# Physics 111: Lecture 3

Mechanics

Nov 14, 2013 (4 years and 8 months ago)

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Physics 111: Lecture 3, Pg
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Physics 111: Lecture 3

Today’s Agenda

Reference frames and relative motion

Uniform Circular Motion

Physics 111: Lecture 3, Pg
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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.

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
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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
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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
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Uniform Circular Motion

What does it mean?

How do we describe it?

What can we learn about it?

Physics 111: Lecture 3, Pg
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What is UCM?

Motion in a circle with:

R

Constant Speed
v =

|
v
|

R

v

x

y

(x,y)

Puck on ice

Physics 111: Lecture 3, Pg
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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
.

For one complete revolution:

2

R

=
R

c

c

=
2

has
period

2

.

1 revolution = 2

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
.

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

frequency

(
f
)
= revolutions / second
(a)

angular velocity
(

)

(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
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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 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
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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
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Centripetal Acceleration

UCM

results in acceleration:

Magnitude
:

a

= v
2

/ R

Direction
:

-

r
(toward center of circle)

R

a

^

Physics 111: Lecture 3, Pg
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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

(a)
500 m

(b)
1000 m

(c)
2000 m

Physics 111: Lecture 3, Pg
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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

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
rot
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

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
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
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