General Physics (PHY 2130)
Lecture 13
•  Rotational kinematics
  Non-uniform circular motion
  Orbits and Kepler’s laws
http://www.physics.wayne.edu/~apetrov/PHY2130/
Lightning Review
Last lecture:
1.  Rotational kinematics
  angular displacement, angular velocity and angular acceleration
  relations between angular and linear quantities
Review Problem: : Your car’s wheels are 65 cm in diameter and are spinning at ω =
101 rads/sec. How fast in km/hour is the car traveling, assuming no slipping?
3
Example: Your car’s wheels are 65 cm in diameter and are spinning at ω = 101
rads/sec. How fast in km/hour is the car traveling, assuming no slipping?
Given:
d = 65 cm = 0.65 m
ω = 101 rads/sec
Find:
v=?
Idea: since the car is
moving with constant
velocity, divide total path
to total time!
v
X
total distance ( 2π r ) N
2π r
v=
=
=
= ωr
total time
T
(T ) N
= (101 rads/sec) (32.5 cm )
= 3.28 ×10 3 cm/sec = 118 km/hr
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Period and Frequency
The time it takes to go one time around a closed path is called the period (T).
total distance 2πr
vav =
=
total time
T
Comparing to v = rω:
2π
ω=
= 2πf
T
f is called the frequency, the number of revolutions (or cycles) per second.
5
Circular Orbits
Consider an object of mass m in a
circular orbit about the Earth.
The only force on the satellite is
the force of gravity:
Gms M e
∑ F = Fg = r 2
∑ F = ms a r
Gms M e
v2
∴
= ms
2
r
r
the speed of the satellite:
GM e
v=
r
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Escape Speed
 The escape speed is the speed needed for an
object to soar off into space and not return
2GM E
vesc =
RE
 For the earth, vesc is about 11.2 km/s
 Note, v is independent of the mass of the object
7
Example: How high above the surface of the Earth does a satellite need to be so
that it has an orbit period of 24 hours?
Given:
T = 24 hours
Idea: relate velocity of a satellite to both r and T!
Previously:
GM e
v=
r
Also need,
= 86400 s
Combine these and solve for r:
Find:
h=?
(
−11
2
2
)(
2πr
v=
T
⎛ GM e 2 ⎞
r = ⎜
T ⎟
2
⎝ 4π
⎠
24
1
3
)
⎛ 6.67 ×10 Nm /kg 5.98 ×10 kg
2 ⎞
(86400 s ) ⎟⎟
r = ⎜⎜
2
4π
⎝
⎠
= 4.225 ×10 7 m
r = Re + h ⇒ h = r − Re = 35,000 km
This orbit is called geosynchronous and is used for TV satellites!
1
3
Kepler’s Laws
 All planets move in elliptical orbits with the Sun at one
of the focal points.
 A line drawn from the Sun to any planet sweeps out
equal areas in equal time intervals.
 The square of the orbital period of any planet is
proportional to cube of the average distance from the
Sun to the planet.
Kepler’s First Law
  All planets move in
elliptical orbits with
the Sun at one focus.
•  Any object bound to
another by an inverse
square law will move in
an elliptical path
•  Second focus is empty
Kepler’s Second Law
 A line drawn from the
Sun to any planet will
sweep out equal areas
in equal times
•  Area from A to B and C to
D are the same
Kepler’s Third Law
 The square of the orbital period of any planet is proportional
to cube of the average distance from the Sun to the planet.
2
T = Kr
3
or
⎛ GM e 2 ⎞
r = ⎜
T ⎟
2
⎝ 4π
⎠
1
3
•  For orbit around the Sun, KS = 2.97x10-19 s2/m3
•  K is independent of the mass of the planet
It can be generalized to:
⎛ GM 2 ⎞
r = ⎜ 2 T ⎟
⎝ 4π
⎠
1
3
where M is the mass of the central body. For example, it would be Msun if
speaking of the planets in the solar system.
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Example: The Hubble Space Telescope orbits Earth 613 km above
Earth’s surface. What is the period of the telescope’s orbit?
Given:
Idea: period T is the time needed to
complete one revolution, i.e. a path of
h = 613 km
length s = 2πr! So,
= 6.13 × 105 m
T=
2π r
v
Need v! To find it, write 2nd Newton’s law:
Find:
T=?
GmHs M e
v2
∑ F = Fg = r 2 = mHS ar = mHS r
Or with numbers (and remembering that r = RE+h):
r3
(613×10 3 m + 6371×10 6 m)3
⇒ T = 2π
= 2π
GM e
(6.674 ×10 −11 Nm 2 / kg 2 )(5.974 ×10 24 kg)
T = 5808s = 1.613h
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13
Tangential and Angular Acceleration
Recall: average angular acceleration:
α av =
ω f − ωi
Δt
Δω
=
Δt
instantaneous angular acceleration:
Δω
α = lim
Δt →0 Δt
SI unit of α is rads/sec2.
tangential component of velocity: vt
tangential acceleration:
= rω
Δvt
Δω
at =
=r
Δt
Δt
at = r α
(in the limit Δt → 0)
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Total Acceleration
  What happens if linear velocity
also changes?
  Two-component acceleration:
•  the centripetal component of the
acceleration is due to changing
direction
•  the tangential component of the
acceleration is due to changing speed
 Total acceleration can be found
from these components:
slowing-down car
2
t
2
C
a = a +a
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Example: A child pushes a
merry-go-round from rest to a
final angular speed of 0.50 rev/s
with constant angular
acceleration. In doing so, the
child pushes the merry-goround 2.0 revolutions. What is
the angular acceleration of the
merry-go-round?
Δv x = v fx − vix = a x Δt
Δω = ω f − ω i = α Δt
1
Δx = (v fx + vix )Δt
2
1
Δx = vix Δt + a x Δt 2
2
2
2
v fx −vix = 2a x Δx
1
(ω f + ω i )Δt
2
1
Δθ = ω i Δt + α Δt 2
2
ω f 2 −ω i 2 = 2αΔθ
vt = rω,
Δθ =
at = rα
Solution: given final and initial angular velocity and angle. Find acceleration.
2 −ω2
ω
i
ω f2 − ω i2 = 2αΔθ , so α = f
2Δθ
2
(0.50 rev s) 2 − 0 # 2π rad &
=
%
( = 0.39 rad s .
2(2.0 rev)
$ rev '
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Example: A disk rotates with constant angular
acceleration. The initial angular speed of the disk is
2 π rad/s. After the disk rotates through 10π
radians, the angular speed is 7π rad/s. (a) What is
the magnitude of the angular acceleration? (b) How
much time did it take for the disk to rotate through
10π radians? (c) What is the tangential acceleration
of a point located at a distance of 5.0 cm from the
center of the disk?
Δv x = v fx − vix = a x Δt
Δω = ω f − ω i = α Δt
1
(v fx + vix ) Δt
2
1
Δx = vix Δt + a x Δt 2
2
2
2
v fx −vix = 2a x Δx
1
(ω f + ω i ) Δt
2
1
Δθ = ω i Δt + α Δt 2
2
2
2
ω f −ω i = 2αΔθ
Δx =
vt = rω,
Δθ =
at = rα
2 −ω 2
ω
(7π rad s) 2 − (2π rad s) 2
2
2
f
i
=
= 7.1 rad s 2 .
(a) ωf − ωi = 2αΔθ , so α =
2Δθ
2(10π rad)
(b)
1
2Δθ
2(10π rad)
Δθ = (ωf + ωi )Δt , so Δt =
=
= 2.2 s .
2
ωf + ωi 7π rad s + 2π rad s
(c)
(7π rad s) 2 − (2π rad s) 2
at = r α = (0.050 m)
= 0.35 m s 2
2(10π rad)
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Artificial Gravity
Nautilus-X (proposed space station)
A large rotating cylinder in deep
space (g≈0).
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FBD for person at the top
position
FBD for person at the
bottom position
y
y
N
x
x
N
Apply Newton’s 2nd Law to each:
2
F
=
N
=
ma
=
m
ω
r
∑ y
r
2
F
=
−
N
=
−
ma
=
−
m
ω
r
∑ y
r
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Example: A space station is shaped like a ring and rotates to simulate gravity.
If the radius of the space station is 120m, at what frequency must it rotate so
that it simulates Earth’s gravity?
Using the result from the previous slide:
2
F
=
N
=
ma
=
m
ω
r
∑ y
r
N
mg
ω=
=
=
mr
mr
g
= 0.28 rad/sec
r
The frequency is f = (ω/2π) = 0.045 Hz (or 2.7 rpm).
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Apparent weight
What would happen on Earth?
Our weight would change!
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Example: What is the minimum speed for the car so that it maintains contact
with the loop when it is in the top (pictured) position?
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Example: What is the minimum speed for the car so that it maintains contact
with the loop when it is in the top (pictured) position?
r
FBD for the car at the top
of the loop:
y
Apply Newton’s 2nd Law:
x
N
w
v2
∑ Fy = − N − w = −mar = −m r
v2
N +w=m
r
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Example continued:
The apparent weight at the top of loop is:
v2
N +w=m
r
⎛ v 2
⎞
N = m⎜⎜ − g ⎟⎟
⎝ r
⎠
The car will just about fall if N = 0!
N = 0 when
⎛ v 2
⎞
N = m⎜⎜ − g ⎟⎟ = 0
⎝ r
⎠
v = gr
This is the minimum speed needed to make it around the loop.
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Example continued:
Consider the car at the bottom of the loop; how does the apparent weight
compare to the true weight?
FBD for the car at the bottom
of the loop:
y
N
x
w
Apply Newton’s 2nd Law:
v2
∑ Fy = N − w = mac = m r
v2
N −w=m
r
⎛ v 2
⎞
⎜
N = m⎜ + g ⎟⎟
⎝ r
⎠
Here,
N > mg
Forces in Accelerating Reference Frames
 Distinguish real forces from fictitious forces
 Centrifugal force is a fictitious force
 Real forces always represent interactions
between objects
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Summary
• A net force MUST act on an object that has circular motion.
• Radial Acceleration ar=v2/r
• Definition of Angular Quantities (θ, ω, and α)
• The Angular Kinematic Equations
• The Relationships Between Linear and Angular Quantities
• Uniform and Nonuniform Circular Motion
vt = rω and at = rα