Yep, that is what I did I expressed in terms of y^2 and then went ahead with that. But I just didn't understand how this solution came about basically.TSny said:
this is the solution. not mine :)hutchphd said:
well.. that is exactly what I don't understand..hutchphd said:
The moment of inertia of a thin, square plate of side length \( a \) and mass \( m \) about an axis perpendicular to the plate and passing through its center is given by \( I = \frac{1}{6}ma^2 \).
The derivation involves integrating the mass distribution over the area of the plate. By dividing the plate into infinitesimal mass elements and summing their contributions, we find that \( I = \frac{1}{6}ma^2 \), where \( m \) is the mass and \( a \) is the side length of the plate.
The moment of inertia of a thin, square plate about an axis along one of its edges is \( I = \frac{1}{3}ma^2 \), where \( m \) is the mass and \( a \) is the side length of the plate.
The moment of inertia depends on the axis of rotation. For a central axis perpendicular to the plate, it is \( \frac{1}{6}ma^2 \). For an axis along one edge, it is \( \frac{1}{3}ma^2 \). The distribution of mass relative to the axis determines the moment of inertia.
Yes, the parallel axis theorem can be used. If the moment of inertia about the center of mass is known (e.g., \( \frac{1}{6}ma^2 \) for a central axis), the moment of inertia about any parallel axis a distance \( d \) away can be found using \( I = I_{cm} + md^2 \).