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Chapter 12: Problem 43

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 from a stable to unstable equilibrium where it's resting on its corner is \( \Delta U = m \cdot g \cdot s ( \sqrt{2/3} - 1/2 )\).

Step by step solution

01

Calculate the initial and final centers of gravity

For the cube in stable equilibrium, its center of gravity is \(g_1 = s/2\). For the cube in unstable equilibrium, its center of gravity is located in the center of the cube and its height from the ground is \(g_2 = s\sqrt{2/3}\).
02

Determine the change in potential energy

This can be calculated as \( \Delta U = m \cdot g \cdot ( g_2 - g_1 )\). \(g\) is the acceleration due to gravity.
03

Substituting for \(g_1\) and \(g_2\)

Substituting \(g_1 = s/2\) and \(g_2 = s\sqrt{2/3}\) in the above equation gives \(\Delta U = m \cdot g \cdot (s \sqrt{2/3} - s/2)\)
04

Simplifying the equation

Simplify the above equation to get the required energy which is \( \Delta U = m \cdot g \cdot s ( \sqrt{2/3} - 1/2 )\)

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Most popular questions from this chapter

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