Dynamics · Chapter 6 of 10 · Advanced
Systems of Particles and Variable Mass
A whole system moves as if its mass were at one point. Apply that to streams of fluid and to bodies that shed mass, and you get jet forces and the rocket equation.
Readiness check
This chapter scales the particle laws up to systems and streams. Tick only what you can do closed-notes.
- Apply impulse-momentum to a single particle.
- State conservation of momentum.
- Find a mass-flow rate ṁ = ρAv.
- Work with natural logarithms.
- Define a system boundary clearly.
The core idea
The net external force on any system equals its total mass times the acceleration of its mass center; for a flow, force equals the rate at which momentum is carried through.
ΣF = m aGΣF = ṁ(vout − vin)Δv = ve ln(m₀/m_f)Internal forces between members of a system cancel in action-reaction pairs, so only external forces move the mass center. Applied to a steady stream, this becomes the momentum-flow form: the force on a vane or pipe is the rate of change of the fluid's momentum. Applied to a body ejecting mass, it gives thrust and the rocket equation.
The skills, taught in order
This chapter generalises the particle laws and applies them to flows. Five skills run from the mass center to the rocket equation.
6.1 The mass center of a system
For a collection of particles, ΣF = m aG, where m is the total mass and aG the acceleration of the mass center. This is why a tumbling wrench's center still follows a clean parabola: the messy internal motion does not affect G.
6.2 Energy and momentum of a system
The work-energy and impulse-momentum principles extend to systems by summing over the members. Internal forces can do net work (a spring between two carts), so kinetic energy is not always conserved, but internal impulses always cancel, so total momentum responds only to external impulse.
6.3 Conservation for systems
With no external force, the mass center moves at constant velocity and total momentum is conserved. An exploding shell is the classic case: the fragments fly apart, yet their mass center continues along the original trajectory as if nothing happened.
6.4 Steady mass flow
For a steady stream entering and leaving a control volume, the force equals the momentum flow rate: ΣF = ṁ(vout − vin), with ṁ = ρAv. This single relation gives the thrust of a jet engine, the force of water on a turbine vane, and the reaction on a pipe bend.
| Situation | Force | Why |
|---|---|---|
| Jet on a flat plate | F = ṁv | all axial momentum removed |
| Jet engine | T = ṁ(ve − vi) | momentum added to the flow |
| Pipe bend | F = ṁ(vout − vin) | direction of momentum changes |
6.5 Variable mass and the rocket equation
When a body ejects mass at relative speed ve, that mass flow produces a thrust ṁve. Integrating over a burn gives the Tsiolkovsky rocket equation, Δv = ve ln(m₀/mf): the velocity gained depends on the exhaust speed and the logarithm of the mass ratio. The logarithm is why staging matters.
| Mass ratio m₀/m_f | Δv (for ve = 2500 m/s) |
|---|---|
| 2 | 1733 m/s |
| 3 | 2747 m/s |
| 5 | 4024 m/s |
| 10 | 5756 m/s |
Engineering connection: jet and rocket propulsion, turbine and pump blading, pipe-bend anchoring, and any process with mass crossing a boundary.
Worked example 1: force of a water jet on a plate
A water jet of cross-section 0.001 m² leaves a nozzle at 25 m/s and strikes a fixed flat plate at right angles, spreading sideways. Take water density 1000 kg/m³ and find the force on the plate.
- ProblemFind the force the jet exerts on the plate in Figure 1.
- Given / findA = 0.001 m², v = 25 m/s, ρ = 1000 kg/m³. Find F.
- AssumptionsSteady flow, the jet leaves the plate with no axial velocity component, plate fixed.
- ModelSteady mass flow: the plate removes all of the jet's axial momentum, so F = ṁ(v − 0).
- Equationsṁ = ρAv F = ṁ(vin − vout) = ṁv
- Solveṁ = 1000 × 0.001 × 25 = 25 kg/s. F = 25 × 25 = 625 N.
- CheckThe force scales as v², since both ṁ and the velocity change grow with v. Doubling the jet speed would quadruple the force to 2500 N, which is why high-pressure jets cut and clean so effectively.
- ConclusionThe momentum-flow form turns a fluid problem into one line of dynamics. The same method, with vout reversed for a curved vane, would nearly double the force.
Worked example 2: the rocket equation
A rocket has an initial mass of 5000 kg, of which 3500 kg is propellant, and an exhaust speed of 2500 m/s. Find the ideal velocity gain (no gravity or drag), and the thrust if the propellant burns in 70 s.
- ProblemFind the ideal Δv and the thrust for the rocket in Figure 2.
- Given / findm₀ = 5000 kg, propellant 3500 kg so mf = 1500 kg, ve = 2500 m/s, burn time 70 s. Find Δv and T.
- AssumptionsConstant exhaust speed, no gravity or drag (ideal Δv), steady burn rate.
- ModelTsiolkovsky equation for Δv; thrust from the mass-flow rate times exhaust speed.
- EquationsΔv = ve ln(m₀/mf) ṁ = (m₀ − mf)/t T = ṁ ve
- SolveMass ratio = 5000/1500 = 3.33. Δv = 2500 × ln(3.33) = 2500 × 1.204 = 3010 m/s. Burn rate ṁ = 3500/70 = 50 kg/s, so thrust T = 50 × 2500 = 125 000 N = 125 kN.
- CheckThe Δv is less than the exhaust speed because the mass ratio is modest; reaching orbital Δv (about 9 km/s) would need either a far higher mass ratio or staging. Thrust comfortably exceeds the initial weight (5000 × 9.81 = 49 kN), so the rocket can lift off.
- ConclusionThe logarithm is the tyranny of the rocket equation: each extra increment of Δv demands a disproportionately larger mass ratio, which is exactly why rockets are staged.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Ignoring ejected mass | Rocket treated as constant-mass | "Is the system losing or gaining mass?" | Use the variable-mass thrust term; mass changes during the burn. |
| Internal forces moving G | Mass center accelerated by internal push | "Is this force external to the system?" | Only external forces move the mass center; internal pairs cancel. |
| Wrong velocity change in flow | Jet force uses speed, not change | "What is vout − vin?" | Force is ṁ times the vector velocity change, including direction. |
| Linear thinking on Δv | Expecting Δv to scale with fuel mass | "Is the relation linear or logarithmic?" | Δv grows with ln(mass ratio); returns diminish sharply. |
Practice ladder
A pipe carries 12 kg/s of water at 8 m/s and discharges it to atmosphere through a nozzle at 30 m/s. Find the axial force needed to hold the nozzle.
Show answer
F = ṁ(vout − vin) = 12(30 − 8) = 264 N. The nozzle must be anchored against this reaction, the same effect that makes a fire hose buck.
For the Worked Example 1 jet, how would the force change if the plate were a deep cup that reversed the flow (vout = −v)?
Show answer
F = ṁ(v − (−v)) = 2ṁv = 2 × 625 = 1250 N. Reversing the flow doubles the momentum change, which is why turbine buckets and Pelton wheels are cupped.
A rocket must gain 4000 m/s with an exhaust speed of 3000 m/s. What mass ratio is required, and what fraction of the launch mass is propellant?
Show answer
m₀/mf = eΔv/ve = e4000/3000 = e1.333 = 3.79. So mf is 1/3.79 = 26% of launch mass, meaning 74% must be propellant, a sobering illustration of the rocket equation.
Pick a real momentum-flow device (a garden hose, a leaf blower, a model rocket, a turbine). Estimate the mass-flow rate and velocities, and compute a thrust or reaction force.
What good work looks like
A clear control volume, ṁ = ρAv computed, the velocity change taken as a vector, and a thrust or anchoring force with a magnitude check.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Analyse one real propulsion or flow device with the momentum-flow or rocket equation, defining the boundary and computing a force or Δv with a stated assumption.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. What governs the motion of a system's mass center?
Only the external resultant: ΣF = m aG; internal forces cancel.
2. Write the steady mass-flow force.
ΣF = ṁ(vout − vin), with ṁ = ρAv.
3. State the rocket equation.
Δv = ve ln(m₀/mf).
4. Why does an exploding shell's center keep its path?
The explosion is internal, so with only gravity external, the mass center follows the original trajectory.
5. Why are rockets staged?
Because Δv grows only logarithmically with mass ratio; dropping empty mass resets the ratio for the next stage.
Textbook mapping
| Item | Mapping |
|---|---|
| Primary source | Meriam and Kraige, Engineering Mechanics: Dynamics (7th ed), Chapter 4 (Kinetics of Systems of Particles) |
| Cross-reference | Hibbeler, Dynamics, Ch. 15 (steady flow, variable mass) · Beer and Johnston, Ch. 14 |
| Core topics | 6.1 Mass center · 6.2 System energy and momentum · 6.3 Conservation · 6.4 Steady mass flow · 6.5 Variable mass and the rocket equation |
| Engineering connection | Jet and rocket propulsion, turbine blading, and pipe-bend forces. |
| Read next | Chapter 7: Plane Kinematics of Rigid Bodies. |