Two skaters of equal mass, holding hands with their arms stretched out. are spinning with an angular velocity about a vertical axis midway between them. What is their new angular velocity when the separation between the skaters is halved?
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Conservation of angular momentum is a fundamental principle in physics. When no external torques act on a rotating system, the angular momentum L equals I omega remains constant. Consider two skaters of equal mass m, holding hands with arms stretched out, spinning about a vertical axis midway between them. Initially, they are separated by distance 2r₁ and rotating with angular velocity omega₁.
To find the moment of inertia for our two-skater system, we treat each skater as a point mass. The moment of inertia for point masses is calculated as I equals the sum of mass times radius squared. Each skater has mass m and is located at distance r₁ from the central rotation axis. Therefore, the total moment of inertia is I₁ equals m times r₁ squared plus m times r₁ squared, which simplifies to I₁ equals 2mr₁ squared.