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Chapter 12: Problem 49

A uniform, solid cube of mass \(m\) and side \(s\) is in stable equilibrium when sitting on a level tabletop. How much energy is required to bring it to an unstable equilibrium where it's resting on its corner?

Short Answer

Expert verified
The energy required to bring the cube to an unstable equilibrium where it's resting on its corner is \((mg*s)((\sqrt{3} -1)/2)\).

Step by step solution

01

Identify stable and unstable equilibrium conditions

We start by understanding the two equilibrium states the cube can be in. In a stable equilibrium, the cube is resting on a face, while in an unstable equilibrium, it is balancing on a corner. The key difference between these two states is the position of the center of mass of the cube relative to the tabletop.
02

Calculate the height of the center of mass in stable equilibrium

When the cube is in stable equilibrium, it is flat on the tabletop. Since the center of mass of a cube is located at the center, the height (h1) of the center of mass from the tabletop in stable equilibrium is \(h1 = s/2\).
03

Calculate the height of the center of mass in unstable equilibrium

When the cube is in unstable equilibrium, it is resting on its corner. Using the Pythagorean theorem for a right-angled triangle (consisting of half a side, the cube's center, and the cube's corner), the height (h2) of the center of mass from the tabletop can be found as \(h2 = \sqrt{(0.5s)^2 + (0.5s)^2 + (0.5s)^2} = s\sqrt{3}/2\).
04

Determine the energy required to go from stable to unstable equilibrium

The work done or energy required (E) to move the cube from stable to unstable equilibrium is equivalent to the change in gravitational potential energy. E = mgh2 - mgh1 = mg((s\sqrt{3}/2) - (s/2)) = (mg*s)((\sqrt{3} -1)/2).

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