Transcript Circular Motion Powerpoint

Circular Motion
What is circular motion?
Anything that
rotates or
revolves around
a central axis is
in circular motion.
Rotation vs. Revolution
An axis is the straight line around which
rotation takes place.
 Rotation: an object turns about an
internal axis. “Spinning”
 Revolution: an object turns about an
external axis, “Turning”
Rotation vs. Revolution
The Ferris wheel
turns about an
axis.
The Ferris wheel
rotates, while the
riders revolve
about its axis.
Rotation vs. Revolution
Where is rotation?
Where is revolving?
Uniform Circular Motion
is the motion of an object in a circle with a
constant or uniform speed.
We have been referring to constant
speed. How would you describe
the velocity?
Remember velocity is a vector quantity so
has magnitude and direction
Velocity of Uniform Circular
Motion
 constant magnitude
changing direction-always
tangential to the circle
Miniature golf: where will the golf ball go?
Over point A, B, or C?
A
C
B
B
As an object moves in uniform circular
motion (remember speed is constant) is
there acceleration?
Yes
Although speed is constant, velocity is
not!!!
Acceleration is a change in velocity
It is accelerating because the direction of
the velocity vector is changing.
Identify the three controls on an
automobile which allow the car to be
accelerated.
In uniform circular motion, what is the
direction of the acceleration?
conditions for uniform circular motion
The velocity vector and the acceleration
vector are always perpendicular to each
other.
The acceleration vector changes only the
direction of the velocity vector not the
magnitude of it.
The acceleration vector is directed inwards
conditions for uniform circular motion
A
V
In circular motion, what is the direction of
the velocity vector?
In circular motion, what is the direction of
the acceleration vector?
Question to think about:
You are riding the carousel at the Woodlands
Mall. How long does it take to make a complete
circle? How many times does it do this in a
second? Are these related?
What does it mean to be periodic?
Period vs Frequency
T = Period
Time for one cycle or revolution (seconds)
f = Frequency
Number of cycles or revolutions per second
(hertz or sec-1)
Formulas
Calculating Period and Frequency
sec onds
T
revolutions
revolutions
f 
sec ond
T = period or time
for
one
revolution (sec)
f = frequency or
revolutions per
second (Hz or
sec-1)
Formulas
The period is inversely proportional to
the frequency
1
T
f
T = period or time for
one revolution
(sec)
f = frequency or
revolutions per
second (Hz or
sec-1)
Example 1
A merry-go-round makes 6
revolutions in 10 seconds. What is
its frequency?
revolutions
The definition of
f 
frequency is the
sec onds
number of
6revolutions
f 
revolutions per
10
sec
onds
second
f  0.6rev / sec
f  0.6 Hz
Example 2
A merry-go-round makes 6
revolutions in 42 seconds. What is
its period?
The definition of
period is time
required to
complete one
revolution
sec onds
T
revolutions
T =revolution
42sec/6rev
s
f 
sec
ond
The period is 7
Example 3
A merry-go-round has a frequency
.5Hz. What is its period?
1
T
f
T = 1rev/.5HZ
Period =2sec
Remember Hz = revolutions/sec
If we know time of 1 revolution (T), How
do we determine velocity of 1 revolution?
We will consider velocity at a point tangent
to the circle
Velocity = d/t
For distance: dependent on the
circumference of the circle
d = 2ᴫr (distance of 1 revolution)
t = T (time of 1 revolution)
Together:
2r
v
T
Calculating speed
2r
v
T
OR
v  2rf
v = speed (m/s)
r = radius of circle
(m)
T = period or time
for
one
revolution (sec)
f = frequency
Example 4:
A lifeguard twirls her whistle on a
string in a horizontal circle with a radius of 0.34 m.
It takes 1.5 second to complete the circle. What is
the average tangential speed of the whistle?
Given:
Example 4:
A lifeguard twirls her whistle on a
string in a horizontal circle with a radius of 0.34m.
It takes 1.5 second to complete the circle. What is
the average tangential speed of the whistle?
Given:
r = 0.34 m
T = (1.5 sec /1rev) = 1.5 sec/rev
2r
v
T
Example 4:
A lifeguard twirls her whistle on a
string in a horizontal circle with a radius of 0.34 m.
It takes 1.5 second to complete the circle. What is
the average tangential speed of the whistle?
Given:
r = 0.34 m
T = (1.5 sec /1rev) = 1.5 sec/rev
2r
v
T
2 (0.34m)
v
1.5 sec/ rev
v  1.42m / sec
the average speed and the radius of the
circle are ________ proportional.
A twofold increase in radius corresponds
to a _______ increase in speed if period
remains the same
Example 5:
A lifeguard twirls her whistle on a
string in a horizontal circle with a radius of 28 cm.
It takes 4.8 second to revolve 5 times. What is the
average tangential speed of the whistle?
Given:
r=
T=
Example 5:
A lifeguard twirls her whistle on a
string in a horizontal circle with a radius of 28 cm.
It takes 4.8 second to revolve 5 times. What is the
average tangential speed of the whistle?
Given:
r = 28 cm = 0.28 m
T = (4.8 sec /5rev) = 0.96 sec/rev
Example 5:
A lifeguard twirls her whistle on a
string in a horizontal circle with a radius of 28 cm.
It takes 4.8 second to revolve 5 times. What is the
average tangential speed of the whistle?
Given:
r = 28 cm = 0.28 m
T = (4.8 sec /5rev) = 0.96 sec/rev
2r
v
T
2 (0.28m)
v
0.96 sec/ rev
v  1.83m / sec
As an object moves in a uniform circle
What happens to its speed?
What happens to its velocity?
What happens to its acceleration?
Formulas
Calculating Centripetal Acceleration
using speed
2
v
ac 
r
ac = centripetal
acceleration (m/s2)
r = radius of circle (m)
v = speed (m/s)
What causes an object to have
Centripetal Acceleration?
Centripetal Force NOT centrifugal force
What’s the difference?
Centripetal = center seeking
Centrifugal = outward seeking
Without a net centripetal force, an object cannot
travel in circular motion. In fact, if the forces are
balanced, then an object in motion continues in
motion in a straight line at constant speed.
Without a centripetal
force, an object in
motion continues
along a straight-line
path.
With a centripetal
force, an object in
motion will be
accelerated and
change its
direction.
Courtesy http://www.physicsclassroom.com/mmedia/circmot/cf.cfm
frame of reference
Centripetal forces are those seen by an
observer in an inertial frame of reference.
Centrifugal forces are those felt by an
observer in an accelerating frame of
reference. As a car goes around a corner,
the passengers think they feel a force
towards the outside of the curve, in reality
this is due to inertia. Centrifugal force is a
misnomer!!!
What causes centripetal acceleration?
 To have acceleration, there must be a net force
towards the center of the circle
 What is the force for the following:
Earth circling the sun
Force of Gravity
Car turning a bend
Force of Friction
Xena warrior princess throwing a ball on a chain
Force of Tension on chain
Formulas
Calculating centripetal force using
centripetal acceleration
Fc  ma c
OR
mv
Fc 
r
2
Fc = centripetal force (N)
m = mass (kg)
ac = centripetal
acceleration (m/s2)
Example 6
A little girl is swinging her 5 kg purse in
horizontal circles using the strap that
allows the purse to swing 20 cm from her
hand. The girl is able to get the purse to
make 10 revolutions in 8 seconds. What
was the speed of the purse? What is the
centripetal acceleration of the purse? How
much tension is in the purse string?
Example 6
Given:
m=
r=
T=
Example 6 determine velocity
Given:
m = 5 kg
r = 20 cm = 0.2 m
T = (8sec / 10rev) = 0.8 sec/rev
2r
v
T
Example 6 determine velocity
Given:
m = 5 kg
r = 20 cm = 0.2 m
T = (8sec / 10rev) = 0.8 sec/rev
2r
v
T
2 (0.2m)
v
0.8 sec/ rev
v  1.57 m / sec
Example 6 determine acceleration
with velocity of 1.57 m/s
Given:
m = 5 kg
r = 20 cm = 0.2 m
T = (8sec / 10rev) = 0.8 sec/rev
2
v
ac 
r
Example 6 determine acceleration
with velocity of 1.57 m/s
Given:
m = 5 kg
r = 20 cm = 0.2 m
T = (8sec / 10rev) = 0.8 sec/rev
2
v
ac 
r
2
(1.57 m / s )
ac 
0 .2 m
2
ac  12.3m / s
Example 6 determine FC with
acceleration of 12,3 m/s2
Given:
m = 5 kg
r = 20 cm = 0.2 m
T = (8sec / 10rev) = 0.8 sec/rev
Fc  ma
Example 6 determine FC with
acceleration of 12.3 m/s2
Given:
m = 5 kg
r = 20 cm = 0.2 m
T = (8sec / 10rev) = 0.8 sec/rev
Fc  ma
Fc  (5kg)(12.3m / s )
2
Fc  61.5 N
Example 7: Shelby twirls her 25g whistle on a 0.450m
lanyard. She twirls the cord at a uniform speed in
a horizontal circle. If the speed is 4.75m/sec, what is the
centripetal acceleration of the whistle? How much tension
is in the lanyard?
Example 7: Shelby twirls her 25g whistle on a 0.450m
lanyard. She twirls the cord at a uniform speed in
a horizontal circle. If the speed is 4.75m/sec, what is the
centripetal acceleration of the whistle? How much tension
is in the lanyard?
a = 50.1 m/s2
F = 1.25 N
Concept questions:
 An object moves in uniform horizontal circular
motion. If the radius of the object triples, what
happens to the speed of the object?
 The speed will triple
 How does doubling velocity affect the ac?
 It increases four fold
 How would calculating Fc change if the uniform
circular motion was vertical instead of
horizontal?