# Moment of Inertia of a Square

• khangaroo
In summary, the moment of inertia of a 4 kg square, with mass concentrated along its perimeter and each side of length 6 m, is 48 kg⋅m² when rotated about an axis perpendicular to the plane of the square at its center of mass. When rotated about an axis perpendicular to the plane of the square at one of its corners, the moment of inertia is 120 kg⋅m². The moment of inertia of a rod rotated about its CM is \large_{I^{CM}_{rod}=\frac{1}{12}md^{2}}, and the parallel-axis theorem is \large_{I=I_{CM}+Md^{2}}.

## Homework Statement

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 $$\large_{kg\cdot m^{2}}$$.

## Homework Equations

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

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

## 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.

They look ok.
For the second one you could have used the perpendicular axis theorem straight from the answer in a). Answer is ok though

## What is the definition of moment of inertia for a square?

Moment of inertia for a square is a measure of its resistance to rotational motion, calculated by multiplying the mass of the square by the square of its distance from the axis of rotation.

## How do you calculate the moment of inertia for a square?

The moment of inertia for a square can be calculated using the formula I = 1/12 * m * s^2, where m is the mass of the square and s is the length of one side.

## What factors affect the moment of inertia for a square?

The moment of inertia for a square is affected by its mass, shape, and distance from the axis of rotation. A square with a larger mass or longer sides will have a larger moment of inertia, while a square with a smaller mass or shorter sides will have a smaller moment of inertia.

## How is moment of inertia related to rotational motion?

The moment of inertia for a square is directly proportional to the square of its distance from the axis of rotation. This means that the further the square is from the axis of rotation, the greater its moment of inertia and the more difficult it is to rotate.

## Why is moment of inertia important in physics?

Moment of inertia is an important concept in physics because it helps us understand and predict the behavior of objects in rotational motion. It is also crucial in engineering and design, as it determines the amount of torque needed to rotate an object and affects its stability and balance.