# Chapter 12 - Exercises and Problems - Page 224: 50

$.333mgh$

#### Work Step by Step

We know that the center of mass of a uniform isosceles triangle on its base is 1/3 of the way to the top, or $.3333h$. We know that when it is flipped onto its apex, its center of mass will be $1h-.3333h=.66667h$. Thus, we find that the energy needed is: $\Delta U = mg(.667-.333)h=.333mgh$

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