In the instance of a rectangular plate, we normally discover the mass minute of inertia once the axis is passing through the centre perpendicular to the aircraft.

You are watching: Moment of inertia of a rectangular plate We usage the adhering to expressions to calculate the minute of inertia of a rectangular plate;

For x-axis;

## Moment Of Inertia Of A Rectangular Plate Derivation

### 1. Line Passing Thturbulent The Base For the derivation of the minute of inertia formula for a rectangular plate, we will certainly think about a rectangular section and also reduced out an elemental component at a distance (y) from the x-axis. Let its thickness be dy and s be the mass per unit volume of the plate.

Mass of the rectangular body = thickness x volume

M = ρ x bdt

Mass of the elepsychological section;

dM =ρ x btdy

Now we will certainly uncover the mass moment of inertia of the elepsychological sexpedition about the x-axis.

I = ∫ y2 dM

I = ∫ (ρ b dy t)y2

In the next action, we will certainly uncover the mass moment of inertia of the rectangular plate.

I = o∫d ρ b dy t y2

I = ρ bt o∫d y2dy

I = ρ bt d3 / 3

(since ρ = M/bdt)

I = Md2 / 3

At this allude, we apply the parallel axis theorem which offers us;

Ibase = ICG + M x h2

ICG = Ibase – Mh2

ICG = Md2 / 3 – M x (d / 2)2

ICG = Md2 / 12

This is how we identify the mass minute of inertia of a rectangular section around a line passing through the base.

### 2. Line Passing Through The Centre of Gravity

Now if we desire to derive the expression for a line passing via the centre of gravity then the actions for derivation are almost the exact same. We just have to make a few transforms.

See more: Irc Botnets Often Make Use Of What Chat Protocol In Order To Receive Commands We will certainly take one rectangular elementary spilgrimage and think about the thickness to be (dY) and also it will certainly be at a distance (Y) from the X-X axis.

The rectangular elementary spilgrimage location will be dA = dY.B

Mass of the rectangular elepsychological part is given by;

dm = ρ x T x dA

dm = ρ x T x dY. B

dm =ρBT. dY

Now we can say that the mass minute of inertia of the rectangular elemental section around the X-X axis ,

(Im)xx = dm.Y2

If we substitute the values for dm we get;

(Im)xx = ρBT. Y2dY

Next off step requires determining the mass minute of inertia of the whole rectangular area. We have to incorporate the above equation in between limit (-D/2) to (D/2). We will create it as;

(Im)xx = -D/2∫D/2 ρBT. Y2dY

(Im)xx = ρBT -D/2∫D/2 Y2dY

(Im)xx = ρBT -D/2D/2

(Im)xx = ρBT / 3 <(D/2)3 – (-D/2)3>Im) XX = ρBT.D3/12

(Im) XX = ρBTD.D2/12

(Im) XX = ρV.D2/12

(Im) XX = M.D2/12

Likewise, we deserve to additionally recognize the mass minute of inertia of the rectangular area around the Y-Y axis. We will certainly obtain;