Moment Of Inertia Of Solid Cone

Moment of inertia of solid cone can be expressed using the given formula;

I = 3 / 10 MR2

However, in this lesson, we will understand how the formula is derived and used in solving the problems. Let us first go through the derivation of the moment of inertia formula for a solid cone.

Solid Cone Moment Of Inertia Formula Derivation

We will take a solid cone where its axis will pass through the centre with radius = r, height = h. We will need to determine the mass though.

Moment Of Inertia Of Solid Cone

We take the elemental disc whose mass is given by;

dm = ρ ⋅ π r2dz

Density is given as;

ρ = M / V = M / ⅓ π R2h

With this, we will calculate the dm.

dm = M / ⅓ π R2h X π r2dz

dm = 3M / R2h X r2dz

If we consider the similarity of the triangle, then we have;

R / r = h / z

r = R X z / h

Now,

dm = 3M / R2h ⋅ R2 / h2 ⋅ z2dz

dm = 3M / h3 X z2dz

If we consider z-axis, the moment of inertia of the elemental disk will be;

dI = ½ dmr2

dI = ½ ⋅ 3M / h3 X z2 ⋅ z2 R2 / h2 dz

dI = 3 / 2 ⋅ M R2 / h5 X z4dz

Now we will follow the integration process. Here;

I = 3 / 2 ⋅ M R2 / h5 oh z4dz

I = 3 / 2 ⋅ M R2 / h5 [z5 / 5 ]oh

I = 3 / 2 ⋅ M R2 / h5 ⋅ h5 / 5

Therefore, I = 3/10 MR2