2. Physics 111: Lecture 3, Pg 2
Inertial Reference Frames:Inertial Reference Frames:
A Reference FrameReference 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 IRFs in this course.
Valid IRFs 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
3. Physics 111: Lecture 3, Pg 3
Relative MotionRelative Motion
Consider a problem with twotwo distinct IRFs:
An airplane flying on a windy dayAn 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 airwith respect to the air.
4. Physics 111: Lecture 3, Pg 4
Relative Motion...Relative Motion...
The plane is moving north in the IRF attached to the air:
VVp, a is the velocity of the plane w.r.t. the air.
AirAir
VVp,a
5. Physics 111: Lecture 3, Pg 5
Relative Motion...Relative Motion...
But suppose the air is moving east in the IRF attached to
the ground.
VVa,g is the velocity of the air w.r.t. the ground (i.e. wind).
VVa,g
AirAir
VVp,a
6. Physics 111: Lecture 3, Pg 6
Relative Motion...Relative Motion...
What is the velocity of the plane in an IRF attached to the
ground?
VVp,g is the velocity of the plane w.r.t. the ground.
VVp,g
7. Physics 111: Lecture 3, Pg 7
Relative Motion...Relative Motion...
VVp,g = VVp,a + VVa,g Is a vector equation relating the airplane’s
velocity in different reference frames.
VVp,g
VVa,g
VVp,a
Tractor
8. Physics 111: Lecture 3, Pg 8
Lecture 3,Lecture 3, Act 1Act 1
Relative MotionRelative 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/s50 m
35250 =
50150 =
9. Physics 111: Lecture 3, Pg 9
Lecture 3,Lecture 3, Act 1Act 1
solutionsolution
The time taken to swim straight across is (distance across) / (vy )
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
10. Physics 111: Lecture 3, Pg 10
Lecture 3,Lecture 3, Act 1Act 1
solutionsolution
So the y component of your velocity with respect to the water is
So the time to get across is
y
x
3m/s
50
3
29
m
m s
s=
50 m
3m/s
11. Physics 111: Lecture 3, Pg 11
Uniform Circular MotionUniform Circular Motion
What does it mean?
How do we describe it?
What can we learn about it?
12. Physics 111: Lecture 3, Pg 12
What is UCM?What is UCM?
Motion in a circle with:
Constant Radius R
Constant Speed v = |vv|
R
vv
x
y
(x,y)
Puck on ice
13. Physics 111: Lecture 3, Pg 13
How can we describe UCM?How can we describe UCM?
In general, one coordinate system is as good as any other:
Cartesian:
» (x,y) [position]
» (vx ,vy) [velocity]
Polar:
» (R,θ) [position]
» (vR ,ω) [velocity]
In UCM:
R is constant (hence vR = 0).
ω (angular velocity) is constant.
Polar coordinates are a natural way to describe UCM!Polar coordinates are a natural way to describe UCM!
RR
vv
x
y
(x,y)
θ
14. Physics 111: Lecture 3, Pg 14
Polar Coordinates:Polar Coordinates:
The arc length s (distance along the circumference) is
related to the angle in a simple way:
s = Rθ, where θ is the angular displacement.
units of θ are called radians.
For one complete revolution:
2πR = Rθc
θc = 2π
θ has period 2π.
1 revolution = 21 revolution = 2ππ radiansradians
RR
vv
x
y
(x,y)
s
θ
15. Physics 111: Lecture 3, Pg 15
Polar Coordinates...Polar Coordinates...
x = R cos θ
y = R sin θ
π/2 π 3π/2 2π
-1
1
0
sincos
RR
x
y
(x,y)
θ
θ
16. Physics 111: Lecture 3, Pg 16
Polar Coordinates...Polar Coordinates...
In Cartesian coordinates, we say velocity dx/dt = v.
x = vt
In polar coordinates, angular velocity dθ/dt = ω.
θ = ωt
ω has units of radians/second.
Displacement s = vt.
but s = Rθ = Rωt, so:
RR
vv
x
y
s
θ=ωt
v = ωR
Tetherball
17. Physics 111: Lecture 3, Pg 17
Period and FrequencyPeriod and Frequency
Recall that 1 revolution = 2π radians
frequency (f) = revolutions / second (a)
angular velocity (ω) = radians / second (b)
By combining (a) and (b)
ω = 2π f
Realize that:
period (T) = seconds / revolution
So T = 1 / f = 2π/ω
RR
vv
s
ω = 2π / T = 2πf
ω
18. Physics 111: Lecture 3, Pg 18
Recap:Recap:
RR
vv
s
θ=ωt
(x,y)
x = R cos(θ) = R cos(ωt)
y = R sin(θ) = R sin(ωt)
θ = arctan(y/x)
θ = ωt
s = v t
s = Rθ = Rωt
v = ωR
19. Physics 111: Lecture 3, Pg 19
Aside: Polar Unit VectorsAside: Polar Unit Vectors
We are familiar with the Cartesian unit vectors: i j ki j k
Now introduce
“polar unit-vectors” rr and θθ :
rr points in radial direction
θθ points in tangential direction
RR
x
y
ii
jj θθ
rr
^^
θθ
^^
^^ ^^
^^
^^
(counter
clockwise)
20. Physics 111: Lecture 3, Pg 20
Acceleration in UCM:Acceleration in UCM:
Even though the speedspeed is constant, velocityvelocity is notnot constant
since the direction is changing: must be some acceleration!
Consider average acceleration in time ∆t aav = ∆vv / ∆t
vv2
ω∆t
vv1
vv1
vv2
∆vv
RR
21. Physics 111: Lecture 3, Pg 21
Acceleration in UCM:Acceleration in UCM:
seems like ∆vv (hence ∆vv/∆t )
points at the origin!
RR
Even though the speedspeed is constant, velocityvelocity is notnot constant
since the direction is changing.
Consider average acceleration in time ∆t aav = ∆vv / ∆t
∆vv
22. Physics 111: Lecture 3, Pg 22
Acceleration in UCM:Acceleration in UCM:
Even though the speedspeed is constant, velocityvelocity is notnot constant
since the direction is changing.
As we shrink ∆t, ∆vv / ∆t dvv / dt = aa
aa = dvv / dt
We see that aa points
in the - RR direction.
RR
23. Physics 111: Lecture 3, Pg 23
Acceleration in UCM:Acceleration in UCM:
This is calledThis is called Centripetal Acceleration.
Now let’s calculate the magnitude:
vv2
vv1
vv1
vv2
∆vv
RR
∆∆RR
∆ ∆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
24. Physics 111: Lecture 3, Pg 24
Centripetal AccelerationCentripetal Acceleration
UCM results in acceleration:
Magnitude: a = v2
/ R
Direction: - rr (toward center of circle)
R
aa
ω
^^
25. Physics 111: Lecture 3, Pg 25
Useful Equivalent:Useful Equivalent:
( )
R
R
a
2
=
We know that and v = ωR
Substituting for v we find that:
a
v
R
=
2
ω
a = ω2
R
26. Physics 111: Lecture 3, Pg 26
Lecture 3,Lecture 3, Act 2Act 2
Uniform Circular MotionUniform 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
27. Physics 111: Lecture 3, Pg 27
Lecture 3,Lecture 3, Act 2Act 2
SolutionSolution
a
v
R
g= =
2
9
2
2
2
2
s
m
8199
s
m
90000
g9
v
R
.×
==
m1000m
819
10000
R ≈=
.
D R m= ≈2 2000
2km
28. Physics 111: Lecture 3, Pg 28
Example: Propeller TipExample: 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 aa here?
29. Physics 111: Lecture 3, Pg 29
Example:Example:
First calculate the angular velocity of the propeller:
so 3500 rpm means ω = 367 s-1
Now calculate the acceleration.
a = ω2
R = (367s-1
)2
x (0.8m) = 1.1 x 105
m/s2
= 11,000 g
direction of aa points at the propeller hub (-rr ).
1 1
1
60
2 0105 0105rpm s-1
= = ≡
rot
x
s
x
rad
rot
rad
smin
min
. .π
^^
30. Physics 111: Lecture 3, Pg 30
Example: Newton & the MoonExample: 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 106
s (period ~ 1 month)
R = 3.84 x 108
m (distance to moon)
RE = 6.35 x 106
m (radius of earth)
R RE
31. Physics 111: Lecture 3, Pg 31
Moon...Moon...
Calculate angular velocity:
So ω = 2.66 x 10-6
s-1
.
Now calculate the acceleration.
a = ω2
R = 0.00272 m/s2
= 0.000278 g
direction of aa points at the center of the Earth (-rr ).
1
27 3
1
86400
2 2 66 10 6
.
.
rot
day
x
day
s
x
rad
rot
xπ = −
s-1
^
32. Physics 111: Lecture 3, Pg 32
Moon...Moon...
So we find that amoon / g = 0.000278
Newton noticed that RE
2
/ R2
= 0.000273
This inspired him to propose that FMm ∝ 1 / R2
(more on gravity later)
R RE
amoon g
33. Physics 111: Lecture 3, Pg 33
Lecture 3,Lecture 3, Act 3Act 3
Centripetal AccelerationCentripetal 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 106
m.)
(a) 0 m/s2
(b) 8.9 m/s2
(c) 9.8 m/s2
34. Physics 111: Lecture 3, Pg 34
First calculate the angular frequency ω:
Realize that:
Lecture 3,Lecture 3, Act 3Act 3
Centripetal AccelerationCentripetal Acceleration
RO
300 km
RO = RE + 300 km
= 6.4 x 106
m + 0.3 x 106
m
= 6.7 x 106
m
RE
1-
s00115.0
rot
rad
2x
s60
1
x
rot
91
1
== πω
min
min
35. Physics 111: Lecture 3, Pg 35
Now calculate the acceleration:
Lecture 3,Lecture 3, Act 3Act 3
Centripetal AccelerationCentripetal Acceleration
a = ω2
R
a = (0.00115 s-1
)2
x 6.7 x 106
m
a = 8.9 m/s2
36. Physics 111: Lecture 3, Pg 36
Recap for today:Recap for today:
Reference frames and relative motion.
Uniform Circular Motion
