1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Conservation of rotational energy