Ph 425 Analytical Mechanics Lecture Notes (R. Egerton, A:Ph425ch7.doc)

Dealing with particles connected by internal forces, not necessarily rigid bodies

7.1 Center of mass: definition of vector and coordinates (Eqs. 7.1.1 & 7.1.2)

Linear momentum of system is total mass times velocity of center of mass (Eq.7.1.4) since total mass m is constant (m’ = 0).

Newton’s 2^nd law: internal forces cancel so S F_ext = m p’ = m a_cm if 3rd law holds.

If F_ext = 0, v_cm = constant

7.2 Angular momentum L

S torque = dL/dt = 0 for an isolated system, if the internal forces are central.

Eq.(7.2.14): L has orbital (v_cm) and spin (about C of M) components,

which are not necessarily parallel to one another

(Examples: earth moving around the sun, electrons around an atomic nucleus)

Likewise, the kinetic energy T can be assigned orbital and spin components

3. Two-body motion

Assume an isolated system (no external forces) in which the bodies are particles or uniform spheres, interacting via central forces.

The system is described at any one instant by 6 quantities, which could be:

The position of particle 1 (3 coordinates)

The position of particle 2 (3 coordinates)

However, it is sometimes more convenient to use:

The relative position of particle 1 relative to particle 2 (3 coordinates)

The position of the center of mass (3 coordinates)

Since the system is isolated, v_cm = constant and is of no interest.

If we now measure coordinates of the two particles relative to the C of M,

m₁r₁ +m₂r₂ = (m_cm) (0) = 0 , as in Eq.(7.3.1)

If R denotes the position of mass 1 relative to mass 2,

R = r₁ – r₂ = r₁(1+m₁/m₂)

Since forces are the same in any inertial system, we can write Newton’s 2^nd law as:

m₁ (r₁)’’ = F₁ = f(R) (R/R) = f(R) e_R

R’’ = (1+m₁/m₂) (r₁)’’ = (1+m₁/m₂) f(R) (R/R) /m₁ = (1/m₁ + 1/m₂) f(R) (R/R)

m R’’ = f(R) (R/R) where 1/m = 1/m₁ + 1/m₂

which is the equation of motion for a single mass m moving in a central field f(R);

m is the reduced mass of the system, ( = m/2 for m₁=m₂=m ; ~ m₁ if m₂ >> m₁)

Our 2-body problem has therefore been reduced to a 1-body problem.

Example: planet(mass m) + sun (mass M).

Assume (for simplicity) a nearly circular orbit (radius r), so that:

m v²/r = GmM/r² , v² = GmM/m r

(Note: only the inertial mass on LHS of equation gets replaced by reduced mass)

The period T is given by T² = (2p r/v)² = 4p ²r^3m /(GmM) = 4p ² r³ / [G(m+M)]

In our solar system, the largest planet (Jupiter) has m ~ M/1000, so T is reduced by 0.05% due to the sun’s motion about the common C or M.

Binary stars or stars rotating with a black hole provide other examples (p.265).

In the case of Rutherford scattering, motion of the nucleus can be included by replacing E = ½ mv² by E = ½ m v² in the formula for ds /dW .

4. Three-body problem

Adding a third body adds so much complexity that analytical solutions are not possible, except under special (resticted 3-body) conditions where two of the bodies (the primaries) are much more massive than the third (the tertiary) and all are moving in near-circular orbits in the same plane.

The tertiary then merely responds to the gravitational attraction of the primaries.

Three forms of solution are known:

1. The tertiary orbits both primaries at a distance >> the primary separation.
2. The tertiary orbits one of the primaries at a distance << the primary separation.

For example, moons of the planets in our solar system.

3. The tertiary partakes in the rotational motion of the primaries because it is located at one of five Lagrangian points. Only two of these, however, are points of stable equilibrium; one example is the Trojan asteroids which orbit the sun close to Jupiter.

The proof of (3) makes use of an effective potential which incorporates a Coriolis term;

see pp. 270 – 274.

3. Two-particle collisions

In simple collision situations, the forces are internal (equal and opposite) and so the system can be regarded as isolated. Linear momentum is conserved.

If kinetic energy is conserved, the collision is termed elastic.

Direct (head-on) collisions constitute a 1-dimensional problem,

And can be characterised by a coefficient of restitution: e = v’ / v where v and v’ are the relative speeds before and after the collision.

Elastic collisions correspond to e = 1, totally-inelastic collisions (where the bodies stay together after the collision) correspond to e = 0.

Some materials provide values of e which increase or decrease with initial velocity v.

Can show (Problem 7.9) that e is related to the energy Q liberated in the collision (by dissipative forces):

Q = ½ m v² (1-e ²)

Collisions usually involve forces of duration, such that we can integrate the force to give a quantity called impulse: P = ò F dt = ò (ma) dt = m (D v)

= change in momentum of one particle relative to the other

In particle-scattering theory, the impulse approximation corresponds to regarding the time interval as approaching zero, so that F is a delta function.

More generally, it is possible to divide the interaction time into two parts: compression phase (impulse P_c) and a restitution phase (impulse P_r). Can show (p. 285) that P_r/ P_c = e

4. Oblique collisions

These are a two-dimensional problems (Fig. 7.6.1, p.288) so conservation of momentum is a vector equation, given by Eq.(7.6.1) if one particle (mass m₂) is initially stationary, often a good approximation in atomic- and particle-physics situations.

If (in addition) we have an elastic collision (Q=0) and equal masses, can show that the two particles emerge at right angles to each other (j ₁ + j ₂ = p /2, independent of the scattering angles j ₁ and j ₂ )

Analysis of collisions often makes use of center-of-mass (CM) coordinates.

Using these coordinates, the particles approach a fixed point (the CM) and then separate, but in a direction which is at an angle q to the line of approach (see Fig.7.6.1).

Eq.(7.16.3) gives the relationships between j ₁ , j ₂ and q , v_cm

j ₁ and j ₂ refer to laboratory coordinates in which measurements are made.

In CM coordinates, the kinetic energy (and momentum) can be zero after the collision.

If we can arrange for the CM to be stationary, this situation corresponds to the maximum amount of released energy Q. This principle is utilized in particle-physics colliders, where similar particles approach in opposite directions at the same speed and the aim is to create new particles with the energy Q.

(In a fixed-target accelerator, the original particles retain much of their KE after the collision, since we must satisfy conservation of momentum as well as conservation of energy).

5. Variable-mass problems

Newton’s 2^nd Law must be expressed in its original form: F = dp/dt where p refers to the momentum of the entire system. Best illustrated by taking examples.

Conveyor belt: mass m(t) moving at velocity v(t) and acquiring particles of velocity u(t). In time dt the change in momentum is:

dP = (m + dm)(v + dv) – mv – u dm = m dv (+ dm dv) – V dm

where V = u - v = velocity of the particles relative to mass m

This change in momentum requires an external force:

F = dP/dt = m(dv/dt) – V(dm/dt)

Consider the horizontal (x-) components of motion:

If F_x = 0 and u_x = 0, dv_x/dt = (0-v_x) dm/dt < 0 ; the belt will decelerate

(forces F_y and F_z would be required to prevent other motion if u_y ¹ 0 ¹ u_z )

F_x >0 is needed to maintain constant v_x .

The rocket is a related case, in which dm < 0 (the rocket loses fuel);

V(dm/dt) is called the thrust and is negative (u itself is opposite to the motion)

If F = 0 (no gravity, rocket in space) and thrust is constant (magnitude V = -V),

m dv/dt = V dm/dt = (-V) dm/dt , giving v – v₀ = -V ò (dm / m) = V ln (m₀/m)

If F = mg = - mg (opposing the motion) and v₀ = 0 (takeoff from earth)

m dv/dt = V dm/dt – mg = (-V) dm/dt - mg, giving v = -V ò (dm / m) - ò g(t) dt

= V ln (m₀/m) – g (t-t₀) = V ln (m₀/m) – g(mass of fuel / rate of burning) in the initial portion of the flight

In general, m₀/m must be large to achieve v >> V (or simply to escape from gravity), in other words a large fuel/payload ratio is needed (typically 20:1 for launching a near-earth satellite; see Example 7.7.1, p.295).