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A Rolling Disk

  1. Apr 26, 2014 #1
    1. The problem statement, all variables and given/known data
    A solid, Uniform disk of radius 5.00 cm and mass 1.50 kg is rolling without slipping a long a horizontal surface. The disk makes 2.00 revolution per second

    a. Find the total kinetic energy (translational + rotational) of the disk

    b. Find the minimum height h of the step (placed in front of the rolling disk) that will prevent the disk from rolling past it. (Hint: assume that the hight h is adjusted so that the disk rolls just up to the top of the step and stops. Conserve Energy)


    2. Relevant equations
    W= 2.00 Revolution x 2pi radian
    V = wr
    I = (1/2)mr^2
    KE_rot = (1/2)Iw^2
    KE = (1/2)mv^2


    3. The attempt at a solution
    a. I assumed that K_total = KE_rot + KE
    K_total = 4.46 x 10^-3 Joules

    b. I have no Idea how to solve this....help?
     
  2. jcsd
  3. Apr 26, 2014 #2
    Correction!
    a. K_Total = 0.444 J
     
  4. Apr 26, 2014 #3
    The height must be such that energy will be conserved between the initial translational and rotational velocity and the final trans/rot velocity. Ktotal(0) + U(0) = Ktotal(1) + U(1)

    I'd begin by drawing a picture. What does the final total Kinetic energy have to be?
     
  5. Apr 26, 2014 #4
    I tried using 1/2 m Vo^2 + mgh(o) = 1/2 m V1^2 + mgh(1)

    and since it says that assuming that it "stops" at the top of the step so I set V1 = 0
    and also h(o) = 0

    cancels all of the m
    so I get h1 = Vo^2 / 2g
    Is that right?
     
  6. Apr 26, 2014 #5
    OK let's stop right here for a sec. 1/2 * Vo^2 is the translational kinetic energy, but you also need to include the rotational kinetic energy because the object is rotating as well as moving forward.
     
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