# Moment of Inertia of a Square

1. Mar 29, 2010

### khangaroo

1. The problem statement, all variables and given/known data

Consider a 4 kg square which has its mass concentrated along its perimeter, with each side of length 6 m.

(a) What is the moment of inertia of the square about an axis perpendicular to the plane of the square at its center of mass? Use the parallel axis theorem and divide the square into parts. The moment of inertia of a rod rotated about its CM is $$\large_{I^{CM}_{rod}=\frac{1}{12}md^{2}}$$.
Answer in units of $$\large_{kg\cdot m^{2}}$$.

(b) What is the moment of inertia of the square about an axis perpendicular to the plane of the square at one of its corners?
Answer in units of $$\large_{kg\cdot m^{2}}$$.

2. Relevant equations

Moment of inertia of a rod rotated about its CM:
$$\large_{I^{CM}_{rod}=\frac{1}{12}md^{2}}$$

Parallel-axis theorem:
$$\large_{I=I_{CM}+Md^{2}}$$

3. The attempt at a solution

(a) I divided the square into 4 rods, and determined the moment of inertia in each of them.
$$\large_{I^{CM}_{rod}=\frac{1}{12}(\frac{m}{4})d^{2}=\frac{1}{48}md^{2}}$$
Then using the parallel-axis theorem:
$$\large_{I=I_{CM}+Md^{2}=\frac{1}{48}md^{2}+\frac{1}{4}m{(\frac{d}{2})} ^{2}=\frac{1}{12}md^{2}}$$
Whole system:
$$\large_{I=4{(\frac{1}{12})}md^{2}=\frac{1}{3}md^{2}}$$

I then plugged in the given values for m and d and got 48 $$\large_{kg\cdot m^{2}}$$ but it wasn't correct.

(b) I divided the square into 4 rods again.
Skipping to the parallel-axis theorem:
For 2 of the rods:
$$\large_{I=I_{CM}+Md^{2}=\frac{1}{48}md^{2}+\frac{1}{4}m{(\frac{d}{2})} ^{2}=\frac{1}{12}md^{2}}$$
For the other 2 rods:
$$\large_{I=I_{CM}+Md^{2}=\frac{1}{48}md^{2}+\frac{1}{4}m{(\frac{d\sqrt{5}}{2})} ^{2}=\frac{1}{3}md^{2}}$$
Whole system:
$$\large_{I=2{(\frac{1}{12})}md^{2}+2{(\frac{1}{3})}md^{2}=\frac{5}{6}md^{2}}$$

Again, I then plugged in the given values for m and d and got 120 $$\large_{kg\cdot m^{2}}$$ for this part but it wasn't correct.

I have no idea what I'm doing wrong.