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PHYSICS 220
Lecture 08
Circular Motion
Textbook Sections 5.3 – 5.5
Lecture 8
Purdue University, Physics 220
1
Overview
• Last Lecture
– Circular
Ci l M
Motion
ti
• θ = angular position radians
• ω = angular velocity radians/second
• α = angular
l acceleration
l ti radians/s
di
/ 2
• Linear to Circular conversions s = r θ
– Uniform Circular Motion
• Speed
S
d iis constant
t t
• Direction is changing
• Acceleration toward center a = v2 / r
– Uniform
U if
Ci
Circular
l A
Acceleration
l ti Ki
Kinematics
ti
• Similar to linear
• Today
– Banked curves
– Circular orbits
Lecture 8
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Circular Motion
Angular velocity
Linear velocity
v = rω
Δθ
ωav =
Δt
Linear velocity is tangent to circle
Centripetal acceleration
R
2
v
2
ac = = ω r
r
a
v
Direction: toward the center
Centripetal Acceleration is radially inward
Lecture 8
Purdue University, Physics 220
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Circular vs. Linear Motion
with comparison to 1-D kinematics
Angular
Linear
α = constant
a = constant
ω = ω0 + α t
v = v0 + at
1
2
1
x = x0 + v0t + at 2
2
θ = θ 0 + ω0t + α t 2
And for a point at a distance R from the rotation axis:
x = Rθ
Lecture 8
v = ωR
R
Purdue University, Physics 220
a = αR
R
4
Kinematics for Circular Motion
with constant acceleration
Linear Variables x,v,a.
x v a Constant a
Δx = x − x0 = vavg Δt
Δθ = θ − θ 0 = ωavg Δt
v − v0 = at
ω − ω0 = α t
ω0 + ω
v0 + v
vavg =
2
1 2
x − x0 = v0t + at
2
1 2
θ − θ 0 = ω0 t + α t
2
v − v = 2aΔx
ω 2 − ω02 = 2αΔθ
2
Lecture 8
A
Angular
l V
Variables
i bl θ,ω,α. Constant
C
α
2
0
ωavg =
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5
Quiz
1) Two children ride on a merry-go-round. Bob is 2 m from the
axis
i off rotation
i and
dM
Mary iis 4 m ffrom the
h axis.
i Whi
Which
h iis true:
A) Mary has larger speed and acceleration but the same
angular speed
B) Bob has larger speed, acceleration, and angular speed
y has larger
g speed,
p
, acceleration,, and angular
g
speed
p
C)) Mary
Lecture 8
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Quiz
2) Which statement is true for a car in uniform circular motion?
A) The car is moving with a constant speed, therefore the car has a net force
of zero
B) For
F the
th car to
t stay
t traveling
t
li around
d a horizontal
h i
t l circular
i l ttrack,
k th
there mustt
be a net force pointing radially inward, toward the center of the circle. If there
wasn't, the car would drive in a straight line.
C) If you turn sharp, you tend to be pulled in the opposite direction of the turn.
FN
f
R
W
ΣF = ma = mv2/R
Lecture 8
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Unbanked Curve
• A car rounding a curve travels
in an approximate circle
• The radius of this circle is
called the radius of curvature
• Forces in the y-direction
– Gravityy and the normal force
• Forces in the x-direction
– Friction is directed toward the
center of the circle
• Since friction is the only force
acting in the xx-direction,
direction it
supplies the centripetal force
Ffriction
Lecture 8
mv 2
=
= μs m g
r
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Unbanked Curve
What is the maximum velocity a car can go around an
unbanked curve?
y : N − mg = 0 N = mg
f s ≤ μs N
mv 2
x : Fc = f s = μ s N = μ s mg =
r
The maximum
velocity to go
v = μ s gr
around an unbanked curve
depends only on μs
(g and r are fixed))
Dry road: μs=0.9
I road:
Icy
d μs=0.1
01
Lecture 8
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Banked Curve
• The maximum speed can be
increased by banking the
curve
• Assume no friction between
the tires and the road
• The car travels in a circle,
so the net force is a
centripetal force
• There are forces due to
gravity and the normal force
acting on the car
Lecture 8
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Banked Curve
• There is a horizontal component of the normal
force
– Letting the horizontal be the x-direction
• The speed at which the car will just be able to
g
the turn without sliding
g up
p or down the
negotiate
banked road is
v = r g tanθ
• Wh
When θ = 0,
0 v = 0 and
d you cannott tturn on a very
icy road without slipping
Lecture 8
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Banked Curve
A car drives around a curve with radius 410 m at a
speed of 32 m/s. The road is banked at 5.0. The
mass of the car is 1400 kg.
A) What is the frictional force on the car?
B) At what speed could you drive around this curve so that
the force of friction is zero?
Lecture 8
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Banked Curve
y
θ
x
θ =5
y-direction
y
Σ F y = ma y = 0
N cos θ − mg − f sin θ = 0
(1)
0
r = 410 m
v = 32 m / s
N
x-direction
ΣF x = max = ma
2
v
N sin θ + f cos θ = ma = m
r
Lecture 8
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((2))
W
f
13
Banked Curve
2 equations and 2 unknown we can solve for N in
(1) and substitute in (2)
f sin θ + m g
v2
N =
N sin θ + f cos θ = m
r
co s θ
2
f
sin
θ
+
mg
mv
⎛
⎞
⎜
⎟ sin θ + f cos θ =
r
cos θ
⎝
⎠
2
mv
2
2
f (sin θ + cos θ ) =
cos θ − mg sin θ
r
⎛ v2
⎞
f = m ⎜ cos θ − g sin θ ⎟ = 2300 N
⎝ r
⎠
Lecture 8
Purdue University, Physics 220
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Banked Curve
A car drives around a curve with radius 410 m at a
speed of 32 m/s. The road is banked at 5.0°. The
mass of the car is 1400 kg.
A) What is the frictional force on the car?
B) At what speed could you drive around this curve so that
the force of friction is zero?
Like an airplane
f =0
2
v
cos θ = g sin θ
r
v = gr tan θ = 19m / s
Lecture 8
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Circular Motion and Forces
• Newton’s Second Law can be applied to circular
motion:
2
G
G
mv
∑ F = m a → ∑ F = m aC = r
• The force must be directed toward the center of
the circle
• The
Th centripetal
t i t l force
f
can be
b supplied
li d b
by a variety
i t
of physical objects or forces
• The “circle” does not need to be a complete circle
Lecture 8
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Centripetal Force Example
• The centripetal
acceleration is
produced by the
tension in the string
• If the string
g breaks,,
the object would move
g
in a direction tangent
to the circle at a
p
constant speed
Lecture 8
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Conical Pendulum
Net force must point towards the center of the circle
x : ∑ Fx = T sin φ = ma x = mω 2 r
r = L sin φ
T sin
i φ = mω 2 L sin
i φ
T = mω 2 L
y : ∑ Fx = T cos φ = mg
mω 2 L cos φ = mg
g
ω =
L cos φ
2
ω=
Lecture 8
Purdue University, Physics 220
g
L cos φ
18
ILQ: Carousel Motion
• Java applet
What is the direction of the acceleration?
A) Along the tension in the cable
B) In a horizontal direction toward the center
C) In a horizontal direction away from the center
Lecture 8
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Circular Orbits of Planets
Ancient Greeks:
The Geocentric Model implies Earth
Earth-Centered
Centered Universe
Lecture 8
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Circular Orbits of Planets
• Copernicus (1500's)
• Developed
D
l
d a mathematical
th
ti l
model for a Sun-centered
solar system
– Published in De Revolutionibus
– Dedicated the book to Paul III
– The De Revolutionibus was placed
on the Index of Prohibited Books in
1616 as a result of the Galileo
affair
– It was taken off the Index in 1758
Lecture 8
Purdue University, Physics 220
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Circular Orbits of Planets
Lecture 8
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Orbits of Planets
Tycho Brahe
(1546-1601)
Johannes Kepler
(1571-1630)
• Made precise
measurements of the
positions of the planets
Lecture 8
• Described the shape of
planetary orbits
as well as their orbital
speeds
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Kepler’s Laws
• First law: The orbit of a
planet
l
t about
b t the
th Sun
S is
i
an ellipse with the Sun
att one focus.
f
• Second law: A line
joining a planet and the
Sun sweeps out equal
areas in equal intervals
of time.
Lecture 8
Purdue University, Physics 220
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Kepler’s Third Law
• The square
q
of a p
planet's orbital p
period is
proportional to the cube of the length of its orbit's
j axis.
semimajor
• Or simply…
simply
T2 = R3
if T is measured in
years and R is measured in astronomical units.
• An AU is the average distance of the Earth from
the Sun. 1 AU = 93,000,000 miles = 8.3 light
minutes
Lecture 8
Purdue University, Physics 220
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Kepler’s Laws
Elliptical
orbits
orbits…
Equal areas in
equal time
T2 = R3
• These were empirical laws
Lecture 8
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Newton’s Law of Universal
Gravitation
• F
From Kepler's
K l ' 3rd
3dL
Law, N
Newton
t d
deduced
d
d iinverse square
law of attraction.
Gm1m2
F=
2
d
• G
G=6
6.67
67 × 10-11 N m2/kg2
Your book derives Kepler’s
Kepler s third law
from Newton’s law for a circular orbit
Lecture 8
Purdue University, Physics 220
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4
π
T2 =
r3
GM sun
27
Satellites and Planets
GmM earth
mv 2
∑ Fr = r 2 = mar = r
GM earth
v=
r
- Speed
p
is independent
p
of mass of satellite
- Satellites in lower orbits have greater speeds
Geostationary orbits:
A circular orbit in Earth’s equatorial
plane
l
whose
h
period
i d iis equall tto
Earth’s rotational period.
~35,786
,
km above ground
g
Lecture 8
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Satellites ACT
Two identical (except for color) satellites are in
circular orbits around the Earth
Earth. The red satellite is
farther from the earth than the green one. Which
satellite has the greater centripetal acceleration?
A) Red
B) Green
C) They are the same
GMm
F = mac = 2
r
ac = ω 2 r
ag =
r3 =
Lecture 8
GM earth
r2
GM earthth
ω2
Purdue University, Physics 220
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Satellites ACT
Two identical (except for color) satellites are in
circular orbits around the Earth
Earth. The red satellite is
farther from the earth than the green one. Which
satellite has the greater velocity?
A) R
Red
d
B) Green
C) Th
They are th
the same
GmM
G
M mv
=
2
r
r
Lecture 8
2
GM
v=
r
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Weightlessness
• Inside the Space Shuttle, the astronauts experience
apparent weightlessness
weightlessness.
• The force of gravity on the Earth pulls on the astronauts to
keep them in a circular orbit around the Earth, so we can’t
can t
accurately say there is “no gravity” there.
• Both the astronauts and the Space Shuttle are in uniform
circular motion, and are continually accelerating towards
the Earth. The are both “falling” at the same rate.
• The apparent weight (how heavy you feel) is defined as
the magnitude of the normal force acting on the body:
G G
W′ = m g −a
where g is the local gravitational field strength:
Lecture 8
Purdue University, Physics 220
G GM
g= 2
r
31
Artificial Gravity
Lecture 8
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Summary of Concepts
• Uniform Circular Motion
– Speed is constant, but velocity is not
– Linear to Circular conversions s = r θ
– Acceleration toward center a = v2 / r
• Kepler
Kepler’s
s Laws
– The orbit of a planet about the Sun is an ellipse with the Sun at
one focus
– A line joining a planet and the Sun sweeps out equal areas in
equal intervals of time
– The square of a planet
planet's
s orbital period is proportional to the cube
of the length of its orbit's semimajor axis T2 = R3
Lecture 8
Purdue University, Physics 220
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