- #1

Sunbodi

- 22

- 0

## Homework Statement

The moment of inertia for a perpendicular axis through the center of a uniform, thin, rectangular metal sheet with sides a and b is (1/12)M(a2 + b2). What is the moment of inertia if the axis is through a corner?

The answer is given as this was a powerpoint lecture and it is: I = 1/3 M(a^2 + b^2)

I'm not looking for how to solve the whole thing however as the moment of Inertia is given in the context of the problem. I'm trying to understand how D was found to be: (a^2/2^2 + b^2 /2^2).

## Homework Equations

I = I (cm) + Md^2

## The Attempt at a Solution

D is meant to be the distance between the the new axis and the center of mass, if I personally were to solve this I'd state that the center of mass is half the distance between the axis that contains a and the axis that contains b. If that's the case, the distance from the center of mass to a diagonal would be a^2 + b^2 because you're increasing both a and b values by the same distance it originally was from the center of mass.