Inertia Of A Hollow Cylinder

Moment of inertia of a hollow cylinder that is rotating on an axis passing through the centre of the cylinder where information technology has an internal radius r₁ and external radius r₂ with mass M tin exist expressed in the following style.

Subsequently, the moment of inertia of a hollow cylinder with a thin wall is determined using the following equation;

Here, the centrality goes through the centre of the cylinder and M = mass and r = radius.

Calculating Moment Of Inertia Of A Hollow Cylinder

If we take a hollow cylinder it will consist of inner radius r₁ and outer radius r_two with mass M, and length L. We volition calculate its moment of inertia about the central centrality.

Moment Of Inertia Of A Hollow Cylinder

However, before we get into the derivation we have to be enlightened of certain things. These are;

The cylinder is split into infinitesimally thin rings.
Each ring will have a thickness of dr with length Fifty.

Meanwhile, the whole adding procedure involves summing up the moments of infinitesimally thin cylindrical shells. Information technology is quite similar to the derivation of the moment of inertia of a solid cylinder.

Let's begin.

1. First, let us recall the moment of inertia equation:

dI = r² dm

Here we have to find dm. It is given as;

dm = ρ dV

Since we take mentioned dV in the above equation, we have to summate it. It will be given every bit;

dV = dA h

Here the dA is considered as the surface area of the ring on top. Now we become;

dA = π (r + dr)ⁱⁱ – π r²

dA = π (r² + 2rdr + (dr)ⁱⁱ) – π rⁱⁱ

Here, (dr)² = 0.

dA = 2 π r dr

We tin can besides find dA past differentiating method.

A = π r²

If we differentiate wrt r,

dA = 2 π r dr

2. We have to substitute dA into dV.

dV = ii π r h dr

If we substitute into dm nosotros go;

dm = ρ 2 π r h dr

We so substitute the dm expression into the dI equation.

dI = r² ρ 2 π r hdr

dl = ρ 2 π r³ h dr

3. Now, we can observe the expression for density. We use the equation;

ρ = M/V

At present,

ρ = Thousand / hπ(r₂ ² – r_one ⁱⁱ)

iv. The next step involves using integration to find the moment of inertia. The integration basically occurs from the inner radius to the outer radius:

dl = ρ 2 π r³ h dr

I = ii ρ π h_r1∫^r2 rⁱⁱⁱ dr = 2 ρ π h [r₂ ⁴/ 4 – r_ane ^four/ 4]

= (ρ π h/2) [r₂ ^iv– r_one ⁴]

I =(π h/2) [ M / π (r₂ ⁱⁱ – r₂ ^two) h] [ (r₂ ⁱⁱ – r_one ²) (r₂ ² + r_one ^two)]

I = ½ Yard (r₂ ² + r_one ²)

Now, if nosotros take a thin hollow cylinder, setting r = r₁ = r_ii and replacing it into the first equation we go;

I = Mr^two

⇒ Cheque Other Object'due south Moment of Inertia:

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