![]() $$I = \int r^2 dm = \int \int (x^2 + y^2) \rho dx dy = \rho\int_ įor (let i = 1 i < points. ![]() In brief, for our right triangle we have: ![]() There is a great YouTube video that walks you through calculating the moment of inertia for a similar triangle, so you can watch that for a more in-depth derivation. We know the area is equal to wh/2 so we can calculate the density if we only have the mass. Let's define the right triangle as having a width w and a height h, rotating about the origin with a uniformly distributed mass of density ρ. The first step is going to be calculating the moment of inertia for a right triangle, since we can get a simple closed formula. Unlike mass, which is a constant for a given body, the moment of inertia depends on the location of the center of rotation. r Distance from the axis of the rotation. Here I'm going to describe how to get the moment of inertia for an arbitrary triangle, and then I'll show a triangulation algorithm to apply this to any polygon. Moment of Inertia Formula Moment of Inertia (I) miri2 where, m Sum of the product of the mass. In order to do proper collision response between rotating objects, we needed to calculate the moment of inertia about their center of mass. In my previous two articles I discussed collision detection and response between rigid bodies.
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