Physics 111: Lecture 13
Today’s 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
Physics 111: Lecture 13, Pg 1
Potential Energy & Force

For a conservative force we define the
potential energy function:
x
U  W    F dx
x
dU
Therefore: F  
U    F dx  C
dx
2
1

Consider some potential energy functions we know, and
find the forces:
1
dU
U x  kx 2  C
Fx  
 kx
Spring:
2
dx

Gravity near earth:
Newton’s Gravity:
U y  mgy  C
UR
GMm

C
R
Fy  
dU
 mg
dy
dU
GMm
FR  

dR
R2
It’s true!!
Physics 111: Lecture 13, Pg 2
Potential Energy Diagrams

Consider a block sliding on a
frictionless surface, attached
to an ideal spring.
1
U s  kx 2
2
m
x
U
0
x
Physics 111: Lecture 13, Pg 3
Potential Energy Diagrams

Consider a block sliding on a
frictionless surface, attached
to an ideal spring.
F
m
x
1
U s  kx 2
2

U
F = -dU/dx = -slope
F
x
0
x
Physics 111: Lecture 13, Pg 4
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
y
m
U
k
x2
2 gm
is the height of an object
in the bowl at position x
0
x
Physics 111: Lecture 13, Pg 5
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 won’t
move.
m
x
U
0
x
Physics 111: Lecture 13, Pg 6
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
0
x
Physics 111: Lecture 13, Pg 7
Equilibrium
Balance
cone
Birds



Suppose U(x) looked U
unstable
like this:
This has two equilibrium
neutral
positions, one is stable
(+ curvature) and one is
unstable (- curvature).
stable
Think of a small object
sliding on the U(x)
0
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.
x
Physics 111: Lecture 13, Pg 8
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!
Physics 111: Lecture 13, Pg 9
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:
N
RCM 
 m i ri
i 1
N
m2
m1
r
r1
y
 mi
i 1
RCM
2
r3
m3
x
m4 r
4
(In this case, N = 4)
Physics 111: Lecture 13, Pg 10
System of Particles: Center of Mass

If the system is made up of only two particles:
N
RCM 
 m i ri
i 1
N
 mi
m1 r1  m2 r2

m1  m2
i 1

So:
m1  m2 r1  m2 r2  r1 
m1  m2 
RCM  r1 
m2
r2  r1 
M
where M = m1 + m2
r2 - r1
m2
m1
r1
RCM
r2
y
x
Physics 111: Lecture 13, Pg 11
System of Particles: Center of Mass

If the system is made up of only two particles:
RCM  r1 
m2
r2  r1 
M
where M = m1 + m2
If m1 = m2
RCM
1
 r1  r2  r1 
2
the CM is halfway between
the masses.
r2 - r1
+
m1
r1
RCM
m2
r2
y
x
Physics 111: Lecture 13, Pg 12
System of Particles: Center of Mass

If the system is made up of only two particles:
RCM  r1 
m2
r2  r1 
M
where M = m1 + m2
If m1 = 3m2
RCM
r2 - r1
1
 r1  r2  r1 
4
the CM is now closer to
the heavy mass.
m2
+
m1
RCM
r1
r2
y
x
Physics 111: Lecture 13, Pg 13
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.
m1
+
m2
m1
+
m2
Physics 111: Lecture 13, Pg 14
System of Particles: Center of Mass

We can consider the components of RCM separately:
 i m i x i i m i y i i m i z i 

( X CM ,YCM , Z CM )  
,
,

M
M
M


m2
m1
r
r1
y
RCM
2
r3
m3
x
m4 r
4
(In this case, N = 4)
Physics 111: Lecture 13, Pg 15
Example Calculation:

X CM 
YCM
Consider the following mass distribution:
 i mi x i m0  ( 2 m )12  m24

 12
M
4m
m0  ( 2 m )12  m0
my
 i i i 
6
M
4m
2m
(12,12)
m
m
(0,0)
(24,0)
RCM = (12,6)
Physics 111: Lecture 13, Pg 16
System of Particles: Center of Mass

For a continuous solid, we have to do an integral.
dm
y
r
x
RCM 
 rdm   rdm
M
 dm
where dm is an infinitesimal
mass element.
Physics 111: Lecture 13, Pg 17
System of Particles: Center of Mass

We find that the Center of Mass is at the “center” of the
object.
y
RCM
x
Physics 111: Lecture 13, Pg 18
System of Particles: Center of Mass

We find that the Center of Mass is at the “center” of the
object.
y
RCM
x
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).
Physics 111: Lecture 13, Pg 19
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
+
+
+
+
+
Physics 111: Lecture 13, Pg 20
Pisa
System of Particles: Center of Mass

Bottle
The center of mass for a combination of objects is the
average center of mass location of the objects:
N
RCM 
+
m2
R2 - R1
+
R2
RCM
 mi
so if we have two objects:
RCM
y
i 1
N
i 1
+
m1 R1
 mi R i
m1R1  m2 R2

m1  m2
 R1 
m2
R2  R1 
M
x
Physics 111: Lecture 13, Pg 21
Lecture 13, Act 1
Center 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
X
CM
(1)
(2)
Physics 111: Lecture 13, Pg 22
Lecture 13, Act 1
Solution

The CM of each half-disk will be closer to the fat end than to the
thin end (think of where it would balance).

The CM of the compound object will be halfway between the
CMs of the two halves.

This is higher than the CM of the disk
X
X
X
CM
X
(1)
(2)
Physics 111: Lecture 13, Pg 23
System of Particles: Center of Mass Double
cone


The center of mass (CM) of an object is where we can
freely pivot that object.
pivot
+
CM
Gravity acts on the CM of an object (show later)
If we pivot the object
somewhere else, it will
orient itself so that the
CM is directly below
the pivot.
pivot
CM
pivot
+
CM

mg

This fact can be used to find
the CM of odd-shaped objects.
Physics 111: Lecture 13, Pg 24
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.
Physics 111: Lecture 13, Pg 25
Lecture 13, Act 2
Center 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
Physics 111: Lecture 13, Pg 26
Lecture 13, Act 2
Solution
The center of mass of the
object is at its center and is
initially directly over the wire
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
CM
CM
mg
mg
(front view)
Physics 111: Lecture 13, Pg 27
Lecture 13, Act 2
Solution

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
Physics 111: Lecture 13, Pg 28
Velocity and Acceleration
of 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:

N
M  
mi 



i 1
RCM
1 N
  m i ri
M i 1
VCM
dRCM 1 N
dri 1 N

  mi
  mi v i
dt
M i 1
dt M i 1
ACM
dVCM 1 N
dv i 1 N

  mi
  m i ai
dt
M i 1
dt
M i 1
The velocity and acceleration of the CM is just the weighted
average velocity and acceleration of all the particles.
Physics 111: Lecture 13, Pg 29
Linear Momentum:

Definition: For a single particle, the momentum p is
defined as:
(p is a vector since v is a
p = mv
vector).
So px = mvx etc.

Newton’s 2nd Law:
F = ma
d
( mv)
 m dv 
dt
dt

F 
dp
dt
Units of linear momentum are kg m/s.
Physics 111: Lecture 13, Pg 30
Linear Momentum:

For a system of particles the total
momentum P is the vector sum of
the individual particle momenta:
N
P
But we just showed that  m i v i  MVCM
i 1
So
N
N
i 1
i 1
 pi   m i v i
1 N


 mi v i 
VCM 
M i 1


P  MVCM
Physics 111: Lecture 13, Pg 31
Linear Momentum:

So the total momentum of a system of particles is just the
total mass times the velocity of the center of mass.
P  MVCM
dV
dP
 M CM  MACM   m i ai   Fi ,net
dt
dt
i
i

Observe:

We are interested in
dP
so we need to figure out
dt
 Fi ,net
i
Physics 111: Lecture 13, Pg 32
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.

 Fi ,NET  F13  F12  F1,EXT
i

m3
F31
 F21  F23 
 F31  F32 
 F1,EXT
(since the other forces
cancel in pairs...Newton’s
3rd Law)
F32
F13
F23
m1 F12
F21
m2
F1,EXT
All of the “internal” forces cancel !!
Only the “external” force matters !!
Physics 111: Lecture 13, Pg 33
Linear Momentum:

Only the total external force matters!
dP
  Fi ,EXT  FNET ,EXT
i
dt
m3
Which is the same as:
FNET ,EXT 
dP
 MACM
dt
m1
m2
F1,EXT
Newton’s 2nd law applied to systems!
Physics 111: Lecture 13, Pg 34
Center of Mass Motion: Recap
Pork chop

We have the following law for CM motion:
FEXT 
Pendulum
dP
 MACM
dt

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.

Physics 111: Lecture 13, Pg 35
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 = 1.5m
m
Physics 111: Lecture 13, Pg 36
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
x=L
x=0
Finding the CM:
If we take the astronaut on the left to be at x = 0:
x cm 
M ( 0 )  m( L ) m( L ) 2

 L
M m
2 .5 m 5
Physics 111: Lecture 13, Pg 37
Lecture 13, Act 3
Center 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
before
20 ft
(b) 15 ft
(c) 16.7 ft
? ft
after
Physics 111: Lecture 13, Pg 38
Lecture 13, Act 3
Solution

Since the man and the canoe have the same mass, the
CM of the man-canoe system will be halfway between
the CM of the man and the CM of the canoe.

Initially the CM of the system is 20 ft from shore.
X
X
x
20 ft
CM of system
Physics 111: Lecture 13, Pg 39
Lecture 13, Act 3
Solution

Since there is no force acting on the canoe in the x-direction, the
location of the CM of the system can’t change!

Therefore, the man ends up 5 ft to the left of the system CM,
and the center of the canoe ends up 5 ft to the right.

He ends up moving 5 ft toward the shore (15 ft away).
15 ft
X
10 ft
X
20 ft
x
5 ft
CM of system
Physics 111: Lecture 13, Pg 40
Recap of today’s lecture

Systems of particles
(Text: 8-1)

Center of mass
(Text: 8-1 & 12-6)

Velocity and acceleration of the center of mass (Text:8-3)

Dynamics of the center of mass
Linear Momentum

Example problems

Look at textbook problems Chapter 8: # 3, 7, 17, 29, 35, 77, 111
(Text: 8-3 to 8-4)
Physics 111: Lecture 13, Pg 41