one on the right side (pulley rotating clockwise) a should be negative (-). We cannot come to a decisive conclusion as to what the final equation to find a for each side should be.

    m = block mass
    M = pulley mass
    r = radius of pulley
    t = tension of string
    a = tangential acceleration (acceleration of the block)
    I = Inertia of pulley
    A = angular acceleration of pulley
    g = Gravity (assume g=10 if needed)

    2. Relevant equations

    Positive "a" Formula Set:


    (+)Formula One: Pulley Rotating CCW

    T = rt, T = IA , I = 1/2M(r^2), and A = a/r

    therefore... rt = 1/2 M(r^2)(a/r)
    (r^2)t = 1/2M(r^2)a
    t = 1/2Ma

    (+)Formula Two: Block Falling to the Left

    t = sum of all forces
    t = mg + ma

    Negative "a" Formula Set:


    (-)Formula One: Pulley Rotating CW

    T = rt, T = IA , I = 1/2M(r^2), and A = -a/r

    therefore... rt = 1/2 M(r^2)(-a/r)
    (r^2)t = 1/2M(r^2)(-a)
    t = -(1/2Ma)

    (-)Formula Two: Block Falling to the Right

    t = sum of all forces
    t = mg + [m(-a)]
    t = mg - ma
    (This part is where we some of the confusion begins. We assume t = mg "-" ma since the block is falling to the right and "a" should be negative)

    3. The attempt at a solution

    Positive, CCW, Left

    t = mg + ma
    t = 1/2Ma
    1/2Ma = mg + ma
    both sides x 2
    Ma = 2mg + 2ma
    both sides -2ma
    Ma - 2ma = 2mg
    factor out a
    a(M - 2m) = 2mg
    both sides / (M-2m)
    a = 2mg / (M - 2m)

    Negative, CW, Right

    t= mg - ma
    t = -(1/2Ma)
    -(1/2Ma) = mg - ma
    both sides x-(2)
    Ma = -2mg + 2ma
    both sides -2ma
    Ma - 2ma = -2mg
    actor out a
    a(M - 2m) = -2mg
    both sides / (M - 2m)
    a = -2mg / (M - 2m)

    I've worked these out to the best of my ability and I believe that they are correct. Can someone please either confirm this or correct me ASAP?
  2. jcsd
  3. Feb 27, 2007 #2


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    Science Advisor
    Homework Helper

    In both cases, you are taking t to be positive when there is a tension in the string. For the lower block, the correct equation is always t = mg + ma if you measure a as positive downwards.

    It is obvious physically that the lower block moves downwards, therefore the acceleration a, measured positive downwards, must be positive.

    Your "negative, CW, right" answer is wrong. The mistake is saying t = mg - ma.
  4. Feb 27, 2007 #3
    Thanks, I'll explain this to the rest of my class!
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