Conservation of Angular Momentum and Central Force Motion

Conservation of Angular Momentum and Central Force Motion

The previous article established the angular-momentum equation

(1)

for a particle system when internal torques cancel and O is a fixed point in an inertial frame.

The immediate consequence is one of the fundamental conservation laws of mechanics:

(2)

For a central force, the force is directed along the radius vector:

(3)

Therefore

(4)

so angular momentum about the force center is conserved.

This single fact leads to several major results:

• central force motion is planar,
• mr²𝜃 is constant,
• equal areas are swept in equal times,
• radial and angular motion can be separated,
• the radial dynamics can be represented with an effective potential.

1 The conservation law

Starting from

(5)

if

(6)

then

(7)

Integrating,

(8)

Because angular momentum is a vector, conservation means that both its magnitude and direction remain constant.

2 Conservation about a chosen point

Angular momentum and torque must be computed about the same point.

A system may have zero torque about one point but nonzero torque about another.

Therefore the statement

(9)

is incomplete unless the reference point or axis is understood.

For a central force, the natural reference point is the force center.

For a rotating isolated system, the center of mass is often the most useful reference point.

3 Angular impulse form

The finite-time relation is

(10)

Thus exact angular-momentum conservation over an interval requires zero net external angular impulse:

(11)

The external torque does not have to vanish at every instant if its total angular impulse over the interval is zero.

4 Redistribution of mass in an isolated rotating system

Angular momentum can remain constant while angular velocity changes.

For fixed-axis rotation about a principal axis,

(12)

If external torque is negligible,

(13)

Therefore

(14)

Reducing the moment of inertia increases angular speed.

Increasing the moment of inertia decreases angular speed.

5 Example 1: contracting rotating system

A rotating system initially has

Ii	= 6.0 kg m2,	(15)
ωi	= 2.0 rad∕s.	(16)

It changes configuration so that

(17)

With negligible external torque,

(18)

Thus

ωf	= ωi	(19)
	= (2.0)	(20)
	= 6.0 rad∕s.	(21)

Therefore

(22)

6 Angular momentum conservation does not imply kinetic-energy conservation

For fixed-axis rotation,

(23)

Using

(24)

we can write

(25)

If L is conserved while I decreases, the rotational kinetic energy increases.

The additional energy can come from internal work, such as muscular work in a person pulling masses inward.

Thus:

(26)

7 Example 2: energy change during contraction

Use the system from Example 1.

Initial rotational kinetic energy:

(27)

Final rotational kinetic energy:

(28)

Therefore

(29)

Angular momentum is conserved, but kinetic energy increases because internal work is done while changing the mass distribution.

8 Central forces

A force is central if it has the form

(30)

where

(31)

The force can point inward or outward, but it must lie along the line joining the particle to the force center.

Examples include:

• Newtonian gravity from a fixed spherical source,
• electrostatic force between point charges,
• an ideal isotropic spring force F = −kr.

9 Central force implies zero torque

Torque about the force center is

(32)

For a central force,

(33)

Therefore

(34)

Hence

(35)

10 Central force motion is planar

Because

(36)

the vectors r and v are perpendicular to L.

If L is constant in direction, the motion remains in the plane perpendicular to L.

Thus:

(37)

If

(38)

the motion is purely radial and lies along a line through the force center.

11 Angular momentum in plane polar coordinates

For planar motion,

(39)

and

(40)

Then

L	= mr × v	(41)
	= m ×.	(42)

The radial term vanishes:

(43)

Since

(44)

we obtain

(45)

Thus the angular-momentum magnitude is

(46)

12 Specific angular momentum

Divide by particle mass:

(47)

The quantity h is called the specific angular momentum.

Its SI units are

(48)

For central force motion,

(49)

This notation is especially common in orbital mechanics.

13 Areal velocity and Kepler’s equal-areas law

During a small angular displacement d𝜃, the radius vector sweeps area

(50)

Divide by dt:

(51)

Using

(52)

we obtain

(53)

If angular momentum is conserved, areal velocity is constant:

(54)

Therefore equal areas are swept in equal times.

14 Example 3: areal velocity

A satellite has specific angular momentum

(55)

Its areal velocity is

(56)

Therefore

(57)

In a time interval

(58)

the swept area is

(59)

15 Polar-coordinate equations of motion

The acceleration in Plane polar coordinates is

(60)

For a central force,

(61)

Therefore the tangential force component is zero:

(62)

Newton’s second law gives

m	= F(r),	(63)
m	= 0.	(64)

16 Deriving angular-momentum conservation from the transverse equation

The transverse equation is

(65)

Multiply by r:

(66)

Recognize the derivative

(67)

Thus

(68)

Therefore

(69)

which is exactly conservation of specific angular momentum.

17 Reducing the radial equation

From

(70)

we have

(71)

The radial equation is

(72)

Substitute the angular-momentum relation:

(73)

Therefore

(74)

The second term behaves like an outward radial contribution associated with the angular motion.

It is not an additional fundamental force. It appears because the radial coordinate is being used to describe motion with conserved angular momentum.

18 Conservative central forces

If the central force is conservative,

(75)

The kinetic energy is

(76)

Use

(77)

Then

(78)

Thus the total energy is

(79)

19 Effective potential

Define the effective potential

(80)

Then the energy equation becomes

(81)

This looks like a one-dimensional energy equation for radial motion.

The term

(82)

is often called the angular-momentum barrier or centrifugal barrier.

20 Turning points and apses

Because

(83)

the allowed radial region satisfies

(84)

A radial turning point occurs when

(85)

Therefore

(86)

at a radial turning point.

In orbital motion, radial turning points are called apses.

For a bound orbit:

• the minimum radius is periapsis,
• the maximum radius is apoapsis.

21 Circular orbit condition

For a circular orbit,

(87)

so

(88)

The radial equation becomes

(89)

Thus

(90)

Equivalently, a circular orbit occurs at a stationary point of the effective potential:

(91)

A stable circular orbit corresponds locally to a minimum:

(92)

22 Newtonian gravity example

For a particle of mass m moving around a fixed mass M,

(93)

The effective potential is

(94)

The attractive gravitational term dominates at large enough radius, while the angular-momentum barrier becomes increasingly important at small radius when L≠0.

For a circular orbit,

(95)

Therefore

(96)

Since

(97)

we obtain

(98)

Thus

(99)

23 Apsis speed relation

At periapsis and apoapsis,

(100)

The velocity is therefore purely transverse.

Angular momentum gives

(101)

Thus for two apses,

(102)

Therefore

(103)

The object moves faster at smaller radius.

24 Example 4: speed change between apses

Suppose an orbit has

rp	= 7.0 × 106 m,	(104)
ra	= 1.4 × 107 m.	(105)

If the apoapsis speed is

(106)

then

vp	= va	(107)
	= 2(4.0)	(108)
	= 8.0 km∕s.	(109)

Thus

(110)

25 Radial motion as a special case

If

(111)

then

(112)

Away from r = 0,

(113)

The motion remains on a fixed radial line.

The effective potential reduces to the ordinary potential:

(114)

Thus the angular-momentum barrier is absent for purely radial motion.

26 Puck pulled through a central hole

Consider a puck moving on a frictionless horizontal table while a string passes through a small hole at the origin.

The string tension is radial:

(115)

Therefore the torque about the hole is zero:

(116)

Hence

(117)

If the string is pulled inward and r decreases, the angular speed increases.

Unlike an isolated skater example, an external agent pulling the string can do work, so the puck’s kinetic energy need not remain constant.

27 Example 5: puck radius change

A puck moves at radius

(118)

with tangential speed

(119)

It is slowly pulled inward until

(120)

At the instant considered, assume the motion is again purely transverse.

Angular momentum gives

(121)

Thus

(122)

Therefore

(123)

The kinetic energy increases by a factor of four because the external pulling agent does work.

28 Two-body central force systems

Consider two particles with masses m₁ and m₂ interacting only through equal-and-opposite central forces.

Define

(124)

as the relative position.

Let

(125)

and define the reduced mass

(126)

In the center-of-mass frame,

ρ1	= r,	(127)
ρ2	= −r.	(128)

The internal angular momentum about the center of mass becomes

(129)

Thus the two-body problem can be represented by a single effective particle of mass μ moving in the relative coordinate.

29 Relative equation of motion

Let F₁₂ be the force on particle 1 due to particle 2.

Then

m1r1	= F12,	(130)
m2r2	= −F12.	(131)

Subtract:

(132)

Since

(133)

we obtain

(134)

This result is developed more fully in M04-09.

30 Conservation law versus symmetry

Angular momentum is conserved whenever the appropriate external torque vanishes.

For a central force, this occurs because the force has rotational symmetry about the center: there is no preferred direction in space, only a dependence on radius.

At a more advanced level, rotational symmetry and angular-momentum conservation are connected by Noether’s theorem.

The present Newtonian derivation reaches the conservation law directly through torque.

31 Common mistakes

1. Conserving angular momentum without stating the point or axis about which torque is zero.
2. Assuming angular-momentum conservation implies kinetic-energy conservation.
3. Using I_iω_i = I_fω_f when a significant external torque acts.
4. Treating the effective angular-momentum term L²∕(mr³) as a new fundamental force.
5. Forgetting that central force motion is planar because the direction of L is constant.
6. Forgetting the factor 1∕2 in the areal-velocity relation.
7. Using rv = constant at arbitrary points of an orbit when the velocity is not perpendicular to r.
8. At arbitrary points, replacing L = mrv sin 𝜃 with L = mrv.
9. Confusing specific angular momentum h = L∕m with ordinary angular momentum.
10. Forgetting that U_eff includes both the physical potential U(r) and the angular-momentum barrier.
11. Setting a circular-orbit condition using F = 0 instead of balancing the radial dynamics.
12. Assuming an external radial force cannot do work. A radial force does no torque about the center, but it can do work when radial displacement occurs.
13. Applying single-particle mass m instead of reduced mass μ when using the relative coordinate for a genuine two-body problem.

32 Practice exercises

1. A rotating system changes its moment of inertia from 8 kg m² to 2 kg m² with negligible external torque. If ω_i = 1.5 rad∕s, find ω_f.
2. For the preceding problem, compare the initial and final rotational kinetic energies.
3. Prove that a central force produces zero torque about the force center.
4. Starting from plane polar velocity, derive L = mr²𝜃.
5. Define specific angular momentum and derive h = r²𝜃.
6. Derive the equal-areas relation dA∕dt = L∕(2m).
7. Starting from the transverse polar equation, derive d(r²𝜃)∕dt = 0.
8. Use angular-momentum conservation to eliminate 𝜃 from the radial equation and derive

   (135)

9. For a conservative central force, derive

   (136)

10. Explain the physical meaning of the effective potential and the angular-momentum barrier.
11. Derive the circular-orbit condition from dU_eff∕dr = 0.
12. For Newtonian gravity, derive v_c = .
13. At two apses of a bound orbit, derive r_pv_p = r_av_a.
14. A puck is pulled from radius 1.2 m to 0.40 m. If its initial transverse speed is 1.5 m∕s and torque about the hole is zero, find the final transverse speed.
15. Derive L_CM = μr ×r for a two-body system.

33 Summary

Angular momentum is conserved when net external torque about the chosen reference point vanishes:

(137)

For a central force,

(138)

Thus central force motion has conserved angular momentum:

(139)

The specific angular momentum is

(140)

The areal velocity is

(141)

For a conservative central force,

(142)

with

(143)

For two-body central force motion,

(144)

M04-09 develops the two-body reduction and relative-coordinate dynamics in detail.

References

References