How Much Energy Is Needed to Tip a Can?

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Discussion Overview

The discussion centers on calculating the minimum energy required to tip a can, focusing on the potential energy involved in the tipping process. Participants explore the theoretical aspects of energy calculations related to the can's center of gravity and its geometry as it approaches the tipping point.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant proposes calculating the potential energy of the can in its stable position and just before tipping to determine the energy required to tip it.
  • Another participant suggests using the formula for potential energy, u1 = mg*h1 and u2 = mg*h2, where h1 and h2 represent the heights of the center of gravity in different positions.
  • A different participant expresses difficulty in determining the new height of the center of gravity during the tipping process, mentioning a known distance related to the can's radius.
  • A participant shares a link to an image that may assist in visualizing the geometry involved in the calculations.

Areas of Agreement / Disagreement

Participants have not reached a consensus on the method for calculating the height of the center of gravity during tipping, indicating that multiple approaches and uncertainties remain in the discussion.

Contextual Notes

There are unresolved geometric considerations regarding the calculation of the center of gravity's height as the can tips, which may depend on specific definitions and assumptions about the can's dimensions.

Juanka
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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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