Physical Sciences 2: Lecture 1b
September 6, 2016
Pre-Reading for Lecture 1b:
Momentum and Center of Mass
•
In physics, the concept of motion is linked with the concept of momentum. In everyday
speech, we say that something has “momentum” if it has a tendency to continue along its
current trajectory. Here are several recent headlines from the New York Times that
include the term “momentum”:
•
U.S. Inflation Tame Despite Economy Gaining Momentum
U.S. Seeks 'Momentum' in Deploying Extra U.N. Troops to South Sudan



In physics, momentum is a vector quantity p defined by: p = mv


⎛  dr ⎞
where m is the mass and v is the velocity ⎜ v = ⎟
⎝
dt ⎠
•
Momentum is important in physics because: Momentum is Conserved.
system
boundary
To define “conserved,” we usually divide the universe into two parts: one part, called the
system, which contains the object(s) of interest, and another part, called the surroundings,
which is simply the rest of the universe. You should envision an imaginary boundary
around the system that separates it from the surroundings:
surroundings
In many instances, we can consider a system to be approximately isolated from the
surroundings: such a system does not interact in any way with the rest of the universe.
Although there are no truly isolated systems, an object in deep space (far away from
gravitational interactions with any planets or stars) would be isolated to a very high
degree of approximation. Given that definition of an isolated system, the Law of
Conservation of Momentum is:
The momentum of an isolated system is constant.
Or, expressed mathematically,

dp
=0
For an isolated system,
dt
According to this formulation, conservation of momentum means that the momentum of
an isolated system does not change with time.
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Physical Sciences 2: Lecture 1b
•
September 6, 2016
For an isolated system containing only one object, the mass of that object will be
constant, and so using our definition of momentum, we know:


d
dv
(mv ) = 0 which implies that
For a single isolated object,
=0
dt
dt
•
So momentum conservation means that the velocity of an isolated object is constant.
This is Newton’s First Law: An isolated object at rest will remain at rest; an isolated
object in motion will remain in motion in a straight line at constant speed.
•
Since objects are never truly isolated, we often study situations in which objects act as if
they are isolated. We can define an object as functionally isolated if it will keep going
in a straight line at constant speed without any intervention. Examples include objects
sliding on ice, rolling without friction, floating on water, or gliding on a cushion of air.
•
If a system contains two or more objects, the total momentum is the vector sum of the
momenta of the individual objects:

 




ptotal = p1 + p2 + p3 +… = m1v1 + m2 v2 + m3v3 +…
•
If an isolated system contains two or more objects, the total momentum is constant, but
momentum can be transferred from one object to another within the system.
•
Momentum is transferred from one object to another by interactions. If a system is not
isolated, that means it must be interacting in some way with its surroundings.
•
There are two basic kinds of macroscopic interactions that we will discuss in this course:
- Contact interactions require direct contact between the objects. Pushing and pulling
are examples of contact interactions. Friction is a more subtle example of a contact
interaction, as is drag (friction in a fluid). The important idea is: any time one object is
in contact with another, there will be some kind of contact interaction.
- Long-range interactions do not require contact. The most important example for this
course is gravity, which is a universal attractive interaction between all objects. Other
examples include electric and magnetic forces, which can act without direct contact.
•
All physical interactions actually arise from one of the four “fundamental interactions,”
which are known to physicists as:
- The electromagnetic force (electricity and magnetism)
- The weak nuclear force (responsible for some kinds of nuclear decay)*
- The strong nuclear force (holds protons and neutrons together inside nuclei)
- The gravitational force (gravity!)
Essentially all macroscopic interactions—pushing, pulling, friction, drag—as well as
microscopic interactions down to the level of atoms and molecules are electromagnetic
interactions. It’s amazing that electromagnetism and gravity describe almost all of the
interactions responsible for everyday physics, chemistry, and biology. *(Indeed,
electromagnetism and the weak force are part of the same “electroweak” interaction.)
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Physical Sciences 2: Lecture 1b
September 6, 2016
The other important idea for this lecture is the center of mass. For any system
 made up
of a whole bunch of particles, we define the position of the center of mass rCM as:
⎛ 1 ⎞




rCM = ⎜
m1r1 + m2 r2 + m3r3 +…)
(
⎟
⎝ M total ⎠

where m1 is the mass of particle 1, r1 is the position of particle 1, etc…, and Mtotal is the
total mass of all the particles. Why do we make such a definition?
If you have only two particles, the center of mass will be somewhere in between the
particles, and if you put a pivot at that point, the two particles would be in balance on a
“see-saw.” For instance:
The center of mass is incredibly useful because it allows us to
define precisely what we mean by the position of a complicated
object. For instance, how would you define the position of a cow?
Would it be the position of its head? its ear? its udder? its tail?
That would be utterly ridiculous. In physics we need a precise
definition of “the position of a cow,” and we choose the center of
mass of the cow to define its position.
If you are standing, your center of mass is about at your belly
button, but if you move around your center of mass will also move.
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Physical Sciences 2: Lecture 1b
September 6, 2016
Why is the center of mass important and useful?
Recall that if a system contains many objects, the total momentum of the system is given
by the vector sum of the momenta of all the objects in the system:

  

ptotal = p1 + p2 + p3 +… = ∑ pk
k




Or, using the definition of momentum for each object: ptotal = m1v1 + m2 v2 + m3v3 +…
Likewise, the total mass of the system is simply: M total = m1 + m2 + m3 +…

dr

We define the velocity of the system as the velocity of the center of mass vCM = CM
dt
Putting all these definitions together, we will find that the total momentum of an arbitrary
system is given by:


ptotal = M totalvCM
In other words, as far as momentum is concerned we can treat a complicated system as
if it were a single particle of mass Mtotal located at the center of mass of the system.
Now, what does this mean for an isolated system? Well, an isolated system will have a
constant total mass, and if it is isolated its momentum must be constant. So, since


ptotal = M totalvCM


if Mtotal is constant and ptotal is constant, then vCM must be constant. This means that the
center of mass of an isolated system always moves at a straight line at constant
speed (or, if the CM is at rest, it will remain at rest).
•
Learning objectives: After this lecture, you will be able to…
1. Identify the interactions between any pair of objects.
2. Find the interactions that could have caused a change in the momentum of an object.
3. Determine whether a system of objects is functionally isolated.
4. Use momentum conservation for a system of several objects to solve for various
unknown quantities (masses, speeds, angles…).
5. Define a particle and explain why “particles” are merely idealizations.
6. Define the center of mass (CM) of an object and explain why the CM is important and
why it is useful.
7. Calculate the CM of a collection of particles.
8. Describe the motion of the CM for an isolated system.
9. Use the CM to make predictions about the motion of an isolated system.
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Physical Sciences 2: Lecture 1b
September 6, 2016
Lecture 1b: Momentum and Center of Mass
•
Highlights from the pre-reading:
If a system contains many objects, the total momentum of the system is given by the
vector sum of the momenta of all the objects in the system:

  

ptotal = p1 + p2 + p3 +… = ∑ pk
k




Or, using the definition of momentum for each object: ptotal = m1v1 + m2 v2 + m3v3 +…
Likewise, the total mass of the system is simply: M total = m1 + m2 + m3 +…
We define the position of the entire system as the position of the center of mass:
⎛ 1 ⎞




rCM = ⎜
m1r1 + m2 r2 + m3r3 +…)
(
⎟
⎝ M total ⎠

drCM

and the velocity of the entire system as the velocity of the center of mass vCM =
dt
Putting all these definitions together, we will find that the total momentum of an arbitrary
system is given by:


ptotal = M totalvCM
In other words, as far as momentum is concerned we can treat a complicated system as
if it were a single particle of mass Mtotal located at the center of mass of the system.
•
Learning objectives: After this lecture, you will be able to…
1. Identify the interactions between any pair of objects.
2. Find the interactions that could be responsible for the change in the momentum of an
object.
3. Determine whether a system of objects is functionally isolated.
4. Use momentum conservation for a system of several objects to solve for various
unknown quantities (masses, speeds, angles…).
5. Define a particle and explain why “particles” are merely idealizations.
6. Define the center of mass (CM) of an object and explain why the CM is important and
why it is useful.
7. Calculate the CM of a collection of particles.
8. Describe the motion of the CM for an isolated system.
9. Use the CM to make predictions about the motion of an isolated system.
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Physical Sciences 2: Lecture 1a
September 1, 2016
Activity 2: Motion
•
This multi-flash photograph shows two pool balls (6 cm in diameter) rolling on a surface
without friction. The grid lines are spaced 10 cm apart, and the balls are photographed
at a rate of 25 frames per second. Initially, ball 2 is at rest (the dark black one), and ball
1 is entering from the left. The two balls collide and move off to the right.
ball 2
ball 2
ball 1
ball 1
1.

Estimate the velocity ( v ) of:
a) ball 1 before the collision
b) ball 1 after the collision
c) ball 2 after the collision
Activity continues on next page…
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Physical Sciences 2: Lecture 1a
September 1, 2016
Activity 2 (continued)
2.
Identify all the interactions involving these balls.
3.
The mass of a pool ball is about 160 g. Considering the two balls together as a single
system, calculate:

a) the initial momentum pi of the system

b) the final momentum p f of the system
c) Is momentum conserved? Why or why not?
4.
Bonus! The kinetic energy of an object is defined as K = 12 mv 2 . Is kinetic energy
conserved for this system of two balls?
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Physical Sciences 2: Lecture 1b
September 6, 2016
Interactions
•
Kinesin is pulling a vesicle along a microtubule:
1.
Identify all the interactions involving the vesicle. Classify each interaction as a contact
interaction or a long-range interaction.
2.
At various times, the vesicle starts and stops moving. Can you identify some interactions
that could be responsible for these changes in its motion?
3.
What are the fundamental interactions responsible for each of the interactions you
identified above?
4.
Bonus! An airplane is flying. Identify all the interactions involving the airplane.
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Physical Sciences 2: Lecture 1b
September 6, 2016
Measuring Mass
•
A cannon rolls freely on wheels on a flat
horizontal surface. When the cannon fires, the
cannonball moves to the right while the cannon
itself recoils to the left.
If the mass of the cannonball is 10 kg, the final
speed of the cannonball is 100 m/s, and the final
speed of the cannon is 1 m/s, what is the mass of
the cannon?
•
Bonus! You repeat the cannonball experiment on the moon, where the strength of gravity
is only 1/6 of that here on Earth. Would anything be different?
•
How can you use momentum conservation to measure the mass of a person?
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Physical Sciences 2: Lecture 1b
September 6, 2016
Am I getting it?
•
The time intervals in each of these multi-flash photos are constant. Answer each question
with the letter (from the photo) that is closest to the location described in the question.
Assume that the x-axis points to the right, and y-axis points up (as is conventional).
Bouncing Basketball
1.
Where is the ball’s speed the fastest?
2.
Where is the speed the slowest?
3.
Where is the y-component of the
ball’s velocity negative?
Swinging on a Rope
4.
Where is the boy’s speed the fastest?
5.
Where is his speed the slowest?
6.
Where is the pole vaulter’s speed the
fastest?
7.
Where is his speed the slowest?
8.
Where is his velocity closest to being
vertical?
9.
Where is his velocity closest to being
horizontal?
Photo by Harold Edgerton, MIT photo pioneer
Pole Vault
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Physical Sciences 2: Lecture 1b
September 6, 2016
Motion can be complicated!
•
On the previous page, in which photo was it easy to determine the velocity of the object?
•
In which photo was it especially difficult?
•
I’m going to walk across the room… what is my velocity?
•
One kind of object has an unambiguous velocity: a particle. What is a particle?
•
It’s tough to calculate my velocity while I’m walking across the room, but it’s easy to
calculate my momentum. Why?
•
In addition, determining my total mass is easy…
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Physical Sciences 2: Lecture 1b
September 6, 2016
Activity 1: The definition of the center of mass
•
•

Recall the definition of center of mass: rCM =
1



m1r1 + m2 r2 + m3r3 +)
(
M total
We want to define the velocity of a system such that the following relationship holds:


ptotal = M totalvsystem
•

Solve the above equation to find an expression for vsystem in terms of the masses and
velocities of all the particles in the system.
•
Now we define the center of mass in order that the following statement is true:
The velocity of a system equals the time derivative of the location of its center of mass.
Write an equation for this statement and show that it is true given the definition (above).
•
For the specific case of two masses (m1 and m2) separated by a
distance R, find a formula for the location of the center of mass of
the system.
1
•
Bonus! Find the CM of the uniform sheet of metal shown at right.
2
3
2
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Physical Sciences 2: Lecture 1b
September 6, 2016
Activity 2: Center of mass for isolated systems
d 




The velocity of a system is vsystem ≡ vCM = rCM , and its momentum is ptotal = M totalvCM .
dt
•
If a system is isolated, what must be true about its total momentum?
•

What must be true about the velocity of the center of mass ( vCM ) of an isolated system?
•
So, if the center of mass of an isolated system is at rest, it will…
•
A person (mass mp = 75 kg) is standing on the Earth (mass mE = 6.0 × 10 24 kg ). The
person and Earth together can be treated as an isolated system at rest. If the person jumps
up 50 cm, how far does the Earth move?
•
Bonus! An object of mass 4m is
isolated and at rest. It breaks apart
into three objects. Your camera
captures the locations of two pieces
of mass m: one is 10 cm to the right
of the original location; the other is 6
cm below. Where is the other piece?
10 cm
Before
After
m
4m
6 cm
at rest
11
m
Where is 2m piece?
Physical Sciences 2: Lecture 1b
September 6, 2016
Motion of the center of mass of an isolated system
•
Let’s take a look at how the center of mass moves for an isolated system…
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Physical Sciences 2: Lecture 1b
September 6, 2016
Activity 3: Center of mass for more complex objects
•
If an object has a plane of symmetry, prove that the center of mass must lie
on that plane. (Hint: choose a coordinate system such that the plane of
symmetry is defined by x = 0.)
•
You can always find the center of mass of a complicated
object by grouping particles together. Consider, for
instance, a water molecule (shown at right). An O atom
is 16 times more massive than an H atom.
- First consider the two H atoms as a group. Where is
the center of mass of this group? Mark that point on the
figure.
- Now treat that point (the CM of the group of H atoms)
as a point particle with a mass equal to the mass of both
H atoms. Find the CM of that fictitious particle and the
O atom. This will be the CM of the entire water molecule.
•
Bonus! Prove that the center of mass can be calculated by grouping particles together.
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Physical Sciences 2: Lecture 1b
September 6, 2016
Putting it together
•
A car (mass 1500 kg) is at rest. What are all the interactions involving the car?
•
The car starts to drive forwards. What interaction is responsible for making the car go?
•
Although the car alone is not an isolated system, the car, road, and Earth together can be
treated as an isolated system that is initially at rest. If the car ends up driving at 60 mph,
what has happened to the road (and the Earth)?
•
What would happen if the road were lighter than the car?
•
Bonus! If everyone in China jumped up at the same time, how far would the Earth move?
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Physical Sciences 2: Lecture 1b
September 6, 2016
One-Minute Paper
Your name: _____________________________
Names of your group members:
TF: _____________________________
_________________________________
_________________________________
_________________________________
•
Please tell us any questions that came up for you today during lecture. Write “nothing”
if no questions(s) came up for you during class.
•
What single topic left you most confused after today’s class?
•
Any other comments or reflections on today’s class?
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