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PHYS 705: Classical Mechanics Introduction and Derivative of Moment of Inertia Tensor 2

L and N depends on the Choice of Origin

Consider a particle moving in a circle with constant v. If we pick the origin O to be at the center of the circle and calculate L wrt this origin, L is a fixed quantity. L v m O r L r m v

In order for the particle to actually maintain this circular motion, there must be a centripetal force acting on it.

v But, notice that this does not give rise m O to any torque since Fc

N rFc 0 and L N 0

(Note: r and F c are collinear) 3

L and N depends on the Choice of Origin

Now, consider the same situation but with the origin O chosen to be below the plane of rotation.

L r m v points as shown v (into page) m Most importantly, it is no longer a dL r constant, i.e, N 0 O dt And, there must be a torque acting on it.

Fc N rFc 0 (out of page) N m (L changes in direction as given by N) r L O L and N depends on the choice of the origin 4

Non-Collinear Relation between L and w

When a rigid object spins about one of its axes of symmetry, the dynamical equations are simple: 1 T Iw 2 L Iω rot 2 ω w nˆ where - the moment of inertia I is simply a scalar - L and w point in the SAME direction

However, this is not generally the case when a rigid body is NOT rotating about one of its Principle Axes of rotation. 5

Non-Collinear Relation between L and w

Let consider the following dumbbells rotating about some axis which is not an axis of symmetry.

v1 - Here, the total angular momentum is, m1 L L1 L2 ω L1 r L m1r 1 v 1 m2r 2 v 2 L 1 2 - Notice that L and w do NOT point in the same direction. r - In fact, L is changing in time so that there 2 must be a torque: m (In general, one just can’t 2 v dL expect an object to rotate 2 N about a particular chosen dt axis in space w/o N.) 1 -The rotational KE doesn’t work out to be quite T I w 2 either. rot 2 6

Derivation of L and T for a Rigid Body

At any given instant of a rigid body motion, we can consider the body to be performing an infinitesimal rotation about some fixed point O on the rigid body

Let say that in the “fixed” frame, O is moving with velocity v0

Now, let consider another point in the body…

r body The point is located in the “body” frame by r ω O and its velocity in the “fixed” frame is given by, fixed The origin of the “body” vf v0 ω r axis is moving with v0 (r is fixed in the “body” frame – rigid body) The “body” frame is also rotating about ω (note: we will also drop ‘ for body frame variables from now on) 7

Derivation of T for a Rigid Body

1 2 A particle (label it by a) has T mv vf v0 ω r a 2 f 1 T mv ωr v ωr a 2 0 0 1 1 2 Tmmv2 v ωr m ωr a 20 0 2

v0 ω m r Now, summing over all particles, we have

12 1 2 Ttotal ma v0 v 0 ω mar a ma ωr a 2a a 2 a

M MRcm 8

Derivation of T for a Rigid Body

Now, let’s pick the origin of the “body” frame to be at the CM so that Rcm 0

Then, we have the splitting of the total kinetic energy

12 1 2 TMtotalv0 ma ωr a TT CM rot 2 2 a

KE of total M KE about the CM (the TCM T as if all at CM interesting part) rot

Writing out the cross product:

22 2 2 ωra ωrωr a aaa w r ωr

Recall AB CD ACBD ADBC 9

Derivation of T for a Rigid Body

a a a T Writing out the components for r a x 1 ,, x 2 x 3 explicitly, we have 1 aa a a Tmrota ww ii xx kk ww ii xx jj (sum rule) 2

Inserting wi w jij 1 T mww xxaa ww xx aa rot2 a ijijkk ijij 1 aa aa Trotw i ma ijkk xxxx ij w j 2 aa aa Defining the Moment of Inertia Tensor by Iij ma ijkk xxxx ij a 1 T w I w rot2 iij j 10

Derivation of T for a Rigid Body

Notes: 1. Summary:

Iij is defined relative to some origin O and a particular choice of axes through the origin.

aa aa Iij ma ijkk xxxx ij a where the position of the ath particle is measured relative to O at

a a a T ra r1,, r 2 r 3 11

Derivation of T for a Rigid Body

2. One could, in general, calculate I ij for some other origin O which might 1 not be the CM of the rigid body and T can still be given as T w I w rot rot2 iij j But, if O is NOT at CM, then the total KE cannot be separated into:

(The M v 0 ω R CM terms in full expression Ttotal T CM T rot for T total might not be zero.)

2+. If O for the body axes is fixed in the fixed frame, then v 0 0 and

1 2 1 2 Ttotal ma v0 v0 ω ma r a maω r a 2 a a 2 a 1 TT w I w total rot2 iij j There are two natural choices for O: CM and a fixed point of rotation. 12

Derivation of T for a Rigid Body 3. There is a generalization of the Parallel Axis Theorem that allows one to relate the Momentum of Inertia Tensor wrt the CM to one relative to another parallelly translated origin. 2 Iij J ij M a ij a ij a where is moment of inertia tensor about the CM of the rigid body Iij

Jij is moment of inertia tensor about some other point O T a a1,, a 2 a 3 is a vector pointing from O to CM

a CM O NOTE: the two sets of axes must be PARALLEL 13

Derivation of L for a Rigid Body

Now, we look at the general expression for the angular momentum of a rigid body… - Here, we again assume that the rigid body is O body r rotating at a constant angular velocity ω with ω one point stationary calling this O.

- The origin of the “body” frame is assumed to fixed be fixed, i.e. v0 0

- Then, the velocity vector v f as measured in the “fixed” frame is: v f ω r

(note: with v 0 0 , contribution for the velocity vector comes from the rotation of the body axes only [the explicit subscript f will be suppressed].) - Then, the total ang momentum about O as measured in the “fixed” frame is:

Lram aa v r aa m ω r a (sum) 14

Derivation of L for a Rigid Body

- Using the vector identity: A BC BAC CAB - We have: 2 L raa mωr a mr aaaa ω rrω (sum)

a a a T -Again, writing out the components for r a x 1 ,, x 2 x 3 explicitly, we have

Lmw rraa rr a a w ia ikk ijj (sum)

- Inserting wi w jij Lmw rraa rr aaw mrraa rr aa w ia j kk ij ijj a kkij ij j - Recognizing the pre-factor as the momentum of inertia tensor, I m xxxxaa aa ij a ijkk ij Li I ijw j a 15

T and L for a Rigid Body

Summary: 1 T w I w ( T is a scalar and the sum is over both i and j) rot2 iij j rot ( L is a vector and the sum is only over j) Li I ijw j

aa aa ( is a tensor and the sum is only over a) Iij ma ijkk xxxx ij Iij a Notes: - The moment of inertia I is no longer a scalar as in the simpler case: L Iω

- As we have seen in our earlier example, L and w are not necessary collinear !

- However, we can still formally write, 1 1 T ωL ωIω rot 2 2 16

Moment of Inertia Tensor

We will see now that our old familiar scalar I is still in the equation.

Let pick nˆ to align with the axis of rotation, then we can write ω w nˆ 1 Substituting w w n into T w I w , we have i i rot2 iij j w 2 T n I n rot2 iijj w 2 aa aa Trot n i ma ijkk x x x ij x n j 2 2 w aa aa ma xkk x n ijij n n ii x x jj n 2 1 2 w 2 T mr 2 r nˆ (still sum over a) rot 2 aa a 17

Moment of Inertia Tensor 2 w 2 T mr 2 r nˆ rot 2 aa a da Now, consider the picture to the right, nˆ O ra Notice that d a is the distance to the

position r a with respect to the chosen axis of rotation nˆ and

2 2 2 ra d a r a nˆ

So, we can write,

2 w 2 Trot mad a (The blue term is our usual definition 2 a for the scalar moment of inertia) 18

Moment of Inertia Tensor

Comments: da nˆ 1. The “projection” n i I ij n j nIn ˆ ˆ is the O r Moment of Inertia about the axis of rotation nˆ a going through the origin O

2. The tensor itself I ij has in it information for ALL I’s in any direction.

3. A particular representation of I ij depends on the choice of origin and axes chosen in calculating it. 4. (later) There will also be a generalization of the parallel axis thm: ω I ICM M a2 a a ij ij ij ij M CM a 19

An Example: Moment of Inertia for a Cube

x3 b b A cube with uniform density r , mass M, and b side b is placed as show with its corner located O at the origin O x2 The finite sum for I ij is now an integral, x1 I r2 x x r dV ij ij ij V As an exercise, let evaluate the components explicitly,

b b b Ir xxx2 2 2 xdxdxdx 2 11 1 2 3 1 1 2 3 0 0 0 b b b r x2 x 2 dx dx dx 2 3 2 3 1 0 0 0 20

An Example: Moment of Inertia for a Cube x b b b 3 Ir x2 x 2 dx dx dx b 11 2 3 2 3 1 b 0 0 0 b O b b b b b r x2 dx dx x 2 dx dx dx 2 2 3 3 3 2 1 x2 0 0 0 0 0 x1 b bb3 b b 3 b b4 2 r dx dx dx r2 dx r b5 3 2 1 1 0 03 0 3 0 3 3 2 Substituting M r b 3 , we then have, I Mb2 11 3

Since the cube is symmetric, similar calculations will result for I22, I 33 2 I Mb2 ii 3 21

An Example: Moment of Inertia for a Cube

Now, consider off-diagonal terms, x3 b b Ir x x dV i j ij ij b V O b b b r x x dx dx dx ijij k (indices don’t sum here) x2 0 0 0 x1 b b b r x dx x dx dx ii jj k 0 0 0 b b b2 b b2 b 2 b 5 r xjj dx dx k rdxk r 2 2 4 0 0 2 0

1 I Mb2 i j ij 4 22

An Example: Moment of Inertia for a Cube

Summarizing, we have, x3 b b 8 3 3 b Mb2 O I 3 8 3 ij x 12 2 3 3 8 x1 23

Principal Axes & Principal Moments of Inertia

Notice that since I ij is always a real symmetric matrix, one can always diagonalize it. 8 3 3 11 0 0 2 2 Mbdiagonalize Mb I 3 8 3 0 11 0 (shown later) ij 12 12 3 3 8 0 0 2

In the language of linear algebra, this corresponds to the rotation of our system (with the same origin) to a set of orthogonal basis corresponding to the eigenvectors of I . ij These axes are called the Principal Axes.

And, the diagonal elements will be the eigenvalues of I ij .

They are the Principal Moment of Inertia. 24

Principal Axes & Principal Moments of Inertia

- If I ij is NOT diagonalized, we will in general have L and w not collinear and the Kinetic energy as a tensor product, i.e., 1 L I w and T w I w i ij j rot2 iij j

- If I ij is diagonalized, then the situation simplifies and we get our regular

simpler results for L and T rot :

1 2 LI1 11w 1,, LI 2 22 w 2 LI 3 33 w 3 and T I w rot2 ii i - The physical significance is that when there is no torque acting on the system, a rigid body when tossed into space will spin around one of these principal axes. It can be shown that rotation about the principal axes with the largest and the smallest principal moments will be stable while rotation about the middle one will not. (HW prob) 25

Principal Axes & Principal Moments

x3 Now, back to our example for a cube b b with its corner at the origin, we had b 8 3 3 O Mb2 I 3 8 3 x2 ij 12 x 3 3 8 1

The characteristic polynomial (without the common pre-factor) is: 8 3 3 3 2 det 3 8 3 0 24 165 242 0 3 3 8

Solving for gives: 11,11,2 26

Principal Axes & Principal Moments

The eigenvectors e ee 1 ,, 2 e 3 can then be evaluated by solving,

8 3 3 e1 e 1 In this basis, the Moment of 3 8 3 e e Inertia tensor is diagonalized: 2 2 3 3 8 e3 e 3 11 0 0 Mb2 For the solved eigenvalues, we have: I 0 11 0 ij 12 12 e 1 1,1,1 0 0 2

2,311 e 2 1,1,0 (degenerate eigenspace)

e3 1,0,1 27

Principal Axes & Principal Moments

-The eigenvector (Principle Axis) e1 1,1,1 x3 corresponding to 1 2 is shown: e1

- The other two eigenvectors can be

any vectors in the plane to e 1 . x2

-Note: x1 S ee e 1. 1 2 3 with the e ' s as

columns is a similarity transform which diagonizes Iij ,

8 3 3 11 0 0 2 2 Mb1 Mb I 3 8 3 ISIS 0 11 0 diag 12 12 3 3 8 0 0 2 28

Principal Axes & Principal Moments

2. For a given origin O, I ij contains ALL the moments of inertia through ANY axes passing through O and for a given axis with unit vector nˆ

Inˆ nInˆ ˆ n iijj I n

3. Among all these possibilities, there are three Principle Axes given by the

eigenvectors of I ij and their associated Principle Moments of Inertia.

4. One can choose to represent I ij using these Principle Axes as the coordinate basis and the resulting Moment of Inertia tensor will be diagonalized.

5. Otherwise, I ij will in general have non-zero off-diagonal elements but all of them with the same origin are related by a similarity transformation. 29

Principal Axes & Principal Moments

(no sum) 6. The Principal Axes are special in the sense that:

- if the RB is spinning along one of the Principal Axis, then Li I iiiw - (later) it can be shown that those are the preferred directions for the object to spin around when there is no external torque ! 30

Parallel Axis Theorem for Inertia Tensor

So, we know that all I ij with the same origin are related by a rotation. Now, how are they related if they are calculated using different origin?

x3 x3 Iij is moment of inertia tensor about the CM of the rigid

r r body a CM x J is moment of inertia tensor about Q Q 2 ij x 1 T x2 a a1,, a 2 a 3 is vector pointing from Q to CM x1 NOTE: the two sets of axes must be PARALLEL Let calculate the inertia tensor in the Q frame

aa aa (sum) Jij ma ijkk xx xx ij a 31

Parallel Axis Theorem for Inertia Tensor

x x 3 3 a a a a - Note, we have, r ar or xi ax i i r r - Plugging it in our expression for , a Jij CM x2 Q aa aa x J m xx xx 1 x ij a ijkk ij 2 a

x1 a a a a Jij ma ijkkkk axax axax ii jj a aaa aa aa ma ijak a k 2 a kkxxkkx aia j a ijx a j x i xx ij a a aa a maijxxk k xx i j m a ij akka a i a j a a a a a ma 2ijkkax ax ij axj i a 32

Parallel Axis Theorem for Inertia Tensor

a x3 x3 - Now, since all the x i are evaluated with the origin at CM,

r r we have, a a ma xk 0 CM x Q 2 a x 1 x - We also have m M being the total mass, 2 a a x1 a a aa Jij ma ij xx k k xix j ma ija k ak ai a j a a a a a 2ijka mxa k ai mxa j aj mxa i a a a - So the big complicated equation reduces simply to,

Jij Iij Mij akka a i a j

wrt Q wrt CM 33

Parallel Axis Theorem for Inertia Tensor

- Example: Cube again… Now, we would like to b b calculate I with origin at CM, ij b a Q - From geometry, the displacement from the corner to the CM

T CM is b b b 3 a ,, and a2 b 2 2 2 2 4 - Also, recall that we have the inertia tensor wrt to Q (corner) as

8 3 3 Mb2 J 3 8 3 ij 12 3 3 8 34

Parallel Axis Theorem for Inertia Tensor

- Applying the Parallel Axis Theorem: b b

Jij I ij M ijkk a a a ij a b a Q Diagonal elements: CM 2 2 2 2 8Mb 3 b b Iii J ii M a a ii a M 12 4 4 8 3 3 8Mb2 Mb 2 Mb 2 Mb2 J 3 8 3 ij 12 2 6 12 3 3 8 Off-diagonal elements: 3Mb2 b 2 Iij J ij M a ij a M 12 4 3Mb2 Mb 2 0 12 4 35

Parallel Axis Theorem for Inertia Tensor

- Putting everything together, we have: b b 1 0 0 b Mb2 a I 0 1 0 Q ij CM 6 0 0 1 Notes:

1. Since I ij is basically proportional to the identity matrix, Iij will be the same by ANY similarity transformation (any rotation about CM). Mb2 Mb 2 ISIS' 1 SS 1 I ij ikkllj6 ikkllj 6 ij ij

2. ANY set of axes through the CM will give the same I ij for this cube.

3. Basically, wrt its CM, the cube looks like a sphere.