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Regular polygons with n = 3, 4, 5,... sides each have their centre of mass at a height h above the ground. Each polygon rolls without slipping on a horizontal floor, pivoting about its leading vertex. As a polygon tips over one vertex, its centre of mass rises and then falls. The greatest rise in height of the centre of mass during this motion is delta. How does delta depend on n and h?

1. delta = h*sin²(pi/n)
2. delta = h*sin(2pi/n)
3. delta = h*(1/cos(pi/n) - 1)
4. delta = h*tan²(pi/2n)

Correct answer: delta = h*(1/cos(pi/n) - 1)

Solution

As the polygon rotates about a vertex, the COM (at fixed distance d from that vertex) is highest when straight above the vertex, lowest when the adjacent side lies flat.

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