Moments of Inertia

This lesson introduces the moment of inertia — the rotational analogue of mass. Just as mass measures resistance to linear acceleration, moment of inertia measures resistance to angular acceleration.

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1 Definition

For a system of particles, the moment of inertia about a given axis is:

$$\boxed{I = \sum m_i r_i^2}$$I=∑mi​ri2​​

where $r_i$ri​ is the perpendicular distance of the $i$i-th particle from the axis.

For a continuous body:

$$I = \int r^2\,\mathrm{d}m$$I=∫r2dm

Property	Detail
SI unit	kg m$^2$2
Always positive	Since $r^2 \ge 0$r2≥0
Depends on the axis	Different axes give different $I$I
Depends on mass distribution	Mass further from the axis contributes more

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2 Standard Results

Body	Axis	Moment of inertia
Thin rod of mass $M$M, length $L$L	Through centre, perpendicular to rod	$\frac{1}{12}ML^2$121​ML2
Thin rod	Through one end, perpendicular to rod	$\frac{1}{3}ML^2$31​ML2
Uniform disc/cylinder of mass $M$M, radius $R$R	Through centre, perpendicular to face	$\frac{1}{2}MR^2$21​MR2
Thin hoop/ring of mass $M$M, radius $R$R	Through centre, perpendicular to plane	$MR^2$MR2
Solid sphere of mass $M$M, radius $R$R	Through centre	$\frac{2}{5}MR^2$52​MR2
Thin spherical shell	Through centre	$\frac{2}{3}MR^2$32​MR2
Rectangular lamina $a \times b$a×b, mass $M$M	Through centre, perpendicular to lamina	$\frac{1}{12}M(a^2 + b^2)$121​M(a2+b2)

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3 Derivation: Uniform Rod About Its Centre

Consider a rod of length $L$L and mass $M$M, with linear density $\rho = M/L$ρ=M/L. Place the centre at the origin.

$$I = \int_{-L/2}^{L/2} x^2 \rho\,\mathrm{d}x = \rho \left[\frac{x^3}{3}\right]_{-L/2}^{L/2} = \frac{M}{L} \cdot \frac{2(L/2)^3}{3} = \frac{M}{L} \cdot \frac{L^3}{12} = \frac{ML^2}{12}$$I=∫−L/2L/2​x2ρdx=ρ[3x3​]−L/2L/2​=LM​⋅32(L/2)3​=LM​⋅12L3​=12ML2​

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4 The Parallel Axis Theorem

If the moment of inertia about an axis through the centre of mass is $I_{\text{cm}}$Icm​, then the moment of inertia about a parallel axis at distance $d$d from the centre of mass is:

$$\boxed{I = I_{\text{cm}} + Md^2}$$I=Icm​+Md2​

Example 1

Find the moment of inertia of a uniform rod of mass $M$M and length $L$L about an axis through one end.

Solution

$I_{\text{cm}} = \frac{1}{12}ML^2$Icm​=121​ML2, $d = L/2$d=L/2.

$I = \frac{1}{12}ML^2 + M(L/2)^2 = \frac{1}{12}ML^2 + \frac{1}{4}ML^2 = \frac{1}{3}ML^2$I=121​ML2+M(L/2)2=121​ML2+41​ML2=31​ML2. $\checkmark$✓

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5 The Perpendicular Axis Theorem