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;