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Two Disks Collide (Torque)

  1. Feb 11, 2017 #1
    1. The problem statement, all variables and given/known data
    Disks A and B each have a rotational inerita ##0.300 kg.m^2## about the central axis and a radius of of ##20.0cm##and are free to rotate on a central rod through both of them.To set them spinning around the rod in the same direction,each wrapped with a string that is pulled for ##10.0s## (the string detaches at the end).The magnitudes of the forces pulling the strings are ##30.0N## for a disk A and ##20.0N## for a disk B.After the strings detach,the disks happen to collide and frictional force between them to the same final angular speed in ##6.00s##
    What are (a) magnitude of the average frictional torque that brings them to the final angular speed and (b) the loss in kinetic energy as the torque acts on them? (c) Where did the "lost energy" go ?

    2. Relevant equations
    ##τ=rFsinθ##
    ##τ=I∝##

    3. The attempt at a solution
    First I found the angular accelerations for both objects
    For A-
    ##τ=rF=I∝##

    ##∝=\frac {rF} {I}##

    ##∝=\frac {0.2m.30N} {0.3 kgm^2}=20\frac {rad} {s^2}##
    For B-
    ##∝=\frac {0.2m.20N} {0.3 kgm^2}=13,3\frac {rad} {s^2}##

    For angular speed (w) which they pulled ##10 sec##, so ##w-w_0=∝t##
    ##w_A=20.10=200\frac {rad} {s}##
    ##w_B=13,3.10=133\frac {rad} {s}##

    They are coming equal in 6 sec so the change in ##w_A## is ##200-(\frac{200+133} {2})=33.5\frac {rad} {s}##

    so angular acceleration is ##∝=\frac {33.5} {6}=5.58\frac {rad} {s^2}##
    so
    ##τ=I∝=0.3.5.58=1.67N.m##

    For (b)

    Initial rotational kinetic energy for A is from ##\frac 1 2Iw^2=(0.5).(0.3) (200^2)##
    Last kinetic energy for A is ##\frac 1 2Iw^2=(0.5).(0.3) (166.5^2)##
    The differance is ##1841,66J## which its not the correct answer..

    Or I tried ##Δ(E_r)=W=τΔθ=1.67Nm.33.5\frac {rad} {s}.6s=335.67J##
    but answer says its ##333J##
    Where I did wrong ?

    And for answer c is of course heat

    Thanks
     
  2. jcsd
  3. Feb 11, 2017 #2

    BvU

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    You stopped looking after considering A. What happens to B ?
    The 333 or 336 comes about because you calculate with numbers instead of symbols. The advice is to work with symbols as long as feasible; then -- in the final expression -- a lot of things cancel and you get a more accurate answer.
    Additional benefit: less typing on the calculator, less chance for typing errors and easier approximate checks on the numerical outcome.
     
  4. Feb 11, 2017 #3
    Ok for B I found ##1504.98J## so the differance between them should give us the result whch its approximatly ##336J##
    I found ##W=τΔθ## I cant use symbols after this ?
     
  5. Feb 11, 2017 #4
    If logic is correct then I dont think 333J or 336J matter so much..?
     
  6. Feb 11, 2017 #5

    BvU

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    You can express both factors in terms of given variables. If you do that, the correct answer is exactly 1000/3 J. You should do that exercise.

    In such a calculation in an exercise, a 1% deviation is considered an error. Loss of score for no purpose.
     
  7. Feb 11, 2017 #6
    Ok I found by ##W=I(w^2)##

    ##0.3.({200-(\frac {1000} {6}))^2}=W##

    from there I found,but in exam or somewhere else I wouldnt do this probably...

    Thanks
     
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