Systems of Particles

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Systems of Particles
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Define the center of mass of a system of discrete
particles or rigid bodies of uniform mass or
density.

• The terms center of mass and center of gravity
  are used synonymously.
• The center of mass is the point at which all
  particles can be considered to be concentrated.
• It is the point of a body at which the force of
  gravity can be considered to act and which
  undergoes no internal motion.

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Define the center of mass of a system of discrete
particles or rigid bodies of uniform mass or
density.

• The center of mass of uniform object is located
  at its geometric center.
• Non-uniform objects which are composed of uniform
  sections or individual particles, the following
  equation can be used.

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• mi mass of the individual particle
• M total mass of the system
• xi i component of the position vector for the
  location of the particle
• yi j component of the position vector for the
  location of the particle
• zi k component of the position vector for the
  location of the particle
• xCM, yCM, and zCM are the x, y, and z coordinates
  of the center of mass

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Define the center of mass of a system of discrete
particles or rigid bodies of uniform mass or
density.

• Three particles, each of mass 2.5 kg are located
  at the corners of a right triangle whose sides
  are 2 m and 1.5 m long as shown below. Locate
  the center of mass.

xCM 1.33 m yCM 0.50 m
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Define the center of mass of a system of discrete
particles or rigid bodies of uniform mass or
density.

• For non uniform objects that are made up of a
  continuous distribution of matter, the this
  equation can be used

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Define the center of mass of a system of discrete
particles or rigid bodies of uniform mass or
density.

• Find the center of mass of a uniform cone of
  height h and radius R made of copper which has a
  density of 8930 kg/m3. Hint Set up your
  coordinate axis so that the z axis is going
  upward from the tip of the cone to the top.

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Define the center of mass of a system of discrete
particles or rigid bodies of uniform mass or
density.

• A tiny slice horizontally through the cone would
  be a tiny cylinder with height dz.
• Using the definition of density and the equation
  for the volume of a cylinder, we can find that
  the mass of that tiny cylinder would be
  dm 8930(p)(r2)dz

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Define the center of mass of a system of discrete
particles or rigid bodies of uniform mass or
density.
r is related to z. The ratio of r / z is the
same as the ratio of R / h. So r is Rz / h.
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Define the center of mass of a system of discrete
particles or rigid bodies of uniform mass or
density.

• So the location of the center of mass would be
• (0,0, )

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Define the center of mass of a system of discrete
particles or rigid bodies of uniform mass or
density.

• When Newtons second law is applied to the center
  of mass, the vector sum of all the forces acting
  on the system is equal to the total mass of the
  system times the acceleration of the center of
  mass.

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Define the center of mass of a system of discrete
particles or rigid bodies of uniform mass or
density.

• A rocket is fired into the air. At the moment it
  reaches its highest point, a horizontal distance
  D from its starting point, it separates into two
  parts of equal mass. Part 1 falls vertically to
  earth. Where does part II land? Assume the
  acceleration due to gravity is constant.

Part II lands at a horizontal distance of 3D
since the center of mass still has to be located
at 2D. Part I is a distance of D before the
center of mass and Part II is a distance of D
beyond the center of mass
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• For projectiles, even if the object itself is
  rotating, the center of mass will follow a
  parabolic path.

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In this movie you will see an irregularly shaped
object as a projectile. The center of mass is
where the light bulb is located. To view the
video, click on the link below http//groups.phys
ics.umn.edu/demo/mechanics/1D4010.html
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• A fisherman stands at the back of a perfectly
  symmetrical boat of length L. The boat is at rest
  in the middle of a perfectly still and peaceful
  lake, and the fisherman has a mass 1/4 that of
  the boat. If the fisherman walks to the front of
  the boat, by how much is the boat displaced?

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• If youve ever tried to walk from one end of a
  small boat to the other, you may have noticed
  that the boat moves backward as you move forward.
  
• Thats because there are no external forces
  acting on the system, so the system as a whole
  experiences no net force.
• The center of mass of the system cannot move
  since there is no net force acting on the system.
• The fisherman can move, the boat can move, but
  the system as a whole must maintain the same
  center of mass.
• Thus, as the fisherman moves forward, the boat
  must move backward to compensate for his movement.

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• Because the boat is symmetrical, we know that the
  center of mass of the boat is at its geometrical
  center, at x L/2.
• We can calculate the center of mass of the system
  containing the fisherman and the boat

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• In the figure below, the center of mass of the
  boat is marked by a dot, while the center of mass
  of the fisherman-boat system is marked by an x.

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• At the front end of the boat, the fisherman is
  now at position L, so the center of mass of the
  fisherman-boat system relative to the boat is

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• The center of mass of the system is now 3 / 5
  from the back of the boat.
• But we know the center of mass hasnt moved,
  which means the boat has moved backward a
  distance of 1/5 L, so that the point 3/ 5 L is
  now located where the point 2 /5 L was before the
  fisherman began to move.

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Define linear momentum and rewrite Newtons
Second Law in terms of momentum.

• Linear Momentum is the product of mass and
  velocity.
• Momentum is a vector since it is the product of a
  scalar (mass) and a vector (velocity).
• The momentum vector is in the direction of the
  velocity vector.
• Momentum is measured in kgm/s

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Define linear momentum and rewrite Newtons
Second Law in terms of momentum.

• p linear momentum
• m mass
• v velocity

26
Define linear momentum and rewrite Newtons
Second Law in terms of momentum.

• A fast moving car has more momentum than a slow
  moving car of the same mass.
• An loaded eighteen wheeler has more momentum than
  a volkswagon beetle traveling at the same speed.
• The more momentum an object has the harder it is
  to stop and the greater effect it will have if
  brought to a stop by collision or impact.

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Define linear momentum and rewrite Newtons
Second Law in terms of momentum.

• A force is required to change the momentum of an
  object whether it is to increase the momentum,
  decrease it, or to change the direction.
• Newtons second law could be rewritten as the
  rate of change of momentum of a particle is
  proportional to the net force applied to it.
• Remember the rate of change of velocity is
  acceleration.

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Define linear momentum and rewrite Newtons
Second Law in terms of momentum.
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Define linear momentum and rewrite Newtons
Second Law in terms of momentum.

• Water leaves a hose at a rate of 5 kg / s with
  a speed of 50 m / s. It strikes a wall, which
  stops it. (We are ignoring any splashing back).
  What is the force exerted by the water on the
  wall?

Final momentum is zero since it is stopped.
Initial momentum is 250 kg m/s each second
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State the law of conservation of linear momentum
for a closed, isolated system and apply this law
to the solution of problems.

• A closed system is one in which objects can
  neither enter nor leave the system.
• An isolated system is one in which there is no
  net external force acting on the system.
• For example billiard balls have the force due to
  gravity acting on them but it is balanced by the
  table pushing up on them. So in that case there
  is no net external force.
• There can be internal forces acting.

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State the law of conservation of linear momentum
for a closed, isolated system and apply this law
to the solution of problems.

• The law of conservation of linear momentum only
  holds true in a closed and isolated system.
• The net external force is zero
• The total mass of the system does not change

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State the law of conservation of linear momentum
for a closed, isolated system and apply this law
to the solution of problems.

• For dP / dt to be equal to zero, the momentum (P)
  must be constant (not changing over time).
• So when there in a closed (mass of system doesnt
  change) and isolated (no net external forces)
  system the total momentum of the system does not
  change (is conserved)

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State the law of conservation of linear momentum
for a closed, isolated system and apply this law
to the solution of problems.

• The Law of conservation of momentum states the
  total momentum before the total momentum after.

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State the law of conservation of linear momentum
for a closed, isolated system and apply this law
to the solution of problems.

• The law of conservation of momentum does NOT say
  that each particle within the system retains its
  same momentum.
• The amount of momentum one particle in the system
  gains is equal to the amount of momentum lost by
  another.

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State the law of conservation of linear momentum
for a closed, isolated system and apply this law
to the solution of problems.

• A 10,000 kg railroad car traveling at a speed of
  24 m / s strikes an identical car at rest. If
  the cars lock together as a result of the
  collisions, what is their common speed afterward?

12 m / s
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State the law of conservation of linear momentum
for a closed, isolated system and apply this law
to the solution of problems.

• Calculate the recoil velocity of a 4 kg rifle
  which shoots a 0.050 kg bullet at a speed of 280
  m / s.

-3.5 m / s
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Define collision and state the impulse-linear
momentum theorem apply this theorem to the
solution of problems.

• A collision is an interaction between two bodies
  if the interaction occurs over a short time
  interval and is so strong that other forces
  acting are insignificant compared to the forces
  the two objects are exerting on each other.

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Define collision and state the impulse-linear
momentum theorem apply this theorem to the
solution of problems.

• An impulse is the product of a net force and time
  over which it acts.
• Remember the net force acting is proportional to
  the rate of change of momentum of an object or
  system.
• So impulses must cause changes in momentum as
  well
• Beginning with the Newtons second law definition
  of momentum, we can derive the impulse momentum
  theorem.

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Define collision and state the impulse-linear
momentum theorem apply this theorem to the
solution of problems.

• J impulse (Ns)
• m mass (kg)
• ?v change in velocity (m/s)

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Define collision and state the impulse-linear
momentum theorem apply this theorem to the
solution of problems.

• Dimensionally, is Ns kgm/s? Hint Remember
  the base units for a Newton.
• 1 N 1 kgm/s2

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Define collision and state the impulse-linear
momentum theorem apply this theorem to the
solution of problems.

• Calculate the impulse suffered when a 70 kg
  person lands on firm ground after jumping from a
  height of 5 m. Then estimate the average force
  on the persons leg if the landing is stiff
  legged where the body moves 1 cm during impact.
• Hint
• Find the impact velocity using linear motion
  equations.
• Find the Impulse
• Find the average velocity during impact
• Find the time of impact using the definition of
  average velocity
• Find the average force acting during impact

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Define collision and state the impulse-linear
momentum theorem apply this theorem to the
solution of problems.
A negative velocity since it is moving downward.
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Define collision and state the impulse-linear
momentum theorem apply this theorem to the
solution of problems.
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Define collision and state the impulse-linear
momentum theorem apply this theorem to the
solution of problems.
46
Define collision and state the impulse-linear
momentum theorem apply this theorem to the
solution of problems.
47
Define collision and state the impulse-linear
momentum theorem apply this theorem to the
solution of problems.
48
Distinguish between elastic and inelastic
collisions in one and two dimensions solve
problems involving these types of problems.

• Elastic collisions are collisions in which both
  momentum and kinetic energy are conserved.
• Inelastic collisions are collisions in which only
  momentum is conserved.

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Distinguish between elastic and inelastic
collisions in one and two dimensions solve
problems involving these types of problems.
Conservation of momentum equation
Conservation of Kinetic Energy Equation
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Distinguish between elastic and inelastic
collisions in one and two dimensions solve
problems involving these types of problems.

• For elastic collisions, you will typically have
  to
• use the conservation of momentum equation to
  solve for one of your two unknown variables in
  terms of the other
• substitute that into the conservation of kinetic
  energy equation
• solve for your unknown
• substitute that value back into your first
  derived equation
• solve for the other unknown

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Distinguish between elastic and inelastic
collisions in one and two dimensions solve
problems involving these types of problems.

• There are some common special cases where it is
  unnecessary to do the extensive algebra. Those
  special cases are on the slides that follow.

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Distinguish between elastic and inelastic
collisions in one and two dimensions solve
problems involving these types of problems.

• In an elastic collision
• If the masses are equal, they will exchange
  velocities.
• For example
• Before the collision
• Ball A is going 5 m/s and ball B is going 3 m/s
• After the collision
• Ball A is going 3 m/s and ball B is going 5 m/s

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Distinguish between elastic and inelastic
collisions in one and two dimensions solve
problems involving these types of problems.

• In an elastic collision
• If object two is initially at rest (v2 0)
• After the collision, the objects will have the
  following velocities.

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Distinguish between elastic and inelastic
collisions in one and two dimensions solve
problems involving these types of problems.

• In an elastic collision
• If object 2 is at rest (v2 0) and object 1 is
  much more massive than object 2
• In other words, object 1 slows down very little
  and object 2 ends up going almost twice the
  velocity of object 1

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Distinguish between elastic and inelastic
collisions in one and two dimensions solve
problems involving these types of problems.

• In an elastic collision
• If object 2 is at rest (v2 0) and object 1 is
  much less massive than object 2

Example a super ball bouncing the massive
earth is object 2 and doesnt move noticeable
after the collision and the super ball rebounds
with about the same velocity it hit with but in
the opposite direction
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Click picture to start video
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Distinguish between elastic and inelastic
collisions in one and two dimensions solve
problems involving these types of problems.

• Think of the interaction of the basketball and
  the floor as the case where object 2 (the earth)
  is at rest and much less massive object 1 (the
  basketball) hits it.
• The basketball will bounce off with approximately
  the same speed it hit with

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Distinguish between elastic and inelastic
collisions in one and two dimensions solve
problems involving these types of problems.

• Think of the interaction between the tennis ball
  and the basketball as object 2 (the tennis ball)
  at rest when it is hit by much more massive
  object 1 (the basketball) so the speed of the
  basket ball is almost unchanged but object 2 (the
  tennis ball) takes off with approximately twice
  the speed of the basketball.
• The same situation is true for the interaction
  between the tennis ball and the ping pong ball.

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Distinguish between elastic and inelastic
collisions in one and two dimensions solve
problems involving these types of problems.

• The conservation of momentum equation can be used
  in 2 and 3 dimensions as well.
• You will just solve for each dimension
  separately.
• Keep in mind that if an object is moving with
  velocity v
• vx vcosT
• vy vsinT
• vz vcosF