Rigid BodiesMonday, 14 November 2011

Changes in angular momentum are caused by external torques.While the momentum of a massive particle is aligned with its ve-locity, the angular momentum of a rotating rigid body typically isnot aligned with its angular velocity. Even in the absence of externaltorques, this means that the angular velocity vector precesses aboutthe angular momentum vector.

Phys

ics11

1

Perhaps it is wise to recap swily the basics from Physics 24.

1. A force F applied at a position r with respect to a designated origin produces a torqueτ = r × F about that origin.

2. e angular momentum of a collection of point particles about a designated origin isL =∑

αr′α ×p

′

α , where the prime indicates measurement with respect to the designated

origin.

3. e angular momentum may be expressed as the sum of two parts: an orbital part(R ×P) expressed in terms of the center-of-mass position and momentum, and a spinpart (∑α rα×pα) expressed in terms of coordinatesmeasuredwith respect to the centerof mass.

4. In the absence of a net external torque on a system, its (total) angularmomentum in aninertial frame is conserved. Choosing an origin wisely can oen simplify the analysisof a problem.

5. e rotational kinetic energy of a spinning rigid body may be expressed as T = 12 Iω2,

where I is the moment of inertia for rotation about the spin axis.

6. e moment of inertia for rotation about an axis is given by the sum I = ∑α mαr2α ,where rα is the perpendicular distance from the axis to mass point α.

e last two points have been highlighted in red because they will need to be modiedhere to handle situations in which the axis of rotation itself changes orientation during themotion or in which the instantaneous axis of rotation does not coincide with an axis of sym-metry of the object.We will not have time for an exhaustive treatment of rotating rigid bodies; their motions

can be quite complicated. We will, however, develop the correct equations of motion andanalyze a few cases of interest and importance.

Physics 111 1 Peter N. Saeta

2. KINETIC ENERGY OF A RIGID BODY

1. e Angular Velocity and Angular Momentum Aren’t Necessarily Parallel

m

a

a

m

Figure 1: An asymmetric rotor rotatesabout a vertical axis. Describe the forcesthat must be supplied by the bearings(shown in blue) to keep the rotor rotatingabout a vertical axis.

Consider the asymmetric rotor illustrated inFig. 1. As the rotor spins about the vertical axis, thebearings (shown in blue triangles) must supply var-ious forces to keep the axis vertical. How do thoseforces vary in time? Do they supply a torque to therotor? If so, what does this imply about the rotor’sangular momentum?

e asymmetric rotor illustrates the perhapssurprising fact that the angular velocity need not beparallel to the angular momentum, and that a rigidbody cannot spin placidly about just any axis weplease. In the absence of external torques, a system’sangular momentum is conserved, of course. Un-like conservation of total momentum, however, theconservation of angular momentum can give rise tosome remarkably complicated and counterintuitivemotions of rigid bodies. As with much of “Newto-nian” mechanics, fundamental advances in describ-ing the rotation of rigid bodies were made by the

Swissmathematician and physicist, Leonhard Euler (1707–1783). Euler found general expres-sions for the kinetic energy and angular momentum of a rigid body, in terms of the momentof inertia tensor, and derived the equations of motion of symmetric and asymmetric tops.We will derive Euler’s equations for a torque-free top and use them to explain the curiousbehavior of tossed tennis rackets.

2. Kinetic Energy of a Rigid Body

We showed that the kinetic energy of a system of mass points mα may be expressed asthe sum of two terms: 12MV 2 and 12 ∑α mα r2α , where M = ∑α mα is the total mass, V is thevelocity of the center of mass, and rα is the position of particle α with respect to the centerof mass. We now apply this for the special case that the mass points compose a rigid body.In this case, the two parts of the kinetic energy may be expressed

Ttrans =12∑α

mαV 2 = 12

MV 2 (1)

Trot =12∑α

mα(ω × rα)2 (2)

Peter N. Saeta 2 Physics 111

2. KINETIC ENERGY OF A RIGID BODY

Dene the Levi-Civita є symbol1 by

єi jk=

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

1 i jk = 123 = x yz or cyclic permutations−1 i jk = 132 = xzy or cyclic permutations0 otherwise

(3)

so that, using the Einstein summation convention, we can write

(A × B)i= єi jk AjBk (4)

We now use an important identity for the Levi-Civita є symbol,

єi jkєi lm= δ jl δkm

− δ jmδkl (5)

which is proved at the end of these notes. Using these denitions and results, we can derive

∣A × B∣2 = (єi jk AjBk)(єi lmAl Bm

) = (δ jl δkm− δ jmδkl

)AjBk Al Bm

= Ai Ai B jB j− Ai Bi AjB j

= A2B2 − (A ⋅ B)2 (6)

is expression allows us to rewrite Eq. (2) as

Trot =12∑α

mα(ωiωirα jrα j− ωiω jrαirα j) =

12

ωiω j∑α

mα(r2αδ i j− rαirα j

) (7)

We dene the sum in this expression to be the inertia tensor (or moment of inertia tensor)I i j,

I i j=∑

αmα(r2αδ i j

− rαirα j) (8)

in terms of which we may write the kinetic energy as

Trot =12

I i jωiω j (9)

Exercise 1 Compute the components of the inertia tensor I i j for the asymmetric rotor shownhere, assuming that themass of the connecting rods is negligible compared to the two spheres.Neglect the spatial extent of the two masses.

1Tullio Levi-Civita (1873–1941) was an Italian mathematician whose 1900 book with Ricci-Curbastro onthe theory of tensors, Methodes de calcul dierentiel absolu et leurs applications, was used by Einstein to learnthe tensor calculus. When asked much later what he liked best about Italy, Einstein said “spaghetti and Levi-Civita.” Levi-Civita corresponded with Einstein, worked on general relativity and later on quantum mechanicsand Dirac’s equation. Because he was Jewish, Levi-Civita was stripped of his professorship in 1938 and died inRome in 1941.

Physics 111 3 Peter N. Saeta

3. ANGULARMOMENTUM

m

a

a

m

z

x

b

b

3. Angular Momentum

Just as the kinetic energy of a system of mass points may be expressed as the sum ofthe translational kinetic energy of a mass point M = ∑α mα at the center of mass and thekinetic energy of each constituent mass point with respect to the center of mass, the angularmomentum about any chosen origin may be similarly factored:

L ≡∑αr′α × p

′

α = R × P

²orbital

+∑αrα × pα

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶spin

(10)

In this expression, R is the position of the center of mass with respect to the chosen ori-gin, and P is the total momentum of the particles. e nal summation involves quantitiesmeasured with respect to the center of mass.

Exercise 2 Prove Eq. (10).

Peter N. Saeta 4 Physics 111

4. TENSORS

We now seek to evaluate the spin angular momentum on the right-hand side of Eq. (10)for a rigid body, for which rα = ω × rα .erefore,

Lspin =∑αrα × pα =∑

αmαrα × (ω × rα)

Once again, we employ the Levi-Civita symbol to simplify this expression:

Li=∑

αmαєi jkr jєklmωl rm

=∑α

mαєki jєklmωl rα jrαm = ωl∑α

mα(δ i l r2α − rαirα j) = I i jω j

at is, the (spin) angular momentum is given by the dot product of the moment of inertiatensor and the angular velocity,

Li= I i jω j or L =

←→I ⋅ ω (11)

Comparing to Eq. (9), we see that we may also represent the rotational kinetic energy as

Trot =12

ωi Li=12ω ⋅ L (12)

4. Tensors

What’s a tensor? A simple answer is that a tensor is an object that eats a vector andreturns a vector. It is linear: if you give it an input vector twice as long, you get out an outputvector twice as long as the original output vector and pointing in the same direction. It is themost general linear beast that maps vectors into other vectors that need not point in the samedirection.

Physics 111 5 Peter N. Saeta

5. INERTIA TENSOR

Another answer is that a tensor is a generalization of a vector. In fact a vector is a rst-rank tensor. So, what’s a vector? A vector has a magnitude and a direction. It may be rep-resented in a particular coordinate system by projecting onto an orthogonal set of basis unitvectors, Ai = A ⋅ ei , but the vector has a life independent of any particular coordinate repre-sentation. Rotating the coordinate axes changes the unit vectors ei and therefore the compo-nents of the vector,Ai , in the rotated coordinate system, but does not change the vector—onlyits representation.is is the passive view of rotation.Alternatively, rotating the vector in the opposite direction while keeping the coordinate

system xed produces the same change in the coordinate representation of the vector (thecomponents Ai) as the passive rotation.is is called an active rotation. Some authors preferto focus on active rotations; others on passive rotations. In either case, a quantity is a vectorif it behaves under rotations as a vector does.And how exactly does a vector behave? Rotations preserve length (Ai iAi = Aj′Aj′),

and they preserve the handedness of the coordinate system. In summation notation, thecomponents of the vector A in the rotated (primed) coordinate system are given in terms ofthe unrotated (unprimed) system by

Ai′= λi′ jAj (13)

where λi′ j = cos(x i′ , x j) = ei′ ⋅ e j. e denition of the direction cosine makes it clear thatλi j′ = λ j′ i .Furthermore, the nature of the inverse transformation is clear: rotate by the same angle in

the opposite direction.e transformationmatrix thus satises the orthogonality condition

λi jλk j= δ ik

= λ ji λ jk (14)

How does a tensor behave?e inertia tensor for a single mass point is I i j = m(r2δ i j −r ir j), which contains two vector components. On rotating the coordinate system, we needto rotate each component. e tensor analog to the transformation equation for vectors,Eq. (13), is thus

Bi′ j′= λi′k λ j′ l Bkl (15)

at is, we need one λ for each tensor index. It is straightforward to generalize this expressionto tensors of arbitrary rank.

5. Inertia Tensor

e inertia tensor dened by Eq. (8) has an important property that greatly simpliesthe description of rotating rigid bodies: we can always nd a body coordinate system thatdiagonalizes the inertia tensor. In that coordinate system, the only nonzero elements of I i j

are those with i = j. is means that if the body rotates about one of these principal axes,the angular velocity and the angular momentum are aligned. e proof follows closely thediscussion of linear algebra in the notes on coupled oscillators. e principal moments ofinertia are obtained by setting

∣I i j− Iδ i j

∣ = 0 (16)

Peter N. Saeta 6 Physics 111

6. EULERIAN ANGLES

where I i j is the inertia tensor in any Cartesian coordinate system. e principle momentsare always real and positive, and the principal axes are automatically orthogonal if the threeprincipal moments are distinct, or may be chosen to be orthogonal if two or more momentsare identical.2 erefore, we may choose to use the principal axes to express the (spin) an-gular momentum (about the center of mass) as

Li= I iωi (17)

where I i ≡ I ii .

6. Eulerian Angles

Weneed three angles to describe the orientation of a rigid body. One common denitionof these angles was provided by Euler and is illustrated in Fig. 2. Because the gure uses lowerindices, I will imitate its conventions in this section. We rst rotate about the x3′ = z axisthrough angle ϕ, which takes Cartesian axes x′ into axes x′′, as shown in Fig. 2(a). is

2In Physics 116 you will see a more general result thatHermitianmatrices, which satisfyH ji = (Hi j)∗, havereal eigenvalues.

Figure 2: Eulerian angles. Figure from Classical Dynamics of Particles and Systems, Fourth Edition.Marion and onton (Harcourt, Fort Worth, 1995) 432.

Physics 111 7 Peter N. Saeta

6.1 Angular velocity in Eulerian angles 6. EULERIAN ANGLES

moves x′′1 into position so that we can rotate through angle θ about x′′1 , tipping the z axisaway from the vertical to produce the x′′′i axes, as shown in Fig. 2(b). Finally, we rotate onceagain about the (local) z axis, this time through angle ψ, to yield the nal (unprimed) axes.e complete rotation may be described by a 3 × 3 rotation matrix λ given by λ = λψλθλϕ,where

λϕ =⎛⎜⎝

cos ϕ sin ϕ 0− sin ϕ cos ϕ 00 0 1

⎞⎟⎠

λθ=⎛⎜⎝

1 0 00 cos θ sin θ0 − sin θ cos θ

⎞⎟⎠

λψ=⎛⎜⎝

cosψ sinψ 0− sinψ cosψ 00 0 1

⎞⎟⎠

yielding

λ =⎛⎜⎝

cosψ cos ϕ − cos θ sin ϕ sinψ cosψ sin ϕ + cos θ cos ϕ sinψ sinψ sin θ− sinψ cos ϕ − cos θ sin ϕ cosψ − sinψ sin ϕ + cos θ cos ϕ cosψ cosψ sin θ

sin θ sin ϕ − sin θ cos ϕ cos θ

⎞⎟⎠

Although rotations through nite angles do not commute, and are therefore not vectors,innitesimal rotations do commute and may be represented by vectors aligned with the axisof rotation. erefore, the angular velocity is a vector. We would like to represent the an-gular velocity vector in the body-centered coordinates [the ones shown without primes inFig. 2(c)].

6.1 Angular velocity in Eulerian angles

Wewould like to use the Eulerian angles as generalized coordinates. Since, T = 12 I i jωiω j =

12 I i(ωi)2, we will need to express ωi in terms of ϕ, θ, and ψ. Referring to Fig. 2(c), we get

ψ = ψ e3 (18)

θ = θ(cosψ e1 − sinψ e2) (19)

ϕ = ϕ[sin θ(sinψ e1 + cosψ e2) + cos θ e3] (20)

erefore, the components of the angular velocity in the body coordinate system are

ω1 = ϕ sin θ sinψ + θ cosψ (21)

ω2 = ϕ sin θ cosψ − θ sinψ (22)

ω3 = ϕ cos θ + ψ (23)

Peter N. Saeta 8 Physics 111

6. EULERIAN ANGLES 6.2 Euler’s equations

Exercise 3 Conrm equations 18 through 20.

6.2 Euler’s equations

In the absence of external forces, L = T , so Lagrange’s equation for ψ becomes

ddt

(∂T∂ψ

) −∂T∂ψ

= 0 (24)

where the kinetic energy isT =

12

I i(ωi

)2 (25)

We can evaluate the partial derivatives using the chain rule,

∂T∂ψ

=∂T∂ωi

∂ωi

∂ψand

∂T∂ψ

=∂T∂ωi

∂ωi

∂ψ

Using Eqs. 21–23, we get

∂ω1

∂ψ= ϕ sin θ cosψ − θ sinψ = ω2 ∂ω1

∂ψ= 0

∂ω2

∂ψ= −ϕ sin θ sinψ − θ cosψ = −ω1 ∂ω2

∂ψ= 0

∂ω3

∂ψ= 0

∂ω3

∂ψ= 1

Substituting into Lagrange’s equation gives

ddt

(I3ω3) − [I1ω1ω2 + I2ω2(−ω1)] = 0

or

ω3 = I1 − I2

I3ω1ω2 (26)

Physics 111 9 Peter N. Saeta

6.3 Tennis Racket 6. EULERIAN ANGLES

Since the choice of which principal axis to call number 3 is entirely arbitrary, we can cyclicallypermute the indices to obtain two more equations,

ω1 = I2 − I3

I1ω2ω3 and ω2 = I3 − I1

I2ω3ω1 (27)

6.3 Tennis Racket

We can now understand the behavior of the thrown tennis racket. In ight, its centerof mass falls with constant acceleration g, but no external torque acts (if we may neglectair resistance).e racket’s principal axes are along the handle (moment I1), perpendicularto the handle and parallel to the face (moment I2), and perpendicular to both the handleand the face (moment I3). We will toss the racket attempting to spin it about one of theprincipal axes—say x1—but will inevitably impart some small angular velocity about theother two principal axes. If the rotation is stable, the magnitude of the angular velocity aboutthe x2 and x3 axes will oscillate and remain small; if it is unstable, their magnitudes will growexponentially.We will take as the initial angular velocity,

ω = ω1 e1 + δ2 e2 + δ3 e3

where ∣δ2∣ ≪ ω1 and ∣δ3∣ ≪ ω1. From Eq. (27),

ω2 = (I3 − I1

I2ω1)ω3

ω3 = (I1 − I2

I3ω1)ω2

Dierentiating the rst equation (treating ω1 as a constant), and using the second equationto replace ω3, we get

ω2 ≈ (I3 − I1

I2ω1)(

I1 − I2

I3ω1)ω2 = (

(I3 − I1)(I1 − I2)I2I3

(ω1)2)ω2

We can also dierentiate the second and use the rst to eliminate ω2. Let

(Ω1)2 =(I3 − I1)(I2 − I1)

I2I3(ω1)2

enω2 + (Ω1)2ω2 = 0 and ω3 + (Ω1)2ω3 = 0

As dened above, I1 < I2 < I3, so (Ω1)2 > 0 and the solutions for ω2(t) and ω3(t) are bothoscillatory. Hence, rotation about the principal axis with the smallest moment of inertia isstable. Of course, we can cyclically permute the indices to investigate stability for rotation

Peter N. Saeta 10 Physics 111

8. PROOF OF EQ. (5)

about the other two axes, as well. We nd that rotation about x3 is stable, but when we seekto spin about the axis with the intermediate moment of inertia,

(Ω2)2 =(I1 − I2)(I3 − I2)

I3I1(ω2)2 < 0

e angular velocity componentsω1 andω3 depend exponentially on the time, and somotionabout x2 is unstable.

7. Summary

• e kinetic energy of a system of mass points may be decomposed as the sum of thecenter of mass kinetic energy (all the mass concentrated at the center of mass andmoving with the center-of-mass velocity) and the kinetic energy with respect to thecenter of mass. For a rigid body, all the particles undergo a coordinated motion withrespect to the center of mass, allowing us to simplify the expression for the kineticenergy.

• Finite rotations do not commute, but innitesimal rotations do commute. Hence, an-gular velocity is a vector.

• In general, the angular momentum of a rigid body is not aligned with its angularvelocity. Strange, but true. Nonetheless, the angular momentum (about the cen-ter of mass) is proportional to the angular velocity, Li = I i jω j. e quantity I i j =

∑α mα(r2αδ i j − rαirα j) is called the inertia tensor.• Because it is a real, symmetric second-rank tensor (which means it is equivalent to amatrix), it is always possible to diagonalize the inertia tensor.is means that one canalways nd an orthogonal coordinate system in which the angular momentum andangular velocity are aligned for pure rotation about each of the coordinate axes.

• A rigid body in torque-free motion rotates stably about two of three principal axes,but unstably about the principal axis of intermediate moment of inertia.

• A symmetric rigid body spinning about its axis of symmetry and subject to an externalgravitational torque precesses about the line of gravitational force. Given the appro-priate initial condition, it precesses uniformly (in the absence of a frictional torque).Under more general initial conditions, the spin axis nutates, tracing curlicues as theaxis precesses about the vertical.

8. Proof of Eq. (5)

We wish to prove thatєi jkєi lm

= δ jl δkm− δ jmδkl

First, note that the expression on the le is antisymmetric under the exchanges j ↔ k andl ↔ m. So is the expression on the right. For the product on the le to be nonzero, each єmust be nonzero, which means that i jk must be some permutation of 123. Let us supposethat j = 2 and k = 3.en the only nonzero term in єi jk is when i = 1. For the second epsilonsymbol to be nonzero, therefore, lmmust be either 2 3 or 3 2. In the former case both epsilons

Physics 111 11 Peter N. Saeta

8. PROOF OF EQ. (5)

are 1 so the product is 1; in the latter, the second is −1, as is the product. Does this work outon the right-hand side?Let’s consider rst when lm = 2 3.en the right-hand side is

δ22δ33 − δ23δ32 = 1 − 0 = 1

as it must be. If lm = 3 2, then the right-hand side becomes

δ23δ32 − δ22δ33 = 0 − 1 = −1

also as it must be.ere is nothing sacred about 2 and 3, of course. Using the permutation properties of the

Levi-Civita symbol, we can generate all possible pairs of 2 distinct indices. We have alreadyconrmed that both sides are antisymmetric under these permutations, so the identity isproven.

Peter N. Saeta 12 Physics 111