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Extra resources for Alphabet Puzzles (Grades K-1)
This is easy enough in this 2D example since we have only one rotational axis, coming out of the paper, and thus need only perform the calculation once. The first step is to calculate the local moment of inertia of each component about its own neutral axis. Given the limited information we have on the geometry and mass distribution of each component, we will make a simplifying approximation by assuming that each component can be rep‐ resented by a rectangular cylinder, and will thus use the corresponding formula for moment of inertia from Figure 1-5.
We can use this relation to rewrite the angular equation of motion in terms of local, or body-fixed, coordinates. Further, the vector to consider is the angular momentum vector Hcg. Recall that Hcg = Iω and its time derivative are equal to the sum of moments about the body’s center of gravity. These are the pieces you need for the angular equation of motion, and you can get to that equation by substituting Hcg in place of V in the derivative transform relation as follows: ∑ Mcg = dHcg/dt = I (dω/dt) + (ω × (I ω)) where the moments, inertia tensor, and angular velocity are all expressed in local (body) coordinates.
Products of inertia Just like the parallel axis theorem, there’s a similar transfer of axis formula that applies to products of inertia: Ixy = Io(xy) + m dx dy Ixz = Io(xz) + m dx dz Iyz = Io(yz) + m dy dz where the Io terms represent the local products of inertia (that is, the products of inertia of the object about axes that pass through its own center of gravity), m is the object’s mass, and the d terms are the distances between the coordinate axes that pass through the object’s center of gravity and a parallel set of axes some distance away (see Figure 1-10).