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Pulley and torque homework

  1. Nov 30, 2008 #1
    1. The problem statement, all variables and given/known data
    The two blocks in the figure (Intro 1 figure) are connected by a massless rope that passes over a pulley. The pulley is 17 cm in diameter and has a mass of 2.0 kg. As the pulley turns, friction at the axle exerts a torque of magnitude 0.46 Nm .

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

    (Intro 1 Figure):http://session.masteringphysics.com/problemAsset/1070551/3/12.P70.jpg


    2. Relevant equations

    torque= r * F
    sum of torques = moment of inertia * angular acceleartion
    angular acceleration * radius = acceleration
    x=v0t + 1/2 a t^2

    3. The attempt at a solution

    I now that v0 is o so the i solve for t in the equation, t = sqrt(2h / a).
    a = [ F / r + (m1 - m2)g ] / (m1 + m2 + m3 / 2)

    then i plug in a into t=sqrt(2h/a)
    = sqrt{ h(2m1 + 2m2 + m3) / [ F / r + (m1 - m2)g ] }
    =t = sqrt{ 1( 2 * 4 + 2 * 2 + 2) / [ 0.46 / 0.17 + (4 - 2)9.81 ] }
    =.79s
    I got it wrong, can anyone help me to see what i did wrong.
     
  2. jcsd
  3. Nov 30, 2008 #2
    The diameter is 17cm, so your radius is half of that. Give that a try. I see you have 17 instead for radius.
     
  4. Nov 30, 2008 #3

    Doc Al

    User Avatar

    Staff: Mentor

    Show how you derived this result.
     
  5. Nov 30, 2008 #4
    This is how i found the equation for acceleration

    Let:
    m1 be the left hand mass,
    T1 be the left hand tension,
    m2 be the right hand mass,
    T2 be the right hand tension,
    g be the acceleration due to gravity,
    a be the acceleration of the masses,
    alpha be the angular acceleration of the pulley,
    r be the radius of the pulley (assumed solid),
    F be the friction couple,
    h be the descent for m1,
    t be the time taken.

    For mass m1:
    m1 g - T1 = m1 a
    T1 = m1(g - a)

    For mass m2:
    T2 - m2 g = m2 a
    T2 = m2(g + a)

    For the pulley:
    (T1 - T2)r - F = I alpha
    = (m3 r^2 / 2)(a / r)

    T1 - T2 = F / r + m3 a / 2

    Eliminating T1 and T2:
    m1(g - a) - m2(g + a) = F / r + m3 a / 2

    (m1 - m2)g - (m1 + m2)a - m3 a / 2 - F / r
    (m1 + m2 + m3 / 2)a = F / r + (m1 - m2)g

    a = [ F / r + (m1 - m2)g ] / (m1 + m2 + m3 / 2)

    Did i derieved it correctly?


    Also did i solve for time corectly.
    h = at^2 / 2
    t = sqrt(2h / a)
    = sqrt{ h(2m1 + 2m2 + m3) / [ F / r + (m1 - m2)g ]


    Ok i see that i plug in the wrong number for the radius.
    =t = sqrt{ 1( 2 * 4 + 2 * 2 + 2) / [ 0.46 / 0.085 + (4 - 2)9.81 ] }
    t=.7478 s

    is this answer correct?
     
  6. Nov 30, 2008 #5

    Doc Al

    User Avatar

    Staff: Mentor

    OK.
    Redo this last step.
     
  7. Nov 30, 2008 #6
    I dont understand wat you mean, did i made a mistake deriving it? i did over and i still got the same formula.
     
  8. Nov 30, 2008 #7
    hmm I got

    2g-Ff=ma and Ff*r=I*(a/r), where: a and Ff is unknown. how is this wrong?

    you can't look at it like one object anymore?
     
    Last edited: Nov 30, 2008
  9. Nov 30, 2008 #8

    Doc Al

    User Avatar

    Staff: Mentor

    Everything you did up to and including where I said "OK" was good. You just made an algebraic mistake where I said "Redo this last step".
     
  10. Nov 30, 2008 #9
    Thanks for your help, i figure out where i made an algebra mistake
     
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