PHY5200 F07

Chapter 3: Momentum and Angular Momentum

Reading

Taylor 3.2-3.3 (today) and 3.3-3.4 for Monday. Friday is the first exam.

Recall

Conservation of Momentum: d_tP = F^ext.

Applied to a rocket, this yields the rocket equation

v(m) = v₀ + v_exln(m₀/m).

Systems of Particles

Up to now we've applied Newton's laws to "point-like" particles only. Even when speaking of objects that are clearly not point-like, such as blocks, cars, trucks, and pendulum bobs, we've somehow managed to ignore the physical extent of the object. Although not yet shown, this treatment is justified to the extent that the body is rigid and not rotating (spinning or twisting). Now we will begin to investigate the case of "rigid" bodies, objects with a fixed, rigid, physical size, or, stated more appropriately for our purposes, a collection of particles where the relative locations of the particles is fixed. To understand how Newton's laws apply in this case, we start with a system of point-like particles, each particle obeying Newton's laws in the ways we have already studied. By summing the effects of all the particles, we will obtain results for rigid bodies.

The result we are after is that the motion of a rigid body can be decomposed into two parts:

The motion of a point [commonly the center of mass (CM)] of the entire system governed by forces external to the system.
Rotation about that point governed by external torques.

This result is immensely useful, and allows us to treat the motion of planets and stars as well as tops and pendula.

The Center of Mass

For a system of n particles, we define the center of mass (CM) of the system as

r_CM = (m₁r₁ + m₂r₂ + ... + m_nr_n) / (m₁ + m₂ + ... + m_n) = (1/M) ∑_i m_i r_i

where m_i and r_i are the mass and position vector of the i^th particle, and M = ∑_i m_i is the total mass of the system. For a continuous object the sum can be taken to an integral over the entire object

r_CM = (1/M) &int_vol r dm = (1/M) ∫_vol rρ(r)dV

where the integrals are taken over the volume of the object, dm is an element of mass, ρ(r) is the density of the object (possibly a function of position), and dV is a volume element (dxdydz in rectangular coordinates). The velocity of the center of mass is found by differentiating with respect to time:

v_CM = dr_CM / dt = (1/M) ∑_i m_i dr_i / dt = (1/M) ∑_i m_i v_i

Logically we define the center of mass momentum as the total mass times of the system times the center of mass velocity. The result is:

p_CM = M v_CM = ∑_i m_i v_i = ∑_i p_i

To summarize, the momentum of the center of mass of a system of particles is equal to the vector sum of the momenta of the particles.

Forces: Internal and External

Suppose each particle is acted on by an external force F_i, and particle j exerts a force F_ij on particle i. By Newton's third law, particle i exerts a force on particle j that is equal in magnitude and opposite in direction, F_ij = -F_ji. The total force acting on particle i is

F^tot_i = F_i + ∑_j F_ij

where we include a term F_ii in the sum for convenience, but we force this term to be zero. The total force acting on the system is the sum of the forces acting on each part of the system.

F_tot = ∑_i F_{tot i} = ∑_i F_i + ∑_i≠j F_ji = ∑_i F_i = ∑_i dp_i / dt = dp_CM / dt

This statement says that we can treat the overall motion of a system of particles as if we had a single particle of mass M located at the center of mass of the system, and subject to the total external force on the system. This is an enormous simplification! No matter how the system of particles is arranged, or how complex is their internal interactions, we can treat the entire system as a single particle located at the center of mass of the system for the purposes of understanding the overall motion -- that is, the motion of the center of mass.