Anyone know how to calculate the gravitational potential energy of a balanced Mass such as an upside down 2d triangle balanced on a single point. Is there any way to calculate the gravitational potential energy of the Tilt motion of the upside down triangle, as it falls to one side.

Thanks guys

  • $\begingroup$ Is it balanced or is there a tilt-motion? I feel those two messaged are counteracting. $\endgroup$ – Steeven Oct 2 '18 at 16:06

You can use the formula for centre-of-mass, CoM, to calculate the location of the "centre", which is the point the total mass "averages down to". Place your coordinate system somewhere and calculate the following x- and y-coordinates:

$$x_{com}=\frac{\sum mx}{\sum m}\quad , \quad y_{com}=\frac{\sum my}{\sum m}\,.$$

Similar for the z-coordinate. If you have individual particles, it is fairly easy to sum them all up. If you have an extended body, you must integrate $\int$ instead of summing $\sum$; it may be tricky depending on the exact geometry.

But if the geometry happens to be fairly simple and symmetric and the mass evenly distributed, you may be able to easily figure out the CoM location with no calculations. If your triangle is equilateral, e.g., (and its mass evenly spread), then you know by symmetry that the CoM is exactly in the geometric centre.

Regardless of how you find the CoM, you can consider this point as where gravity pulls. As if the entire object's mass was actually only concentrated at this point. This is therefore easy to input into a gravitational potential energy formula, such as $$U=mgh\, ,$$

depending on your purpose. Just remember that the height $h$ in this particular formula is not the distance of the CoM from the ground, but the distance between where the CoM starts out and where it ends after tilting over. Only this amount of potential energy is released.


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