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Mass suspended from a pulley

  1. Jun 22, 2016 #1
    1. The problem statement, all variables and given/known data
    A mass of 0.5kg is suspended from a flywheel, if the mass is released from rest and falls a distance of 0.5m in 1.5s calculate:

    A, the Linear acceleration of the mass
    B, the angular acceleration of the wheel
    C, the tension in the rope
    D, the friction torque.

    Known information
    Mass of wheel 3 kg Mp
    Outside radius of wheel 0.3m
    Radius of gyration 0.212 m

    2. Relevant equations


    3. The attempt at a solution

    A,
    mg - T = ma
    TR=Ia/R
    mg-(Ia/R^2) = ma
    a= mg/(m+(I/R^2))
    Swapping I out for I=(MpR^2)/2
    gives
    a=mg/(m+((MpR^2)/R^2)
    a=mg/(m+(Mp/2)
    a=2.4525
    2.45 m/sec^-2

    B,
    Angular acceleration = a/R
    2.4525/0.3 = 8.175
    8.175 rads/sec^-2

    C,
    T=mg-ma
    T=0.5*9.81-0.5*2.4525
    T=3.68
    3.68N

    D, - The problem!
    Torque (TNET) = acceleration torque (TA) + frictional torque (TF)
    TF = TR
    3.68*0.3
    1.103625

    TA = Angular acceleration*I
    I = MpR^2/2
    I = 3*0.3^2/2
    I = 0.135
    0.135*8.175
    = 1.103625

    Thus suggestion there is no frictional Torque,

    I have spent a few weeks on an off with this question, I have looked at out threads on here, the lesson provided with the module and YOUTUBE. I am still not able to work out were I am going wrong, hope someone can help, Thanks,
     
  2. jcsd
  3. Jun 22, 2016 #2
    Funny how posting that that prompted thought,

    Should net torque use the radius of gyration instead of radius, thus giving a TNET of 0.779895, apply the same to the TA calculation it gives 0.551258
    thus
    TF = TNET - TA
    TF = 0.228637

    Or have I gone wrong further up the chain?
     
  4. Jun 22, 2016 #3

    gneill

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    Staff: Mentor

    Hi SWINEFLU, Welcome to PF.

    You have a problem with part A. You're looking for the linear acceleration of the mass and you're given distance and time, so why not avoid all the rotational motion involvement to begin with? Simple kinematics will yield the linear acceleration.
     
  5. Jun 22, 2016 #4

    gneill

    User Avatar

    Staff: Mentor

    The radius of gyration will let you compute the moment of inertia using the given mass of the wheel. What's the relationship between radius of gyration, moment of inertia, and mass? This should be one of your Relevant Equations.
     
  6. Jun 22, 2016 #5
    Thanks gneill, I understand were you are coming from as this is what I have read online in several places, but surely the rotating of the mass of which the string is attached to affects the overall acceleration of the mass?
     
  7. Jun 22, 2016 #6

    gneill

    User Avatar

    Staff: Mentor

    It does, yes, but whatever affect it has results in the observed motion which is that the mass travels a given distance in a given time, hence a deducible acceleration (assuming linear motion and constant acceleration).
     
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