Of the thin bars (in the form of a square), the member along the rotation axis has (approximately) no rotational inertia about that axis (since it is thin), and the member farthest from it is very much like it (by being parallel to it) except that it is displaced by a distanceh; it has rotational inertia given by the parallel axis theorem: IImhmRmR22220=+=+=com.Now the two members of the square perpendicular to the axis have the same rotational inertia (that is I3= I4). We find
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