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Gravitational potential energy of a body

An interesting application of the above formulas is in calculating the gravitational potential energy of a body. Suppose you have a turnip and wish to calculate its potential energy. Well for every atom i of mass tex2html_wrap_inline1211 in the turnip, the potential energy, is just tex2html_wrap_inline1255 times the vertical position of the atom tex2html_wrap_inline1257 . To get the total potential energy, we sum over all atoms:

equation224

By some fancy foot work we see this is

equation226

But tex2html_wrap_inline1259 is just the total mass tex2html_wrap_inline1261 , and the term in parenthesis is just the definition of the center of mass in the z-direction, tex2html_wrap_inline1263 . So finally we obtain the simple formula:

equation229

In other words, to compute the potential energy of a turnip, we just have to find its center of mass, and its total mass. If we replace the turnip with a point of mass tex2html_wrap_inline1261 located at the center of mass, it will have the same potential energy.

This gives us a nifty way to experimentally determine the center of mass. Take the right icoscoles triangle we looked at earlier, see 1.2.1.

If we hang it from the right angled corner, the hypotenuse will lie horizontally.

{

  figure233

Why? We can replace the triangle by the center of mass (the red spot), and we're hanging it from the top (the + symbol). A single mass will always dangle vertically (dashed purple line) from where it's being held.

We could also hang it from another corner and the same thing should happen. The red spot should dangle vertically from the point its being held.

{

  figure237

If we didn't know where the center of mass was, this method would allow us to determine its position. It's just the intersection of the two purple lines

{

  figure242

This is a neat way of determining the center of mass of a body.


next up previous
Next: F=ma revisited Up: center of mass for Previous: solution

Joshua Deutsch
Fri Jan 17 12:19:41 PST 1997