Angular Momentum of Particle Systems

Angular Momentum of Particle Systems

Linear momentum measures translational motion:

(1)

Angular momentum measures how that motion is distributed about a chosen point.

For a particle at position r relative to an origin O, with linear momentum p, define the angular momentum about O as

(2)

For a Newtonian particle,

(3)

Its magnitude is

(4)

where 𝜃 is the angle from r to p.

Angular momentum is a vector. Its direction is determined by the right-hand rule.

1 Angular momentum depends on the chosen origin

Linear momentum of a particle does not depend on the coordinate origin.

Angular momentum generally does.

If the position vector changes, then

(5)

changes even when p is unchanged.

The subscript O is therefore important.

Whenever angular momentum is used, the reference point or axis should be stated.

2 Magnitude and perpendicular distance

The magnitude

(6)

can be rewritten using the perpendicular distance from the origin to the line of motion.

Define

(7)

Then

(8)

This form is often geometrically clearer than the full cross product.

3 Units and dimensions

Since

(9)

the SI unit is

(10)

The dimensions are

(11)

Angular momentum is not energy, even though both frequently contain products of mass, length, and velocity.

4 Cartesian components

Let

r	= xex + yey + zez,	(12)
p	= pxex + pyey + pzez.	(13)

Then

(14)

so

Lx	= ypz − zpy,	(15)
Ly	= zpx − xpz,	(16)
Lz	= xpy − ypx.	(17)

For planar motion in the xy plane,

(18)

and therefore

(19)

5 Example 1: particle angular momentum

A 2.0 kg particle is at

(20)

and moves with

(21)

Its momentum is

(22)

Therefore

LO	= (3ex + 4ey) × (10ex)	(23)
	= 40	(24)
	= −40ez kg m2∕s.	(25)

Thus

(26)

The negative z direction is consistent with the clockwise sense of the motion about the origin.

6 Torque

Torque about the same origin is

(27)

Its magnitude is

(28)

where ϕ is the angle from r to F.

Torque is the rotational analogue of force in the angular-momentum equation.

7 Derivation of the angular-momentum equation for one particle

For a particle,

(29)

Differentiate:

	= × p + r ×	(30)
	= v × mv + r × F.	(31)

Because

(32)

the first term vanishes.

Therefore

(33)

for a particle when O is fixed in an inertial frame.

8 Zero torque and constant angular momentum

If

(34)

then

(35)

Hence

(36)

This means both:

• the magnitude of L_O remains constant,
• its direction remains constant.

A zero torque about one point does not imply zero torque about every point.

9 Angular impulse

Integrate the torque equation from t_i to t_f:

(37)

Define the angular impulse

(38)

Then

(39)

This is the angular counterpart of

(40)

10 Example 2: constant torque

A particle has initial angular momentum

(41)

A constant torque

(42)

acts for

(43)

The angular impulse is

(44)

Since

(45)

the final angular momentum is

(46)

11 Angular momentum of a particle system

For N particles, define total angular momentum about O:

(47)

Equivalently,

(48)

Each particle contributes angular momentum about the same chosen origin.

12 Rate of change for a particle system

Differentiate the total:

(49)

For each particle,

(50)

Thus

(51)

Separate external and internal forces:

(52)

13 Cancellation of internal torques

For a pair of particles i and j,

(53)

by Newton’s third law.

The pair’s internal torque about O is

τijpair	= r i × Fij + rj × Fji	(54)
	= (ri − rj) × Fij.	(55)

If the internal force is central, meaning

(56)

then

(57)

Thus ordinary central Newtonian internal forces produce no net internal torque.

14 System angular-momentum equation

When internal torques cancel,

(58)

Integrating,

(59)

If the net external torque about O vanishes,

(60)

This is the system-level conservation law developed further in M04-08.

15 Important caveat about internal-force cancellation

Equal-and-opposite internal forces are not enough by themselves to guarantee zero pair torque about an arbitrary origin.

The pair torque is

(61)

It vanishes when the force lies along the line joining the particles.

Thus the familiar particle-system result

(62)

assumes the ordinary Newtonian central-force model for internal interactions, or a more complete description in which all relevant angular momentum is included.

16 Center of mass decomposition

Let

(63)

and

(64)

Write each particle position as

(65)

where ρ_i is measured from the center of mass.

Similarly,

(66)

where

(67)

The center of mass definitions imply

∑ imiρi	= 0,	(68)
∑ imivi′	= 0.	(69)

Now expand

LO	= ∑ i(R + ρi) × mi(V + vi′)	(70)
	= R × MV + ∑ iρi × mivi′.	(71)

Therefore

(72)

The first term is the orbital angular momentum of the center of mass about O.

The second term is angular momentum about the center of mass:

(73)

17 Physical meaning of the decomposition

The decomposition

(74)

separates two different motions:

• motion of the system as a whole around the origin,
• motion of particles relative to the center of mass.

A system can therefore have:

• nonzero orbital angular momentum but no internal rotation,
• zero orbital angular momentum but nonzero internal angular momentum,
• both at the same time.

18 Example 3: translating and spinning system

Suppose a compact system of total mass

(75)

has center of mass position

(76)

and center of mass velocity

(77)

Its orbital angular momentum about the origin is

Lorb	= RCM × MVCM	(78)
	= (2ex) × (12ey)	(79)
	= 24ez kg m2∕s.	(80)

If its internal angular momentum about the center of mass is

(81)

then

(82)

19 Change of origin

Let O′ be displaced from O by a constant vector a:

(83)

Then

LO′	= ∑ i(ri − a) × pi	(84)
	= LO − a ×∑ ipi.	(85)

Therefore

(86)

Angular momentum about different origins differs by a term involving total linear momentum.

If

(87)

then angular momentum is the same about all origins related by a constant displacement.

20 Central force and angular momentum

Suppose a particle experiences a central force:

(88)

Then F is parallel or antiparallel to r.

Therefore

(89)

Hence

(90)

21 Planar motion under a central force

If

(91)

is constant in direction, both r and v remain perpendicular to the same fixed vector L.

Therefore the motion remains in a plane perpendicular to L.

Thus central-force motion is planar.

This result is fundamental in orbital mechanics.

22 Angular momentum in polar coordinates

For planar motion,

(92)

Then

L	= m ×	(93)
	= mr2𝜃 .	(94)

Hence

(95)

The radial velocity does not contribute because

(96)

23 Areal velocity

During time dt, the radius vector sweeps an approximate triangular area

(97)

Therefore

(98)

Using

(99)

we obtain

(100)

If angular momentum is conserved,

(101)

This is the mechanical basis of Kepler’s equal-areas law.

24 Example 4: orbital angular momentum

A satellite of mass

(102)

moves instantaneously perpendicular to its radius vector at

r	= 7.0 × 106 m,	(103)
v	= 7.5 × 103 m∕s.	(104)

Because

(105)

the angular-momentum magnitude is

L	= mrv	(106)
	= (500)(7.0 × 106)(7.5 × 103).	(107)

Thus

(108)

25 Bridge to rigid body angular momentum

For a collection of particles forming a rigid body, angular momentum is still

(109)

For rotation about a fixed axis z,

(110)

The z component becomes

Lz	= ∑ imiri⊥vi	(111)
	= ∑ imi 2ω.	(112)

Define

(113)

Then

(114)

26 Why L = Iω requires care

For fixed-axis rotation about a symmetry or principal axis, the scalar relation

(115)

is often sufficient.

For general three-dimensional rigid body motion, however, angular momentum need not be parallel to angular velocity.

The correct vector relation is

(116)

where I is the inertia tensor.

That full rigid body treatment belongs later in the rotational-dynamics sequence.

27 Angular momentum versus linear momentum

Linear momentum is

(117)

Angular momentum about O is

(118)

Thus angular momentum combines:

• translational momentum,
• geometry relative to a chosen origin.

A particle can have:

• nonzero linear momentum but zero angular momentum about a point on its line of motion,
• nonzero linear momentum and nonzero angular momentum about another point,
• zero linear momentum and therefore zero particle orbital angular momentum.

28 Common mistakes

1. Treating angular momentum as a scalar when its direction matters.
2. Forgetting to state the origin about which angular momentum is computed.
3. Using L = rp without the sin 𝜃 factor when r and p are not perpendicular.
4. Confusing torque with angular momentum.
5. Assuming zero net force implies zero net torque.
6. Assuming zero torque about one point implies zero torque about every point.
7. Forgetting that internal torque cancellation requires the appropriate central-force structure in the particle model.
8. Mixing angular momentum about the center of mass with angular momentum about a laboratory origin.
9. Forgetting the orbital term R_CM × P.
10. Using L = Iω as a general Vector Identity for arbitrary three-dimensional rigid body motion.
11. Forgetting that central force means force along the radius vector from the force center.
12. Confusing angular impulse with ordinary linear impulse.
13. Forgetting that a particle moving radially has zero angular momentum about the force center even if its speed is nonzero.

29 Practice exercises

1. A particle has r = (2e_x + 3e_y) m and p = (4e_x − 1e_y) kg m∕s. Find L_O.
2. A 3.0 kg particle moves at 5.0 m∕s along a straight line whose perpendicular distance from the origin is 2.0 m. Find the angular-momentum magnitude about the origin.
3. Derive dL_O∕dt = τ_O for a particle about a fixed inertial origin.
4. A constant torque of 6 N m acts for 0.50 s. Find the angular impulse.
5. Show that the internal torque of a central Newton’s-third-law force pair vanishes.
6. Derive the total system equation dL_O∕dt = τ_O,ext under the central internal-force assumption.
7. Derive

   (119)

8. A system has R_CM = 4e_x m, total momentum P = 3e_y kg m∕s, and L_CM = 2e_z kg m²∕s. Find L_O.
9. Show how angular momentum changes when the origin is shifted by a constant vector a.
10. Prove that a central force produces zero torque about the force center.
11. Derive L = mr²𝜃 for planar polar motion.
12. Derive the areal-velocity relation dA∕dt = L∕(2m).
13. Explain why a particle moving directly toward the origin has zero angular momentum about that origin.
14. For fixed-axis rigid body rotation, derive L_z = I_zω from the particle definition of angular momentum.
15. Explain why the general vector relation for a rigid body is L = I ⋅ω rather than simply L = Iω.

30 Summary

For a particle,

(120)

Its magnitude is

(121)

Torque is

(122)

About a fixed inertial origin,

(123)

For a particle system with central internal forces,

(124)

The total angular momentum separates as

(125)

For a central force,

(126)

For fixed-axis rigid body rotation,

(127)

M04-08 develops conservation of angular momentum and central-force motion in greater depth.

References

References