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Homework Help: Conservation of rotational energy

  1. Nov 13, 2006 #1
    A cylindrical can of paint with (I = .5MR^2) starts from rest and rolls down a roof as shown in the link

    Determine how far the can lands from the edge of the house.
    This is my work, but I am not getting the correct answer. Please tell me where I messed up. The correct answer is 6.18m

    mgH = (1/2)MV_cm^2 (1+B)
    gH = (1/2)V_cm^2 (1+B)
    3g = (1/2)V_cm^2(1+.5)
    3g = (3/4)v_cm^2
    4g = v_cm^2
    v = 2SQRT(g)

    Now, I know that x = x_0 + v_o*cos(theta)*t

    x = 2SQRT(g)*cos(30)*t

    To solve for t, I used y

    y = 10 - 2SQRT(g)*sin(30)*t - (1/2)gt^2
    0 = 20 - 4SQRT(g)*sin(30)*t - gt^2
    0 = gt^2 + 2SQRT(g)*t - 20

    t = 2SQRT(g) +/- SQRT[4g-4(g)(-20)]/2g
    t = 7.72 seconds or 4.79

    x = 2SQRT(g)*cos(30)*7.72
    x = 41.8

    x = 2SQRt(g)*cos(30)*4.79
    x = 26.0

    The correct answer is 6.18. My answer is too far off, but I don't know where I went wrong.
    Last edited: Nov 13, 2006
  2. jcsd
  3. Nov 13, 2006 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    I'm not sure what your initial equation is meant to represent, but note that total energy must be conserved. i.e,

    gravitational potential = rotational kinetic energy + linear kinetic energy
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