Derive, for both the single particle and system of particles, the Work-Energy Principle or Work-Energy Theorem

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Derive, for both the single particle and system of particles, the Work-Energy Principle or Work-Energy Theorem. 2. Draw a system of particles and write all the basic equations that describe the mechanics of a System of Particles. 3. Compare and contrast between Hamilton’s H and Lagrange’s L.

The Correct Answer and Explanation is:

1. Derivation of the Work-Energy Theorem

Single Particle:

Start with Newton’s Second Law:F⃗=ma⃗\vec{F} = m \vec{a}F=ma

Multiply both sides by the displacement dr⃗\vec{dr}dr:F⃗⋅dr⃗=ma⃗⋅dr⃗\vec{F} \cdot \vec{dr} = m \vec{a} \cdot \vec{dr}F⋅dr=ma⋅dr

Since a⃗=dv⃗dt\vec{a} = \frac{d\vec{v}}{dt}a=dtdv​, and v⃗=dr⃗dt\vec{v} = \frac{d\vec{r}}{dt}v=dtdr​, then:a⃗⋅dr⃗=dv⃗dt⋅v⃗dt=v⃗⋅dv⃗\vec{a} \cdot \vec{dr} = \frac{d\vec{v}}{dt} \cdot \vec{v} dt = \vec{v} \cdot d\vec{v}a⋅dr=dtdv​⋅vdt=v⋅dv

Thus:F⃗⋅dr⃗=mv⃗⋅dv⃗\vec{F} \cdot \vec{dr} = m \vec{v} \cdot d\vec{v}F⋅dr=mv⋅dv

Integrating both sides:∫F⃗⋅dr⃗=∫mv⃗⋅dv⃗=12mv2−12mv02\int \vec{F} \cdot \vec{dr} = \int m \vec{v} \cdot d\vec{v} = \frac{1}{2} m v^2 – \frac{1}{2} m v_0^2∫F⋅dr=∫mv⋅dv=21​mv2−21​mv02​

So,W=ΔKW = \Delta KW=ΔK

System of Particles:

Sum of external work on all particles equals change in total kinetic energy:∑Wext=ΔKtotal\sum W_{\text{ext}} = \Delta K_{\text{total}}∑Wext​=ΔKtotal​

Here, internal forces cancel due to Newton’s Third Law. So:∑i=1nF⃗ext,i⋅dr⃗i=Δ(∑i=1n12mivi2)\sum_{i=1}^{n} \vec{F}_{\text{ext},i} \cdot \vec{dr}_i = \Delta \left( \sum_{i=1}^{n} \frac{1}{2} m_i v_i^2 \right)i=1∑n​Fext,i​⋅dri​=Δ(i=1∑n​21​mi​vi2​)

2. System of Particles and Basic Equations

Diagram (You can draw this by sketching several particles labeled m1,m2,…m_1, m_2, \ldotsm1​,m2​,… with external forces F⃗ext\vec{F}_{\text{ext}}Fext​ and internal forces between them)

Basic Equations:

  • Total external force:

F⃗ext=Ma⃗CM\vec{F}_{\text{ext}} = M \vec{a}_{\text{CM}}Fext​=MaCM​

  • Momentum conservation:

P⃗=∑miv⃗i\vec{P} = \sum m_i \vec{v}_iP=∑mi​vi​

  • Angular momentum about origin:

L⃗=∑r⃗i×miv⃗i\vec{L} = \sum \vec{r}_i \times m_i \vec{v}_iL=∑ri​×mi​vi​

  • Torque:

τ⃗ext=dL⃗dt\vec{\tau}_{\text{ext}} = \frac{d\vec{L}}{dt}τext​=dtdL​

  • Work-energy for system:

∑F⃗ext⋅dr⃗=ΔKtotal\sum \vec{F}_{\text{ext}} \cdot \vec{dr} = \Delta K_{\text{total}}∑Fext​⋅dr=ΔKtotal​

3. Comparison: Hamilton’s H vs Lagrange’s L

Lagrangian L=T−VL = T – VL=T−V

  • It is the difference between kinetic and potential energy.
  • Leads to Lagrange’s equations:

ddt(∂L∂q˙i)−∂L∂qi=0\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_i} \right) – \frac{\partial L}{\partial q_i} = 0dtd​(∂q˙​i​∂L​)−∂qi​∂L​=0

  • Best for systems with constraints and generalized coordinates.

Hamiltonian H=T+VH = T + VH=T+V (in many cases)

  • Represents total energy if potential energy does not depend on velocity.
  • Used in Hamilton’s equations:

q˙i=∂H∂pi,p˙i=−∂H∂qi\dot{q}_i = \frac{\partial H}{\partial p_i}, \quad \dot{p}_i = -\frac{\partial H}{\partial q_i}q˙​i​=∂pi​∂H​,p˙​i​=−∂qi​∂H​

  • More powerful for analytical mechanics and quantum mechanics.

Contrast:

  • Lagrangian depends on positions and velocities; Hamiltonian uses positions and momenta.
  • Lagrangian is better suited to derive equations of motion; Hamiltonian is more symmetric and useful in phase space analysis.

Both are reformulations of Newtonian mechanics and lead to the same physical predictions.

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