Physics 111: Lecture 13 Todays Agenda

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Physics 111 Lecture 13Todays Agenda

• Potential Energy Force
• Systems of Particles
• Center of mass
• Velocity and acceleration of the center of mass
• Dynamics of the center of mass
• Linear Momentum
• Example problems

2
Potential Energy Force

• For a conservative force we define the
  potential energy function
• Therefore
• Consider some potential energy functions we know,
  and find the forces
• Spring
• Gravity near earth
• Newtons Gravity

Its true!!
3
Potential Energy Diagrams

• Consider a block sliding on a frictionless
  surface, attached to an ideal spring.

m
x
U
x
0
4
Potential Energy Diagrams

• Consider a block sliding on a frictionless
  surface, attached to an ideal spring.
• F -dU/dx -slope

F
x
U
F
x
x
0
5
Potential Energy Diagrams

• The potential energy of the block is the same as
  that of an object sliding in a frictionless
  bowl
• Ug mgy 1/2 kx2 Us

m
U
x
is the height of an object in the bowl at
position x
0
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Equilibrium

• F -dU/dx -slope
• So F 0 if slope 0.
• This is the case at the minimum or maximum of
  U(x).
• This is called an equilibrium position.
• If we place the block at rest at x 0, it wont
  move.

m
x
U
x
0
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Equilibrium

• If small displacements from the equilibrium
  position result in a force that tends to move the
  system back to its equilibrium position, the
  equilibrium is said to be stable.
• This is the case if U is a minimum at the
  equilibrium position.
• In calculus language, the equilibrium is stable
  if the curvature (second derivative) is positive.

F
m
x
U
F
x
0
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Equilibrium
Balance cone
Birds
U

• Suppose U(x) looked like this
• This has two equilibrium positions, one is
  stable ( curvature) and one is unstable (-
  curvature).
• Think of a small object sliding on the U(x)
  surface
• If it wants to keep sliding when you give it a
  little push, the equilibrium is unstable.
• If it returns to the equilibrium position when
  you give it a little push, the equilibrium is
  stable.
• If the curvature is zero (flat line) the
  equilibrium is neutral.

unstable
neutral
stable
x
0
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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
Ice table

• 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

• If the system is made up of only two particles

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System of Particles Center of Mass

• If the system is made up of only two particles

where M m1 m2
r2 - r1

m2
m1
RCM
r2
r1
y
x
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System of Particles Center of Mass

• If the system is made up of only two particles

where M m1 m2
If m1 3m2
r2 - r1

m2
m1
RCM
r2
the CM is now closer to the heavy mass.
r1
y
x
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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.

15
System of Particles Center of Mass

• We can consider the components of RCM separately

16
Example Calculation

• Consider the following mass distribution

2m
(12,12)
m
m
(0,0)
(24,0)
RCM (12,6)
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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.

y
RCM
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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System of Particles Center of Mass
Pisa
Bottle

• The center of mass for a combination of objects
  is the average center of mass location of the
  objects

m2
R2 - R1

R2
so if we have two objects
RCM
R1
m1
y
x
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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
Fork, spoon, and match

• 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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Velocity and Accelerationof the Center of Mass

• If its particles are moving, the CM of a system
  can also move.
• Suppose we know the position ri of every
  particle in the system as a function of time.

So
And

• The velocity and acceleration of the CM is just
  the weighted average velocity and acceleration of
  all the particles.

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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

• For a system of particles the total momentum P
  is the vector sum of the individual particle
  momenta

But we just showed that
So
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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

• Suppose we have a system of three particles as
  shown. Each particle interacts with every other,
  and in addition there is an external force
  pushing on particle 1.

m3
F31
F32
F13
F23
F12
m1
F21
(since the other forces cancel in
pairs...Newtons 3rd Law)
m2
F1,EXT
All of the internal forces cancel !! Only the
external force matters !!
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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
Pork chop

• 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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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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Recap of todays lecture

• Systems of particles (Text 8-1)
• Center of mass (Text 8-1 12-6)
• Velocity and acceleration of the center of mass
  (Text8-3)
• Dynamics of the center of mass (Text 8-3 to
  8-4)
• Linear Momentum
• Example problems
• Look at textbook problems Chapter 8 3, 7, 17,
  29, 35, 77, 111