Transcript Physics - Circular Motion

Uniform Circular Motion
http://www.physicsclassroom.com/mmedia/circmot/circmotTOC.html
1
Uniform circular motion
• motion of an
object in a circle
with a constant
or uniform
speed
• constant change
in direction –
which means
changing
velocity
2
Big Bang Theory Physics Lesson
Centrifugal vs. Centripetal Force
• Leonard Explains Centripetal Force to
Penny
• Link
(go to 2:35 minute mark)
3
Centrifugal vs Centripetal Forces
• In common language we have mistakenly
been using centrifugal force when we
should correctly refer to the phenomenon as
centripetal force
• Centripetal force is a force that makes a
body follow a curved path towards a point
as to a center
4
Centrifugal vs Centripetal Forces
• Sometimes an outward force is attributed to
circular motion.
• Centrifugal force represents the effects of
inertia that arise in connection with rotation
and which are experienced as an outward
force away from the center of rotation.
5
Centrifugal vs Centripetal Forces
Example of the Lady Bug in a Can
6
Definitions
• Period (T in secs - s) : The time it takes for
one full rotation or revolution of an object
T= 1
f
• Frequency (f in hertz - Hz): The number of
revolutions or rotations per unit time.
f= 1
T
7
When an object spins in a circle, the distance it
travels in one revolution is the circumference of
the circle, 2πr
The time it takes for one revolution is the period, T.
If v = distance traveled
t
Therefore circular velocity will be….
vc = 2πr
T
8
Uniform Circular Motion: Period
Object repeatedly
finds itself back where
it started.
distance = rate  time
distance 2r
time =

rate
v
2r
T=
v
The time it takes to
travel one “cycle” is
the “period”.
9
Quantifying Acceleration: Magnitude
v1
v2
Similar Triangles:
v x

v
x
10
Quantifying Acceleration: Magnitude
v v  t

v
r
v2  t
v 
r
v v
a

t
r
2
Centripetal Acceleration
11
Applying Newton’s 2nd Law:
F  ma
mv
F
r
2
Centripetal Force
Always points toward center of circle.
(Always changing direction!)
Centripetal force is the magnitude of the force
required to maintain uniform circular motion.
12
Direction of Centripetal Force,
Acceleration and Velocity
With a centripetal
force, an object in
motion continues along
a straight-line path.
Without a centripetal
force, an object in
motion continues along
a straight-line path.
13
Direction of Centripetal Force,
Acceleration and Velocity
14
What if velocity decreases?
15
What if mass decreases?
16
What if radius decreases?
17
What provides the centripetal force?
• Tension (with or without elasticity)
• Gravity
• Friction (on ground) – air resistance negligible
• Normal Force
Centripetal force is NOT a new “force”. It is simply a
way of quantifying the magnitude of the force
required to maintain a certain speed around a circular
path of a certain radius.
18
Relationship Between Variables of Uniform
Circular Motion
Suppose two identical objects go around in
horizontal circles of identical diameter but one
object goes around the circle twice as fast as the
other. The force required to keep the faster object
on the circular path is
The answer is E. As the
A. the same as
velocity increases the
B. one fourth of
centripetal force required to
maintain the circle increases
C. half of
as the square of the speed.
D. twice
E. four times
the force required to keep the slower object on the path.19
Relationship Between Variables of Uniform
Circular Motion
Suppose two identical objects go around in
horizontal circles with the same speed. The
diameter of one circle is half of the diameter of
the other. The force required to keep the object
on the smaller circular path is
A. the same as
The answer is D. The centripetal force needed
B. one fourth of to maintain the circular motion of an object is
inversely proportional to the radius of the circle.
C. half of
Everybody knows that it is harder to navigate a
D. twice
sharp turn than a wide turn.
E. four times
the force required to keep the object on the larger
path.
20
Relationship Between Variables of Uniform
Circular Motion
Suppose two identical objects go around in horizontal circles of
identical diameter and speed but one object has twice the
mass of the other. The force required to keep the more
massive object on the circular path is
A. the same as
B. one fourth of
Answer: D.The mass is directly
C. half of
proportional to centripetal force.
D. twice
E. four times
21
Tension Can Yield a Centripetal Acceleration:
If the person doubles the
speed of the airplane,
what happens to the
tension in the cable?
mv
F = ma 
r
2
Doubling the speed, quadruples the force (i.e.
tension) required to keep the plane in uniform circular
22
motion.
Friction Can Yield a Centripetal Acceleration:
23
Car Traveling Around a Circular Track
Friction provides the centripetal acceleration
24
Friction Can Yield a Centripetal Acceleration
FN
fs
Force
X
Y
W
0
-mg
FN
0
FN
fs
-sFN
0
Sum
ma
0
W
What is the maximum
speed that a car can use
around a curve of radius
“r”?
25
Maximum Velocity
F
y
 0   mg  FN
FN  mg
F
x
 ma   
max
s
mv 2
max

  s mg
r
2
max
v  s  g  r
v  smax  g  r
mg
Force
X
Y
W
0
-mg
FN
0
FN
FC
-sFN
0
Sum
ma
0
26
Centripetal Force: Question
A car travels at a constant
speed around two curves.
Where is the car most likely to
skid? Why?
mv
F = ma 
r
2
Smaller radius: larger force
required to keep it in uniform
circular motion.
27
Gravity Can Yield a Centripetal Acceleration:
Hubble Space Telescope
orbits at an altitude of 598 km
(height above Earth’s surface).
What is its orbital speed?
28
Gravity and Centripetal Acceleration:
Centripetal acceleration provided by gravitational force
G  m  ME m  v

2
R
R
2
29
Gravity and Centripetal Acceleration:
G  m  ME m  v

2
R
R
2
Solve for the velocity….
G  m  ME  R
v 
2
m R
G  ME
2
v 
R
G  ME
v
R
2
30
Hubble Space Telescope:
v
GM E
RE  598km
(6.67  1011 m3 kg -1s-2 )  (5.974  1024 kg)
v
6,976,000 m
v  7,600 m / s
31
Banked Curves
Why exit ramps in highways are banked?
32
Banked Curves
Q: Why exit ramps in highways are banked?
33
Banked Curves
Q: Why exit ramps in highways are banked?
A: To increase the centripetal force for the higher exit speed.
34
The Normal Force Can Yield a Centripetal Acceleration:
Engineers have learned to “bank” curves so that
cars can safely travel around the curve without
relying on friction at all to supply the centripetal
acceleration.
How many forces are
acting on the car (assuming
no friction)?
35
Banked Curves
Why exit ramps in highways are banked?
FN cosq = mg
36
Banked Curves
Why exit ramps in highways are banked?
FN cosq = mg
37
The Normal Force as a Centripetal Force:
Two: Gravity and Normal
Force
X
Y
W
0
-mg
FN
Sum
FNsinq FNcosq
ma
0
38
The Normal Force as a Centripetal Force:
F
y
  mg  FN cosq  0
mg
FN 
cosq
mv 2
 Fx  FN sin q  ma  r
mg
mv 2
 sin q 
cosq
r
v2
tan q 
gr
39
The Normal Force and Centripetal Acceleration:
How to bank a curve…
2
v
tan q 
gr
…so that you don’t rely on friction at all!!
40
Artifical Gravity
41
Vertical Circular Motion
42
Vertical Circular Motion
43
Calculations
• Practice Set 1 – link
• Practice Set 2 - link
44
Acknowledgements
• http://www.physicsclassroom.com/mmedia/
circmot/ circmotTOC.html
45
The End!
46