Physics 106P: Lecture 3 Notes - ieu.edu.trhomes.ieu.edu.tr/hozcan/Phys100/Lect03.pdfPhysics 111:...
Transcript of Physics 106P: Lecture 3 Notes - ieu.edu.trhomes.ieu.edu.tr/hozcan/Phys100/Lect03.pdfPhysics 111:...
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Physics 111: Lecture 3, Pg 1
Physics 101: Lecture 3
Today’s Agenda
Reference frames and relative motion
Uniform Circular Motion
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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 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
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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.
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Physics 111: Lecture 3, Pg 4
Relative Motion...
The plane is moving north in the IRF attached to the air:
Vp, a is the velocity of the plane w.r.t. the air.
Air
Vp,a
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Physics 111: Lecture 3, Pg 5
Relative Motion...
But suppose the air is moving east in the IRF attached to the ground.
Va,g is the velocity of the air w.r.t. the ground (i.e. wind).
Va,g
Air
Vp,a
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Physics 111: Lecture 3, Pg 6
Relative Motion...
What is the velocity of the plane in an IRF attached to the ground?
Vp,g is the velocity of the plane w.r.t. the ground.
Vp,g
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Physics 111: Lecture 3, Pg 7
Relative Motion...
Vp,g = Vp,a + Va,g Is a vector equation relating the airplane’s velocity in different reference frames.
Vp,g
Va,g
Vp,a
Tractor
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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
35250
50150
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Physics 111: Lecture 3, Pg 9
Lecture 3, Act 1 solution
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
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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
329
m
m ss
50 m
3 m/s
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Physics 111: Lecture 3, Pg 11
Uniform Circular Motion
What does it mean?
How do we describe it?
What can we learn about it?
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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
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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]
» (vx ,vy) [velocity]
Polar:
» (R,) [position]
» (vR ,) [velocity]
In UCM:
R is constant (hence vR = 0).
(angular velocity) is constant.
Polar coordinates are a natural way to describe UCM!
R
v
x
y
(x,y)
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Physics 111: Lecture 3, Pg 14
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:
2R = Rc
c = 2
has period 2.
1 revolution = 2 radians
R
v
x
y
(x,y)
s
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Physics 111: Lecture 3, Pg 15
Polar Coordinates...
x = R cos
y = R sin
/2 3/2 2
-1
1
0
sin cos
R
x
y
(x,y)
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Physics 111: Lecture 3, Pg 16
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 = Rt, so:
R
v
x
y
s t
v = R
Tetherball
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Physics 111: Lecture 3, Pg 17
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 = 2f
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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 = Rt
v = R
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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)
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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 aav = v / t
v2
t
v1
v1 v2
v
R
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Physics 111: Lecture 3, Pg 21
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 aav = v / t
v
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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 dv / dt = a
a = dv / dt
We see that a points
in the - R direction.
R
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Physics 111: Lecture 3, Pg 23
Acceleration in UCM:
This is called Centripetal Acceleration.
Now let’s calculate the magnitude:
v2
v1
v1 v2
v
R
R
v
v
R
RSimilar triangles:
But R = vt for small t
So:
v
t
v
R
2 v
v
v t
R
av
R
2
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Physics 111: Lecture 3, Pg 24
Centripetal Acceleration
UCM results in acceleration:
Magnitude: a = v2 / R
Direction: - r (toward center of circle)
R a
^
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Physics 111: Lecture 3, Pg 25
Useful Equivalent:
R
Ra
2
We know that and v = R
Substituting for v we find that:
av
R
2
a = 2R
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Physics 111: Lecture 3, Pg 26
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
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Physics 111: Lecture 3, Pg 27
Lecture 3, Act 2 Solution
av
Rg
2
9
2
2
2
2
s
m8199
s
m90000
g9
vR
.
m1000m819
10000R
.D R m 2 2000
2km
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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?
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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 = 2R = (367s-1)2 x (0.8m) = 1.1 x 105 m/s2 = 11,000 g
direction of a points at the propeller hub (-r ).
1 11
602 0105 0105 rpm s-1
rotx
sx
rad
rot
rad
smin
min. .
^
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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 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
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Physics 111: Lecture 3, Pg 31
Moon...
Calculate angular velocity:
So = 2.66 x 10-6 s-1.
Now calculate the acceleration.
a = 2R = 0.00272 m/s2 = 0.000278 g
direction of a points at the center of the Earth (-r ).
1
27 3
1
864002 2 66 10 6
..
rot
dayx
day
sx
rad
rotx s-1
^
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Physics 111: Lecture 3, Pg 32
Moon...
So we find that amoon / g = 0.000278
Newton noticed that RE2 / R2 = 0.000273
This inspired him to propose that FMm 1 / R2
(more on gravity later)
R RE
amoon g
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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 106 m.)
(a) 0 m/s2
(b) 8.9 m/s2
(c) 9.8 m/s2
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Physics 111: Lecture 3, Pg 34
First calculate the angular frequency :
Realize that:
Lecture 3, Act 3 Centripetal 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-s 00115.0 rot
rad2 x
s60
1 x
rot
91
1
min
min
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Physics 111: Lecture 3, Pg 35
Now calculate the acceleration:
Lecture 3, Act 3 Centripetal Acceleration
a = 2R
a = (0.00115 s-1)2 x 6.7 x 106 m
a = 8.9 m/s2
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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: 3-5, 5-2, also 9-1)
Look at Textbook problems Chapter 3: # 67, 69, 113, 117