JEE Main 2024 Question Paper Solution Discussion Live JEE Main 2024 Question Paper Solution Discussion Live

Moment Of Inertia Of An Ellipse

Moment of inertia of ellipse is usually determined by the following expression;

I = M (a2 + b2) / 4

We will further understand how this equation is derived in this article.

Derivation of Moment of Inertia of Ellipse

Moment of Inertia of Ellipse

1. The first step involves determining the parametric equation of an ellipse.

It is given as;

(a cos t, b sin t)

Now we need to modify the parametric equation to get the transformation equation. We get;

(r cos θ, λ r sin θ)

Where

λ= b / a

Here, we have to consider a and b as the semi-major and semi-minor axis of the ellipse.

a = semi-major axis

b = semi-minor axis

We get the transformation equation as;

x = r cos θ

y = λ r sin⁡ θ

After this, we will need to find the area element using the Jacobian. It is given as;

Find the area element using the Jacobian

2. Finding the area

While calculating the area we have to remember that r will be integrated from 0 to semi-major axis a.

A = oa o λ r d θ

A = λ a2 π

A = π ab

3. Calculating moment of inertia

In this case, the moment of inertia formula will be;

I = ρ ∫ (x2 + y2) dA

I = ρ oa o λ r3 (cos2 θ + λ2 sin2⁡ θ) drdθ

I = ρλ a4 / 4 o (cos2 θ + λ2 sin2⁡ θ) dθ

By symmetry we get;

I = ρλ a4 ½ 2 o (cos2 θ + λ2 sin2⁡ θ) dθ

I = ρλ a4 ½ [B(½ , 3/2) + λ2b (3/2, ½ ) ]

B = beta function.

Now if we make use of the beta-gamma relation we get;

I = ρ λab / 4 (a2 + b2)

I = M (a2 + b2) / 4

Meanwhile, we also have the following equation for a filled ellipse whose radius along the x-axis is a and whose radius along the y-axis is b. It is given as;

Ix = π / 4 ab3

Iy = π / 4 a3b

⇒ Check Other Object’s Moment of Inertia:

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