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Conservation of Energy and Angular Speed.

  1. Dec 16, 2013 #1
    1. The problem statement, all variables and given/known data
    Use the conservation of energy law to find the speed of the blocks and the angular speed of the pulley when the heavier block hits the ground. The system starts at rest. (going to include picture)
    http://i.imgur.com/CjB3gzK.jpg


    2. Relevant equations
    Moment of inertia for a disk (I believe): I = (1/2)mR^2
    KE = (1/2)m*v^2
    KE of pulley = (1/2)I(v/R)^2


    3. The attempt at a solution
    R = radius .2 m
    m1 = 10 kg rock
    m2 = 4 kg rock
    mp = mass of pulley

    Tried reading some things out of my textbook but that didn't help much =(. Decided to let the PE of each block at it's initial position be 0, so the amount of GPE that is converted to KE is:
    KE = (10 kg - 4 kg)(9.8 m/s^2)(2 m) = 117.6 J

    calculate moment of inertia:
    I = (1/2)mpR^2 = (1/2)*2*.2^2 = .04 kg*m^2

    then the KE of the system:

    K = (1/2)(m1 + m2)v^2 + (1/2)I(v/R)^2

    K = v^2(1/2)[10 kg + 4 kg + (.04 kg*m^2 / (0.20 m)^2)] = v^2(7.5 kg)

    Set v^2(7.5) equal to the available energy that was converted from PE and solve for v:

    v = √(117.6 J / 7.5 kg) = 3.96 m/s

    Is this stuff even right? I basically just derived it from my textbook and am having a hard time completely understanding it. Really not sure how I would get angular speed .. would it just be ω = v / R?

    Rereading this gives me a headache so if anyone helps, I commend you! ;D
     
    Last edited: Dec 16, 2013
  2. jcsd
  3. Dec 16, 2013 #2
    The key for the angular speed is seeing that the speed of the blocks has to equal the speed of a point on the pulley's surface as it rotates, seeing as the rope doesn't slip.
     
  4. Dec 16, 2013 #3
    care to elaborate? lol, not sure how I would go about doing that
     
  5. Dec 16, 2013 #4
    You've already found the velocity of the block. The velocity of the block must equal the velocity of a point on the pulley's surface. Using that, apply the equation that relates translational speed to rotational speed.
     
  6. Dec 16, 2013 #5
    so it is ω = v / R ?
    ω = 3.96 m/s / .20 m = 19.8 rad/s?

    Btw, do you happen to know if all the other stuff is right?
     
  7. Dec 16, 2013 #6
    I didn't analyze your work equation by equation but the method did seem correct. And yep, your method to find the angular velocity is correct :)
     
  8. Dec 16, 2013 #7
    Sweet, thanks a lot =D
     
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