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Physics 207 WebPages - Fall 1999
Chapter 9

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Chapter 16

CHAPTER 9: Systems of Particles

Brandon Anis, William Pruett, George Hitt, Lynette Penn, Calvin Williams, Rudy Corbett

*Denotes problem that may be on the Exam

 

Main topics:

  • Center of Mass
  • Newton’s 2nd Law For a System of Particles
  • Linear Momentum/Conservation of
  • Systems of Varying Mass

Center of Mass

The Center of mass a point at which you can treat any size object as a point mass when applying external forces.

In a Boomerang, the center of mass is located in it’s internal concavity.

 

The formula for the center of mass is given by

rcm = mir

  • Tips: When working with center of mass problems, break r down into it’s x, y, and z components and remember that the center of mass is not always in the object.

Newton’s 2nd Law for a System of Particles

Fext = Macm

Where

1. Fext is the vector sum of all the external forces acting on the system. Always be sure not to include internal forces, or forces that one part of a system exerts on another part.

2. M is the total mass of the system and is constant.

3. acm is the acceleration of the center of mass of the system and not any other point in the system.

Linear Momentum/Conservation of

Linear momentum (a vector quantity) of a particle is defined as

p = mv

Or momentum is equal to the mass times the velocity. Its SI units are kilogram-meter per second.

For a system of particles, it simply changes to

P = Mvcm

Using momentum, Newton’s second law can be written as

Fext =

In any closed, isolated system, the law of conservation of linear momentum states that

Pi=Pf

Question: If a system of particles has zero momentum does the system necessarily have zero kinetic energy?

If P = Mvcm = 0 does KE=mv² = 0 ?

NO! Momentum is a vector quantity, while kinetic energy is a scalar.

  • Tips: Make sure the system is closed and isolated. Be careful that boundaries of the system exclude objects that exert non-conservative forces. In other words, in that case Pi would not equal Pf.

Systems with Varying Mass

Things get a little more complicated when the mass begins to vary, like in a rocket for example. The book gives two equations for working with rockets.

First Rocket Equation says

Ru = Ma

Where R is the rate at which the fuel is consumed and u is the speed at which it is being ejected. Thrust, T, is defined as Ru, giving you

T = Ma

Where a is the acceleration at whatever time it’s mass is M.

Second Rocket Equation says

vf – vi = u ln

With Mi as the initial mass and Mf as the final mass

  • Tips: Systems with varying mass are characteristic of rocket problems.

The shuttle uses thousands of tons of fuel

to place only a few tons of material into orbit.

Links:

Test your skills with momentum conservation with a game of pool:

http://members.aol.com/jiping/pool.html

University of Wisconsin discussion of Linear Momentum/C.M.

http://lupine.physics.wisc.edu/207/lecture10/image-index.html

University of Illinois at Urbana-Champaign - Physics notes

http://courses.physics.uiuc.edu/cyberprof-docs/physics/phys101/lect/

 

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