Georgi, Howard. Physics 16 - Mechanics and Special Relativity (англ.)

lectur

e 17

Topics:

Where are we now?

Impulse and rigid bodies

The moment of inertia tensor

~

L and ~ω

The body frame

Where are we now?

We are poised to take the next big step into understanding the motion of rigid bodies. By putting

together the ideas of impulsive torque and the angular velocity vector, we can solve some more

interesting problems. To do this, however, we must introduce and understand the moment of

inertia tensor.

Impulse and rigid bodies

So to follow up the notion that thinking about impulsive torques is easier to understand, we are

going to spend some time dealing with the following question. Suppose that we are out in space,

and we come upon a light rigid rectangular frame with masses at the corners, as shown:

....................................................

.

..

.

y

x

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

m

.

.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

×

~

F

(1)

It is ﬂoating at rest, say in the x-y plane, when we hit one of the masses with a hammer, applying

a large force in the −z direction for a very short time, as shown. This produces an impulsive force

and torque on the rigid frame. The question is, immediately after the hammer blow, what is the

velocity of each of the masses? We will not actually solve this problem until next time, but today,

1

8

we will start to explore a crucial component of the answer — we will discuss the moment of inertia

tensor and the relation between the angular momentum

~

L and the angular velocity ~ω.

But ﬁrst you might ask — Why have I formulated the problem in this peculiar way? Why not

just ask for the trajectory of each of the masses for all time after the hammer blow? The answer is

that these trajectories are MUCH harder to ﬁnd and to understand than the velocities I have asked

about. I will begin by explaining why this is so. You may guess that the answer has something to do

with impulse — the fact that we have applied the force in a very short time so that the frame does

not have a chance to move while the force is being applied. That is correct, but it is only part of the

difference. Complications arise because the direction of the angular momentum after the hammer

blow does not coincide with the instantaneous axis of rotation of the body. When we calculate

~

L

in terms of ~ω and the parameters m

j

and ~r

j

that describe the rigid body, it just turns out that except

in very special circumstances,

~

L and ~ω are not in the same direction. This fact will cause us lots

of grief when we try to calculate the actual trajectories, and I want to postpone the worst of it. In

fact, in this course, we will not actually ever solve for the full trajectories in this case, although

we will do so in some interesting and very non-trivial examples. But just ﬁnding the velocities

of the masses right after the hammer strike is not so bad. We do this by studying carefully the

vectors ~p (linear momentum),

~

L (angular momentum), and ~ω (angular velocity) and understanding

the relationships between them in detail. We will ﬁnd a peculiar relationship depending on a

complicated object called the moment of inertia tensor. We will explain this as cleverly as we can,

but this is one of those cases in which no amount of cleverness can make it look really simple.

There are times when you just have to be very careful and let the mathematics carry you along. I

hope that going over it in several different ways will help you get a feeling for this difﬁcult subject.

The moment of inertia tensor

Let’s go back to our expression for the angular momentum of a rigid body about the point

~

R,

~

L =

X

j

(~r

j

−

~

R ) × ~p

j

=

X

j

m

j

(~r

j

−

~

R ) × (

˙

~r

j

−

˙

~

R ) (2)

where the sum runs over the various massive parts of the system, labeled by the index j. Now we

can calculate this for a rigid body rotating with angular velocity vector ~ω around

~

R by using

(

˙

~r

j

−

˙

~

R ) = ~ω × (~r

j

−

~

R ) (3)

Putting (3) into (2) gives

~

L =

X

j

m

j

(~r

j

−

~

R ) ×

³

~ω × (~r

j

−

~

R )

´

(4)

Notice that the components of

~

L are just linear combinations of the components of ~ω, but the

coefﬁcients are some complicated looking sums.

To get a feel for these scary looking sums, let us begin by looking at a case when there is just

a single term.

~

L = m

1

(~r

1

−

~

R ) ×

³

~ω × (~r

1

−

~

R )

´

(5)

2

This describes the angular momentum of a single particle of mass m

1

at the point ~r

1

rotating about

the point

~

R with angular velocity ~ω. And let’s simplify further by putting

~

R = 0 (this is just a

choice of origin anyway, so it doesn’t cost anything). Then (5) becomes

~

L = m

1

~r

1

× (~ω × ~r

1

) = −m

1

~r

1

× (~r

1

× ~ω) (6)

We can understand what (6) is by decomposing ~ω into pieces perpendicular to and parallel to ~r

1

,

~ω = ~ω

⊥

+ ~ω

k

(7)

Now the parallel part does not contribute at all because of the cross products, so

~

L = −m

1

~r

1

× (~r

1

× ~ω

⊥

) (8)

But now each of the cross products rotates the vector ~ω

⊥

by 90

◦

in the plane perpendicular to ~r

1

and multiplies the magnitude by |~r

1

|. Thus if we do this twice we get −|~r

1

|

2

~ω

⊥

, so (8) becomes

~

L = −m

1

~r

1

× (~r

1

× ~ω

⊥

) = m

1

|~r

1

|

2

~ω

⊥

= m

1

(~r

1

· ~r

1

) ~ω

⊥

(9)

Thus the angular momentum is not so complicated in this case. It is just m

1

r

2

1

times the perpen-

dicular component of ~ω.

The notation is (9) is not good enough to allow us to generalize this to the case of more than

one j. The problem is that ~ω

⊥

depends implicitly on the direction of ~r

1

. To put this back into (4)

we have to make this dependence explicit. To do that, note that the parallel component of ~ω can be

written as

~ω

k

= ˆr

1

(ˆr

1

· ~ω) (10)

and thus from (7), we can write

~ω

⊥

= ~ω − ˆr

1

(ˆr

1

· ~ω) (11)

Combining (9) and (11) gives

~

L = m

1

(~r

1

· ~r

1

) ~ω

⊥

= m

1

(~r

1

· ~r

1

)

³

~ω − ˆr

1

(ˆr

1

· ~ω)

´

= m

1

(~r

1

· ~r

1

) ~ω − m

1

~r

1

(~r

1

· ~ω) (12)

Putting the arbitrary reference point

~

R back in and factoring out the factor of m

1

gives

m

1

h³

(~r

1

−

~

R ) · (~r

1

−

~

R )

´

~ω − (~r

1

−

~

R )

³

(~r

1

−

~

R ) · ~ω

´i

(13)

Now all the dependence on ~r

1

is completely explicit, and we can apply the same procedure to each

of the terms in the sum in (4). The result is

~

L =

X

j

m

j

h³

(~r

j

−

~

R ) · (~r

j

−

~

R )

´

~ω − (~r

j

−

~

R )

³

(~r

j

−

~

R ) · ~ω

´i

(14)

Equation (14) is the desired relation between

~

L and ~ω. However, it is useful to write it in a

different form. The ﬁrst thing to notice is that both sides of (14) are vectors, and that the right hand

3

side is proportional to the vector ~ω, but is not, in general, in the same direction as ~ω. The second

thing to notice is that the second term in (14) involves a dot product. We can think of this as the

sum that appears in a matrix multiplication. This allows us to write (14) in a slightly more useful

form in matrix notation. First deﬁne a matrix

~

B as follows:

~

B =

X

j

m

j

(~r

j

−

~

R ) (~r

j

−

~

R ) =







B

xx

B

xy

B

xz

B

yx

B

yy

B

yz

B

zx

B

zy

B

zz







(15)

=

X

j

m

j







[~r

j

−

~

R ]

x

[~r

j

−

~

R ]

x

[~r

j

−

~

R ]

x

[~r

j

−

~

R ]

y

[~r

j

−

~

R ]

x

[~r

j

−

~

R ]

z

[~r

j

−

~

R ]

y

[~r

j

−

~

R ]

x

[~r

j

−

~

R ]

y

[~r

j

−

~

R ]

y

[~r

j

−

~

R ]

y

[~r

j

−

~

R ]

z

[~r

j

−

~

R ]

z

[~r

j

−

~

R ]

x

[~r

j

−

~

R ]

z

[~r

j

−

~

R ]

y

[~r

j

−

~

R ]

z

[~r

j

−

~

R ]

z







(16)

We have used a notation of two vector signs to indicated this matrix (Dave, in his book, just uses

bold face and you have to ﬁgure out from the context whether you are dealing with a vector or a

matrix). In general, this is not a good thing to do, but for our purposes in this course, it doesn’t

do any harm to use a slightly sloppy notation in which we think of our matrices as “bivectors.”

In fact, it will be kind of helpful. The reason is that the matrices we will have to deal with are

symmetric, so we don’t have to be careful to distinguish which vector index is which. You can see

the symmetry from the deﬁnition of

~

B, because

B

xy

= B

yx

, B

xz

= B

zx

, B

yz

= B

zy

. (17)

Then we can think of the dot product in (14) as matrix multiplication, and just indicate it as a

dot product (this is actually the notation that Mathematica uses). Then the second term in (14) can

be written as

−

~

B · ~ω (18)

The ﬁrst term in (14) involves the trace of the matrix

~

B

X

j

m

j

(~r

j

−

~

R )

2

= Tr

~

B = B

xx

+ B

yy

+ B

zz

(19)

So we can combine the two terms in (14) as a single matrix equation,

~

L =

~

I · ~ω (20)

where

~

I is the matrix







B

xx

+ B

yy

+ B

zz

0 0

0 B

xx

+ B

yy

+ B

zz

0

0 0 B

xx

+ B

yy

+ B

zz







−







B

xx

B

xy

B

xz

B

yx

B

yy

B

yz

B

zx

B

zy

B

zz







(21)

or

~

I =







B

yy

+ B

zz

−B

xy

−B

xz

−B

yx

B

xx

+ B

zz

−B

yz

−B

zx

−B

zy

B

xx

+ B

yy







(22)

4

Like the matrix

~

B, the matrix

~

I is symmetric —

I

xy

= I

yx

, I

xz

= I

zx

, I

yz

= I

zy

. (23)

Thus

~

I

is completely determined by 6 numbers,

I

xx

, I

yy

, I

zz

, I

xy

, I

xz

, and I

yz

. (24)

The object

~

I is a thing called the “moment of inertia tensor.” The term “tensor” refers to a

large class of objects in mathematics that are generalizations of vectors. We are not going to get

into tensors in general. This particular tensor is fairly simple because it can be thought of as a

machine for taking linear combinations of the components of one vector (~ω) to get another vector

(

~

L). Equations (20) and (22) show that you can also think of

~

I as a 3 × 3 matrix, and think of the

right hand side of (20) as a matrix product of the matrix

~

I and the vector ~ω. Because (20) is an

equation for a vector, it has a meaning independent of the particular coordinate system in which

we describe the vectors. But like a vector, the tensor

~

I will have different components in different

coordinate systems. What makes the moment of inertia tensor a tensor is that when we go from one

coordinate system to another, the components of

~

I transform in the just the right way so that (20)

is still a correct vector equation. That is the meaning of the two vector signs on

~

I. This is implicit

in (20), in which one of the vector signs on

~

I is associated with the vector nature of

~

L on the left

hand side, while the other vector index is combined by the dot product with the vector index of ~ω

so that the right hand side behaves like a vector. This will be very important in our understanding

of rigid bodies.

The general point about tensors is that one can actually go backwards through this argument

and argue that

~

I (because it takes a vector into another vector) behaves under rotations like a

product of two vectors. That doesn’t mean that it is a product of two vectors, just that it rotates the

way a product of two vectors rotates when we go from one coordinate system to another.

In term of

~

I, we can quickly derive for rotation about a ﬁxed axis, ˆn, the component of angular

momentum in the ˆn direction:

ˆn ·

~

L = ˆn ·

~

I · ~ω = ˆn ·

~

I · ωˆn = ω ˆn ·

~

I · ˆn (25)

But we showed last week that ω times the moment of inertia about the axis ˆn was the component

of angular momentum in the ˆn direction. Thus (25) implies that

I

ˆn

= ˆn ·

~

I · ˆn (26)

Notice that this means that

~

I contains all the information about the moment of inertia about any

possible axis!

Let’s see what

~

I looks like for the example system if we take the reference point

~

R to be the

center of mass. If the rectangular frame has sides of length ` and `

0

, and we center the object in the

5

z = 0 plane, the positions of the mass are as shown below:

....................................................

.

..

.

y

x

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

m

j = 1

(`/2, `

0

/2, 0)

j = 2

(`/2, −`

0

/2, 0)

j = 4

(−`/2, `

0

/2, 0)

j = 3

(−`/2, −`

0

/2, 0)

.

.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

(27)

The center of mass (the vector

~

R) is at the origin, by symmetry or explicit calculation

~

R =

1

M

X

j

m

j

~r

j

= (0, 0, 0) (28)

Then we have

m

j

= m , ~r

j

−

~

R = (±`/2, ±`

0

/2, 0) (29)

where the ± signs in (29) are independent, running over all four possibilities as j runs from 1 to 4.

Now we can compute the various components of the moment of inertia tensor: For example

~

I

xx

=

X

j

m

j

³

(~r

j

−

~

R )

2

− (r

j

− R)

2

x

´

=

X

j

m

j

³

(r

j

− R)

2

y

+ (r

j

− R)

2

z

´

= m `

0

2

~

I

yy

=

X

j

m

j

³

(r

j

− R)

2

x

+ (r

j

− R)

2

z

´

= m `

2

~

I

xy

=

~

I

yx

= −

X

j

m

j

³

(r

j

− R)

x

(r

j

− R)

y

´

= 0

~

I

zz

=

~

I

xx

+

~

I

yy

~

I

xz

=

~

I

yz

=

~

I

zx

=

~

I

zy

= 0

(30)

6

In matrix form, this looks like

~

I =







m`

0

2

0 0

0 m`

2

0

0 0 m(`

2

+ `

0

2

)







(31)

Notice that the vanishing of

~

I

xy

is slightly of non-trivial. It follows from the form of (29), but

requires a cancelation between different terms in the sum. This happened only because we were

clever (or at least not stupid) in our choice of coordinate system.

~

L and ~ω

The important thing about the moment of inertia tensor is equation (20), which describes the con-

nection between the angular momentum vector and the angular velocity vector. Because

~

I is a

tensor, and not simply a number, (20) implies that these two vectors are generally not in the same

direction. There is a very nontrivial relation between them that depends on the form of

~

I.

One example of the non-trivial connection between

~

L and ~ω is the fact that if one of these

vectors is constant, the other is generally changing with time. For example, imagine that we take

the rectangular frame we have been discussing and run a frictionless rod through the centers of

masses 1 and 3, as shown below

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

m

j = 1

(`/2, `

0

/2, 0)

j = 2

(`/2, −`

0

/2, 0)

j = 4

(−`/2, `

0

/2, 0)

j = 3

(−`/2, −`

0

/2, 0)

.

.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

(32)

If we now set the frame rotating about the ﬁxed rod by pushing mass 2 into the plane, the system

7

will have an angular velocity vector along the rod, as shown below:

.

..

.

..

.

..

.

..

.

~ω

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

m

j = 1

(`/2, `

0

/2, 0)

j = 2

(`/2, −`

0

/2, 0)

j = 4

(−`/2, `

0

/2, 0)

j = 3

(−`/2, −`

0

/2, 0)

.

.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

(33)

Here ~ω is in the direction (`, `

0

, 0), and in the ﬁgure, ` > `

0

. But to ﬁnd the x and y components

of

~

L, we multiply ~ω by

~

I,

~

L =







L

x

L

y

L

z







=







m`

0

2

0 0

0 m`

2

0

0 0 m(`

2

+ `

0

2

)













ω

x

ω

y

ω

z







=







m`

0

2

ω

x

m`

2

ω

y

m(`

2

+ `

0

2

)ω

z







(34)

In this case, ω

z

= 0, so this has the effect of multiplying the x and y components of ~ω by different

constants,

L

x

=

~

I

xx

ω

x

= m`

0 2

ω

x

, L

y

=

~

I

yy

ω

x

= m`

2

ω

y

, (35)

8

and this implies that

~

L is in a different direction, proportional to (`

0

, `, 0), as shown below:

.

..

.

..

.

..

.

..

.

..

.

~

L

~ω

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

m

j = 1

(`/2, `

0

/2, 0)

j = 2

(`/2, −`

0

/2, 0)

j = 4

(−`/2, `

0

/2, 0)

j = 3

(−`/2, −`

0

/2, 0)

.

.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

(36)

Now as the frame rotates about the frictionless rod, the angular momentum vector rotates with

it! The component of

~

L in the direction of the frictionless rod, L

k

, does not change. Because

there is no friction, there can be no torque in the direction of the rod, and thus the rate of change

of this component of

~

L vanishes. That in turn means that the magnitude of ~ω remains constant

as well. But the rod can and does produce forces on the frame perpendicular to the rod. These

forces, shown below, produce the torque that causes the perpendicular component of the angular

momentum vector,

~

L

⊥

, to change. This component executes uniform circular motion with angular

frequency ω = |~ω| in the plane perpendicular to ~ω. Thus the magnitude of the torque is

|~ω ×

~

L| = ω|L

⊥

| (37)

This business of angular momentum and angular velocity being in different directions may not

make obvious sense to you. But you should be able to feel this torque in your bones. You should

be able to see that the accelerations required to keep this system in uniform circular motion are

in different planes, and that this must produce a torque. Since torque is rate of change of angular

momentum, you can go backwards and conclude that the angular momentum does not point along

the angular velocity. I hope this will help you develop intutition for the meaning of the moment of

inertia tensor.

9

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

~

L

~ω

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

m

j = 1

(`/2, `

0

/2, 0)

j = 2

(`/2, −`

0

/2, 0)

j = 4

(−`/2, `

0

/2, 0)

j = 3

(−`/2, −`

0

/2, 0)

.

.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

(38)

A practical example of the torque that result from angular momentum that is not in the direction

of the angular velocity is the process of balancing tires for your car. Here is a ﬁgure from the web

(at http://www.discounttire.com/dtc/brochure/info/tireBalance.jsp):

a vibration to be felt. The illustration below shows how an imbalance creates vibration.

Static Imbalance:

Occurs when there is a

heavy or light spot in the

tire so that the tire won't

roll evenly and the tire

Dynamic Imbalance:

Occurs when there is

unequal weight on one

or both sides of the

tire/wheel assembly's

The reason that a mechanic spins your tire during the balancing process is check for dynamic im-

balance, which results from mass that is balanced around the center of mass but is not symmetrical

with respect to the axis. When a tire with this imbalance spins on the axle, the angular momentum

10

Georgi, Howard. Physics 16 - Mechanics and Special Relativity (англ.)

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