Transcript PowerPoint Presentation - Chapter 3 Kinematics in 2d

Chapter 5
Uniform Circular Motion
ac=v2/r
Uniform circular motion:
Motion in a circular path with constant speed
v
s
r
1) Speed and period
• Period, T: time for one revolution
• Speed is related to period:
– Path for one revolution: s = 2r
– Speed: v = s/T = 2r/T
2) Centripetal Acceleration
v
s
r
• Directed toward centre
2
v
• Magnitude: a 
c
r
ac
3) Centripetal force
v
s
ac
r
Fc
• Toward centre (parallel to acceleration) since F = ma
• Magnitude:
mv 2
Fc  mac 
r
4) Driving around circular curves
a) Friction only:
vmax  s gr
b) Banked curve, no friction:
v2
tan  
rg
4) Driving around circular curves
c) Banked curve with friction
fs(ii)
fs(i)
Net force F is vector sum of
gravity, the normal force, and
friction:
r
r r
F  FN  mg  fs
Since the ideal (zero friction)
angle is given by
v2
tan  
rg
i) If v2 > gr tan friction prevents sliding up
ii) If v2 < gr tan friction prevents sliding down
4) Driving around circular curves
c) Banked curve with friction
Example: Find maximum velocity
for µ = 0.80, = 47º, and r = 60 m
FN
fs(ii)
y
fs
(i)
x
Fc


r
r r
Fc  FN  mg  fs
fs
y-motion
x-motion
0  FN cos  mg  fs sin 
When v  vmax , fs   FN so
Fc  FN sin   fs cos
FN 
mg
cos   sin 
mg
mv 2
When v  vmax , fs   FN , and using Fc =
,
r
mvmax 2
FN 
r(sin    cos )
4) Driving around circular curves
c) Banked curve with friction
Example: Find maximum velocity
for µ = 0.80, = 47º, and r = 60 m
FN
fs(ii)
y
fs
x
(i)
Fc


r
r r
Fc  FN  mg  fs
fs
mg
Eliminate FN , and solve for vmax :
vmax
2
 sin    cos 
 gr 
 cos   sin  
vmax  88m/s  317km/h
5) Satellites in circular orbits
a) Speed and radius
For a distance r, the gravitational force is
mM
FG 2E
r
For circular motion with radius r,
the centripetal force is
mv 2
Fc 
r
• Radius of orbit determines speed
(independent of mass)
For circular motion around the earth, F  Fc , so
• Accelerations of all objects at the
same radius are equal (no
acceleration relative to each other)
mM E mv 2
G 2 
r
r
GM E
v
r
or
==> No apparent force between
them: apparent weightlessness in
orbiting satellite
5) Satellites in circular orbits
b) Period, Synchronous orbits
(i) Period and radius
Time for one revolution: T
Speed is constant, so v 
For earth orbit, v 
2 r
GM E

or
T
r
3
2 r 2
T
GM E
2 r
T
GM E
, so
r
5) Satellites in circular orbits
b) Period, Synchronous orbits
(ii) Radius of synchronous orbit:
Definition: satellite is stationary above earth’s surface
Conditions:
• T = 1 sidereal day
• above equator
Radius is given by r
3
2

T GM E
2
Using T  24h  4 min = 86160 s
gives r  4.22  107 m = 6.61RE
6) Centrifugal force* and artificial gravity
a) Artificial gravity
A rotating cylinder exerts a centripetal
force on objects on the inside surface: e.g.
if r = 1700 m, and Fc = mg (usual force of
gravity), then mg = mv2/r, giving
v  gr  130 m/s = 468 km/h
Centripetal force acts toward the centre
(up, like the normal force on earth) but
gravity acts down. What has taken the
place of gravity, from the perspective of
the cylinder-dwellers?
Centrifugal force…
does not exist
is a fictitious, pseudo, virtual force
is an inertial force
* The concept of centrifugal force is not required material.
6) Centrifugal force and artificial gravity
a) Artificial gravity
A rotating cylinder exerts a
centripetal force on objects
on the inside surface: e.g. if
r = 1700 m, and Fc = mg
(usual force of gravity),
then mg = mv2/r, giving
v  gr  130 m/s = 468 km/h
Centripetal force acts toward the centre (up, like the normal force on
earth) but gravity acts down. What has taken the place of gravity,
from the perspective of the cylinder-dwellers?
Centrifugal force…
does not exist (in inertial frames)
is a fictitious, pseudo, virtual force
is an inertial force
6) Centrifugal force and artificial gravity
This episode of Quirks & Quarks on CBC radio
illustrates the confusion that can exist about artificial
gravity.
http://www.cbc.ca/quirks/archives/01-02/mp3/qq131001f.mp3
The question was: “What happens when you jump inside a rotating cylinder
like the one on the movie 2001: A Space Odysey”
Contrary to the answer given, you do not float to the other side and hit your
head. The simulated gravity does not require continuous contact with the
surface. From your perspective (the jumper’s), you will be pulled back to the
surface just as you would in a gravitational field, apart from a small deviation
depending on the radius of the cylinder. To understand this, we will consider a
simpler inertial force first.
6) Centrifugal force and artificial gravity
b) Inertial force -- Apparent force resulting from an
accelerating reference frame.
e.g. An accelerating spaceship (far from planets)
m
FN
a
From an inertial frame: FN  ma  ma0
a0
6) Centrifugal force and artificial gravity
b) Inertial force -- Apparent force resulting from an
acceleration reference frame.
e.g. An accelerating spaceship (far from planets)
m
ma0 FN
F  0
a0
FN  ma0  0
From an inertial frame: FN  ma  ma0
Inside the spaceship, acceleration is zero, so net force should
be zero. Same equation, new interpretation:
Inside the ship an inertial force equal to ma0 is acting toward the
floor.
6) Centrifugal force and artificial gravity
b) Inertial force -- Apparent force resulting from an
acceleration reference frame.
e.g. An accelerating spaceship (far from planets)
a0
From an inertial frame: FN  ma  ma0
Inside the spaceship, acceleration is zero, so net force should
be zero. Same equation, new interpretation:
Inside the ship an inertial force equal to ma0 is acting toward the
floor. A dropped ball falls to the floor just as it would if a constant
force acted on it.
6) Centrifugal force and artificial gravity
b) Inertial force -- Apparent force resulting from an
acceleration reference frame.
e.g. An accelerating spaceship (far from planets)
a0
From an inertial frame: FN  ma  ma0
Inside the spaceship, acceleration is zero, so net force should
be zero. Same equation, new interpretation:
Inside the ship an inertial force equal to ma0 is acting toward the
floor. A dropped ball falls to the floor just as it would if a constant
force acted on it.
6) Centrifugal force and artificial gravity
b) Inertial force -- Apparent force resulting from an
acceleration reference frame.
e.g. An accelerating spaceship (far from planets)
a0
From an inertial frame: FN  ma  ma0
Inside the spaceship, acceleration is zero, so net force should
be zero. Same equation, new interpretation:
Inside the ship an inertial force equal to ma0 is acting toward the
floor. A dropped ball falls to the floor just as it would if a constant
force acted on it.
6) Centrifugal force and artificial gravity
c) Centrifugal force: the inertial force in a rotating reference frame
Consider a cup of water swung around in a vertical circle. What
keeps the water in the cup when it is upside-down?
In an inertial frame, only the normal force and
gravity act to give centripetal force which
produces centripetal acceleration (no such thing
as centrifugal force)
Answer: Inertia. The cup pulls the water down faster
than its natural falling rate.
Gravity plus the normal force provide centripetal force
that produces circular motion:
mv 2
FN  mg 
r
mg
FN
6) Centrifugal force and artificial gravity
c) Centrifugal force: the inertial force in a rotating reference frame
F=mg+FN=mv2/r
Inside the cup, the water is at rest, as is an
observer in a boat inside the cup; i.e. they do not
accelerate in this reference frame.
What keeps water in the cup from
the perspective of the cup-dweller?
mg
Answer: Invent centrifugal force to balance
real forces and ensure zero acceleration. So
F  0
mv 2
FN  mg 
0
r
FN
6) Centrifugal force and artificial gravity
c) Centrifugal force: the inertial force in a rotating reference frame
F=mg+FN
What happens if a ball is dropped?
If the force holding it Fh is removed, it
should fall radially (down to the cupdweller), but in an inertial frame it moves
tangentially.
How can both be true?
Fh
mg
FN
6) Centrifugal force and artificial gravity
This animation shows what happens when a ball is dropped
inside a rotating cylinder.
6) Centrifugal force and artificial gravity
In the rotating reference
frame it moves radially
outward (down to the
observer), apart from a small
curvature which decreases
for larger cylinder radii.
An object released moves with constant velocity
on the tangent following to Newton’s first law.
6) Centrifugal force and artificial gravity
An object that leaves the floor with a small
radial velocity (like a jumping person) will
have velocity with both tangential and radial
components and will move at a small angle to
the the tangent. It will leave the floor and then
return to it again very close the point of
departure.
To an observer inside the
cylinder, it moves up and
down (except for small
displacement due to finite
radius of the cylinder).
6) Centrifugal force and artificial gravity
c) Centrifugal force: the inertial force in a rotating reference frame
An accelerating reference frame can duplicate the effect
of gravity (not only when in contact). Centrifugal force is
introduced within the reference frame to account for the
observed acceleration.
However, when analyzing the situation from an
inertial frame (as we do in this course), inertial forces
are not present. “Centrifugal force does not exist.”
6) Centrifugal force and artificial gravity
d) Equivalence principle
- Inertial mass (F = ma) is the same as gravitational mass (F = GmM/r2)
- It is, however, clear that science is fully justified in assigning such a numerical equality
only after this numerical equality is reduced to an equality of the real nature of the two
concepts.” -- Einstein
- The happiest thought of my life…
The gravitational field has only a relative existence... Because for an observer freely
falling from the roof of a house - at least in his immediate surroundings - there exists no
gravitational field. -- Einstein
A uniform gravitational field is completely equivalent to a
uniformly accelerated reference frame.
• No local experiment can distinguish them
• Gravity is geometrical
7) Vertical circular motion
• Not usually uniform motion since the speed is changing
• Net force not always toward the centre
• Component of acceleration toward the centre (the
centripetal component) is still v2/r, so
mv 2
Fc 
r
7) Vertical circular motion
mv32
FN 3  mg 
r
FN 2
FN 4
mv4 2

r
mv12
FN1  mg 
r
mv2 2

r
7) Vertical circular motion
mv32
FN 3  mg 
r
Minimum speed at top:
FN 3  0
mv32
mg 
r
v3  rg
For r = 6 m, v = 7.6 m/s or 27 km/h