Moment of inertia of a square of side 1 unit about an axis through its center of gravity, is .
A)1
B)1/12
C)1/3
D)1/4

Question

Quizplus · Accepted Answer

The moment of inertia of a square with side length $a$ about an axis through its center of gravity and perpendicular to its plane is given by $I = \frac{1}{6}ma^2$, where $m$ is the mass. If we consider a uniform square (mass distributed evenly) and normalize the mass to 1 for simplicity, and since the side length $a = 1$ unit, the formula simplifies to $I = \frac{1}{6}(1)(1)^2 = \frac{1}{6}$. However, this formula is for a thin plate, and the correct formula for a square about its center is $I = \frac{1}{12}m(2a)^2$ when considering it as a rectangle with sides $a$ and $a$. Substituting $m = 1$ and $a = 1$, we get $I = \frac{1}{12}(1)(2)^2 = \frac{1}{12}(4) = \frac{1}{3}$. My explanation contained an error in the calculation and formula application. The correct moment of inertia of a square of side 1 unit about an axis through its center of gravity, considering it as a solid object and not a thin plate, and correcting my mistake, is indeed $\frac{1}{12}$ when considering the mass and dimensions properly. This is derived from the general formula for a rectangle, adjusting for the square's dimensions. My initial explanation mistakenly combined aspects of different formulas and contexts. The correct answer, considering the square's geometry and uniform mass distribution, is $\frac{1}{12}$ of the square's mass times the square of its side length, with the mass normalized to 1 and side length also being 1.

Moment of Inertia of a Square of Side 1 Unit