Physics 111: Lecture 3, Pg
1
Physics 111: Lecture 3
Today’s Agenda
Reference frames and relative motion
Uniform Circular Motion
Physics 111: Lecture 3, Pg
2
Inertial Reference Frames:
A
Reference Frame
is the place you measure from.
It’s where you nail down your
(x,y,z)
axes!
An Inertial Reference Frame (
IRF
) is one that is
not
accelerating
.
We will consider only
IRF
s in this course.
Valid
IRF
s can have fixed velocities with respect to each other.
More about this later when we discuss forces.
For now, just remember that we can make measurements
from different vantage points.
Cart on
track on
track
Physics 111: Lecture 3, Pg
3
Relative Motion
Consider a problem with
two
distinct IRFs:
An airplane flying on a windy day
.
A pilot wants to fly from Champaign to Chicago. Having
asked a friendly physics student, she knows that Chicago
is 120 miles due north of Urbana. She takes off from
Willard Airport at noon. Her plane has a compass and an
air

speed indicator to help her navigate.
The compass allows her to keep the nose of the plane
pointing north.
The air

speed indicator tells her that she is traveling at
120 miles per hour
with respect to the air
.
Physics 111: Lecture 3, Pg
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Relative Motion...
The plane is moving north in the IRF attached to the air:
V
p, a
is the velocity of the plane w.r.t. the air.
Air
V
p,a
Physics 111: Lecture 3, Pg
5
Relative Motion...
But suppose the air is moving
east
in the IRF attached to
the ground.
V
a,g
is the velocity of the air w.r.t. the ground (i.e.
wind
).
V
a,g
Air
V
p,a
Physics 111: Lecture 3, Pg
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Relative Motion...
What is the velocity of the plane in an IRF attached to the
ground?
V
p,g
is the velocity of the plane w.r.t. the ground.
V
p,g
Physics 111: Lecture 3, Pg
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Relative Motion...
V
p,g
=
V
p,a
+
V
a,g
Is a vector equation relating the airplane’s
velocity in different reference frames.
V
p,g
V
a,g
V
p,a
Tractor
Physics 111: Lecture 3, Pg
8
Lecture 3,
Act 1
Relative Motion
You are swimming across a
50m
wide river in which the
current moves at
1 m/s
with respect to the shore. Your
swimming speed is
2 m/s
with respect to the water.
You swim across in such a way that your path is a straight
perpendicular line across the river.
How many seconds does it take you to get across ?
(a)
(b)
(c)
50
3
29
2 m/s
1 m/s
50 m
35
2
50
50
1
50
Physics 111: Lecture 3, Pg
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Lecture 3,
Act 1
solution
The time taken to swim straight across is
(distance across) / (v
y
)
Choose
x
axis along riverbank and
y
axis across river
y
x
Since you swim straight across, you must be tilted in the water so that
your
x
component of velocity with respect to the water exactly cancels
the velocity of the water in the
x
direction:
2 m/s
1m/s
y
x
1 m/s
2
1
3
2
2
m/s
Physics 111: Lecture 3, Pg
10
Lecture 3,
Act 1
solution
So the
y
component of your velocity with respect to the water is
So the time to get across is
y
x
3
m/s
50
3
29
m
m
s
s
50 m
3
m/s
Physics 111: Lecture 3, Pg
11
Uniform Circular Motion
What does it mean?
How do we describe it?
What can we learn about it?
Physics 111: Lecture 3, Pg
12
What is UCM?
Motion in a circle with:
Constant Radius
R
Constant Speed
v =

v

R
v
x
y
(x,y)
Puck on ice
Physics 111: Lecture 3, Pg
13
How can we describe UCM?
In general, one coordinate system is as good as any other:
Cartesian:
»
(x,y)
[position]
»
(v
x
,v
y
)
[velocity]
Polar:
»
(R,
)
[position]
»
(v
R
,
)
[velocity]
In UCM:
R
is constant (hence
v
R
= 0
).
(angular velocity) is constant.
Polar coordinates are a natural way to describe UCM!
R
v
x
y
(x,y)
Physics 111: Lecture 3, Pg
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Polar Coordinates:
The arc length
s
(distance along the circumference) is
related to the angle in a simple way:
s
=
R
,
where
is the
angular displacement
.
units of
are called
radians
.
For one complete revolution:
2
R
=
R
c
c
=
2
has
period
2
.
1 revolution = 2
radians
R
v
x
y
(x,y)
s
Physics 111: Lecture 3, Pg
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Polar Coordinates...
x
=
R
cos
y
=
R
sin
/
2
3
/2
2

1
1
0
sin
cos
R
x
y
(
x
,
y
)
Physics 111: Lecture 3, Pg
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Polar Coordinates...
In Cartesian coordinates, we say velocity
dx/dt = v
.
x = vt
In polar coordinates, angular velocity
d
/dt =
.
=
t
has units of
radians/second
.
Displacement
s = vt
.
but
s = R
= R
t,
so:
R
v
x
y
s
t
v
=
R
Tetherball
Physics 111: Lecture 3, Pg
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Period and Frequency
Recall that
1 revolution = 2
radians
frequency
(
f
)
= revolutions / second
(a)
angular velocity
(
)
= radians / second
(b)
By combining
(a)
and
(b)
= 2
f
Realize that
:
period
(
T
) =
seconds / revolution
So
T = 1 / f = 2
/
R
v
s
= 2
/ T =
2
f
Physics 111: Lecture 3, Pg
18
Recap:
R
v
s
t
(x,y)
x
=
R
cos(
)
=
R
cos(
t)
y
=
R
sin(
)
=
R
sin(
t)
= arctan
(y/x)
=
t
s
=
v
t
s
=
R
=
R
t
v
=
R
Physics 111: Lecture 3, Pg
19
Aside: Polar Unit Vectors
We are familiar with the Cartesian unit vectors:
i j k
Now introduce
“polar unit

vectors”
r
and
:
r
points in radial direction
points in tangential direction
R
x
y
i
j
r
^
^
^
^
^
^
(counter clockwise)
Physics 111: Lecture 3, Pg
20
Acceleration in UCM:
Even though the
speed
is constant,
velocity
is
not
constant
since the direction is changing:
must be some acceleration
!
Consider average acceleration in time
t
a
av
=
v
/
t
v
2
t
v
1
v
1
v
2
v
R
Physics 111: Lecture 3, Pg
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Acceleration in UCM:
seems like
v
(hence
v
/
t
)
points at the origin!
R
Even though the
speed
is constant,
velocity
is
not
constant since
the direction is changing.
Consider average acceleration in time
t
a
av
=
v
/
t
v
Physics 111: Lecture 3, Pg
22
Acceleration in UCM:
Even though the
speed
is constant,
velocity
is
not
constant
since the direction is changing.
As we shrink
t
,
v
/
t
d
v
/ dt =
a
a
= d
v
/ dt
We see that
a
points
in the

R
direction.
R
Physics 111: Lecture 3, Pg
23
Acceleration in UCM:
This is called
Centripetal Acceleration.
Now let’s calculate the magnitude:
v
2
v
1
v
1
v
2
v
R
R
v
v
R
R
Similar triangles:
But
R =
v
t
for small
t
So:
v
t
v
R
2
v
v
v
t
R
a
v
R
2
Physics 111: Lecture 3, Pg
24
Centripetal Acceleration
UCM
results in acceleration:
Magnitude
:
a
= v
2
/ R
Direction
:

r
(toward center of circle)
R
a
^
Physics 111: Lecture 3, Pg
25
Derivation:
R
R
a
2
We know that and
v
=
R
Substituting for v we find that:
a
v
R
2
a =
2
R
Physics 111: Lecture 3, Pg
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Lecture 3,
Act 2
Uniform Circular Motion
A fighter pilot flying in a circular turn will pass out if the
centripetal acceleration he experiences is more than about
9
times the acceleration of gravity
g
. If his F18 is moving
with a speed of
300 m/s
, what is the approximate diameter
of the tightest turn this pilot can make and survive to tell
about it ?
(a)
500 m
(b)
1000 m
(c)
2000 m
Physics 111: Lecture 3, Pg
27
Lecture 3,
Act 2
Solution
a
v
R
g
2
9
2
2
2
2
s
m
81
9
9
s
m
90000
g
9
v
R
.
m
1000
m
81
9
10000
R
.
D
R
m
2
2000
2km
Physics 111: Lecture 3, Pg
28
Example: Propeller Tip
The propeller on a stunt plane spins with frequency
f = 3500 rpm
. The length of each propeller blade is
L = 80cm
.
What centripetal acceleration does a point at the tip of a
propeller blade feel?
f
L
what is
a
here?
Physics 111: Lecture 3, Pg
29
Example:
First calculate the angular velocity of the propeller:
so
3500 rpm
means
=
367 s

1
Now calculate the acceleration.
a
=
2
R
=
(367s

1
)
2
x (0.8m)
=
1.1 x 10
5
m/s
2
=
11,000 g
direction of
a
points at the propeller hub (

r
).
1
1
1
60
2
0
105
0
105
rpm
s
1
rot
x
s
x
rad
rot
rad
s
min
min
.
.
^
Physics 111: Lecture 3, Pg
30
Example: Newton & the Moon
What is the acceleration of the Moon due to its motion
around the Earth?
What we know (Newton knew this also):
T
= 27.3 days = 2.36 x 10
6
s
(period ~ 1 month)
R
= 3.84 x 10
8
m
(distance to moon)
R
E
= 6.35 x 10
6
m
(radius of earth)
R
R
E
Physics 111: Lecture 3, Pg
31
Moon...
Calculate angular velocity:
So
= 2.66 x 10

6
s

1
.
Now calculate the acceleration.
a
=
2
R
=
0.00272 m/s
2
= 0
.000278
g
direction of
a
points at the center of the Earth (

r
).
1
27
3
1
86400
2
2
66
10
6
.
.
rot
day
x
day
s
x
rad
rot
x
s
1
^
Physics 111: Lecture 3, Pg
32
Moon...
So we find that
a
moon
/
g
= 0.000278
Newton noticed that
R
E
2
/ R
2
= 0.000273
This inspired him to propose that
F
Mm
1 / R
2
(more on gravity later)
R
R
E
a
moon
g
Physics 111: Lecture 3, Pg
33
Lecture 3,
Act 3
Centripetal Acceleration
The Space Shuttle is in Low Earth Orbit (LEO) about
300 km
above the surface. The period of the orbit is about
91 min
.
What is the acceleration of an astronaut in the Shuttle in the
reference frame of the Earth?
(The radius of the Earth is
6.4 x 10
6
m
.)
(a)
0 m/s
2
(b)
8.9 m/s
2
(c)
9.8 m/s
2
Physics 111: Lecture 3, Pg
34
First calculate the angular frequency
:
Realize that:
Lecture 3,
Act 3
Centripetal Acceleration
R
O
300 km
R
O
= R
E
+ 300 km
= 6.4 x 10
6
m + 0.3 x 10
6
m
= 6.7 x 10
6
m
R
E
1

s
00115
.
0
rot
rad
2
x
s
60
1
x
rot
91
1
min
min
Physics 111: Lecture 3, Pg
35
Now calculate the acceleration:
Lecture 3,
Act 3
Centripetal Acceleration
a
=
2
R
a
=
(0.00115 s

1
)
2
x
6.7 x 10
6
m
a
=
8.9 m/s
2
Physics 111: Lecture 3, Pg
36
Recap for today:
Reference frames and relative motion.
(Text: 2

1, 3

3, & 4

1)
Uniform Circular Motion
(Text: 5

2, also 9

1)
Look at Textbook problems
Chapter 3: # 47, 49, 97, 105
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