Title: Physics 211: Lecture 13 Todays Agenda
1Physics 211 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
2Potential 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!!
3Potential Energy Diagrams
- Consider a block sliding on a frictionless
surface, attached to an ideal spring.
m
x
U
x
0
4Potential 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
5Potential 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
6Equilibrium
- 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
7Equilibrium
- 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
8Equilibrium
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
9System 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!
10System 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
11System of Particles Center of Mass
- If the system is made up of only two particles
12System 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
13System 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
14System 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.
15System of Particles Center of Mass
- We can consider the components of RCM separately
16Example Calculation
- Consider the following mass distribution
2m
(12,12)
m
m
(0,0)
(24,0)
RCM (12,6)
17System 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
18System of Particles Center of Mass
- We find that the Center of Mass is at the
center of the object.
y
RCM
x
19System 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
20System 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
21System 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
22Lecture 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)
23Lecture 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)
24System 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
25System 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.
26Lecture 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
27Lecture 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)
28Lecture 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
29Velocity 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.
30Linear Momentum
- Definition For a single particle, the momentum
p is defined as
(p is a vector since v is a vector).
p mv
F ma
dv
- Units of linear momentum are kg m/s.
31Linear 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
32Linear 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
33Linear 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 !!
34Linear Momentum
- Only the total external force matters!
m3
Which is the same as
m1
m2
F1,EXT
Newtons 2nd law applied to systems!
35Center 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
36Example 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
37Example 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
38Lecture 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
39Lecture 13, Act 3Solution
x
40Lecture 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
41Recap of todays lecture
- Systems of particles
- Center of mass
- Velocity and acceleration of the center of mass
- Dynamics of the center of mass
- Linear Momentum
- Example problems