Derivation of Moment of Inertia of Annular Disc
Derivation of the Formula
- Mass Distribution:
- Dividing the Disc:
- Mass of an Elemental Ring:
- Moment of Inertia of a Thin Ring:
- Integrating Over the Whole Annular Disc:
- Evaluate the Integral:
Applications of the Formula
Moment of Inertia of Annular Disc FAQs

Moment of Inertia of Annular Disc

The moment of inertia of a uniform annular disc with mass (m), thickness (t), inner radius (R1), and outer radius (R2) is expressed as;

I = ½ M (R₂²+ R₁²)

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In this case, we must consider that the disc is rotating about an axis that runs through the centre.

Also Check: Absorptive & Emissive Power

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Derivation of Moment of Inertia of Annular Disc

(1) We’ll begin by recalling the moment of inertia expression, which is as follows:

dI = r² dm

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In just this case, we can think of dm as the mass of volume dV. The moment of inertia about the z-axis is then considered, yielding the expression;

I_zz = _O∫^R r²dm

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The mass dm of the volume element dV, just like relation in a disc, is related to volume and density by;

dm = ρ dV

Then we should compute dV. We would then assume that the disc has a uniform density in this case. In this case, we’ll also consider a ring with radius r, width dr, and thickness t. We might well obtain;

dV = (2 π r dr)t

Once we substitute the values for dm we get;

dm = ρ (2 πt ) r dr

(2) So we can create a new expression for I_zz. It certainly will be;

I_zz = _R1∫^R2 r²dm

I_zz = 2 π ρ t _R1∫^R2 r³dr

Then, we integrate R₁ and R₂. We will now have;

I_zz = 2 π ρ t [R₂⁴/4 – R₁⁴ / 4]

I_zz = (½) π ρ t [ R₂⁴ – R₁⁴] ……………(1)

Even so, we can also express it as [ R₂² + R₁²][ R₂² – R₁²].

Thus, the equation now becomes;

I_zz = ½ πρt [ R₂² + R₁²][ R₂² – R₁²] ……………..(2)

(3) The total mass of the annular disc must now be determined. We would then subtract the total mass (M₂) of a disc of radius R₂ from the mass (M₁) of a disc of radius R₁.

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We have, Mass = Density x Area x Thickness

So,

M₂ = π ρ t R₂²

M₁ = π ρ t R₁²

Thus, the mass of the annular disc is given as;

M_a = M₂ – M₁

M_a = π ρ t R₂² – πpt R₁²

M_a = π ρ t (R₂² – R₁²)

(4) The following step will be to substitute the above values in equation 2, which really is.

I_zz = ½ π ρ t [ R₂² + R₁²][ R₂² – R₁²]

After adding the values and integrating them, we could indeed write the expression for an annular disc’s moment of inertia as;

I_zz = ½ M_a (R₂² + R₁²)

Also Check: The Moment of Inertia

Derivation of the Formula

Mass Distribution:
- Consider the mass $M$ $M$ of the annular disc to be uniformly distributed over its area.
- The area of the annular disc is: $A = \pi (R_2^2 - R_1^2)$ $A$ $=$ $π$ $($ $R$ $22$ $$ $-$ $R$ $12$ $$ $)$
The mass per unit area (surface density) is:

$\sigma = \frac{M}{\pi (R_2^2 - R_1^2)}$ $σ$ $=$ $π$ $($ $R$ $22$ $$ $-$ $R$ $12$ $$ $)$ $M$ $$

Dividing the Disc:

Divide the annular disc into infinitesimally thin concentric rings.
Each ring has a radius $r$ $r$ and a thickness $dr$ $dr$ .

Mass of an Elemental Ring:

The area of a thin ring is $dA = 2\pi r \, dr$ $dA$ $=$ $2$ $πrdr$ .
The mass of the ring is: $dm = \sigma \, dA = \sigma (2\pi r \, dr)$ $dm$ $=$ $σdA$ $=$ $σ$ $($ $2$ $πrdr$ $)$

Moment of Inertia of a Thin Ring:

The moment of inertia of a thin ring about the central axis is: $dI = dm \cdot r^2$ $dI$ $=$ $dm$ $\cdot$ $r$ $2$
Substituting $dm$ $dm$ : $dI = \sigma (2\pi r \, dr) \cdot r^2 = 2\pi \sigma r^3 \, dr$ $dI$ $=$ $σ$ $($ $2$ $πrdr$ $)$ $\cdot$ $r$ $2$ $=$ $2$ $πσr$ $3$ $dr$

Integrating Over the Whole Annular Disc:

Integrate $dI$ $dI$ from $R_1$ $R$ $1$ $$ to $R_2$ $R$ $2$ $$ : $I = \int_{R_1}^{R_2} 2\pi \sigma r^3 \, dr$ $I$ $=$ $\int$ $R$ $1$ $$ $R$ $2$ $$ $$ $2$ $πσr$ $3$ $dr$

Substituting $\sigma = \frac{M}{\pi (R_2^2 - R_1^2)}$ $σ$ $=$ $π$ $($ $R$ $22$ $$ $-$ $R$ $12$ $$ $)$ $M$ $$ :

$I = \frac{2M}{R_2^2 - R_1^2} \int_{R_1}^{R_2} r^3 \, dr$ $I$ $=$ $R$ $22$ $$ $-$ $R$ $12$ $$ $2$ $M$ $$ $\int$ $R$ $1$ $$ $R$ $2$ $$ $$ $r$ $3$ $dr$

Evaluate the Integral:

The integral of $r^3$ $r$ $3$ is: $\int r^3 \, dr = \frac{r^4}{4}$ $\int$ $r$ $3$ $dr$ $=$ $4$ $r$ $4$ $$

Substituting the limits:

$I = \frac{2M}{R_2^2 - R_1^2} \left[ \frac{R_2^4}{4} - \frac{R_1^4}{4} \right]$ $I$ $=$ $R$ $22$ $$ $-$ $R$ $12$ $$ $2$ $M$ $$ $[$ $4$ $R$ $24$ $$ $$ $-$ $4$ $R$ $14$ $$ $$ $]$

Simplify:

$I = \frac{2M}{4(R_2^2 - R_1^2)} (R_2^4 - R_1^4)$ $I$ $=$ $4$ $($ $R$ $22$ $$ $-$ $R$ $12$ $$ $)$ $2$ $M$ $$ $($ $R$ $24$ $$ $-$ $R$ $14$ $$ $)$

Using the difference of squares: $R_2^4 - R_1^4 = (R_2^2 - R_1^2)(R_2^2 + R_1^2)$ $R$ $24$ $$ $-$ $R$ $14$ $$ $=$ $($ $R$ $22$ $$ $-$ $R$ $12$ $$ $)$ $($ $R$ $22$ $$ $+$ $R$ $12$ $$ $)$ ,

$I = \frac{M}{2} (R_1^2 + R_2^2)$ $I$ $=$ $2$ $M$ $$ $($ $R$ $12$ $$ $+$ $R$ $22$ $$ $)$

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Applications of the Formula

Engineering Design: Helps calculate rotational properties of ring-shaped components like flywheels.
Physics Problems: Used in rotational dynamics to compute torque and angular acceleration.
Astronomy: Used to model celestial rings or circular objects.