Physics 211: Lecture 14 Todays Agenda

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Physics 211 Lecture 14Todays Agenda

• Systems of Particles
• Center of mass
• Linear Momentum
• Example problems
• Momentum Conservation
• Inelastic collisions in one dimension
• Ballistic pendulum

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System of Particles

• Until now, we have considered the behavior of
  very simple systems (one or two masses).
• But real life is usually much more interesting!
• For example, consider a simple rotating disk.
• An extended solid object (like a disk) can be
  thought of as a collection of parts. The motion
  of each little part depends on where it is in the
  object!

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System of Particles Center of Mass

• How do we describe the position of a system
  made up of many parts?
• Define the Center of Mass (average position)
• For a collection of N individual pointlike
  particles whose masses and positions we know

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System of Particles Center of Mass

• We can consider the components of RCM separately

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Example Calculation

• Consider the following mass distribution

2m
(12,12)
m
m
(0,0)
(24,0)
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System of Particles Center of Mass
Baton

• The center of mass is where the system is
  balanced!
• Building a mobile is an exercise in finding
  centers of mass.

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System of Particles Center of Mass

• For a continuous solid, we have to do an integral.

dm
r
y
where dm is an infinitesimal mass element.
x
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System of Particles Center of Mass

• We find that the Center of Mass is at the
  center of the object.

The location of the center of mass is an
intrinsic property of the object!! (it does not
depend on where you choose the origin or
coordinates when calculating it).
RCM
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System of Particles Center of Mass

• We can use intuition to find the location of the
  center of mass for symmetric objects that have
  uniform density
• It will simply be at the geometrical center !

CM




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Lecture 13, Act 1Center of Mass

• The disk shown below (1) clearly has its CM at
  the center.
• Suppose the disk is cut in half and the pieces
  arranged as shown in (2)
• Where is the CM of (2) as compared to (1)?

(a) higher (b) lower
(c) same
XCM
(1)
(2)
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Lecture 13, Act 1Solution

• The CM of each half-disk will be closer to the
  fat end than to the thin end (think of where it
  would balance).

X
X
(1)
(2)
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System of Particles Center of Mass
Double cone

• The center of mass (CM) of an object is where we
  can freely pivot that object.
• Gravity acts on the CM of an object (show later)
• If we pivot the objectsomewhere else, it
  willorient itself so that theCM is directly
  below the pivot.
• This fact can be used to findthe CM of
  odd-shaped objects.

pivot
CM
pivot
pivot
CM

CM
mg
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System of Particles Center of Mass
Odd shapes

• Hang the object from several pivots and see where
  the vertical lines through each pivot intersect!

pivot
pivot
pivot
CM

• The intersection point must be at the CM.

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Lecture 13, Act 2Center of Mass
3 pronged object

• An object with three prongs of equal mass is
  balanced on a wire (equal angles between prongs).
  What kind of equilibrium is this position?

a) stable b) neutral c) unstable
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Lecture 13, Act 2Solution
If the object is pushed slightly to the left
or right, its center of mass will not be above
the wire and gravity will make the object fall off
The center of mass of the object is at its
center and is initially directly over the wire

CM

CM
mg
mg
(front view)
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Lecture 13, Act 2Solution

• Consider also the case in which the two lower
  prongs have balls of equal mass attached to them

CM

CM
mg
mg
In this case, the center of mass of the
object is below the wire
When the object is pushed slightly, gravity
provides a restoring force, creating a stable
equilibrium
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Example Astronauts Rope

• Two astronauts at rest in outer space are
  connected by a light rope. They begin to pull
  towards each other. Where do they meet?

m
M 1.5m
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Example Astronauts Rope...
m
M 1.5m

• They start at rest, so VCM 0.
• VCM remains zero because
• there are no external forces.
• So, the CM does not move!
• They will meet at the CM.

CM
L
xL
x0
Finding the CM
If we take the astronaut on the left to be at x
0
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Lecture 13, Act 3Center of Mass Motion

• A man weighs exactly as much as his 20 foot long
  canoe.
• Initially he stands in the center of the
  motionless canoe, a distance of 20 feet from
  shore. Next he walks toward the shore until he
  gets to the end of the canoe.
• What is his new distance from the shore. (There
  no horizontal force on the canoe by the water).

20 ft
(a) 10 ft (b) 15 ft (c) 16.7
ft
before
20 ft
? ft
after
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Lecture 13, Act 3Solution
x
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Lecture 13, Act 3Solution

• Since there is no force acting on the canoe in
  the x-direction, thelocation of the CM of the
  system cant change!

X
X
x
20 ft
CM of system
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Linear Momentum

• Definition For a single particle, the momentum
  p is defined as

(p is a vector since v is a vector).
p mv

• So px mvx etc.

• Newtons 2nd Law

F ma
dv

• Units of linear momentum are kg m/s.

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Linear Momentum

• So the total momentum of a system of particles is
  just the total mass times the velocity of the
  center of mass.
• Observe
• We are interested in so we need to figure
  out

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Linear Momentum

• Only the total external force matters!

m3
Which is the same as
m1
m2
F1,EXT
Newtons 2nd law applied to systems!
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Center of Mass Motion Recap

• We have the following law for CM motion
• This has several interesting implications
• It tells us that the CM of an extended object
  behaves like a simple point mass under the
  influence of external forces
• We can use it to relate F and A like we are used
  to doing.
• It tells us that if FEXT 0, the total momentum
  of the system can not change.
• The total momentum of a system is conserved if
  there are no external forces acting.

Pendulum
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Momentum Conservation

• The concept of momentum conservation is one of
  the most fundamental principles in physics.
• This is a component (vector) equation.
• We can apply it to any direction in which there
  is no external force applied.
• You will see that we often have momentum
  conservation even when energy is not conserved.

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Elastic vs. Inelastic Collisions

• A collision is said to be elastic when kinetic
  energy as well as momentum is conserved before
  and after the collision.
  Kbefore Kafter
• Carts colliding with a spring in between,
  billiard balls, etc.

• A collision is said to be inelastic when kinetic
  energy is not conserved before and after the
  collision, but momentum is conserved.
  Kbefore ?
  Kafter
• Car crashes, collisions where objects stick
  together, etc.

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Inelastic collision in 1-D Example 1

• A block of mass M is initially at rest on a
  frictionless horizontal surface. A bullet of
  mass m is fired at the block with a muzzle
  velocity (speed) v. The bullet lodges in the
  block, and the block ends up with a speed V. In
  terms of m, M, and V
• What is the initial speed of the bullet v?
• What is the initial energy of the system?
• What is the final energy of the system?
• Is kinetic energy conserved?

x
V
before
after
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Example 1...

• Consider the bullet block as a system. After
  the bullet is shot, there are no external forces
  acting on the system in the x-direction.
  Momentum is conserved in the x direction!

x
V
initial
final
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Example 1...

• Now consider the kinetic energy of the system
  before and after
• Before
• After
• So

Kinetic energy is NOT conserved! (friction
stopped the bullet) However, momentum was
conserved, and this was useful.
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Example 1...

• What if the bullet goes through?

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Inelastic Collision in 1-D Example 2
M
m
ice
v 0
(no friction)
V
M m
v ?
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Example 2...
Air track
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Lecture 14, Act 4Momentum Conservation

• Two balls of equal mass are thrown horizontally
  with the same initial velocity. They hit
  identical stationary boxes resting on a
  frictionless horizontal surface.
• The ball hitting box 1 bounces back, while the
  ball hitting box 2 gets stuck.
• Which box ends up moving faster?

(a) Box 1 (b) Box 2 (c)
same
2
1
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Lecture 14, Act 4Momentum Conservation

• Since the total external force in the x-direction
  is zero, momentum is conserved along the x-axis.
• In both cases the initial momentum is the same
  (mv of ball).
• In case 1 the ball has negative momentum after
  the collision, hence the box must have more
  positive momentum if the total is to be
  conserved.
• The speed of the box in case 1 is biggest!

x
V1
V2
2
1
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Lecture 14, Act 4Momentum Conservation
mvinit (Mm)V2
mvinit MV1 - mvfin
V2 mvinit / (Mm)
V1 (mvinit mvfin) / M
x
V1
V2
2
1
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Ballistic Pendulum
L
L
V0
L
L
H
m
v
M m
V
M

• A projectile of mass m moving horizontally with
  speed v strikes a stationary mass M suspended by
  strings of length L. Subsequently, m M rise
  to a height of H.

Given H, what is the initial speed v of the
projectile?
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Ballistic Pendulum...

• Two stage process

1. m collides with M, inelastically. Both M and
m then move together with a velocity V (before
having risen significantly).
2. M and m rise a height H, conserving KU
energy E. (no non-conservative forces acting
after collision)
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Ballistic Pendulum...

• Stage 1 Momentum is conserved

in x-direction

• Stage 2 KU Energy is conserved

Eliminating V gives
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Ballistic Pendulum Demo
L
L
L
L
H
m
v
M m
M
d

• In the demo we measure forward displacement d,
  not H

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Ballistic Pendulum Demo...
Ballistic pendulum
for
for d ltlt L
Lets see who can throw fast...
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Recap of todays lecture

• Systems of particles (Text 8-1)
• Center of mass (Text 8-1 12-6)
• Linear Momentum (Text 8-3 to 8-4)
• Inelastic collisions in one dimension (Text
  8-6)
• Ballistic pendulum (Ex. 8-14)