Moments of Inertia

The rotational kinetic energy is the sum of the kinetic energy of each of the constituents of the rigid body. We can rewrite the rotational kinetic energy in terms of the angular velocity vector and certain aggregate quantities determined by the distribution of mass in the rigid body.

Substituting our representation of the relative velocity vectors into the rotational kinetic energy we obtain

X

We introduce an arbitrary rectangular coordinate system with ori-gin at the center of rotation and with basis vectors ˆe₀, ˆe₁, and ˆe₂, with the property that ˆe₀×eˆ₁ = ˆe₂. The components of~ω on this coordinate system areω⁰,ω¹, andω². Rewriting~ω in terms of its components, the rotational kinetic energy becomes

X

The quantities I_ij are the components of theinertia tensor with respect to the chosen coordinate system. Note what a remarkable form the kinetic energy has taken. All we have done is interchange the order of summations, but now the kinetic energy is written as a sum of products of components of the angular velocity vector, which completely specify how the orientation of the body is chang-ing, and the quantityIij, which depends solely on the distribution of mass in the body relative to the chosen coordinate system.

We will deduce a number of properties of the inertia tensor.

First, we find a somewhat simpler expression for it. The

compo-nents of the vector ξ~α are (ξα, ηα, ζα).³ Rewriting ~ξα as a sum over its components, and simplifying the elementary vector prod-ucts of basis vectors, the components of the inertia tensor can be arranged in the inertia matrixI, which looks like:

" P The inertia tensor has real components and is symmetric: Ijk = I_kj.

We define the moment of inertiaI about a line by I =X

α

m_α(ξ_α^⊥)², (2.16)

where ξ_α^⊥ is the perpendicular distance from the line to the con-stituent with index α. The diagonal components of the inertia tensorI_iiare recognized as the moments of inertia about the lines coinciding with the coordinate axes ˆei. The off-diagonal compo-nents of the inertia tensor are called products of inertia.

The rotational kinetic energy of a body depends on the distri-bution of mass of the body solely through the inertia tensor. Re-markably, the inertia tensor involves only second order moments of the mass distribution with respect to the center of mass. We might have expected the kinetic energy to depend in a complicated way on all the moments of the mass distribution, interwoven in some complicated way with the components of the angular ve-locity vector, but this is not the case. This fact has a remarkable consequence: for the motion of a free rigid body the detailed shape of the body does not matter. If a book and a banana have the same inertia tensor, that is, the same second order mass moments, then if they are thrown in the same way the subsequent motion will be the same, however complicated that motion is. The fact that the book has corners and the banana has a stem do not affect the motion except for their contributions to the inertia tensor. In general, the potential energy of an extended body is not so simple

3Here we avoid the more consistent notation (ξ⁰_α, ξ¹_α, ξ_α²) for the components of~ξ_αbecause it is awkward to write expressions involving powers of the com-ponents written this way.

and does indeed depend on all moments of the mass distribution, but for the kinetic energy the second moments are all that matter!

Exercise 2.1: Rotational kinetic energy

An interesting alternate form for the rotational kinetic energy can be found by decomposing ~ξ_α into components parallel and perpendicular to the rotation axis ˆω. Show that the rotational kinetic energy can also be written

T_R= ¹₂Iω², (2.17)

where I is the moment of inertia about the line through the center of mass with direction ˆω, andω is the instantaneous rate of rotation.

Exercise 2.2: Steiner’s theorem

LetIbe the moment of inertia of a body with respect to some given line through the center of mass. Show that the moment of inertia I⁰ with respect to a second line parallel to the first is

I⁰=I+M R² (2.18)

where M is the total mass of the body and R is the distance between the lines.

Exercise 2.3: Some useful moments of inertia

Show that the moments of inertia of the following objects are as given:

a. The moment of inertia of a sphere of uniform density with massM and radiusRabout any line through the center is ²₅M R².

b. The moment of inertia of a spherical shell with massM and radius Rabout any line through the center is ²₃M R².

c. The moment of inertia of a cylinder of uniform density with massM and radiusRabout the axis of the cylinder is ¹₂M R².

c. The moment of inertia of a thin rod of uniform density per unit length with mass M and length L about an axis perpendicular to the rod through the center of mass is ₁₂¹M L².

Exercise 2.4: Jupiter

a. The density of a planet increases toward the center. Provide an argument that the moment of inertia is less than that of a sphere of uniform density of the same mass and radius.

b. The density as a function of radius inside Jupiter is well approxi-mated by

ρ(r) = M R³

sin(πr/R) 4r/R ,

where M is the mass and Ris the radius of Jupiter. Find the moment of inertia of Jupiter in terms ofM andR.