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Energy required to tip a can?

  1. Feb 13, 2012 #1
    I want to calculate the minimum energy required to tip a can. (I have a mouse trap and have calculated the potential energy and I want to verify that it indeed can tip over my can.)

    I am thinking I have to calculate the potential energy of the can just as it is tipping and that is the potential energy required to tip the can is that correct?
  2. jcsd
  3. Feb 13, 2012 #2
    how about this

    calculate the potential energy of the untouched can (the can, sitting on level ground, in a stable equilibrium position)
    u1 = mg*h1
    h1 = height of center of gravity of can

    calculate the potential energy of the can just barely about to tip
    u2 = mg*h2
    h2 = height of center of gravity of can as it's about to tip

    E = u2-u1
  4. Feb 13, 2012 #3
    I thought about that however I cannot figure out the geometry to calculate the new height of the center of gravity.

    I know the distance (x value) it will move is equal to (1/2*r), however I do not know how to calculate the height (i am looking at a triangle with only two knowns (90 deg angle and x dist of 1/2*r)
  5. Feb 13, 2012 #4
    http://desmond.imageshack.us/Himg651/scaled.php?server=651&filename=23925225.png&res=medium [Broken]

    http://desmond.imageshack.us/Himg651/scaled.php?server=651&filename=23925225.png&res=medium [Broken]

    this should help
    Last edited by a moderator: May 5, 2017
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