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Conservation of Angular Momentum of Cart Problem

  1. Oct 27, 2011 #1
    1. The problem statement, all variables and given/known data
    Diameter = 20 cm, Massplywood = 8.8kg, Masswheel = 1.86 kg
    Length of incline = 34 m Masstotal = 16.24 kg

    2. Relevant equations

    Energybefore = Energyafter

    KEtranslational = 1/2mtotalv2
    KErotational = 1/2 I ω2
    Isolid disk = 1/2 M R2
    Ug = mgh
    v = rω


    3. The attempt at a solution
    Attached is a copy of my work at the solution.
    1. After drawing a sketch of my problem I set my energies from the top of the incline equal to the energies when at the bottom.
    2. The only energies acting in "before" is gravitational potential energy Ug = mtotalgh. I can find the high using trig h = Lsin(15). Note the gravitational potential energy of the carts center of mass mtotalgd will cancel out on the "left" side of the equation.
    3. The energies acting in "after" is the translational KE, rotational KE, and gravitational potential energy (that cancels out mentioned in 2).
    - KEt = 1/2mtotalv2
    - KEr = 1/2(1/2*4MwheelR2)(v/r)2
    ++ the moment of inertia for a solid disk is I = 1/2 M R2 but there are 4 wheels hence multiplied by 4. Also v = rω, substituting (v/r) for ω.
    4. Simplified and solve for velocity.

    I am doing this homework assignment on webssign and 18.8 m/s 18.77 m/s, whichever, does not seem to be correct.

    muWe3.jpg
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 28, 2011 #2

    gneill

    User Avatar

    Staff: Mentor

    Operator error: Make sure that your calculator is set to use degrees rather than radians.
     
  4. Oct 28, 2011 #3
    Of course, that was it. Thank you.
     
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