Center of Mass Motion and System Dynamics

Center of Mass Motion and System Dynamics

A many-particle system can contain complicated internal motion, yet one special point often moves according to a remarkably simple equation.

That point is the center of mass.

For particles with masses m_i at positions r_i, define the total mass

(1)

and the center of mass position

(2)

For a fixed-mass system, differentiating gives

(3)

and therefore

(4)

Differentiating once more,

(5)

Thus the center of mass moves as though the entire system mass were concentrated there and acted on by the net external force, as far as translational motion is concerned.

1 Discrete-particle definition

For N particles,

(6)

In Cartesian components,

XCM	= ∑ imixi,	(7)
Y CM	= ∑ imiyi,	(8)
ZCM	= ∑ imizi.	(9)

Each coordinate is a mass-weighted average.

Particles with greater mass contribute more strongly to the center of mass location.

2 Two-particle center of mass

For two particles,

(10)

In one dimension,

(11)

If

(12)

the center of mass lies closer to particle 2.

3 Example 1: two particles on a line

Let

m1	= 2.0 kg,	x1	= 0,	(13)
m2	= 3.0 kg,	x2	= 10 m.	(14)

Then

XCM	=	(15)
	= 6.0 m.	(16)

Thus

(17)

The center of mass lies closer to the 3.0 kg particle.

4 The center of mass need not lie inside matter

The center of mass is a weighted geometric point, not necessarily a material point.

Examples include:

• a ring, whose center of mass lies at its empty geometric center,
• a hollow sphere,
• two separated particles,
• a curved wire,
• a rigid body with a cavity.

The center of mass can lie in empty space.

5 Center of mass velocity

For constant particle masses,

(18)

Differentiate:

VCM	=	(19)
	= ∑ imi.	(20)

Therefore

(21)

Multiply by M:

(22)

The right side is the total momentum:

(23)

Hence

(24)

6 Center of mass acceleration

Differentiate the velocity:

(25)

Multiply by M:

(26)

Using Newton’s second law for each particle,

(27)

Thus

(28)

For ordinary Newtonian internal-force pairs,

(29)

Therefore the internal forces cancel in the total:

(30)

Hence

(31)

7 System dynamics equation

The center of mass equation can be written in two equivalent forms:

(32)

and, for fixed total mass,

(33)

This is the system analogue of

(34)

It describes the translational motion of the system as a whole.

8 Internal forces cannot accelerate the center of mass

Suppose a system is isolated:

(35)

Then

(36)

Therefore

(37)

Internal forces can:

• accelerate individual particles,
• cause vibration,
• cause rotation,
• cause recoil,
• convert internal energy into kinetic energy,
• change the shape of a nonrigid system.

But they cannot change the center of mass velocity of an isolated fixed-mass system.

9 Internal rearrangement and center of mass motion

Consider two masses connected by an internal spring.

When the spring expands or contracts, both masses accelerate.

Their individual velocities change, yet

(38)

remains constant if no external force acts.

Thus the masses can move relative to the center of mass while the center of mass moves uniformly.

10 Example 2: recoil and the center of mass

Two skaters initially stand at rest on frictionless ice.

Let

m1	= 50 kg,	(39)
m2	= 75 kg.	(40)

Initially,

(41)

They push apart.

If skater 1 moves at

(42)

then

(43)

Thus

v2	= −v1	(44)
	= −(3.0)	(45)
	= −2.0 m∕s.	(46)

The skaters separate, but

(47)

throughout.

11 Explosion and center of mass trajectory

Suppose a projectile follows a parabolic trajectory under gravity and then explodes internally into fragments.

Neglect air resistance.

The explosion produces large internal forces, but the only external force remains gravity.

Therefore

(48)

Hence

(49)

The center of mass continues along the same parabolic trajectory that the unexploded object would have followed.

The fragments spread apart around that trajectory.

12 Uniform gravity and the center of mass

For particles in a uniform gravitational field,

(50)

The total gravitational force is

Fg	= ∑ imig	(51)
	= g.	(52)

Thus

(53)

The center of mass therefore accelerates as

(54)

if gravity is the only external force.

13 Gravitational potential energy in a uniform field

The total gravitational potential energy is

(55)

Using

(56)

we obtain

(57)

Therefore

(58)

This is another example of the center of mass capturing the translational behavior of a system.

14 The center of mass and rigid body translation

For a rigid body, the center of mass is fixed relative to the body’s material geometry.

The rigid body can:

• translate,
• rotate about its center of mass,
• translate and rotate simultaneously.

Its total linear momentum is still

(59)

Rotation about the center of mass does not change the total linear momentum.

This separation between center of mass translation and motion relative to the center of mass is fundamental to rigid body mechanics.

15 Example 3: person walking on a free cart

A person of mass

(60)

stands on a cart of mass

(61)

The cart can roll freely on a horizontal frictionless track.

The person walks

(62)

to the right relative to the cart.

Let the cart move a distance x to the left relative to the ground.

Then the person’s displacement relative to the ground is

(63)

With no horizontal external force, the center of mass position remains fixed:

(64)

Thus

(65)

Therefore

(66)

Numerically,

x	= (3.0)	(67)
	= 1.2 m.	(68)

Thus the cart moves

(69)

to the left.

16 Why the person-cart problem is not a momentum-only problem

Momentum conservation implies

(70)

at every instant if the system starts from rest and no horizontal external force acts.

However, the requested quantity is usually a displacement.

The position form of the center of mass relation,

(71)

is therefore more direct.

Momentum conservation and fixed center of mass position are two related descriptions of the same isolated translational dynamics.

17 Continuous mass distributions

For a continuous body,

(72)

becomes an integral:

(73)

In Cartesian coordinates,

XCM	= ∫ xdm,	(74)
Y CM	= ∫ y dm,	(75)
ZCM	= ∫ z dm.	(76)

The mass element can be written using an appropriate density.

For a line,

(77)

For a surface,

(78)

For a volume,

(79)

18 Example 4: uniform rod

A uniform rod has length L and linear density

(80)

Place the rod along the x axis from

(81)

to

(82)

Then

(83)

The center of mass is

XCM	= ∫ 0Lxdm	(84)
	= ∫ 0Lxλdx	(85)
	= 0L	(86)
	= .	(87)

Therefore

(88)

Symmetry predicts the same result immediately.

19 Symmetry as a center of mass tool

If a mass distribution has a symmetry plane, the center of mass lies in that plane.

If it has a symmetry axis, the center of mass lies on that axis.

Examples:

• uniform rod: midpoint,
• uniform disk: geometric center,
• uniform sphere: geometric center,
• uniform rectangular block: geometric center.

Symmetry can eliminate one or more integrals.

20 Composite systems

A complicated object can often be divided into simpler pieces.

If piece k has mass M_k and center of mass R_k, then the complete center of mass is

(89)

Each piece can therefore be treated as if its mass were concentrated at its own center of mass for the purpose of locating the overall center of mass.

21 Cavities and negative-mass bookkeeping

A cavity can be handled mathematically by superposition.

One may:

1. begin with the complete solid body,
2. represent the removed region as a piece with negative mass,
3. compute the combined mass-weighted position.

The negative mass is only a bookkeeping device. It does not represent physical negative matter.

22 Center of mass frame

Define the center of mass frame as the inertial frame moving with velocity

(90)

relative to the laboratory frame.

For particle i,

(91)

Then

∑ imivi′	= ∑ imi(vi − VCM)	(92)
	= P − MVCM	(93)
	= 0.	(94)

Thus

(95)

in the center of mass frame.

23 Kinetic-energy decomposition

The particle-system kinetic energy can be decomposed into center of mass and relative parts:

(96)

The relative part is

(97)

This relation was developed in M03-08.

It shows that system kinetic energy naturally separates into:

• translation of the center of mass,
• internal or relative motion about the center of mass.

24 External work versus center of mass motion

The center of mass equation determines

(98)

through

(99)

This center of mass pseudowork is not generally the same as the total actual external work on a multi-particle system.

Actual external work can also change:

• relative kinetic energy,
• rotation,
• vibration,
• deformation,
• internal energy.

Thus system dynamics and system energy bookkeeping answer related but distinct questions.

25 Example 5: constant external force on a system

A system has total mass

(100)

The net external force is constant:

(101)

Then

(102)

If the center of mass starts at rest,

(103)

The individual particles can have complicated internal motions, but those motions do not alter this center of mass acceleration.

26 Example 6: external impulse and center of mass velocity

A system of total mass

(104)

receives external impulse

(105)

Since

(106)

we obtain

(107)

Thus

(108)

27 System boundary remains essential

The center of mass equations apply to the chosen system.

If a person and cart are both inside the system boundary, their mutual forces are internal.

If only the cart is chosen as the system, the person’s force on the cart is external.

The same physical interaction can therefore be internal or external depending on the system definition.

A correct system boundary is part of the physics, not merely bookkeeping.

28 Fixed-mass limitation

The simple form

(109)

used here assumes a fixed set of particles with constant total mass.

If mass crosses the system boundary, momentum is transported with that mass.

Then an open-system momentum balance is required.

Rockets, leaking containers, and jets therefore need additional terms beyond the fixed-mass center of mass equation.

That is the subject of M04-06.

29 Common mistakes

1. Using an ordinary arithmetic average instead of a mass-weighted average.
2. Assuming the center of mass must lie inside the material.
3. Forgetting that P = MV_CM is a vector equation.
4. Allowing internal forces to accelerate the center of mass of an isolated fixed-mass system.
5. Confusing motion relative to the center of mass with motion of the center of mass.
6. Forgetting that an explosion can change fragment velocities without changing the center of mass trajectory.
7. Treating the center of mass as a physical particle that must coincide with material.
8. Forgetting that gravity gives F_g = Mg only when the field is effectively uniform over the system.
9. Using laboratory-frame velocities in some terms and center of mass-frame velocities in others.
10. Confusing actual external work with center of mass pseudowork.
11. Applying fixed-mass formulas directly to an open variable-mass system.
12. Using “rigid body” motion to imply that the center of mass cannot translate and rotate simultaneously with the body.

30 Practice exercises

1. Two particles of masses 2 kg and 5 kg are located at x = 1 m and x = 8 m. Find X_CM.
2. Three particles have masses m, 2m, and 3m at positions (0, 0), (3, 0), and (0, 4). Find the center of mass coordinates.
3. Derive P = MV_CM from the discrete center of mass definition.
4. Derive F_ext,net = MA_CM for a fixed-mass particle system.
5. Explain why internal forces can change individual particle momenta while leaving total momentum unchanged.
6. Two skaters push apart from rest. Show directly that their center of mass position remains fixed.
7. A projectile explodes in flight. Explain why its center of mass follows the original ballistic trajectory when air resistance is neglected.
8. Prove that the total gravitational potential energy in a uniform field is MgY _CM.
9. A 50 kg person walks 4.0 m relative to a 100 kg free cart. Find the cart displacement relative to the ground.
10. Derive the center of mass of a uniform rod using integration.
11. Find the center of mass of two uniform rods treated as composite pieces with known individual centers of mass.
12. Show that total momentum is zero in the center of mass frame.
13. Explain why a rotating rigid body can have zero center of mass velocity but nonzero kinetic energy.
14. A 12 kg system experiences a constant external force 36e_x N. Find its center of mass acceleration.
15. A 4 kg system receives impulse (−6e_x + 8e_y) N s. Find the change in center of mass velocity.

31 Summary

The center of mass of a discrete system is

(110)

For a continuous distribution,

(111)

For a fixed-mass system,

(112)

The translational system dynamics are governed by

(113)

If the net external force vanishes,

(114)

Internal forces can change relative motion but cannot accelerate the center of mass of an isolated fixed-mass system.

In the center of mass frame,

(115)

These results provide the foundation for the next topic: systems in which mass crosses the chosen boundary.

References

References