Transcript Lecture13-10

Announcements
• EXAM: October 13 (through Chapter 9)
- calculator (no stored functions, etc)
- #2 pencil
Reading and Review
Elastic Collisions
Carefully place a small rubber ball (mass m) on
a) zero
top of a much bigger basketball (mass M) and
b) v
drop these from the same height h so they
arrive at the ground with the speed v. What is
c) 2v
the velocity of the smaller ball after the
d) 3v
basketball hits the ground, reverses direction,
e) 4v
and then collides with the small rubber ball?
m
v
v
M
Elastic Collisions
Carefully place a small rubber ball (mass m) on
a) zero
top of a much bigger basketball (mass M) and
b) v
drop these from the same height h so they
c) 2v
arrive at the ground with the speed v. What is
the velocity of the smaller ball after the
d) 3v
basketball hits the ground, reverses direction,
e) 4v
and then collides with the small rubber ball?
• Remember that relative
velocity has to be equal
v
before and after collision!
Before the collision, the
basketball bounces up with v
v and the rubber ball is
coming down with v, so
their relative velocity is –2v.
After the collision, it
therefore has to be +2v!!
m
M
(a)
3v
v
v
v
(b)
(c)
Recoil Speed
A cannon sits on a stationary
a) 0 m/s
railroad flatcar with a total
b) 0.5 m/s to the right
mass of 1000 kg. When a 10-kg
c) 1 m/s to the right
cannonball is fired to the left at
d) 20 m/s to the right
a speed of 50 m/s, what is the
e) 50 m/s to the right
recoil speed of the flatcar?
Recoil Speed
A cannon sits on a stationary
a) 0 m/s
railroad flatcar with a total
b) 0.5 m/s to the right
mass of 1000 kg. When a 10-kg
c) 1 m/s to the right
cannonball is fired to the left at
d) 20 m/s to the right
a speed of 50 m/s, what is the
e) 50 m/s to the right
recoil speed of the flatcar?
Because the initial momentum of the
system was zero, the final total momentum
must also be zero. Thus, the final
momenta of the cannonball and the flatcar
must be equal and opposite.
pcannonball = (10 kg)(50 m/s) = 500 kg-m/s
pflatcar = 500 kg-m/s = (1000 kg)(0.5 m/s)
Elastic Collisions in 2-D
Two-dimensional collisions can only be solved if
some of the final information is known, such as the
final velocity of one object
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E
A proton collides elastically with another proton that is initially at rest.
The incoming proton has an initial speed of 3.5x105 m/s and makes
a glancing collision with the second proton. After the collision one
proton moves at an angle of 37o to the original direction of motion,
the other recoils at 53o to that same axis. Find the final speeds of the
two protons.
v1
v0 = 3.5x105 m/s
initial
37o
53o
v2
final
A proton collides with another proton that is initially at rest. The
incoming proton has an initial speed of 3.5x105 m/s and makes a
glancing collision with the second proton. After the collision one
proton moves at an angle of 37o to the original direction of motion,
the other recoils at 53o to that same axis. Find the final speeds of the
two protons.
Momentum conservation:
v1
v0 = 3.5x105 m/s
initial
37o
53o
v2
final
if we’d been given only 1 angle, would have needed
the other angle or conservation of energy also!
Is energy conserved in this collision?
? 1
1
1
2
2
2
K i  mv0  mv1 f  mv2 f
2
2
2
?
 v v
2
0
2
1f
v
 3.5 10
2
2f
5
m s    2.8  10 m s    2.1 10 m s 
2 ?
5
2
  3.5  10 m s 
5
5
2
2
In an elastic collision between a moving mass
incident on a stationary identical mass, the angle
between the final momenta is always 90 degrees.
Center of Mass
Treat extended mass as a bunch of small masses:
Fg  mi g   mi  g  M g
In a uniform gravitational field you can treat gravitational force as
if it acts at the “Center of Mass”
Center of Mass
The center of mass of a system is the point where the
system can be balanced in a uniform gravitational field.
For two objects:
The center of mass is closer to the more massive object.
Center of Mass
In general:
RCM
m1r1  m1r1  m1r1 

m1  m1  m1 
mi ri mi ri


mi
M
Symmetry often very useful in determining the
Center of Mass
Center of Mass
The center of mass need not be within the object
Motion about the Center of Mass
The center of mass of a complex or composite object follows a
trajectory as if it were a single particle - with mass equal to the
complex object, and experiencing a force equal to the sum of all
external forces on that complex object
Motion of the center of mass
Action/Reaction pairs inside the system cancel out
The total mass multiplied by the acceleration of the
center of mass is equal to the net external force
The center of mass
accelerates just as
though it were a point
particle of mass M acted
on by
Momentum of a composite object
1 ptot

M t
MAcm   Fext
ptot

t
Recoil Speed II
A cannon sits on a stationary
railroad flatcar with a total
a) 0 m/s
mass of 1000 kg. When a 10-kg
b) 0.5 m/s to the right
cannonball is fired to the left at
c) 1 m/s to the right
a speed of 50 m/s, what is the
d) 20 m/s to the right
speed of the center of mass (of
e) 50 m/s to the right
the flatcar + cannonball)?
Recoil Speed II
A cannon sits on a stationary
railroad flatcar with a total
a) 0 m/s
mass of 1000 kg. When a 10-kg
b) 0.5 m/s to the right
cannonball is fired to the left at
c) 1 m/s to the right
a speed of 50 m/s, what is the
d) 20 m/s to the right
speed of the center of mass (of
e) 50 m/s to the right
the flatcar + cannonball)?
Because the initial momentum of the
system was zero, the final total momentum
must also be zero, regardless of the
release of internal energy, internal forces,
etc.
If no external forces act, the motion of the
center of mass does not change
Center of Mass
The disk shown below in (1) clearly has its
center of mass at the center.
a) higher
b) lower
Suppose the disk is cut in half and the
pieces arranged as shown in (2).
Where is the center of mass of (2) as
compared to (1) ?
c) at the same place
d) there is no definable CM
in this case
(1)
X
CM
(2)
Center of Mass
The disk shown below in (1) clearly has its
center of mass at the center.
a) higher
b) lower
Suppose the disk is cut in half and the
pieces arranged as shown in (2).
c) at the same place
d) there is no definable CM
in this case
Where is the center of mass of (2) as
compared to (1) ?
The CM of each half is closer
to the top of the semicircle
than the bottom. The CM of the
whole system is located at the
midpoint of the two semicircle
CMs, which is higher than the
yellow line.
(1)
X
CM
(2)
CM
A charging bull elephant with a mass of 5240 kg comes directly
toward you with a speed of 4.55 m/s. You toss a 0.150 kg rubber
ball at the elephant with a speed of 7.81 m/s. When the ball
bounces back toward you, what is its speed?
A charging bull elephant with a mass of 5240 kg comes directly
toward you with a speed of 4.55 m/s. You toss a 0.150 kg rubber
ball at the elephant with a speed of 7.81 m/s. When the ball
bounces back toward you, what is its speed?
Our simplest formulas for speed after an elastic
collision relied on one body being initially at rest.
So lets try a frame where one body (the ball) is at rest!
What is the speed of the
elephant relative to the ball?
In the frame where this is the elephant’s speed, what is the final speed of the ball?
Back in the frame of the ground:
A charging bull elephant with a mass of 5240 kg comes directly
toward you with a speed of 4.55 m/s. You toss a 0.150 kg rubber
ball at the elephant with a speed of 7.81 m/s. When the ball
bounces back toward you, what is its speed?
Our simplest formulas for speed after an elastic
collision relied on one body being initially at rest.
So lets try a frame where one body (the ball) is at rest!
What is the
speed ofFormulas
the
NOTE:
elephant relative to the ball?
for 1-D elastic scattering
with non-zero initial velocities are given in
In the frame
where this is the elephant’s
speed,
what is the final speed of the ball?
end-of-chapter
problem
88.
Back in the frame of the ground:
Rotational Kinematics
Angular Position
 x̂ 
Angular Position
θ>0
θ<0
Degrees and revolutions:
Arc Length
Arc length s, from angle
measured in radians:
s = rθ
- What is the relationship between the circumference
of a circle and its diameter?
C/D=π
C =2πr
- Arc length for a full rotation (360o) of a radius=1m circle?
s = 2 π (1 m) = 2 π meters
 x̂ 
1 complete revolution = 2 π radians
1 rad = 360o / (2π) = 57.3o
Angular Velocity
Instantaneous Angular Velocity
Period
= How long it takes
to go 1 full revolution
Period T:
SI unit: second, s
Linear and Angular Velocity
Greater translation for same rotation
Bonnie and Klyde
Bonnie sits on the outer rim of a
merry-go-round, and Klyde sits
midway between the center and the
rim. The merry-go-round makes one
revolution every 2 seconds. Who has
the larger linear (tangential) velocity?
a) Klyde
b) Bonnie
c) both the same
d) linear velocity is zero for
both of them

Klyde
Bonnie
Bonnie and Klyde
Bonnie sits on the outer rim of a
merry-go-round, and Klyde sits
midway between the center and the
rim. The merry-go-round makes one
revolution every 2 seconds. Who has
the larger linear (tangential) velocity?
a) Klyde
b) Bonnie
c) both the same
d) linear velocity is zero for
both of them
Their linear speeds v will be
different because v = r  and
Klyde

Bonnie is located farther out
(larger radius r) than Klyde.
Bonnie
Angular Acceleration
Instantaneous Angular Acceleration
Rotational Kinematics, Constant Acceleration
If the acceleration
is constant:
v = v0 + at
If the angular
acceleration is
constant:
Analogies between linear and rotational kinematics:
Angular Displacement I
An object at rest begins to rotate with a
constant angular acceleration. If this
object rotates through an angle  in the
time t, through what angle did it rotate
in the time ½ t?
a) ½ 
b) ¼ 
c) ¾ 
d) 2 
e) 4 
Angular Displacement I
An object at rest begins to rotate with a
constant angular acceleration. If this
object rotates through an angle  in the
time t, through what angle did it rotate
in the time ½ t ?
The angular displacement is  =
a) ½ 
b) ¼ 
c) ¾ 
d) 2 
e) 4 
 t 2 (starting from rest), and there
is a quadratic dependence on time. Therefore, in half the time, the
object has rotated through one-quarter the angle.
Which child experiences a greater acceleration?
(assume constant angular speed)
Larger r:
- larger v for same ω
- larger ac for same ω
ac is required for circular motion.
An object may have at as well, which implies angular acceleration
Angular acceleration and total linear
acceleration
Angular and linear acceleration
Rolling Motion
If a round object rolls without slipping, there is a
fixed relationship between the translational and
rotational speeds:
Rolling Motion
We may also consider rolling motion to be a
combination of pure rotational and pure
translational motion:
+
=
Truck Speedometer
Suppose that the speedometer of a
truck is set to read the linear speed
a)
of the truck but uses a device that
actually measures the angular
speed of the tires. If larger
b)
diameter tires are mounted on the
truck instead, how will that affect
the speedometer reading as
c)
compared to the true linear speed
of the truck?
speedometer reads a higher
speed than the true linear speed
speedometer reads a lower speed
than the true linear speed
speedometer still reads the true
linear speed
Truck Speedometer
Suppose that the speedometer of a
truck is set to read the linear speed
a)
of the truck but uses a device that
actually measures the angular
speed of the tires. If larger
b)
diameter tires are mounted on the
truck instead, how will that affect
the speedometer reading as
c)
compared to the true linear speed
of the truck?
speedometer reads a higher
speed than the true linear speed
speedometer reads a lower speed
than the true linear speed
speedometer still reads the true
linear speed
The linear speed is v =  R. So when the speedometer measures
the same angular speed  as before, the linear speed v is actually
higher, because the tire radius is larger than before.
Jeff of the Jungle swings on a vine that is 7.20 m long. At the
bottom of the swing, just before hitting the tree, Jeff’s linear
speed is 8.50 m/s.
(a) Find Jeff’s angular speed at this time.
(b) What centripetal acceleration does Jeff experience at the
bottom of his swing?
(c) What exerts the force that is responsible for Jeff’s
centripetal acceleration?
Jeff of the Jungle swings on a vine that is 7.20 m long. At the
bottom of the swing, just before hitting the tree, Jeff’s linear
speed is 8.50 m/s.
(a) Find Jeff’s angular speed at this time.
(b) What centripetal acceleration does Jeff experience at the
bottom of his swing?
(c) What exerts the force that is responsible for Jeff’s
centripetal acceleration?
a)
b)
c)
This is the force that is responsible for
keeping Jeff in circular motion: the vine.
Rotational Kinetic Energy
For this mass m, and a massless
rod
Provided in lecture
notes on: 10/6
PHYS2010 midterm 1, Fall 2008
13) If the net work done on an object is zero, then the object's kinetic energy
A) is zero.
B) decreases.
C) increases.
D) remains the same.
E) cannot be determined without knowing the object mass.
suggested time: 1 minute
Please do not ask questions about this problem
at discussion sessions before 10/7
Provided in lecture
notes on: 10/6
PHYS2010 midterm 1, Fall 2008
14) A horizontal force of 40 N pushes an object of mass 5.0 kg up an inclined plane
through a distance of 1.6 m measured along the plane. The plane is inclined at an
angle of 30o to the horizontal. Neglect friction and use g=10 m/s2, What is the work
done by the normal force on the object?
A) 53 J
B) 15 J
C) 0 J
D) -25 J
E) 37 J
suggested time: 1 minute
Please do not ask questions about this problem
at discussion sessions before 10/7
Provided in lecture
notes on: 10/6
PHYS2010 midterm 2, Fall 2008
5) In a particle accelerator, protons are fired at a stationary target and collide
elastically with the nuclei of the atoms in the target. In one such experiment, a beam of
protons is fired with a speed of 2.10 x 107 m/s. The protons that undergo a collision
rebound with a speed of 1.68 x 107 m/s. What is the mass of the nuclei that the
protons are colliding against?
A) 9.00 proton masses
B) 10.0 proton masses
C) 7.00 proton masses
D) 8.00 proton masses
E) 6.00 proton masses
suggested time: 3-4 minutes
Please do not ask questions about this problem
at discussion sessions before 10/7
Provided in lecture
notes on: 10/6
PHYS2010 midterm 2, Fall 2008
6) A Ferris wheel with a radius of 8.00 m rotates at a constant rate, completing one
revolution in 30.0 s. What is the apparent weight of a 60.0-kg passenger when she is
at the top of the wheel?
A) 615 N
B) 589 N
C) 568 N
D) 325 N
E) 432 N
suggested time: 4-5 minutes
Please do not ask questions about this problem at
discussion sessions before 10/19