Dynamics of a System of Particles Prof. Claude A Pruneau Notes compiled by L. Tarini Physics and Astronomy Department Wayne State University PHY 6200 Theoretical Mechanics

Introduction So far dealt with point-like objects, or at least assume we could… Let’s begin to consider objects with finite size or extent. First discuss a system of “n” particles. Then discuss continuous limit for rigid bodies.

o CM

Center of Mass Discrete System of “n” masses: Continuous System: Total mass:

Problem: Calculate the CM position of a solid (uniform density) half-sphere. Solution: By virtue of symmetry Density: x y z dy a

Along “y” : Answer:

Definitions

Newton’s 3rd Law Newton’s 2nd Law Sum over  : Conclusion:

Conclusion (1) The CM moves as if it were a single particle, of mass equal to the total mass of the system, acted on by the total external force and independent of the nature of the internal forces provided:

Conclusions (2) Linear momentum of the system is the same as if a single particle of mass M were located at the position of the center of mass and moving in the manner the CM moves.

The total linear momentum for a system free of external forces is constant and equal to the linear momentum of the CM. Conclusions (3)

Example: A chain of uniform mass density , length b, and mass M hangs from the ceiling. End b is released at t=0. Solution: Density: A B CM A B x t = 0 t > 0 Equation of motion: Right side in motion: Free fall

Solve for

Energy

Angular momentum of a system of n particles

thus The total angular momentum about an origin is the sum of the angular momentum of the CM about that origin and the angular momentum of the system about the position of the CM.

But Thus

Assumeis along Finally If the net resultant external torques about a given axis vanish, then the total angular momentum of the system about that axis remains constant in time.

Energy of the System Consider the work done on a system in moving all particles from a certain configuration (1) to another (2) when all positions are specified.

The total energy of the system is equal to the sum of the kinetic energy of a particle of mass M moving with the velocity of the center of mass and the kinetic energy of motion of the individual particles relative to the center of mass. =0

If andare conservative forces,

with

corresponds to internal potential energy. The total energy for a conservative system is constant. The term

Example: A rope of uniform linear density and mass wrapped one complete turn around a hollow cylinder of mass M and radius R. Cylinder rotates freely about its axis. Rope’s end is at. When P is at, system is slightly displaced. Find angular velocity. Solution: