How Much Energy Is Needed to Tip a Can?

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To calculate the minimum energy required to tip a can, one must determine the potential energy of the can in both its stable position and just before tipping. The potential energy of the untouched can is calculated using the formula u1 = mg*h1, where h1 is the height of the can's center of gravity. The energy required to tip the can is found by calculating the potential energy just before tipping, u2 = mg*h2, where h2 is the new height of the center of gravity. The challenge lies in accurately determining the geometry to calculate h2 as the can tips, particularly using the known distance related to the radius of the can. Understanding these calculations is essential for verifying if the mouse trap can generate enough energy to tip the can.
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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?
 
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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
 
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)
 
http://desmond.imageshack.us/Himg651/scaled.php?server=651&filename=23925225.png&res=medium

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

this should help
 
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