Rotational Motion - Chapter 4 · 1st term is energy associated with motion of center of mass of system. 2nd term is energy associated with rotational motion of system about center

Chapter 04Rotational Motion

P. J. Grandinetti

Chem. 4300

P. J. Grandinetti Chapter 04: Rotational Motion

Angular MomentumAngular momentum of particle with respect toorigin, O, is given by

𝓁 = r⃗ × p⃗

x

y

z

m

r

O

p

Unlike linear momentum, angular momentumdepends on origin choice.

Rate of change of angular momentum is givenby cross product of r⃗ with applied force.

d𝓁dt

= r⃗ ×dp⃗dt

= r⃗ × F⃗ = 𝜏

Cross product is defined as applied torque, 𝜏.


Conservation of Angular MomentumConsider system of N Particles

m1

m2

m3

m4

m5

x

y

z

Total angular momentum is

L⃗ =N∑

𝛼=1𝓁𝛼 =

N∑𝛼=1

r⃗𝛼 × p⃗𝛼

Rate of change of angular momentum is

dL⃗dt

=N∑

𝛼=1

d𝓁𝛼dt

=N∑

𝛼=1r⃗𝛼 ×

dp⃗𝛼dt

which becomes

dL⃗dt

=N∑

𝛼=1r⃗𝛼 × F⃗net𝛼


Conservation of Angular Momentum

dL⃗dt

=N∑

𝛼=1r⃗𝛼 × F⃗net𝛼

Taking an earlier expression for a system of particles from chapter 1

F⃗net𝛼 = F⃗ext𝛼 +

N∑𝛽=1𝛽≠𝛼

f⃗𝛼𝛽

we obtaindL⃗dt

=N∑

𝛼=1r⃗𝛼 × F⃗ext𝛼 +

N∑𝛼=1


r⃗𝛼 × f⃗𝛼𝛽

and then obtain

dL⃗dt

=N∑

𝛼=1r⃗𝛼 × F⃗ext𝛼 +

��>

0N∑

𝛼=1


r⃗𝛼 × f⃗𝛼𝛽 double sum disappears from Newton’s 3rd law (f⃗12 = −f⃗21)


Conservation of Angular Momentum

dL⃗dt

=N∑

𝛼=1r⃗𝛼 × F⃗ext𝛼 =

N∑𝛼=1

𝜏𝛼 = 𝜏total

If there is no net external torque on system of particles then system’s total angularmomentum, L⃗, is constant,

if 𝜏total = 0, thendL⃗dt

= 0, and L⃗ = constant

This is the principle of conservation of angular momentum.

True in quantum mechanics as well as in classical mechanics.


Orbital and Spin Angular Momentum


Orbital and Spin Angular MomentumConsider system of particles again.Total angular momentum relative to origin is

L⃗ =N∑

𝛼=1𝓁i =

N∑𝛼=1

r⃗𝛼 × m𝛼dr⃗𝛼dt

Defining position of each particle relative to center of mass,

r⃗𝛼 = R⃗ + r⃗◦𝛼

r⃗𝛼 is position of particle position relative to origin,R⃗ is center of mass relative to origin,r⃗◦𝛼 is particle position relative to center of mass.

O

COM

Expression for L⃗ becomes (see notes for derivation)

L⃗ = R⃗ × p⃗ total⏟⏞⏞⏟⏞⏞⏟

relative to origin

+N∑

𝛼=1r⃗◦𝛼 × m𝛼

dr⃗◦𝛼dt

⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟relative to CM


Orbital and Spin Angular Momentum

L⃗ = R⃗ × p⃗ total⏟⏞⏞⏟⏞⏞⏟

relative to origin

+N∑

𝛼=1r⃗◦𝛼 × m𝛼

dr⃗◦𝛼dt

⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟relative to CM

Identify total angular momentum as split into orbital and spin angular momentum terms:

L⃗ = L⃗orbital + L⃗spin

Imagine L⃗orbital as angular momentum of Earth as it moves around sun as the origin,and L⃗spin as angular momentum of Earth as it spins about its center of mass.

Often true, to a good approximation, that orbital and spin parts are separately conserved.


Rotational energyTotal kinetic energy of system of particles is

K =N∑

𝛼=1

12

m𝛼

(dr⃗𝛼dt

)2Written in terms of R⃗ and r⃗◦𝛼 gives

K = 12

M(

dR⃗dt

)2⏟⏞⏞⏞⏞⏟⏞⏞⏞⏞⏟center of mass

+ 12

N∑𝛼=1

m𝛼

(dr⃗◦𝛼dt

)2⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

rotation about center of mass

1st term is energy associated with motion of center of mass of system.

2nd term is energy associated with rotational motion of system about center of mass.


Rigid Bodies


Rigid Bodies

DefinitionWhen all particles in system are rigidly connected we have a rigid body.This ideal model assumes no relative movement of composite particles.

m1

m2

m3

m4

m5

x

y

z


How do we specify the orientation of a rigid body?

DefinitionEuler’s rotation theorem states that orientation of rigid body in a given coordinate system canbe described by rotation through an angle about a single axis.

Only 2 angles needed to define orientation of rotation axis so full orientation of any rigid bodyabout fixed point can be described by just 3 parameters: e.g., polar, 𝜃, and azimuthal, 𝜙,angles and angle of rotation, 𝜒 :

Sign of 𝜒 is determined by right-hand rule: thumbpoints along rotation axis and right-hand fingerscurl in direction of positive rotation.Set of 𝜙, 𝜃, and 𝜒 are called Euler angles


Euler anglesAnother example of a set of Euler angles is convention for giving orientation of an airplane by3 parameters called yaw, pitch, and roll:

Yaw Axis(z)

Roll Axis(x)

Pitch Axis(y)

Yaw, pitch, and roll are just one of many ways of defining the 3 angles implied by Euler’srotation theorem.


Angular velocity vector, �⃗�

DefinitionAngular velocity vector, �⃗�, is vector passing through origin along axis of rotation and whosemagnitude equals magnitude of angular velocity,

�⃗� = 𝜔 e⃗r

𝜔 = d𝜒∕dt and e⃗r is unit vector defined by

e⃗r = sin 𝜃 cos𝜙 e⃗x + sin 𝜃 sin𝜙 e⃗y + cos 𝜃 e⃗z

Direction of �⃗� is determined by right-hand rule: Curl right-hand fingers in direction ofrotation, then thumb points in direction of �⃗�.

Keep in mind when describing motion of rigid body that magnitude and orientation of �⃗� canchange with time.


Angular velocity vector, �⃗�Best origin choice through which �⃗� passes depends on rigid body motion being described.

For molecular rotations : natural origin choice is molecule’s center of mass.

For top spinning on table surface : origin is better located at fixed point where tip of topmeets table surface.

x

x

y

y

z


Relationship between angular momentum and angular velocity vectorsTotal angular momentum of rigid body relative to origin is

J⃗ =N∑

𝛼=1(r⃗𝛼 × m𝛼 v⃗𝛼) =

N∑𝛼=1

m𝛼(r⃗𝛼 × v⃗𝛼)

Note: notation change, J⃗ instead of L⃗ for rigid body.Linear velocity vector, v⃗𝛼, of 𝛼th particle is related to its angular velocity, �⃗�, by

v⃗𝛼 = �⃗� × r⃗𝛼r⃗𝛼 is particle position.�⃗� is identical for all particles in rotating rigid body since all inter-particle distances areconstant.

Combining these two expressions ...

J⃗ =N∑

𝛼=1m𝛼

[r⃗𝛼 × (�⃗� × r⃗𝛼)

]⏟⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏟

vector triple productP. J. Grandinetti Chapter 04: Rotational Motion

Relationship between angular momentum and angular velocity vectorsTotal angular momentum of rigid body relative to the origin is

J⃗ =N∑

𝛼=1m𝛼

[r⃗𝛼 × (�⃗� × r⃗𝛼)

]⏟⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏟

vector triple product

Vector triple product has a well known expansion

a⃗ ×(

b⃗ × c⃗)= (a⃗ ⋅ c⃗) b⃗ − (a⃗ ⋅ b⃗) c⃗

With triple product expansion we get

J⃗ =N∑

𝛼=1m𝛼

[(r⃗𝛼 ⋅ r⃗𝛼

)�⃗� −

(r⃗𝛼 ⋅ �⃗�

)r⃗𝛼]


Relationship between angular momentum and angular velocity vectorsMoment of Inertia Tensor

J⃗ =N∑

𝛼=1m𝛼

[(r⃗𝛼 ⋅ r⃗𝛼

)�⃗� −

(r⃗𝛼 ⋅ �⃗�

)r⃗𝛼]

can be rewritten as matrix equation

J⃗ =⎛⎜⎜⎝

JxJyJz

⎞⎟⎟⎠ =⎛⎜⎜⎝

Ixx Ixy IxzIyx Iyy IyzIzx Izy Izz

⎞⎟⎟⎠⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

I

⎛⎜⎜⎝𝜔x𝜔y𝜔z

⎞⎟⎟⎠⏟⏟⏟

�⃗�

= I ⋅ �⃗�

matrix I is called the moment of inertia tensor (see next slide).

J⃗ = I ⋅ �⃗�When rigid body rotates about origin with angular velocity �⃗� the size and direction of body’s angularmomentum, J⃗, is determined by its moment of inertia tensor, I.


Moment of Inertia Tensor of a rigid bodyElements of Moment of Inertia Tensor about the center of mass

If rigid body rotates about its center of mass then relevant moment of inertia tensor–relativeto center of mass—is calculated according to

Ixx =N∑

𝛼=1m𝛼

(y2𝛼 + z

2𝛼)− M(Y2 + Z2)

Iyy =N∑

𝛼=1m𝛼

(x2𝛼 + z

2𝛼)− M(X2 + Z2)

Izz =N∑

𝛼=1m𝛼

(x2𝛼 + y

2𝛼)− M(X2 + Y2)

Ixy = Iyx = −N∑

𝛼=1m𝛼x𝛼y𝛼 + MXY

Iyz = Izy = −N∑

𝛼=1m𝛼y𝛼z𝛼 + MYZ

Ixz = Izx = −N∑

𝛼=1m𝛼x𝛼z𝛼 + MXZ

M is total massX, Y, Z are coordinates of the center of mass.

Notice that off-diagonal elements are symmetric about diagonal, Ixy = Iyx, Ixz = Izx, etc.Tensors with this property are called symmetric tensors.


The Principal Axis System of a Moment of Inertia TensorOff-diagonal elements are symmetric about the diagonal

I =⎛⎜⎜⎝

Ixx Ixy IxzIxy Iyy IyzIxz Iyz Izz

⎞⎟⎟⎠For any symmetric tensor one can always find an axis system in which it is diagonal.

IPAS = R(𝜙, 𝜃, 𝜒) ⋅⎛⎜⎜⎝

Ixx Ixy IxzIxy Iyy IyzIxz Iyz Izz

⎞⎟⎟⎠ ⋅ RT (𝜙, 𝜃, 𝜒) =⎛⎜⎜⎝

Ia 0 00 Ib 00 0 Ic

⎞⎟⎟⎠Diagonal elements, Ia, Ib, and Ic, called the principal moments of inertia.R(𝜙, 𝜃, 𝜒) is a rotation matrix and RT (𝜙, 𝜃, 𝜒) is its transpose.

DefinitionPrincipal Axis System (PAS) is a coordinate system in which I is diagonal.Convention is to assign axes of body-fixed frame, a, b, and c, such that Ic ≥ Ib ≥ Ia.


Moment of Inertia TensorPrincipal Axis System

When moment of inertia tensor is calculated in the PAS we have

IPAS =⎛⎜⎜⎝

Ia 0 00 Ib 00 0 Ic

⎞⎟⎟⎠Symmetry can help identify the principal axis system. A 2-fold or higher symmetry axis ina molecule is a principal axis.

If 3 principal moments of inertia are equal then any axis in space can be a principal axis.

If 2 principal moments are equal then any axis in plane of the 2 equal principal momentscan be a principal axis.


Moment of Inertia TensorPrincipal Axes

For freely rotating object, i.e.,with no external torque applied, J⃗ is conserved, dJ⃗∕dt = 0.

If freely rotating object has constant (i.e., time independent) angular velocity vector, �⃗�,then �⃗� points along a principal axis, and J⃗ is aligned with �⃗� (recall J⃗ = I ⋅ �⃗�).

If �⃗� of freely rotating object is not aligned with J⃗, then �⃗� is time dependent sinceconservation of angular momentum requires J⃗ to remain time-independent.

When viewed (calculated) in body-fixed frame I is time-independent.

When viewed (calculated) in space-fixed frame I may be time dependent as rigid bodyrotates.


Intermediate Axis Theoremaka Tennis Racket Theorem aka the Dzhanibekov1 effect

DefinitionIntermediate axis theorem states that the rotation of an object with Ia ≠ Ib ≠ Ic around its Iaand Ic principal axes is stable, while rotation around the Ib axis (or intermediate axis) is not.

Dzhanibekov effect

Intermediate Axis Theorem

Ic

Ia

Ib

1Soviet CosmonautP. J. Grandinetti Chapter 04: Rotational Motion

https://www.youtube.com/watch?v=1n-HMSCDYtMhttps://www.youtube.com/watch?v=-Si6iRL5Fj8

Moment of Inertia Tensor

ExampleCalculate moment of inertia tensor of water molecule. Take OH bond length as d = 0.957 Å, H-O-Hangle as Ω = 105◦, and O and H masses as 16.0 u and 1.0 u (1 u = 1.66053904 × 10−27 kg).

z

x105°

# atom mass x y z1 H 1 u +0.5826 Å 0 +0.7592 Å2 H 1 u +0.5826 Å 0 −0.7592 Å3 O 16 u 0 0 0




We know M = 18 u, and X = 0.06473 Å, and Y = Z = 0.

# atom mass x y z1 H 1 u +0.5826 Å 0 +0.7592 Å2 H 1 u +0.5826 Å 0 −0.7592 Å3 O 16 u 0 0 0

Diagonal moment of inertia components are

Ixx =N∑

𝛼=1m𝛼(�

��0

y2𝛼 + z2𝛼) −��

��:0M(Y2 + Z2)

Iyy =N∑

𝛼=1m𝛼

(x2𝛼 + z

2𝛼)− M(X2 +��

0Z2 )

Izz =N∑

𝛼=1m𝛼(x2𝛼 + �

��0

y2𝛼 ) − M(X2 +��

0Y2 )




We know M = 18 u, and X = 0.06473 Å, and Y = Z = 0. Diagonal moment of inertia components are

Ixx = mHz21 + mHz22 = (1 u)(0.7592Å)

2 + (1 u)(−0.7592Å)2 = 1.914 × 10−47m2⋅kg

Iyy = mH(x21 + z

21)+ mH

(x22 + z

22)− MX2

Iyy = (1 u)((0.5826 Å)2 + (0.7592Å)2

)+ (1 u)

((0.5826 Å)2 + (−0.7592Å)2

)− (18 u)(0.06473Å)2

Iyy = 2.916 × 10−47m2⋅kg

Izz = mHx21 + mHx22 − MX

2

Izz = (1 u)(0.5826 Å)2 + (1 u)(0.5826 Å)2 − (18 u)(0.06473Å)2 = 1.002 × 10−47m2⋅kg



ExampleCalculate moment of inertia tensor of water molecule. Take OH bond length as d = 0.957 Å,H-O-H angle as Ω = 105◦, and O and H masses as 16.0 u and 1.0 u.

And the off-diagonal components are

Ixy = Iyx = 0

Iyz = Izy = 0

Ixz = Izx = −mHx1z1 − mHx2z2Ixz = Izx = −(1 u)(0.5826 Å)(0.7592Å) − (1 u)(0.5826 Å)(−0.7592Å)

= 0



ExampleCalculate moment of inertia tensor of water molecule. Take OH bond length as d = 0.957 Å,H-O-H angle as Ω = 105◦, and O and H masses as 16.0 u and 1.0 u.

In the coordinate system for this calculation we obtain a diagonal moment of inertia tensorgiven by

I∕10−47m2⋅kg =⎛⎜⎜⎝

1.914 0 00 2.916 00 0 1.002

⎞⎟⎟⎠Not done yet! Must follow convention for labeling the axes in principal axis system. In thisexample we identify (Iyy = Ic) > (Ixx = Ib) > (Izz = Ia).



ExampleCalculate moment of inertia tensor of water molecule.

Final answerWater molecule has moment of inertia tensor in the PAS given by

IPAS∕10−47m2⋅kg =⎛⎜⎜⎝

1.002 0 00 1.914 00 0 2.916

⎞⎟⎟⎠

z

x105°

When labeling axes we need to maintain right-handed coordinate system where a⃗ × b⃗ = c⃗.



ExampleCalculate moment of inertia tensor of water molecule.

Final answerWater molecule has moment of inertia tensor in the PAS given by

IPAS∕10−47m2⋅kg =⎛⎜⎜⎝

1.002 0 00 1.914 00 0 2.916

⎞⎟⎟⎠ 105°a

b

When labeling axes we need to maintain right-handed coordinate system where a⃗ × b⃗ = c⃗.


Rotational Energy of Rigid Bodiesand

Classification of molecules


Rotational energy of a rigid bodyIn PAS kinetic energy only comes from rotational motion. Define v⃗◦𝛼 as velocity of 𝛼th particle relativeto center of mass.

K = 12

N∑𝛼=1

m𝛼 v⃗◦2𝛼 =

12

N∑𝛼=1

m𝛼 v⃗◦𝛼 ⋅ v⃗◦𝛼

Since v⃗◦𝛼 = �⃗� × r⃗◦𝛼 we can write

K = 12

N∑𝛼=1

m𝛼 v⃗◦𝛼 ⋅ (�⃗� × r⃗◦𝛼) =12�⃗� ⋅

N∑𝛼=1

m𝛼(r⃗◦𝛼 × v⃗◦𝛼) =12�⃗� ⋅

N∑𝛼=1

r⃗◦𝛼 × p⃗𝛼

Recognizing final sum as total angular momentum, J⃗ =∑N

𝛼=1 r⃗◦𝛼 × p⃗𝛼, we have

K = 12�⃗� ⋅ J⃗

Since J⃗ = I ⋅ �⃗� we obtain our expression for the kinetic energy of the rotational motion of a rigid body,

K = 12�⃗� ⋅ I ⋅ �⃗�


Rotational energy of a rigid body

K = 12�⃗� ⋅ I ⋅ �⃗�

Kinetic energy reduces to convenient form in PAS (i.e., a body-fixed frame) in terms of principalmoments:

K = 12(𝜔2aIa + 𝜔

2bIb + 𝜔

2cIc

)In PAS we also have

Ja = Ia𝜔a, Jb = Ib𝜔b, and Jc = Ic𝜔cwhich we use to rewrite kinetic energy as

K =J2a2Ia

+J2b2Ib

+J2c2Ic


Classification of molecules

Molecules grouped into 5 classes based on principal components.

Name Diagonal values ExamplesSpherical Ia = Ib = Ic = I CH4

Prolate Symmetric I|| = Ia < Ib = Ic = I⟂ CH3FOblate Symmetric I⟂ = Ia = Ib < Ic = I|| CHF3

Asymmetric Ia < Ib < Ic CH2Cl2, CH2CHClLinear Ia = 0, Ib = Ic = I OCS, CO2

Molecules with two or more 3-fold or higher rotational symmetry axes are spherical tops.

All molecules with one 3-fold or higher rotational symmetry axis are symmetric tops becauseprincipal moments about two axes normal to n-fold rotational symmetry axis (n ≥ 3) are equal.


Spherical: Ia = Ib = Ic = IMethane is an example of a molecule with this rotational symmetry.

H C

H

H

H

Ia = Ib = Ic = 5.3 × 10−40 g⋅cm2

Spherical molecules have more than one 3-fold axis of symmetry and all 3 principal moments of inertiaare equal.Principal moments of inertia of spherical tetrahedral and octahedral molecules:

In spherical symmetry case kinetic energy takes form

K =J2a + J

2b + J

2c

2I= J

2

2I


Symmetric molecules: two principal moments are equal

Symmetric molecules have a three-fold or higher axis of symmetry.

Presence of this symmetry axis leads to two of the principal moments of inertia being equal.

3rd principal moment is associated with rotation about axis with highest rotational symmetry,called the figure axis.

In prolate and oblate symmetric tops the figure axis is associated with the smallest and largestprincipal moment of inertia, respectively.


Prolate Symmetric: I|| = Ia < Ib = Ic = I⟂think “cigar-shaped”

Label perpendicular moment of inertia component I⟂ for Ib = Ic and parallel component I|| for Ia.Fluoromethane is an example of a molecule with prolate symmetry.

F C

H

H

H

a → Ia = 5.3 × 10−40 g⋅cm2

Ib = Ic = 32.9 × 10−40 g⋅cm2

In prolate symmetric case kinetic energy expression takes form

K =J2a2I|| +

J2b + J2c

2I⟂= J

2

2I⟂+(

12I|| −

12I⟂

)J2a


Oblate Symmetric: I⟂ = Ia = Ib < Ic = I||think “frisbee-shaped”

Label perpendicular moment of inertia component I⟂ for Ia = Ib and parallel component I|| for Ic.Trifluoromethane is an example of a molecule with oblate symmetry.

H C

F

F

F

c → Ia = Ib = 81.1 × 10−40 g⋅cm2

Ic = 149.1 × 10−40 g⋅cm2

In oblate symmetric case kinetic energy expression takes form

K =J2a + J

2b

2I⟂+

J2c2I|| =

J22I⟂

+(

12I|| −

12I⟂

)J2c


Symmetric Rotor Generic Cases


Asymmetric: Ia < Ib < IcMolecules with Ia < Ib < Ic have lowest symmetry.

Dichloromethane is example of molecule with asymmetric symmetry.

C

H

H

Cl

Cl

a↑

→ b

Ia = 26.41 × 10−40 g⋅cm2Ib = 254.51 × 10−40 g⋅cm2Ic = 275.71 × 10−40 g⋅cm2

Asymmetric molecules can possess two-fold axes and planes of symmetry but cannot have anythree-fold or higher symmetry axes.

In asymmetric case kinetic energy expression cannot be simplified from general form

K =J2a2Ia

+J2b2Ib

+J2c2Ic


Linear: Ia = 0, Ib = Ic = ILinear molecules have an axis with infinite rotational symmetry (C∞) since all the atoms lie on one axis.

Carbon dioxide and OCS are two examples of a molecule with linear symmetry.

O C S a → Ia = 0,Ib = Ic = I = 138.0 × 10−40 g⋅cm2

O C O a → Ia = 0,Ib = Ic = I = 71.3 × 10−40 g⋅cm2


Linear Molecule: General Expressions


Linear Molecule: Kinetic EnergyTo obtain kinetic energy expression in linear case we need to reconsider derivation.Because Ia = 0 there can be no rotational motion around the line of atoms. All rotation occurs aboutan axis perpendicular to the a axis.Since Ia = 0 and 𝜔a = 0 we modify

K = 12(𝜔2aIa + 𝜔

2bIb + 𝜔

2cIc

)to be

K = 12(𝜔2b + 𝜔

2c)

I = 12𝜔2I

Similarly we see that J2 = J⃗ ⋅ J⃗ = I2b𝜔2b + I

2c𝜔

2c = I

2𝜔2 and we have

K = J2

2I

as kinetic energy of linear rigid rotor.Warning: looks the same as spherical case, but not the same. Linear case derived with Ia = 0.


Diagonalizing a moment of inertia tensor


Diagonalizing a moment of inertia tensor

Previously, we worked in body-fixed coordinate systems with axes chosen along molecular axes ofhighest rotational symmetry where I is diagonal.

If different coordinate system was used for water:

z

x105°

we would have calculated

I∕10−47m2⋅kg =⎛⎜⎜⎝

1.576 0 0.4410 2.916 0

0.441 0 1.340

⎞⎟⎟⎠

How do we find principal components of I and transformation back to coordinate system where I isdiagonal?


Diagonalizing a moment of inertia tensorWhen angular velocity vector, �⃗�, is aligned with a principal axis we have

J⃗ = I �⃗�c =⎛⎜⎜⎝

Ia 0 00 Ib 00 0 Ic

⎞⎟⎟⎠⎛⎜⎜⎝

00𝜔

⎞⎟⎟⎠ = Ic�⃗�cIn this case, �⃗�c is vector aligned with a principal axis with principal component value Ic.We can write 3 equations:

I�⃗�a = Ia�⃗�a, I�⃗�b = Ib�⃗�b, I�⃗�c = Ic�⃗�c

Divide both sides by corresponding magnitude |�⃗�i|, and rearrange to(I − Ii1

)e⃗i = 0

e⃗i is unit vector along direction of �⃗�i, and 1 is identity matrix.This is a set of 3 simultaneous linear equations for each principal component.


Diagonalizing a moment of inertia tensor(I − Ii1

)e⃗i = 0

1 To find principal components we expand determinant

det(I − Ii1

)=|||||||

Ixx − Ii Ixy IxzIyx Iyy − Ii IyzIzx Izy Izz − Ii

||||||| = 0to get a 3rd-order polynomial equation in Ii. The 3 principal components Ia, Ib, Ic are roots of this3rd-order polynomial equation

2 PAS orientation obtained by substituting each principal component back into top equation(I − Ia1

)e⃗a = 0,

(I − Ib1

)e⃗b = 0,

(I − Ic1

)e⃗c = 0,

and solving for corresponding e⃗i which points along direction for Ii in original coordinate systemwhere I was non-diagonal.


Diagonalizing a moment of inertia tensorExampleGiven the inertia tensor calculated in an arbitrary body-fixed frame shown below

I∕10−47m2⋅kg =⎛⎜⎜⎝

1.576 0 0.4410 2.916 0

0.441 0 1.340

⎞⎟⎟⎠determine (a) its principal components of the inertia tensor and (b) the orientation of the PAS.

(a) For the general form of this tensor in this problem we can create the determinant,||||||Ixx − Ii 0 Ixz0 Iyy − Ii 0Ixz 0 Izz − Ii

|||||| = 0and expand to obtain the polynomial

(I2i − (Ixx + Izz)Ii + IxxIzz − I2xz) (Iyy − Ii) = 0


Diagonalizing a moment of inertia tensorFinding principal components from roots of polynomial

(I2i − (Ixx + Izz)Ii + IxxIzz − I2xz)(Iyy − Ii) = 0

As expected, one root is I1 = Iyy = 2.916 × 10−47m2⋅kg, leaving us to find other 2 roots from

I2i − (Ixx + Izz)Ii + IxxIzz − I2xz = 0

Applying quadratic formula we find 2 solutions

I2 =12

[Ixx + Izz +

√(Ixx − Izz)2 + 4I2xz

]= 1.914 × 10−47m2⋅kg

andI3 =

12

[Ixx + Izz −

√(Ixx − Izz)2 + 4I2xz

]= 1.002 × 10−47m2⋅kg

Finally, follow convention and assign Ia = I3, < Ib = I2, < Ic = I1P. J. Grandinetti Chapter 04: Rotational Motion

Diagonalizing a moment of inertia tensorFinding principal axis system(a) Following convention assign Ia = I3, < Ib = I2, < Ic = I1, and obtain

I∕10−47m2⋅kg =⎛⎜⎜⎝

1.002 0 00 1.914 00 0 2.916

⎞⎟⎟⎠(b) To find orientation of each principal axis substitute principal component values back into(

I − Ia1)

e⃗a = 0,(I − Ib1

)e⃗b = 0,

(I − Ic1

)e⃗c = 0,

Obtain 3 simultaneous equations for each principal axis direction.

These simultaneous equations are usually redundant and only relationships among xi, yi, and zi areobtained, rather than unique values for them.

Normalization of unit vectors gives additional constraint to pin down values for xi, yi, and zi.


Diagonalizing a moment of inertia tensorFinding the principal axis system : I1 case

Starting with I1 = Ic = 2.916 × 10−47m2⋅kg, we obtain

⎛⎜⎜⎝1.576 − 2.916 0 0.441

0 2.916 − 2.916 00.441 0 1.340 − 2.916

⎞⎟⎟⎠⎛⎜⎜⎝

xcyczc

⎞⎟⎟⎠ =⎛⎜⎜⎝−1.34 0 0.441

0 0 00.441 0 −1.576

⎞⎟⎟⎠⎛⎜⎜⎝

xcyczc

⎞⎟⎟⎠ = 0Obtain two equations:

−1.34xc + zc = 00.441xc − 1.576zc = 0

Satisfy two equations with xc = 0, zc = 0 while letting yc take on any possible value.

In other words, solution to these equations is any point on line along y axis.

Define unit vector along direction associated with Ic as e⃗c = e⃗y


Diagonalizing a moment of inertia tensorFinding the principal axis system : I2 caseSwitching to I2 = Ib = 1.914 × 10−47m2⋅kg, we obtain⎛⎜⎜⎝

1.576 − 1.914 0 0.4410 2.916 − 1.914 0

0.441 0 1.340 − 1.914

⎞⎟⎟⎠⎛⎜⎜⎝

xbybzb

⎞⎟⎟⎠ =⎛⎜⎜⎝−0.338 0 0.441

0 1.002 00.441 0 −0.574

⎞⎟⎟⎠⎛⎜⎜⎝

xbybzb

⎞⎟⎟⎠ = 0−0.338xb + 0.441zb = 0

1.002yb = 00.441xb − 0.574zb = 0

Satisfy equations by setting yb = 0, xb = 1 and solving for zb = 0.76644.From this solution we make vector e⃗x + 0.76644 e⃗z pointing in direction associated with Ib.

Using√

x2b + z2b = 1.256 we normalize vector to make unit vector pointing in Ib direction

e⃗b =1

1.256(e⃗x + 0.76644 e⃗z

)= 0.7937 e⃗x + 0.6083 e⃗z


Diagonalizing a moment of inertia tensorFinding the principal axis system : I3 caseFinally for I3 = Ia = 1.002 × 10−47m2⋅kg, we obtain⎛⎜⎜⎝

1.576 − 1.002 0 0.4410 2.916 − 1.002 0

0.441 0 1.340 − 1.002

⎞⎟⎟⎠⎛⎜⎜⎝

xayaza

⎞⎟⎟⎠ =⎛⎜⎜⎝

0.574 0 0.4410 1.914 0

0.441 0 0.338

⎞⎟⎟⎠⎛⎜⎜⎝

xayaza

⎞⎟⎟⎠ = 00.574xa + 0.441za = 0

1, 914ya = 00.441xa + 0.338za = 0

Satisfy equations by setting ya = 0, xa = 1 and solving for za = −1.30159.From this solution make vector e⃗x − 1.30159 e⃗z pointing in direction associated with Ia.

Using√

x2a + z2a = 1.6414 normalize vector to make unit vector pointing in Ia direction

e⃗a =1

1.6414(e⃗x − 1.30159e⃗z

)= 0.6092e⃗x − 0.7930e⃗z


Diagonalizing a moment of inertia tensorFinding the principal axis system

Directions of e⃗a and e⃗b vectors are draw in figure below.

105°

z

x

e⃗a = 0.6092e⃗x − 0.7930e⃗ze⃗b = 0.7937 e⃗x + 0.6083 e⃗z

e⃗c = −e⃗y

Note: sign change in e⃗c = −e⃗y needed to obtain right-handed coordinate system. Direction of e⃗c vectoris directed out of and perpendicular to plane of page.Consistent with our earlier moment of inertia calculation on water.


Rotating frame transformation



When rigid body rotates the “body-fixed” frame becomes a non-inertial frame.

Newton’s laws suggest that they would no longer apply in such a non-inertial (i.e., accelerating orrotating) frame.

There are approaches to obtain correct equations of motion in rotating frame which can be relatedback to some inertial “space-fixed” frame.


Rotating frame transformationIn inertial “space-fixed” frame a vector Q⃗ can be decomposed into its projections onto unit vectors

Q⃗ = Qxe⃗x + Qye⃗y + Qze⃗z

Rate of change of Q⃗ in inertial “space-fixed” frame is(dQ⃗dt

)space

=dQxdt

e⃗x +dQydt

e⃗y +dQzdt

e⃗z

Q⃗ can also be can be decomposed into its projections onto unit vectors in a non-inertial rotating frame

Q⃗ = Qae⃗a + Qbe⃗b + Qce⃗c

Rate of change of Q⃗ in rotating frame is(dQ⃗dt

)rot

=dQadt

e⃗a +dQbdt

e⃗b +dQcdt

e⃗c

How are(

dQ⃗∕dt)

spaceand

(dQ⃗∕dt

)rot

related?


Rotating frame transformationCalculate rate of change of Q⃗ in space-fixed frame in terms of its decomposition in rotating frame.(

dQ⃗dt

)space

=dQadt

e⃗a +dQbdt

e⃗b +dQcdt

e⃗c⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟(

dQ⃗∕dt)

rot

+Qade⃗adt

+ Qbde⃗bdt

+ Qcde⃗cdt

We identify first 3 terms on right as(

dQ⃗∕dt)

rot

Unit vectors in rotating frame, e⃗a, e⃗b, and e⃗c, are time dependent in space-fixed frame fromrotation, so last 3 derivatives are non-zero.

Linear velocity of unit vector as it rotates about a given direction with angular velocity, Ω⃗, is givenby cross product

v⃗i =(

de⃗idt

)space

= Ω⃗ × e⃗i



Last term can be rewritten

Qade⃗adt

+ Qbde⃗bdt

+ Qcde⃗cdt

= QaΩ⃗ × e⃗a + QbΩ⃗ × e⃗b + QcΩ⃗ × e⃗c = Ω⃗×Qae⃗a + Ω⃗×Qbe⃗b + Ω⃗×Qce⃗c = Ω⃗ × Q⃗

Leaving us with (dQ⃗dt

)space

=

(dQ⃗dt

)rot

+ Ω⃗ × Q⃗

This relates rate of change of Q⃗ in space-fixed frame to its rate of change in frame rotating with Ω⃗.


Euler’s equations of motion


Euler’s equations of motionRelationship between rate of change of J⃗ in space- and body-fixed frames is(

dJ⃗dt

)space

=(

dJ⃗dt

)body

+ �⃗� × J⃗

In inertial space-fixed frame (dJ⃗dt

)space

= 𝜏

Combining top 2 equations gives (dJ⃗dt

)body

+ �⃗� × J⃗ = 𝜏

This is Euler’s equation of motion for a rigid body.


Euler’s equations of motion (dJ⃗dt

)body

+ �⃗� × J⃗ = 𝜏

Working in body-fixed frame we relate J⃗ and �⃗� by

J⃗ = 𝜔aIae⃗a + 𝜔bIbe⃗b + 𝜔cIce⃗c

Substituting this into Euler’s equations of motion (top of slide) and get 3 coupled differential equations:

Ia

(d𝜔adt

)body

− (Ib − Ic)𝜔b 𝜔c = 𝜏a

Ib

(d𝜔bdt

)body

− (Ic − Ia)𝜔c 𝜔a = 𝜏b

Ic

(d𝜔cdt

)body

− (Ia − Ib)𝜔a 𝜔b = 𝜏c

What happens if there is no external torques (𝜏 = 0), that is, free rotation?P. J. Grandinetti Chapter 04: Rotational Motion

Euler’s equations of motion – Free rotationWhen there is no external torque, that is, 𝜏 = 0, Euler’s equations describe free rotation of rigid body(

d𝜔adt

)body

=(Ib − Ic)

Ia𝜔b 𝜔c,

(d𝜔bdt

)body

=(Ic − Ia)

Ib𝜔c 𝜔a,

(d𝜔cdt

)body

=(Ia − Ib)

Ic𝜔a 𝜔b

What happens when there’s no external torque and rigid body is rotating about a principal axis?If rigid body with Ia ≠ Ib ≠ Ic is initially rotating about a principal axis, say e⃗c, then 𝜔a = 𝜔b = 0.(

d𝜔adt

)body

=(

d𝜔bdt

)body

=(

d𝜔cdt

)body

= 0

Rate of change of all 3 components of �⃗� is zero:→ magnitude and orientation of �⃗� remains constant in rigid body frame.

Since J⃗ = Ic𝜔ce⃗c:→ magnitude and orientation of J⃗ remains constant in rigid body frame and aligned with �⃗�.

Since J⃗ is conserved for freely rotating rigid body (i.e., 𝜏 = 0):→ magnitude and orientation of J⃗ remains remains constant in space-fixed inertial frame,

and �⃗� is aligned with J⃗P. J. Grandinetti Chapter 04: Rotational Motion

Euler’s equations of motionWhen there is no external torque, that is, 𝜏 = 0, Euler’s equations describe free rotation of rigid body(

d𝜔adt

)body

=(Ib − Ic)

Ia𝜔b 𝜔c,

(d𝜔bdt

)body

=(Ic − Ia)

Ib𝜔c 𝜔a,

(d𝜔cdt

)body

=(Ia − Ib)

Ic𝜔a 𝜔b

What happens when there’s no external torque and rigid body is rotating about an arbitrary axis?

�⃗� becomes time dependent in body-fixed frame.

J⃗ becomes time dependent in body-fixed frame since J⃗ = I ⋅ �⃗�. Remember, I is time-independentin body-fixed frame.

Since J⃗ is conserved for freely rotating rigid body (i.e., 𝜏 = 0):→ J⃗ is time independent in space-fixed inertial frame.→ �⃗� is NOT time independent in space-fixed frame. Remember, I may be time

dependent in space-fixed frame.


Free rotation of oblate symmetric rigid body about an arbitrary axiswhere Ia = Ib = I⟂ < Ic = I||


Euler’s equations of motionFree rotation of oblate symmetric rigid body (Ia = Ib = I⟂ < Ic = I||) about an arbitrary axis(

d𝜔adt

)body

=(Ib − Ic)

Ia𝜔b 𝜔c,

(d𝜔bdt

)body

=(Ic − Ia)

Ib𝜔c 𝜔a,

(d𝜔cdt

)body

=(Ia − Ib)

Ic𝜔a 𝜔b

In case of oblate symmetric body, where Ia = Ib = I⟂ < Ic = I||, Euler’s equations simplify to�̇�a =

(I⟂ − I||)I⟂

𝜔c 𝜔b �̇�b = −(I⟂ − I||)

I⟂𝜔c 𝜔a �̇�c = 0

Using Newton’s dot notation for time derivatives. Define new constant angular frequency, Ω0, as

Ω0 =I⟂ − I||

I⟂𝜔c

and further simplify two coupled differential equations to:

�̇�a = Ω0𝜔b �̇�b = −Ω0𝜔a


Euler’s equations of motionFree rotation of oblate symmetric rigid body (Ia = Ib = I⟂ < Ic = I||) about an arbitrary axis

�̇�a = Ω0𝜔b �̇�b = −Ω0𝜔a

To solve these coupled differential equations we take time derivative again to obtain

�̈�a = Ω0�̇�b and �̈�b = −Ω0�̇�a

Substitute 1st-derivative expressions into 2nd-derivative expressions to get 2 uncoupled 2nd-orderhomogeneous differential equations:

�̈�a + Ω20𝜔a = 0 and �̈�b + Ω20𝜔b = 0

For each equation we can propose a solution of the form

𝜔i(t) = Ai cos kit + Bi sin kit


Euler’s equations of motionFree rotation of oblate symmetric rigid body about an arbitrary axisSubstituting

𝜔i(t) = Ai cos kit + Bi sin kit

into�̈�a + Ω20𝜔a = 0 and �̈�b + Ω

20𝜔b = 0

givesAi(Ω20 − k

2i ) cos kit + Bi(Ω

20 − k

2i ) sin kit = 0

To make equation true for all values of t we set ki = Ω0 and obtain

𝜔a(t) = Aa cosΩ0t + Ba sinΩ0t and 𝜔b(t) = Ab cosΩ0t + Bb sinΩ0t

Choose initial condition: 𝜔a(t = 0) = 𝜔⟂ and 𝜔b(t = 0) = 0 leads toAa = 𝜔⟂, Ba = 0, and Ab = 0, Bb = 𝜔⟂, and finally to

𝜔a(t) = 𝜔⟂ cosΩ0t and 𝜔b(t) = 𝜔⟂ sinΩ0t

Projection of �⃗� onto a-b plane has constant length, 𝜔⟂, and rotates in a-b plane with frequency of Ω0.P. J. Grandinetti Chapter 04: Rotational Motion

Euler’s equations of motionFree rotation of oblate symmetric rigid body about an arbitrary axis

Since 𝜔c is constant then total length of �⃗� isconstant,

𝜔 =√

𝜔2a + 𝜔2b + 𝜔

2c

Define angle �⃗� makes with e⃗c with

tan 𝛼 = 𝜔⟂∕𝜔c

and get

�⃗�(t) = 𝜔[sin 𝛼 cosΩ0t e⃗a − sin 𝛼 sinΩ0t e⃗b + cos 𝛼 e⃗c

]In body-fixed frame �⃗� precesses about e⃗c withangle of 𝛼 at frequency Ω0.

body-fixed frame


Euler’s equations of motionFree rotation of oblate symmetric rigid body about an arbitrary axisMotion of J⃗ in body-fixed frame is

J⃗ = I ⋅ �⃗� = 𝜔[I⟂ sin 𝛼 cosΩ0t e⃗a − I⟂ sin 𝛼 sinΩ0t e⃗b + I|| cos 𝛼 e⃗c]

J⃗ has constant length and precesses around e⃗c at angular frequency Ω0.

body-fixed frame

Angle between J⃗ and e⃗c is

tan 𝜃 =JaJc

=I⟂I|| tan 𝛼

In oblate case I⟂ < I|| and 𝜃 is smaller than 𝛼As both �⃗� and J⃗ rotate around e⃗c all 3 vectorsremain in same plane.


Euler’s equations of motionFree rotation of oblate symmetric rigid body about an arbitrary axis

In space-fixed frame we know that J⃗ remains unchanged, thus �⃗� and e⃗c must precess about J⃗.

body-fixed frame space-fixed frame


Angular MomentumConservation of Angular MomentumOrbital and Spin Angular MomentumOrbital and Spin Angular MomentumRotational energy

Rigid BodiesMoment of inertia tensorRotational energy Classification of moleculesEuler's equations of motion

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