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Mass suspended from rotating flywheel

  1. Oct 21, 2015 #1
    1. The problem statement, all variables and given/known data
    A mass of 0.5kg is suspended from a flywheel,
    The mass is released from rest and falls a distance of 0.5m in 1.5s
    Mass of wheel = 3kg
    Outside radius of wheel = 300mm
    Radius of gyration = 212mm

    2. Relevant equations
    a- Linear acceleration of wheel
    b- Angular acceleration of wheel
    c- Tension in rope
    d- Frictional torque

    3. The attempt at a solution

    a) A = 2 x S / t^2
    = 2x 0.5 / 1.5^2
    = 0.444m/s

    b) Inertia = mass x radius of gyration^2
    = 3kg x 0.212^2
    = 0.134832 KG M 2

    Force = mass x acceleration
    = 0.5kg x 0.444
    = 0.222N

    Torque = Force x Distance
    = 0.222N x 0.3
    = 0.0666 N

    ANGULAR ACCELERATION = TORQUE / INERTIA
    = 0.0666N / 0.134832
    = 0.4944 Rad / s

    c) Tension in rope
    tension = Mg - Ma
    = (0.5kg x 9.81) - (0.5 x 0.444)
    = 4.683N

    d) Frictional torque

    Torque in rope = force x dist
    = 4.683 x 0.3m
    = 1.4049NM

    Frictional Torque = Torque - Accel. Torque
    = 1.4049 - ??????
    = ???????

    Im not too sure how to find the acceleration torque! I solved these questions (or thought I did a while ago and now im writing up my solutions to submit them some bits don't make sense, any advice or guidance would be greatly appreciated)

    Thanks
    A
     
  2. jcsd
  3. Oct 21, 2015 #2

    BvU

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    Hi Andy, welcome to PF :smile: !

    How do you distinguish friction from inertia ?
    There is nothing on friction in the problem statement.
    There is no question in the problem statement.

    Is the radius of gyration a given or did you round off ##0.3\sqrt{1\over 2}## ?
     
  4. Oct 21, 2015 #3
    Hi BvU, thanks for the warm welcome!

    The Radius of gyration is given in question = 212mm or 0.212m
    The last question (d) asks to calculate frictional torque (resisting motion)
     
  5. Oct 21, 2015 #4

    SteamKing

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    These are the wrong units for acceleration.
    These are the wrong units for torque. Remember, the number 0.3 has units.
    These are the wrong units for angular acceleration.
     
  6. Oct 21, 2015 #5

    BvU

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    Ok, got it.

    A picture is nice, I suppose you made one ?

    Acceleration of the 0.5 kg weight is due to the net force on the thing. one force is mg, the other T (tension in wire)
    So reconsider the torque on the wheel/motor assembly.
     
  7. Oct 21, 2015 #6

    BvU

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    Next: you don't know the inertia of wheel/motor assembly, so you'll have to find the angular acceleration in a different way.

    [edit] have to run now. good luck
     
  8. Oct 21, 2015 #7
    Think ill retype this; sorry guys

    a) A=2xS/t²
    = 2x0.5/1.5²
    =0.444 m/s²

    b) a = acceleration / radius
    = 0.444m/s² / 0.3m
    = 1.481 rad/s²

    c) tension = Mg - Ma
    = (0.5kg x 9.81m/s²) - (0.5kg x 0.444m/s²)
    = 4.683N

    d) Calculate frictional torque

    torque in rope = force x radius
    = 4.683N x 0.3M
    = 1.4049NM

    acceleration torque = Mass x radius of gyration² x angular acceleration
    = 3 x 0.212² x 1.481
    = 0.199NM

    Frictional torque = torque in rope - acceleration torque
    = 1.4049 - 0.199
    = 1.205NM



     
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