Phys 207
Announcements
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Hwk2 submission deadline = 1st thing Monday morning
TA office hours
ÍR: Philip Castro / 7:00 – 9:00 pm / 227 SHL
ÍF: Dan Pajerowski / 1:30 – 3:30 pm / 131B SHL
ÍF: Xing Chen / 5:00 – 7:00 pm / 225 SHL (Commons Room)
Today’s Agenda
Uniform Circular Motion
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Phys 207
Uniform Circular Motion
(UCM)
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What does it mean?
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How do we describe it?
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What can we learn about it?
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Page 1
What is UCM?
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Motion in a circle with:
ÍConstant Radius R
y
Í Constant Speed v = |v|
v
(x,y)
R
x
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How can we describe UCM?
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In general, one coordinate system is as good as any other:
Í Cartesian:
» (x,y) [position]
y
» (vx ,vy) [velocity]
v
(x,y)
Í Polar:
R
» (R,θ) [position]
θ
x
» (vR ,ω) [velocity]
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In UCM:
Í R is constant (hence vR = 0).
Í ω (angular velocity) is constant.
Í Polar coordinates are a natural way to describe UCM!
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Page 2
Polar Coordinates:
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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.
Í units of θ are called radians.
y
z
For one complete revolution:
2πR = Rθc
Í θc = 2π
θ has period 2π.
v
θ
(x,y)
R s
x
1 revolution = 2π radians
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Polar Coordinates...
y
x = R cos θ
y = R sin θ
R
(x,y)
θ
x
1
cos
sin
π/2
π
θ
0
3π/2
2π
-1
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Page 3
Polar Coordinates...
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z
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In Cartesian coordinates, we say velocity dx/dt = v.
Í x = vt
In polar coordinates, angular velocity dθ/dt = ω.
Í θ = ωt
y
Íω has units of radians/second.
v
Displacement s = vt.
R
but s = Rθ = Rωt, so:
θ=ωt
s
x
v = ωR
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Period and Frequency
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Recall that 1 revolution = 2π radians
Ífrequency (f) = revolutions / second
Íangular velocity (ω) = radians / second
By combining (a) and (b)
Íω = 2π f
Realize that:
Íperiod (T) = seconds / revolution
ÍSo T = 1 / f = 2π/ω
(a)
(b)
v
R
s
ω
ω = 2π / T = 2πf
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Page 4
Recap:
x = R cos(θ) = R cos(ωt)
y = R sin(θ) = R sin(ωt)
θ = arctan (y/x)
v
R
θ=ωt
θ = ωt
s=vt
s = Rθ = Rωt
(x,y)
s
v = ωR
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Aside: Polar Unit Vectors
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We are familiar with the Cartesian unit vectors: i j k
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Now introduce
^
“polar unit-vectors” r^ and θ :
Ír^ points in radial direction
^
Íθ points in tangential direction
(counter clockwise)
^
θ
^
r
y
R
θ
j
x
i
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Page 5
Acceleration in UCM:
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Even though the speed is constant, velocity is not constant
since the direction is changing: must be some acceleration!
Í Consider average acceleration in time ∆t
aav = ∆v / ∆t
∆v
v2
R
v2
v1
v1
ω∆t
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Acceleration in UCM:
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Even though the speed is constant, velocity is not constant
since the direction is changing.
Í As we shrink ∆t, ∆v / ∆t
dv / dt = a
R
a = dv / dt
We see that a points
in the - R direction.
Let’s take a look – link to Active Figure 4-18
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Page 6
Acceleration in UCM:
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This is called Centripetal Acceleration.
Now let’s calculate the magnitude:
∆v
v2
v1
But ∆R = v∆t for small ∆t
v2
R
∆R
∆v ∆R
=
v
R
Similar triangles:
So:
v1
∆ v v ∆t
=
v
R
∆v v 2
=
R
∆t
a=
v2
R
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Centripetal Acceleration
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UCM results in acceleration:
ÍMagnitude:
a = v2 / R
ÍDirection:
-^
r (toward center of circle)
R
a
ω
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Page 7
Useful Equivalent:
We know that
a=
v2
R
and
v = ωR
Substituting for v we find that:
a=
( ω R )2
R
a = ω2R
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Lecture 5, Act 1
Uniform Circular Motion
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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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Page 8
Lecture 5, Act 1
Solution
a=
m2
v2
s2
R =
=
m
9g
9 × 9 . 81 2
s
90000
2
v
= 9g
R
R=
10000
m ≈ 1000 m
9 .81
D = 2 R ≈ 2000 m
2km
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Example: Propeller Tip
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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?
what is a here?
f
L
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Page 9
Example:
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First calculate the angular velocity of the propeller:
1 rpm = 1
rot
1 min
rad
rad
x
x 2π
= 0105
.
≡ 0105
.
s-1
min 60 s
rot
s
Í so 3500 rpm means ω = 367 s-1
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Now calculate the acceleration.
Ía = ω2R = (367s-1)2 x (0.8m) = 1.1 x 105 m/s2
= 11,000 g
^ ).
Í direction of a points at the propeller hub (-r
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Example: Newton & the Moon
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What is the acceleration of the Moon due to its motion
around the Earth?
What we know (Newton knew this also):
ÍT = 27.3 days = 2.36 x 106 s
(period ~ 1 month)
ÍR = 3.84 x 108 m
(distance to moon)
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ÍRE = 6.35 x 10 m
(radius of earth)
R
RE
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Page 10
Moon...
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Calculate angular velocity:
1 rot
1 day
rad
x
x 2π
= 2.66 x10 −6 s-1
27 .3 day 86400 s
rot
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So ω = 2.66 x 10-6 s-1.
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Now calculate the acceleration.
Í a = ω2R = 0.00272 m/s2 = 0.000278 g
^ ).
Ídirection of a points at the center of the Earth (-r
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Moon...
z
z
So we find that amoon / g = 0.000278
Newton noticed that RE2 / R2 = 0.000273
amoon
g
R
z
RE
This inspired him to propose that FMm ∝ 1 / R2
Í(more on gravity later)
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Page 11
Lecture 5, Act 2
Centripetal Acceleration
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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)
(b)
(c)
0 m/s2
8.9 m/s2
9.8 m/s2
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Lecture 5, Act 2
Centripetal Acceleration
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First calculate the angular frequency ω:
ω=
z
1 rot
1 min
rad
x
= 0.00115 s
x 2π
91 min 60 s
rot
Realize that:
-1
RO
RO = RE + 300 km
= 6.4 x 106 m + 0.3 x 106 m
= 6.7 x 106 m
300 km
RE
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Page 12
Lecture 5, Act 2
Centripetal Acceleration
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Now calculate the acceleration:
a = ω2R
a = (0.00115 s-1)2 x 6.7 x 106 m
a = 8.9 m/s2
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Recap of Lectures 4 and 5
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2-D Kinematics
(Text: 4-1 thru 4-3)
ÍIndependence of x and y components
ÍProjectile motion
ÍBaseball problem
ÍShoot the monkey
(example 4.4)
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Uniform circular motion
(Text: 4.4)
Reminder
Homework 2 is posted on web
Submission deadline is Monday, Feb. 21 at the BEGINNING of lecture
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