Download Ch33

Chapter 8. Dynamics II: Motion in a Plane
Chapter Goal: To learn
how to solve problems about
motion in a plane, especially
circular motion.
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Student Learning Objectives – Ch. 8
• To understand the dynamics of uniform
circular motion.
• To learn the basic ideas of orbital motion.
• To answer “How does the water stay in the
bucket?” and related questions.
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Uniform Circular Motion
The acceleration of uniform
circular motion points to the
center of the circle. Thus the
acceleration vector has only a
radial component ar. This
acceleration is conveniently
written in the rtz-coordinate
system as
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Uniform Circular Motion
The acceleration of uniform
circular motion points to the
center of the circle.
Why? – Let’s investigate…
with a ball on a string, and on a
track (of sorts)
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Dynamics of Uniform Circular Motion
Newton’s second law in terms
of the r-, t-, and z-components,
is as follows:
There is no such thing as a
“centripetal force”. It is a
normal, tension or other
kind of force that results in
a centripetal acceleration.
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Mass on a turntable
Is there a force on the
mass?
What kind of force?
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Mass on a turntable
The turntable slowly
speeds up. Which
mass will slide off
first?
a. m because r > r2
b. m2 because r > r2
c. same time, since ω is
the same for both
masses.
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m2
•
Mass on a turntable
The turntable slowly
speeds up. Which
mass will slide off
first?
a. m because r > r2
b. m2 because r > r2
c. same time, since ω is
the same for both
masses.
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m2
•
Car at constant speed in a straight line – No net
force on the car.
y
n
x
fr
fs
FG
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Car turning a corner at constant speed
Is there a net force on the
car as it negotiates the
turn?
Where did it come from?
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Car turning a “circular” corner at constant speed
When the driver turns the
wheel the tires turn. To
continue along a straight
line, the car must
overcome static friction
and slide. If the static
friction force is less than
the maximum, the tire
cannot slide and so has no
choice but to roll in the
direction of the turn.
fs-max = μs |n|
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skid….
blue arrow
represents
direction car
would slide in the
absence of friction
End of chapter problem #33
A 5 g coin is placed 15 cm
from the center of a
turntable. The table
speeds up to 60 rpm.
a. Does the coin slide
off?
b. If not, at what speed
(in rpms) will it slide?
Us = .80, uk = .50
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End of chapter problem #33
A 5 g coin is placed 15 cm
from the center of a
turntable. The table
speeds up to 60 rpm.
a. Does the coin slide
off? - No
b. If not, at what speed
(in rpms) will it slide?
– 69 rpms
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The banked curve (EOC #8)
• A highway curve of
radius 500 m is
designed for traffic
moving at 90 km/hour.
What is the correct
banking angle for the
road?
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The banked curve (EOC #8)
• A highway curve of
radius 500 m is
designed for traffic
moving at 90 km/hour.
What is the correct
banking angle for the
road?
• The radial axis is in the
true horizontal
direction, relative to the
angle of bank.
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The banked curve (EOC #8)
• A highway curve of
radius 500 m is
designed for traffic
moving at 90 km/hour.
What is the correct
banking angle for the
road? Answer is 7.30
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Comparison of level curve and banked curve
Level curve with friction:
fs = mv2/r. If the speed
increases, the frictional
force does the same, until
fs max. Any v < vmax
works.
Banked curve:
mg tan θ = mv2/r. There is
only one value of speed
where this works without
some help from friction.
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Comparison of level curve and banked curve
For v >v0, mg tan θ is too
small and car will slide
up the bank, without
friction. Static friction
points towards the circle
For v<v0, mg tan θ is too
large and car will slide
down the bank without
friction. Static friction
points away from the
circle.
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A ride at a carnival consists of chairs that are swung in a circle by
15.0-m cables attached to a vertical rotating pole, as shown in
the drawing. The total mass of a chair and occupant is 179 kg.
a) Draw a free body diagram for this situation.
b) Determine the Tension in cable c) speed (v) of chair
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
A ride at a carnival consists of chairs that are swung in a circle by
15.0-m cables attached to a vertical rotating pole, as shown in
the drawing. The total mass of a chair and occupant is 179 kg.
a) Draw a free body diagram for this situation.
b) Determine the Tension in cable c) speed (v) of chair
60°
b. 3510 N
c.
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14.9 m/s
Circular Orbits
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Circular Orbits
An object moving around the earth in a circular orbit of
radius r at speed vorbit will have centripetal acceleration of
That is, if an object moves parallel to the surface with the
speed
then the free-fall acceleration provides exactly the
centripetal acceleration needed for a circular orbit of radius
r. An object with any other speed will not follow a circular
orbit.
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Forces in a non-inertial reference frame
• You are riding in a car
that makes a sudden
stop, and you feel as if a
force throws you
forward towards the
windshield.
• A force is an
interaction between two
objects. What object
caused the force?
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Forces in a non-inertial reference frame
• You can describe your
experience in terms of
what are called
fictitious forces.
• These are not real
forces because no agent
is exerting them, but
they describe your
motion relative to a
noninertial reference
frame.
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Stop to Think – Newtonian Analysis
Is the normal force at
the cars (and people)
at the bottom of the
roller coaster greater
than, less than or
equal to FG? What
would the scale say?
Use a Newtonian
analysis to support
your answer.
Think about sitting on a
scale in the roller coaster
– I hate when that
happens!
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Stop to Think – Newtonian Analysis
Is the normal force of the cars
(and people) at the bottom
of the roller coaster greater
than, less than or equal to
FG?
Fr-net = n – FG = mv2/r
n
Greater than, the roller coaster
is in circular motion, so
there has to be a net force
into the circle.
FG
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Stop to think more
Fr-net = n – FG = mv2/r
What changes (n, FG, both,
neither) if the carts are
going faster?
n
FG
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Stop to think more
Fr-net = n – FG = mv2/r
What changes (n, FG, both,
neither) if the carts are
going faster?
The right side of the equation
has changed, and one of the
left hand terms is fixed so…
n
n must increase!
FG
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Stop to think more
Fr-net = n – FG = mv2/r
What does it mean physically if
n = FG in this situation?
Can n ever be less than FG for
this situation? In what
case(s)?
n
FG
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Tension (Workbook #8)
A ball swings around in a
vertical circle on a
string. Is the tension at
the bottom of the swing
greater than, less than
or equal to FG? Use a
Newtonian analysis to
support your answer.
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Tension
T greater than FG , the ball
is in circular motion, so
there has to be a net
force into the circle.
Fr-net = T – FG = mv2/r
If T=FG, what does that
mean physically?
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What about at the top?
Is the tension at the top of
the swing greater than,
less than or equal to
FG? Use a Newtonian
analysis to support your
answer.
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What about at the top?
For this situation:
FG + T = mv2/r
Can T be 0 for this situation?
Can T be less than zero? (i.e.
away from the circle)?
• What happens physically?
• What happens
mathematically?
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What about at the top?
Derive an expression for
minimum speed the ball can
have and reach the top of
the circle without falling in
terms of T, m, r, and any
constants:
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What about at the top?
The mimimum speed occurs
when T = 0, since T cannot
be negative
• mg = mv2/r
• v = rg
• This value is known as the
vc, the critical speed.
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Why Does the Water Stay in the Bucket?
For this situation, what happens to the normal force and
the gravitational force, as the speed decreases?
Is is physically possible for n< 0 (for the track to pull
on the cart)?
What is the minimum speed necessary for the roller
coaster to stay on the track?
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Why Does the Water Stay in the Bucket?
•mg = mv2/r
rg
•v =
•This value is known as
the vc, the critical
speed.
•If the track isn’t
pressing against the
cart, the cart can’t press
against the track, i.e. it
loses contact and falls.
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A car is rolling over the top of a
hill at speed v. At this instant,
A. n > FG.
B. n < FG.
C. n = FG.
D.We can’t tell about n without knowing v.
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A car is rolling over the top of a
hill at speed v. At this instant,
Fr-net = FG - n = mv2/r
n < FG
If you were sitting on a
scale now, what would
happen as you went
faster?
How fast could you go?
Explain.
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It is easy to show that the
maximum speed you
could attain and still stay
on the road is rg
What is the physical
significance of n
becoming zero for this
instance?
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Why Does the Water Stay in the Bucket?
•The critical angular velocity ωc is that at which gravity
alone is sufficient to cause circular motion at the top.
•Since ω = v/r:
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Why Does the Water Stay in the Bucket?
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EOC #14
• A roller coaster car with
a scale built into the seat
(I hate when that
happens!) goes through
a dip with a radius of 30
m. The scale reading
increases by 50% at the
bottom of the dip. What
is car’s speed?
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EOC #14 - answer
The scale reading is the normal
force:
Fr-net = n – FG = mv2/r
v = 12.1 m/s
n
FG
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Nonuniform Circular Motion
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Nonuniform Circular Motion
• The force component (Fnet)r creates a centripetal
acceleration and causes the particle to change
directions.
• The component (Fnet)t creates a tangential
acceleration and causes the particle to change speed.
• Force and acceleration are related to each
other through Newton’s second law.
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EOC # 18
A new car is tested on a 200-m diameter track.
If the car speeds up at a constant 1.5 m/s each
second, how long after starting is the
magnitude of the car’s centripetal acceleration
equal to that of tangential acceleration?
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EOC # 18
at = ar = 1.5 m/s2
ar = v2/r so v = 12.2 m/s
Is time important?
∆t = ∆v/at = 8.13 s
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EOC #19
A toy train is released with an angular speed of
30 rpm on a 1.0 m-diameter track. The
coefficient of rolling friction is 0.10 (This is a
resistive friction).
a. What is the magnitude of angular acceleration
immediately after being released?
b. How long does it take the train to stop?
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EOC #19 - answer
a. What is the magnitude
of angular acceleration
immediately after being
released?
• Use Newton’s 2nd Law
to solve for Fnet in the t
direction, then calculate
angular accleration:
α = Fnet/mr = 1.96 rad/s2.
Note this opposes
motion.
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EOC #19 - answer
b.
How long does it take to
stop?
Determine whether time is
important (!) and use the
appropriate kinematic
equation for radial
components after
converting ω into SI units
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Banked Turn with friction
A 100-m radius curve in a road is constructed
with a bank angle of 100. On a rainy day, the
static coefficient of friction can be as low as
0.10. What are the highest and lowest speeds
at which a car can make this turn without
sliding in those conditions?
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Banked Curve
For v >v0, normal force
component is too small
and car will slide up the
bank. Static friction
points toward the circle.
For v<v0, normal force
component is too large
and car will slide down
the bank. Static friction
points away from the
circle.
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Comparison of level curve and banked curve
Greatest speed is 16.6 m/s
Slowest speed is 8.57 m/s
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Vertical Circular Motion Problem
An circular clothes dryer (radius = 0.32 m)
is designed so that the clothes tumble
gently as they dry. This means that
when a piece of clothing reaches some
angle above the horizontal, it loses
contact with the cylinder wall and falls.
How many revolutions per second
should the dryer make so this occurs at
θ = 700 ?
Note: the axes shown are not necessarily
the best choice for this problem. They
simply serve to show the orientation of
θ with true horizontal and vertical.
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Vertical Circular Motion Problem
Once an appropriate
free body diagram is
chosen, all that is left
is to write Newton’s
2nd Law for circular
motion for the radial
axis:
mg sin θ = mω2r
ω = .85 rps
r
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θ
FG
t