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

below:

.

..

.

..

.

..

.

..

.

...

..

.

θ

(23)

Suppose that at some angle θ, the torque is τ — into the paper in the ﬁgure. We will actually

ignore the dependence of the torque on θ and simply assume that the magnitude of the torque is

constant and equal to τ . This will allow us to understand what is happening more easily, and it

does not change the qualitative nature of the physics, just some boring details.

Now we will ﬁrst consider a motion in which the top is started with exactly the right preces-

sional velocity to keep it at the angle θ. In this case, the angular velocity has two contributions.

There is a large contribution, ω

s

, in the direction of the symmetry axis of the top — because the top

is rapidly spinning. There is a vertical contribution, ω

p

, due to the precession caused by the torque.

Typically, for a rapidly rotating top, the torque is small in the sense that the precession frequency

ω

p

is much smaller than the angular velocity ω

s

associated with the spin of the top.

.

..

.

ω

p

ω

s

.

..

.

....

.

θ

(24)

The motion described by this ~ω is very similar to what we saw animated for the symmetric top —

a precession of the plane of ˆe

3

on top of a rotation around the ˆe

3

axis. But here, the precession of

the plane is much smaller than the spin.

From (24), we see that the component of angular momentum in the direction of the symmetry

axis is

L

k

= I

3

ω

3

(25)

6

where ω

3

is the total component of angular velocity along the symmetry axis.

ω

3

= ω

s

+ ω

p

cos θ (26)

The quantity ω

3

is actually somewhat more interesting than ω

s

because it is conserved if the tip of

the top is frictionless. The component of angular momentum perpendicular to the symmetry axis

is

L

⊥

= I ω

p

sin θ (27)

What we care about for the precession is the horizontal component of angular momentum,

which is

L

x

= sin θ L

k

− cos θ L

⊥

= sin θ I

3

ω

3

− sin θ cos θ I ω

p

(28)

Normally, we ignore the small second term on the right hand side of (28), and do not distinguish

between ω

s

and ω

3

. But I just wanted to show you that it is possible to take into account the con-

tribution from the precession if we want to. At any rate, the rate of change of angular momentum

for uniform precessional motion is ω

p

L

x

, and thus the condition for uniform precession is

τ = ω

p

L

x

= ω

p

(sin θ I

3

ω

3

− sin θ cos θ I ω

p

) (29)

This is a quadratic equation for ω

p

with solutions

ω

p

=

I

3

ω

3

2I cos θ

±

s

µ

I

3

ω

3

2I cos θ

¶

2

−

τ

sin θ cos θ I

(30)

The plus sign here gives what is called fast precession, which is precession of the order of the

rotational velocity of the top. If the torque is small, we can write it as

ω

p

=

I

3

ω

3

2I cos θ

+

s

µ

I

3

ω

3

2I cos θ

¶

2

−

τ

sin θ cos θ I

≈

I

3

ω

3

I cos θ

(31)

This is just a slightly perturbed version of the free rotation of a symmetric top that we have just

understood in detail — the torque plays very little role in it — and (31) reproduces (15) with

|

~

L| = I ω

p

. This is not what we usually see, because the top usually doesn’t get started moving

that fast. Instead, we are interested in slow precession, the minus sign solution, which gives

ω

p

=

I

3

ω

3

2I cos θ

−

s

µ

I

3

ω

3

2I cos θ

¶

2

−

τ

sin θ cos θ I

=

I

3

ω

3

2I cos θ

Ã

1 −

s

1 −

4Iτ cos θ

sin θ ω

2

3

I

2

3

!

≈

I

3

ω

3

2I cos θ

Ã

1 − 1 +

2Iτ cos θ

sin θ ω

2

3

I

2

3

!

=

τ

sin θ I

3

ω

3

(32)

This is the usual result in which the precession frequency is proportional to the torque and inversely

proportional to the angular momentum.

7

Nutation

So if the top is moving at exactly the right angular velocity, it precesses at constant angular velocity

and stays at the same angle. But what happens if things are not quite right? In particular, suppose

that we are holding the top at some ﬁxed angle and suddenly drop it. This is a typical situation.

What happens?

The best way to understand what happens is to go to an accelerated coordinate system that is

rotating around the pivot point of the top with angular velocity ω

p

. This is not an inertial frame,

of course, which means that there are so-called ﬁctitious forces that make up for the fact that the

frame is accelerating. We will talk about these in detail later. At any rate, in this frame, the motion

with constant θ and precession rate ω

p

just looks like a top with its symmetry axis sitting still in

space. This means that in this frame there must be a ﬁctitious force that exactly cancels the torque,

so that the angular momentum is conserved. I presume that this a basically a coriolis force, but it

doesn’t matter what it is because it must be there.

Now what does the process look like in this rotating frame if we drop the top from rest in the

space frame? In the moving frame, because the symmetry axis of the top was initially at rest in

the space frame, it is initially rotating with angular velocity −ω

p

in the rotating frame. But that

means that there is a small additional component to the angular velocity in the vertical direction,

and thus the angular velocity and the angular momentum of the top are slightly displaced from the

symmetry axis. Because there is no torque in this special frame, this just reduces the free rotation

problem that we have already analyzed. The symmetry axis of the top precesses rapidly around the

angular momentum with frequency L/I. Back in the space frame, this motion is superimposed on

the much slower precession of the angular momentum produced by the torque. The rapid motion

is called nutation.

Not only is the nutational motion very rapid, but the amplitude is usually very small. In the

case we discussed in which the top is dropped from rest, the angle of the angular velocity in the

special frame to the symmetry axis is of order ω

p

/ω

3

. For a rapidly rotating top, the motion is often

to small and too rapid to see. But you can feel it or hear it under the right circumstances.

As I warned you I would, I have ignored the dependence of the torque on θ in discussing

nutation. I hope that you can now see why this doesn’t make much difference. The torque is a

very small effect that produces the slow precession of the plane of ˆe

3

. The nutation results if the

angular momentum is not quite lined up with the symmetry axis. These two effects have very little

to do with one another. Small changes in the torque as ˆe

3

precesses rapidly about

~

L will have a

very small effect on the motion.

Vectors in the body frame

The derivation above of the motion of the free symmetric top in the space frame is slick and (I

hope) easy to understand, but limited to the case I

1

= I

2

. While we will not solve the more general

problem completely, it is interesting to set it up mathematically in the body frame, rotating along

with this rigid body. At least, in this frame, the moments of inertia do not change. We will take the

8

origin to be the center of mass in both the space frame and the body frame. We will also take our

coordinate axes in the body frame to correspond to the three principal axes of the rigid body. Then

the equations that govern the motion of the body take a fairly simple form. Of course, even if we

can solve these equations, it doesn’t tell us everything, because it will still take some more work to

understand how the body is moving in the space frame. But at least this is a start.

Our starting point will be the relation between the change of a vector with respect to the space

frame and the change with respect to the rotating body frame. Consider a vector

~

A. We will call

the rate of change of the vector with respect to the body frame

δ

~

A

δt

. (33)

If

δ

~

A

δt

= 0, the vector is ﬁxed with respect to the body frame, and then we know that its change with

respect to the space frame is simply ~ω ×

~

A where ~ω is the angular velocity of the body. If

δ

~

A

δt

6= 0,

then the change with respect to the space frame is the sum of this and the contribution from the

body’s rotation:

d

~

A

dt

=

δ

~

A

δt

+ ~ω ×

~

A . (34)

What I ﬁnd confusing about (34) is that it seems to refer to two different frames — the space frame

and the body frame. In which frame is it true? The answer is in either one! The quantities on

both sides of the equation are vectors that can be expressed either in the space frame or in the body

frame. The explicit coordinates of

~

A and its derivatives and ~ω will change when we go from one

frame to another, but (34) is true in either. This is why I was careful to use the words “with respect

to” rather than the word “in” in deﬁning

d

~

A

dt

and

δ

~

A

δt

. These objects are the change of the vector

~

A with respect to the space frame and the body frame respectively, but they can be described in

either frame (or for that matter, any other coordinate system).

Let’s discuss an example. Consider the vector ~ω itself. Here (34) implies

d~ω

dt

=

δ~ω

δt

+ ~ω × ~ω =

δ~ω

δt

. (35)

The rate of change of ω with respect to the body frame is the same vector as the rate of change of

omega with respect to the space frame. First consider the situation in which ~ω = ωˆn is constant in

space. This means that the body is rotating about the ﬁxed axis ˆn with constant angular velocity ω.

In the body frame, this axis is ﬁxed in the body, and again ~ω is constant. Thus

d~ω

dt

= 0 ⇒

δ~ω

δt

= 0 (36)

But the relation (35) may seem a bit odd in general, so let’s see how it works in a nontrivial

example. Considering the following object will give us something speciﬁc to think about and will

9

also illustrate some of the curious features of angular velocity.

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

.

..

.

..

.

~ω

2

rotation of axis

rotation

around axis

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

.

~ω

1

t = 0

space frame

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

.

..

.

...

......

..

.

..

.

..

.

(37)

This is intended to show a rigid body in the space frame rotating about its long axis ˆe

1

with angular

velocity

~ω

1

= ω

1

ˆe

1

(38)

which is constant in the body frame and with the axis ˆe

1

rotating about the vertical with angular

velocity

~ω

2

= ω

2

ˆz (39)

constant in the space frame. The total angular velocity is the sum ~ω = ~ω

1

+ ~ω

2

and the component

of the angular velocity in the direction of the axis is executing uniform circular motion and the rate

of change of ~ω is

d

dt

~ω = ~ω

2

× ~ω = ~ω

2

× (~ω

1

+ ~ω

2

) = ~ω

2

× ~ω

1

(40)

In the space frame, as the long axis rotates around with angular velocity ~ω

2

, you see the shorter

crosspiece rotating around the long axis with angular velocity ~ω

1

. This is animated in CROSS.EXE,

where the long axis of the cross is rendered in white and the short axis is yellow and blue (the two

halves are colored differently to make it easier to keep track of the sense of the rotation).

The animation is based on the fact that our description in words of the angular velocity de-

scribes not just what ω looks like at a particular moment, but also how it evolves with time (at least

if we impose an initial position). For example, if the axis ˆe

1

is in the ˆx direction at time t = 0, then

at an arbitrary time

ˆe

1

= ˆx cos ω

2

t + ˆy sin ω

2

t (41)

~ω

1

= ω

1

ˆx cos ω

2

t + ω

1

ˆy sin ω

2

t (42)

~ω = ω

1

ˆx cos ω

2

t + ω

1

ˆy sin ω

2

t + ω

2

ˆz (43)

From this and (39), you can explicitly verify (40):

d

dt

~ω = −ω

1

ω

2

ˆx sin ω

2

t + ω

1

ω

2

ˆy cos ω

2

t = ω

2

ˆz × (ω

1

ˆx cos ω

2

t + ω

1

ˆy sin ω

2

t) (44)

10

Now what does this look like in the body frame? Stare at the animation and imagine that you

are moving around with the cross and try to get a feeling for how you would see the green line

representing the angular velocity. Naturally, ˆe

1

and therefore ~ω

1

doesn’t change, because ˆe

1

in

deﬁned to be ﬁxed in the body frame. But ~ω

2

, which was ﬁxed in the space frame, is not ﬁxed in

the body frame, precisely because of the nonzero ~ω

1

, as shown below.

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

.

..

.

~ω

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

.

−~ω

1

t = 0

body frame

.

..

.

(45)

In the body frame, at t = 0, the angular velocity vector looks the same as it did in the space

frame. But now at later times, the angular velocity vector rotates around the long axis with angular

velocity −~ω

1

, so the perpendicular component, ~ω

2

executes uniform circular motion about the long

axis with angular velocity −~ω

1

, and the rate of change of the vector ~ω is

δ

δt

~ω = −~ω

1

× ~ω

2

(46)

which is the same as (40).

11

lecture 21

Topics:

Where are we now

The free symmetric top in the body frame

Euler’s equations

The free symmetric top ala Euler’s

The tennis racket theorem

Where are we now

Last time we solved the problem of the motion of the free symmetric top in the space frame. This

allows us to understand what the system looks like. But it is sometimes useful and important to

understand what the motion looks like with respect to the body frame. Here we will do this in

two ways - directly, using what we know about the relation between the space frame and the body

frame, and more formally, by deriving general mathematical relations (Euler’s equations) for the

angular velocity in the body frame. Euler’s equations can then be applied to more general system

(in which I

1

6= I

2

).

Before we get started, I wanted to review once more the interesting trick we used last time to

solve the free symmetric top in the space frame without even solving a differential equation because

it is such a beautiful example of the power of thinking and understanding what you are doing rather

than using brute-force mathematics. We used two things. One was the relation between the angular

velocity vector ~ω, the angular momentum vector

~

L, and the unit vector along the symmetry axis of

the top ˆe3:

~ω =

~

L/I

⊥

− Ωˆe

3

(1)

where

Ω =

(I

3

− I

⊥

) ω

3

I

⊥

(2)

The other is the cross-product formula for circular motion about an axis. We know that the velocity

of every point ~r on the rigid body with respect to the body’s center of mass is given by the cross

product with the angular velocity vector

d~r

dt

= ~ω × ~r (3)

Unfortunately, this by itself doesn’t do us much good because if I

⊥

6= I

3

, the angular velocity vec-

tor is constantly changing, so that while (3) is correct at any instant of time, it doesn’t immediately

tell us what happens next. The key to making progress is to not worry about what every point on

the rigid body is doing, but just what the points on the symmetry axis are doing. For ~r = ˆe

3

, we

can use (1), so that (3) becomes

dˆe

3

dt

= ~ω × ˆe

3

=





~

L

I

⊥

− Ωˆe

3





× ˆe

3

=

~

L

I

⊥

× ˆe

3

(4)

1

Notice that the second term in the parentheses vanishes only because we have restricted our at-

tention to ˆe

3

, rather than asking about the motion of a general point. We are using the fact that

rotation of a single line doesn’t uniquely determine the axis because we can always add a compo-

nent of the angular velocity vector along the line itself. So we can choose to consider the rotation

of ˆe

3

along the

~

L direction. The reason that this is an improvement is because

~

L is ﬁxed in space.

Therefore (4) has a simple interpretation at all times — ˆe

3

is undergoing uniform circular motion

about the ﬁxed angular momentum with angular velocity vector

~

L/I

⊥

. Now that we know what is

happening to ~e

3

, we can interpret (6) to tell us what every point on the symmetric top is doing. The

top is spinning around the axis with angular velocity −Ω, and axis is spinning around the angular

momentum according to (4).

The free symmetric top in the body frame

I hope that it will not surprise you that our principal tool in the analyzing the free symmetric top

in the body frame will be the general relation between the change of a vector with respect to the

space frame and its change with respect to the body frame

d

~

A

dt

=

δ

~

A

δt

+ ~ω ×

~

A . (5)

The other thing we need is the vector relation we derived last time between

~

L, ~ω and ˆe

3

,

~ω =

~

L/I

⊥

− Ωˆe

3

(6)

where

Ω =

(I

3

− I

⊥

) ω

3

I

⊥

(7)

This vector relation is true in any coordinate system.

We can now calculate how ~ω and

~

L change in the body frame. We will do this in two ways.

First we can use (6) and the fact we derived last time that with respect to the space frame the plane

of ~ω and ˆe

3

is rotating about

~

L with angular velocity

~

L/I

⊥

to write

δ~ω

δt

=

d~ω

dt

=

~

L

I

⊥

× ~ω (8)

Again this is a vector equation true in any coordinate system, so we can use it in the space frame,

together with (6) to write

δ~ω

δt

=

³

~ω + Ωˆe

3

´

× ~ω (9)

= ~ω × ~ω + Ωˆe

3

× ~ω = Ωˆe

3

× ~ω (10)

But because ˆe

3

is ﬁxed in the body frame, (10) implies that with respect to the body frame, ~ω is

rotating about ˆe

3

with angular velocity Ωˆe

3

. This is the same trick we used in the space frame,

2

except that now it is ˆe

3

that is ﬁxed, so we want to think of the rotation as about ˆe

3

. Note that (10)

implies that ω

3

is constant just as in the space frame because

δ

δt

(ˆe

3

· ~ω) = ˆe

3

·

δ~ω

δt

= ˆe

3

·

³

Ωˆe

3

× ~ω

´

= 0 , (11)

so we can independently derive the fact that the coefﬁcients in in the vector equations (6) are ﬁxed.

Alternatively, we can use the fact that

~

L is ﬁxed in the space frame and write

0 =

δ

~

L

δt

+ ~ω ×

~

L , (12)

so that

δ

~

L

δt

= −~ω ×

~

L = −

³

~

L/I

⊥

− Ωˆe

3

´

×

~

L = Ωˆe

3

×

~

L (13)

This expresses the same physics. With respect to the body frame,

~

L, like ~ω must rotate about the

ˆe

3

axis with angular velocity Ωˆe

3

because the entire plane in which ~ω and

~

L are ﬁxed is rotating.

There is a nice way of understanding all this that may help to cement one of the important

lessons of the last few weeks — while rotations are complicated and do not add like ordinary

vectors, inﬁnitesimal rotations and angular velocities do add. In this case, what we have with

respect to the body frame is that the plane formed by the three vectors

~

L, ~ω and ˆe

3

rotates with

angular velocity

~ω

pb

= Ωˆe

3

. (14)

With respect to the space frame, the plane formed by the three vectors

~

L, ~ω and ˆe

3

rotates with

angular velocity

~ω

ps

=

~

L/I

⊥

(15)

The relation between the space frame and the body frame is that the angular velocity of the plane

with respect to the space frame, ~ω

ps

, is the angular velocity of the plane with respect to the body

frame, ~ω

pb

, plus the angular velocity of the body, ~ω:

~ω

ps

= ~ω

pb

+ ~ω (16)

Indeed, because of (6), this works. Angular velocities add like ordinary vectors.

Euler’s equations

note that this is an approach to equations of motion in an accelerating frame

There is another approach to the problem of the free symmetric top, which is also more gener-

ally useful. We can derive what are called “Euler’s equations” by considering (5) where

~

A is the

angular momentum,

~

L. We can then use the fact that the torque, ~τ , is the rate of change of angular

momentum with respect to the space frame. Using (5), we can write this in terms of the rate of

change of

~

L with respect to the body frame. Then we can write

~τ =

d

~

L

dt

=

δ

~

L

δt

+ ~ω ×

~

L (17)

3

As we discussed last time, we can evaluate a vector equation in any coordinate system. It is

interesting to consider (17) in the body frame in which

~

I is ﬁxed. Then in the usual basis in which

we use the principal axes of our body, ˆe

1

ˆe

2

and ˆe

3

, the moment of inertia tensor is

~

I =







I

1

0 0

0 I

2

0

0 0 I

3







(18)

and we can express the components of

~

L in this basis in terms of the components of ~ω as

L

1

= I

1

ω

1

L

2

= I

2

ω

2

L

3

= I

3

ω

3

(19)

We now expand the cross product in (17) to get

τ

1

=

δL

1

δt

+ ω

2

L

3

− ω

3

L

2

τ

2

=

δL

2

δt

+ ω

3

L

1

− ω

1

L

3

τ

3

=

δL

3

δt

+ ω

1

L

2

− ω

2

L

1

(20)

Now use (19), L

1

= I

1

ω

1

, L

2

= I

2

ω

2

, L

3

= I

3

ω

3

, to write

τ

1

= I

1

δω

1

δt

+ ω

2

I

3

ω

3

− ω

3

I

2

ω

2

τ

2

= I

2

δω

2

δt

+ ω

3

I

1

ω

1

− ω

1

I

3

ω

3

τ

3

= I

3

δω

3

δt

+ ω

1

I

2

ω

2

− ω

2

I

1

ω

1

(21)

or simplifying slightly

τ

1

= I

1

δω

1

δt

− (I

2

− I

3

) ω

2

ω

3

τ

2

= I

2

δω

2

δt

− (I

3

− I

1

) ω

3

ω

1

τ

3

= I

3

δω

3

δt

− (I

1

− I

2

) ω

1

ω

2

(22)

These are Euler’s equations, relating the rate of change of the angular velocity in the body frame to

the torque and angular velocity in the body frame and the principal moments of inertia. We might

have expected that the right hand side would be proportional to products of two components of ~ω.

This ensures that if the angular velocity is in the direction of one of the principle axes, the right

hand terms of all the Euler equations vanish because two of the components of ~ω are zero. And

then if the torque vanishes, δ~ω/δt = 0, because as we know, the body can rotate freely about a

principle axis.

4

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

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