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.
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We usage the adhering to expressions to calculate the minute of inertia of a rectangular plate;
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.
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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
(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;