Let’s try to learn the Moment of inertia of a Rectangle along with many solved problems, examples, calculations, and a clear understanding of the derivation, etc.

## What is Moment of Inertia of Rectangle?

Let’s try to concentrate on the moment of inertia of rectangle basics.

### Moment of Inertia of Rectangle Basics

Following are some basics while calculating the moment of inertia of the rectangle.

- If we want to calculate the moment of inertia of a large body such as a rectangle, we use the method of integration. Firstly, we find the moment of inertia of a very small or infinitesimal section. Then we use that value to calculate the moment of inertia of the whole body, for instance, a rectangle.
- The body’s axis of rotation is external or internal. Otherwise, it may be supposed to be a constraint or a change. In simpler conditions, the moment of inertia is often associated with the body’s axis of rotation.

### Moment of Inertia of Rectangle Definition

The rectangle’s moment of inertia is defined as: **The summation of products is obtained from the entire mass of every attached element of the rectangle and then multiplied the value by the square of the particles with respect to its distance from the central point.**

- A simple mathematical equation can determine the rectangle’s moment of inertia in the blog below.
- There are three cases while calculating the moment of inertia of the rectangle, which is given below.

## Moment of Inertia of Rectangle

### Associate in Axis Passing through Its center of mass

When we take a state of affairs once the axis passes through the center of mass, the instant moment of inertia of a rectangle is given as:

- I = bh
^{3}/ 12

Here, b is employed to denote the rectangle breadth (the dimension parallel to the axis), and ‘h’ is alleged to be the peak (dimension perpendicular to the axis).

### Associate in Axis Passing Through Its Base

If we have a tendency to point out the associate in the axis passing through the bottom, the instant of inertia of a rectangle is expressed as:

- I = bh
^{3}/ 3

This can be simply determined or calculated with the help of the Parallel Axis Theorem. In this consideration, it is already considered that the rectangle’s center of mass is at a distance of h/2 from the bottom.

### Parallel Axis Theorem

Any shape’s moment of inertia in relation to an arbitrary, non-centroid axis can be calculated if its moment of inertia in relation to a centroidal axis parallel to the first one is known.

The following equation expresses the so-called Parallel Axes Theorem:

where I’ is the moment of inertia in terms of an arbitrary axis, I is the moment of inertia in terms of a centroidal axis parallel to the first, and A (=bh) is the area of the form and d is the distance between the two parallel axes.

### A center of mass Axis Perpendicular to Its Base

If the center of mass axis is perpendicular to its base, the instant of inertia of a rectangle is determined by alternating the scale b and h, from the primary equation that is, I = bh^{3} / 3 . We’ll get the subsequent equation;

- Iy = hb
^{3}/ 12

Look, here b changed into the place of h and vice versa.

## Moment of Inertia of Rectangle Calculation

Let’s try to understand how to find out the moment of inertia of a rectangle with respect to its baseline with a detailed explanation. First, we will consider a rectangle section PQRS as per the below image.

- b = the breadth of the rectangle
- h = height of the rectangle

The area of the rectangle is,

Area = Breadth x Height

- A =b x h
- A =bh

Now, we have already learned how to derive the moment of inertia of different components like a rod, sphere, triangle, etc., and understand that it is required to consider one infinitesimal section and find out the moment of inertia of that section.

Once the MI of the small section is able to calculate, it is easy to calculate the moment of inertia of the whole rectangle. Let us consider a small strip of height dy at y distance from the bottom line (PQ) and need to calculate the moment of inertia of that rectangular section.

Now, we can write for the small section,

dA = b x dy ——————— (1)

Where,

- dA = Area of the samll section
- b = bredth of that small section
- dh = height of the small section

Now, based on the formula of the moment of inertia,

dI = dA y^{2} ——————— (2)

Putting the value of Eqn. 1 into Eqn. (2), we get

- dI = dA y
^{2} - dI = b x dy y
^{2}

By integrating,

- ∫ dl = ∫
_{0}^{h}b x dy y2 - I = b ∫
_{0}^{h}dy y2 - I = b [ y
^{3}/3 ]_{0}^{h} - I = b [ h
^{3}/3 -0] - I = bh
^{3}/3

Hence, the moment of inertia of rectangle with respect to its base can be written as

I = bh^{3}/3

In the same way, other conditions can also be derived. We will calculate the same in some other blogs.

## Moment of Inertia of Rectangle Solved Problems

### Question#1 Calculating the moment of inertia of rectangle when width and depth are given

**Problem**: Calculate the moment of inertia of a rectangle having a width of 15 mm and depth of 40 mm.

**Solution:** Using the parallel axes theorem:

I = I_{x} + AH^{2}

H shows the vertical distance, while A is the area of the rectangle.

H= depth/2

And, I= bh^{3}/12

Area of rectangle= bh

Now, we will substitute values in the parallel axis theorem, we get:

- I= I
_{x}+ AH^{2} - I = [(bh
^{3}) / 12] + [bh (h/2)^{2}] - I = bd
^{3}/ 3

Now putting the given values in the equation, we obtain:

- I = bh
^{3}/ 3 - I= (15 x 40
^{3}) / 3 = 320000 mm^{4}

### Question#2 Calculating moment of inertia of rectangle having sides a and b.

**Problem**: Calculate the moment of inertia of a rectangle having sides a and b in respect to an axis passing through the side b considering the figure given below.

**Solution:** The small black rectangular strip has a width dx while its distance from the axis of rotation is x. hence the moment of inertia is given by:

d I= x^{2 }dm = x^{2}p b d x

p is the density, p= 1. Hence, we can write:

d I = bx^{2 }d x

integration from x= 0 to x= a, we get the following expression:

I= **ʃ _{0}**

^{a}bx

^{2}dx = b

**ʃ**

_{0}^{a }x

^{2}dx = bx

^{3}/3 I

_{0}

^{a }= ba

^{3}/3

### Question#3 Calculating moment of inertia about the centroid of the given element A

**Problem**: Calculate “I” for the given element A, having a height of 10 units and a base of 150 units.

**Solution:** For rectangles, we use the following expression:

I= bh^{3}/12

I_{centroid}= (150 units) (10 units)^{3}/ 12

= 12500 units^{4}

^{ }= 12.5*10^{3 }units^{4}

### Question#4 Calculating moment of inertia of the given element about line “x” using the parallel axes theorem

**Problem**: Calculate “I” for the given element A about line x, having a height of 10 units and a base of 150 units.

**Solution:** Using the parallel axes theorem:

I_{x} = I_{centroid }* Ad^{2}

where A shows the area and d is the distance to the centroid of the element from x.

now inserting the values. The value of I_{centroid }is taken from the previous numerical.

I_{x}= 12500 + (10) (150) (106.7 – 5)^{2}

= 15526835 units ^{4}

= 15.5 * 10^{6} units^{4}

## Conclusion

The moment of inertia is the calculation of the force necessary to rotate an object. The value can be manipulated to increase or decrease inertia. In sports such as skating, diving, and gymnastics, the body structure of athletes is constantly changing. In this blog, we have studied the moment of inertia of rectangle & parallelogram and its different aspects. Check out our few most interesting articles, Refer our YouTube Videos. Check out our few most interesting articles,

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