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

If we then deﬁne the torque as

~τ

r

0

=

X

j

(~r

j

− ~r

0

) ×

~

F

j

(3)

Then

d

dt

~

L

r

0

=

X

j

m

j

˙

~r

j

×

˙

~r

j

+

X

j

m

j

(~r

j

− ~r

0

) ×

¨

~r

j

(4)

=

X

j

(~r

j

− ~r

0

) ×

~

F

j

= ~τ

r

0

(5)

The ﬁrst term on the right hand side of (4) vanishes because of the antisymmetry of the cross

product.

It is also worth noting that the the forces in (5) need only include the “external” forces that

prevent the body from being isolated and that break rotation invariance for the non-isolated system.

This must be true for general reasons, and we can also see it explicitly if we make some more

explicit assumptions.

The explicit calculational reason that Morin describes assumes that we can break up the force

into forces between “point particles” and that the force between point particles is a central force.

The central force assumption is not unreasonable because the rotation invariance of the underlying

theory implies that the force between two point particles in the system must be along the line

between them because there is no other vector along which it could point. Then we can use

Newton’s third law. Thus if

~

F

ab

is the force on particle a from particle b,

~

F

ab

= −

~

F

ba

∝ ~r

a

− ~r

b

(6)

The ﬁrst equality is Newton’s third law, and the second comes from the assumption of central force.

But then the contribution to the torque from this pair of forces vanishes because

(~r

a

− ~r

0

) ×

~

F

ab

+ (~r

b

− ~r

0

) ×

~

F

ba

∝ (~r

a

− ~r

b

) × (~r

a

− ~r

b

) = 0 (7)

This is the argument in Morin (though he relegates some of it to problem 8).

But we really don’t need to make such explicit assumptions. The general reason the torque from

internal forces vanishes is that, as we noted, we believe that in any really isolation system, rotation

invariance holds and angular momentum is really concerved. Any violation of angular momentum

conservation arises simply because we have not isolated the system. The internal forces that hold

the body together do not contribute to the torque, because what “internal” means is that they don’t

depend on what is going on outside the system. They are invariant under rotations if as we believe

the full theory of the system in isolation is invariant under rotations. The forces that hold the body

together simply rotate along with the body when it is rotated. If the external forces went away,

the internal forces would still be there holding the body together, but there would be not torque

because the body would be isolated and rotation invariance would not be broken.

2

In pictures, what we are doing is separating a part of our system that call “internal.”

.

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

.

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

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

internal

.

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

.

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

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

.

external

If we rotate the internal and the external parts of the system together, that has no effect on the

Lagrangian. The energy doesn’t change. There is no torque.

.

..

.

..

.

..

.

internal

.

..

.

..

.

..

.

external

But if we rotate the internal parts of the system, while keeping the external parts of the system

ﬁxed, then the energy can change and there be a torque. But this depends only on the relative

3

orientation of what we have called internal and external. We haven’t changed the internal system

itself.

.

..

.

..

.

..

.

internal

.

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

.

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

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

.

external

In fact, we could get the same torque by keeping the internal system completely ﬁxed, and rotating

the external system in the opposite direction because this doesn’t change the relative orientation

of internal and external, and because of overall rotation invariance, that is all that matters.

.

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

.

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

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

internal

.

..

.

..

.

..

.

external

4

Example — oscillations of a hanging rod

Here’s a simple example of using torque and moment of inertia — consider the small oscillations of

a system of a solid rod of length ` with uniform linear mass density ρ pivoted at the top and allowed

to rotate in a vertical plane in the earth’s gravitational ﬁeld, as shown (with the rod constrained to

rotate in the plane of the paper about the pivot at the top):

•

.

θ

(8)

We can take θ to be the angle of the rod from the vertical. The moment of inertia I is given by

I =

X

j

m

j

|ˆn × ~r

j

|

2

(9)

To ﬁnish the job we must actually compute the moment of inertia I. We can write the “sum” in (9)

as

I =

Z

`

0

dx ρ x

2

=

1

3

ρ `

3

(10)

Thus the angular momentum is (out of the paper)

L = I

˙

θ =

1

3

ρ `

3

˙

θ (11)

To ﬁnd the torque, we can use the fact that we can assume that the graviational force from the

uniform ﬁeld acts on the center of mass. This is because torque is linear in the position, so the

integral would do to ﬁnd the torque by adding up the contribution of all the little masses in the rod

is the same integral we would do to ﬁnd the center of mass.

Z

~r × d

~

F =

Z

~r × dm (−gˆz) =

µ

Z

~r dm

¶

× (−gˆz) = M

~

R × (−gˆz) =

~

R ×

~

F (12)

This is

τ = −ρ`g (`/2) sin θ = −

1

2

ρ`

2

g sin θ (13)

again out of the paper. Thus using the fact the torque is rate of change of angular momentum, we

can write

d

dt

I

˙

θ = I

¨

θ =

1

3

ρ `

3

¨

θ = −

1

2

ρ`

2

g sin θ (14)

or

¨

θ = −

3

2

g

`

sin θ (15)

5

For small θ, this is a harmonic oscillator with angular frequency

ω =

s

3g

2`

(16)

Notice, however, that the result is quite different from what we would get for a simple pendulum

with a mass at the center of mass of the rod. That would have angular frequency

ω =

s

2g

`

(17)

The effect of the moment of inertia is to add some extra inertia to the system which decreases the

angular frequency and increases the period of oscillation.

We could do this somewhat more easily using the Lagrangian, without mentioning torque. The

kinetic energy is

1

2

I

˙

θ

2

(18)

The potential energy can be taken to be

ρ`g

`

2

(1 − cos θ) (19)

(the constant 1 is not important, but I guess it is nice to have the energy deﬁned to be 0 at θ = 0).

Then the Lagrangian of the system is

L =

ρ`

3

6

˙

θ

2

−

ρ`

2

(1 − cos θ) (20)

and the Euler-Lagrange equation is

ρ`

3

¨

θ = −

ρg`

2

sin θ (21)

Torque - moving reference point

Things are slightly more complicated when the reference point is moving - when

˙

~r

0

6= 0. But only

a little. The ﬁrst thing to note is that this absolutely has to work the same way if

¨

~r

0

= 0 (22)

because this is what we would get if we looked at a ﬁxed reference point in a frame of refer-

ence moving with constant velocity. In the moving reference frame, positions and velocities are

different, but forces and accelerations are all the same, and Newton’s laws work the same way.

The angular momentum in the moving frame looks like

~

L

r

0

=

X

j

m

j

(~r

j

− ~r

0

) × (

˙

~r

j

−

˙

~r

0

) (23)

6

We get the second terms in the parentheses because we are measuring the position and velocity in

the moving frame. If the reference point is accelerating,

¨

~r

0

6= 0, this may be a dangerous thing to

do, because Newton’s laws don’t work quite the smae way in an accelerating frame, but we will

worry about that in a moment.

Mathematically,

d

dt

~

L

r

0

=

X

j

m

j

(

˙

~r

j

−

˙

~r

0

) × (

˙

~r

j

−

˙

~r

0

) +

X

j

m

j

(~r

j

− ~r

0

) × (

¨

~r

j

−

¨

~r

0

) (24)

Again the ﬁrst term on the right hand side vanishes because of the antisymmetry of the cross

product and this looks just like (4) except for the

¨

~r

0

term. There are a couple of important situations

in which we can simply ignore this term. One, as we expected, is when the second derivative

vanishes,

¨

~r

0

= 0 — that is if ~r

0

is not accelerating. (25)

The other is more interesting. We can rewrite the

¨

~r

0

term using the deﬁnition of the center of mass

as

−

X

j

m

j

(~r

j

− ~r

0

) ×

¨

~r

0

= −M (

~

R − ~r

0

) ×

¨

~r

0

(26)

Evidently this vanishes if

~r

0

=

~

R — that is if ~r

0

is the center of mass; (27)

or if

(~r

0

−

~

R) is parallel to

¨

~r

0

so the cross product vanishes. (28)

Then

d

dt

~

L

r

0

= ~τ

r

0

(29)

where the torque τ

r

0

~τ

r

0

=

X

j

(~r

j

− ~r

0

) ×

~

F

j

(30)

is computed in terms of the forces in the lab frame.

The slightly surprising thing about this derivation is that we can continue to use ~τ = d

~

L/dt if

we take the reference point to be the center of mass even if the center of mass is accelerating!

The best way of understanding this is something we will discuss in more detail next month. We can

actually go to an accelerating coordinate system without changing the physics if we also introduce

what is called a “ﬁctitious force” to make up for the effect of the acceleration. Next month, we

will see that the

¨

~r

0

is associated with such “ﬁctitious force.” This force behaves exactly like a

uniform gravitational force (indeed, Einstein would say that it is not really ﬁctitious - it is entirely

equivalent to a gravitational force - this is the starting point for general relativity - but we won’t go

there right now). But a uniform gravitational force just acts on the center of mass, and if produces

no torque about the center of mass. That is why this works.

Thus if we take the reference point ~r

0

to be the center of mass

~

R, the rate of change of angular

momentum about ~r

0

is equal to the torque about ~r

0

, even if ~r

0

is moving. This is a good thing, so I

7

will assume that the reference point is the center of mass unless explicitly stated. From now on, I

will drop the subscript R.

Example — Rolling ring

Here is another example — a ring of mass m and radius r rolling without slipping down an inclined

plane in a gravitational ﬁeld.

.

..

.

~

F

g

= −mgˆz

~r

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

.

..

.

φ

•

••

••••••

••

•

••

••••

••

•

.

..

.

(31)

This is a system with one degree of freedom. We can specify the position of the ring either by the

distance ` that the ring rolls, or by the angle θ through which the ring rolls, and these are related

by

` = rθ (32)

Let’s do this three ways. First let’s consider the torque and angular momentum about the

reference point deﬁned by the point where the ring touches the plane. The nice thing about this

reference point is that the torque on the ring is independent of the force that the plane exerts on the

ring. It comes entirely from the gravitational force, which as usual we can take to act at the center

of mass. This point is instantaneously at rest, and the motion of the ring at the instant it is at rest is

a rotation about this reference point Then the torque is

mgr sin φ (33)

The moment of inertia is given by the parallel axis theoren as

I = mr

2

+ I

CM

= 2mr

2

(34)

because the mass is all a distance r away from the center of mass. Thus we have

˙

L = I

¨

θ = 2mr

2

¨

θ = mgr sin φ (35)

8

or

¨

θ =

g

2r

sin φ (36)

This describes uniform angular acceleration. In terms of rθ, the distance down the inclined plane,

it is uniform acceleration with g sin φ/2. Note that there is an extra factor of 1/2 compared with

mass sliding down a frictionless plane with the same angle. This is the effect of the extra inertia

associated with the moment of inertia.

I glossed over an important point here, and let me now emphasize it. Our reference point in this

analysis is not moving. We are looking at the system at a particular time, t, and the reference point

is the ﬁxed point where the ring touches the plane at that one time — not the moving point where

the ring touches the plane as a function of time. The way this shows up in the calculation is that

we computed the angular momentum in the lab frame. If we had been using a moving reference

frame, we would have had to measure the velocities of the parts of the system with respect to that

moving frame. That would have been more complicated - and also useless in this case, because the

point at which the ring touches the plane is accelerating as a function of time.

Now let’s do the problem using the Lagrangian. As the ring rolls without slipping, its center

of mass moves down the plane, and it also spins. Thus its kinetic energy is a sum of a linear and

a rotational term. If we describe the conﬁguration of the system by the angle θ that the ring rolls,

the linear distance it moves is r θ. Thus the kinetic energy is

1

2

m (r

˙

θ)

2

+

1

2

I

CM

˙

θ

2

(37)

and I

CM

in this case is just m r

2

because all the mass of the ring is a distance r from the center of

mass. The gravitational potential is

− mgrθ sin φ (38)

Thus the Lagrangian is

m r

2

˙

θ

2

+ mgrθ sin φ (39)

and the Euler-Lagrange equation is

2m r

2

¨

θ = mgr sin φ (40)

which again gives

¨

θ =

g

2r

sin φ (41)

Another way to do the problem is in term of torque and rate of change of angular momentum

about the center of mass. But to do it this way, we need to consider the force

~

F

i

that the inclined

9

plane exerts on the ring, as shown below.

~

F

g

= −mgˆz

mg sin φ

mg cos φ

T

N

~

F

i

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

.

..

.

φ

•

••••••

•

••

•••••

•

.

..

.

..

.

..

.

..

.

..

.

..

.

(42)

The normal component of

~

F

i

(labeled by N in the diagram) just cancels the normal component of

the gravitational force, mg cos φ, because the ring doesn’t accelerate off the plane. The tangential

component of

~

F

i

(labeled by T in the diagram) is the frictional force that keeps the ring from

sliding on the incline. Now the acceleration of the ring down the incline is determined by the

difference between the tangential component of the gravitational force and T ,

ma = mg sin φ − T . (43)

But the angular acceleration of the ring is governed by the moment of inertia, I = mr

2

, and the

torque about the center of mass, which is out of the plane of the paper with magnitude rT ,

mr

2

¨

θ = rT (44)

Because the ring is rolling without slipping

a = r

¨

θ (45)

so (43) becomes

mr

¨

θ = mg sin φ − T . (46)

Multiplying (46) by r and adding it to (44), we can cancel the unknown T and ﬁnd

2mr

2

¨

θ = mgr sin φ (47)

which agrees with (40) as it should.

As we have seen before in other contexts, the Lagrangian technique simpliﬁes things. It allows

us to solve the problem without introducing and then eliminating the frictional force, T . But we

were able to do it just as easily by choosing the reference point appropriately

10

The three dumbbells

A very simple system that shows some of the bewildering complexity of rotational motion is a

system of two equal masses m attached to one another by a string and free to rotate without friction

on a horizontal plane about their common center of mass. The conﬁguration of this system for ﬁxed

R just depends on the angle with angular velocity ω =

˙

θ.

If the distance between the objects is 2R, so that both masses are a distance R from the center

of mass, the angular momentum is perpendicular to the plane of the rotation with magnitude

L = |

~

L| = |~r × ~p | = 2mR

2

ω (48)

so in this case the moment of inertia is

I = 2mR

2

(49)

Now suppose the distance between the masses is variable - R(t) - where the t dependence is

imposed by pulling in or letting out the string. It is probably obvious that angular momentum is

conserved in this case, but let’s be pedantic about it and think about the Lagrangian. It is just the

kinetic energy

L = mR(t)

2

˙

θ

2

+ m

˙

R(t)

2

(50)

Sure enough, the Lagrangian does not depend on θ, so the momentum associated with θ is con-

served - and this is just L

L =

∂L

∂

˙

θ

= 2mR(t)

2

˙

θ = 2mR(t)

2

ω (51)

So if, for example, R(t) is decreasing, ω must be increasing to keep the angular momentum con-

stant.

This is very familiar. This is the way an ice-skater gets spinning very fast. But it is easy to get

confused. When you pull on the rope, you are pulling perpendicular to the direction of motion of

the masses. How is it that this speeds up the masses? In this simple case, it is easy to see how

everything works. While the rope is being pulled in, there is a component of the motion of the

mass in the direction of the centripetal force, and thus the centripetal force feeds energy into the

kinetic energy of the masses, increasing their speed. But still this simple example is a taste of some

of the bizarre behavior of angular momentum in three dimensions. We will see a lot more of it in

the next few weeks.

A simple way to implement this system is to hold a weight in each hand and stand on a fric-

tionless turntable.

11

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

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