Download Uniform circular motion

Today: (Ch. 5)

Circular Motion and Gravitation
Tomorrow: (Ch. 5)

Circular Motion and Gravitation
Free Fall Example!
Water is dripping from a water facet, each
drop coming one second apart. Two drops
are shown here. At t=0s the first drop is
released. At t=1s, the second drop is
released.
(a) What is the velocity of the first drop when
the second is released?
(b) Draw the velocity-time curve for both
drops on a single graph.
(c) Draw the acceleration-time curve for
both drops on a single graph.
Velocity-time graph for drops
v (m/s)
30
20
10
0
1
-10
-20
-30
2
3
4
5
6
7
t (s)
Acc.-time graph for drops
a (m/s2)
30
20
10
0
1
-10
-20
-30
2
3
4
5
6
7
t (s)
Introduction
• Circular motion
– Acceleration is not constant
– Cannot be reduced to a one-dimensional problem
• Examples
– Car traveling around a turn, Centrifuge, Earth orbiting
the Sun
• Gravitation
– Explore gravitational force in more detail
– Look at Kepler’s Laws of Motion
– Further details about g
Vocabulary
Uniform Circular Motion: Motion along a circular path
with a constant speed.
Since the direction changes, the velocity changes,
and there is acceleration, even though the speed is
constant!
Period, T: The time an object takes to complete one
complete revolution. (Time to go around once.)
Description of uniform circular
motion

r
θ
  2  1

av 
t
av  lim
t 0

t
2

r
1
Angular displacement
Average angular velocity
Instantaneous angular velocity
Uniform Circular Motion
• Uniform circular motion assumes constant speed
• The distance traveled in one
cycle is the circumference of
the circle and time taken is
period of the motion, T
T 
2 r
v
r is the radius of the circle
v is the speed of the motion
Centripetal Acceleration
• Speed is constant, the
velocity is not constant
• Direction of acceleration
– Always directed toward
the center of the circle
– This is called the
centripetal acceleration
• Centripetal means
“center-seeking”
• Magnitude of acceleration
Circular Motion and Forces
• Newton’s Second Law to circular motion:
mv
 F  m a   F  maC  r
2
• The force must be directed toward the center of the
circle
• The centripetal force can be supplied by a variety of
physical objects or forces
• The “circle” does not need to be a complete circle
Centripetal Force Example
• The centripetal
acceleration is produced
by the tension in the
string
• If the string breaks, the
object would move in a
direction tangent to the
circle at a constant speed
Problem Solving Strategy –
Circular Motion
• Recognize the principle
– If the object moves in a circle, then there is a
centripetal force acting on it
• Sketch the problem
– Include the path the object travels
– Identify the circular part of the path
– Include the radius of the circle
– Show the center of the circle
– Selecting a coordinate system that assigns the
positive direction toward the center of the circle is
often convenient
• A free body diagram is generally useful
Problem Solving Strategy, cont.
• Identify the principles
– Find all the forces acting on the object
– Find the components of the forces that are directed
toward the center of the circle
– Find the components of the forces perpendicular to the
center
– Apply Newton’s Second Law for both directions
• The acceleration directed toward the center of the
circle is a centripetal acceleration
• Solve for the quantities of interest
• Check your answer
– Consider what the answer means
– Does the answer make sense
Centripetal Acceleration – Car
• A car rounding a curve travels in an approximate circle
• The radius of this circle is called the radius of curvature
• Forces in the y-direction
– Gravity and the normal
force
• Forces in the x-direction
– Friction is directed toward
the center of the circle
• Since friction is the only force
acting in the x-direction, it
supplies the centripetal force
Ffriction
mv 2

 s m g
r
v  s r g
Cornering
Top View
Rear View
r
v
r
Car on Banked Curve Example
• The maximum speed can be
increased by banking the curve
• Assume no friction between the
tires and the road
• The car travels in a circle, so the
net force is a centripetal force
• Forces acting on the car gravity
and normal
• The speed at which the care will
just be able to negotiate the turn
without sliding up or down the
banked road is
v  r g tan
• When θ = 0, v = 0 and you
cannot turn on a very icy road
without slipping
Banked Curves
Highway engineers use banked curves to lessen the
reliance on friction for the centripetal force.
A properly designed banked curve uses part of the
Normal Force to provide the centripetal force.
r
2
v
1
  tan
rg
Examples of Circular Motion
• When the motion is uniform, the total
acceleration is the centripetal acceleration
– Remember, this meant that the speed is
constant
• The motion does not need to be uniform
– Then there will be a tangential acceleration
included
• Many examples can be analyzed by
looking at the two components
Non-Uniform Circular Motion
• If the speed is also
changing, there are two
components to the
acceleration
• One component is tangent
to the circle, at
• The other component is
directed toward the center
of the circle, ac
Circular Motion, Vertical Circle
Example
• The speed of the rock varies with time
• At the bottom of the circle:
– Tension and gravity are in opposite
directions
–
 v2

T  m  g 
 r

• The tension supports the rock (mg)
and supplies the centripetal force
Circular Motion, Vertical Circle
Example, Cont.
• At the top of the circle:
– Tension and gravity are in the same direction
• Pointing toward the center of the circle
–
Ttop
 v2

 m  g 
 r

• There is a minimum value of v needed to keep
the string taut at the top
– Let T = 0
– v  rg
Circular Motion, Roller Coaster
Example
• The roller coaster’s path
is nearly circular
• There is a maximum at
which the coaster will not
leave the top of the track:
– v  rg
– If the speed is greater
than this, N would
have to be negative
– This is impossible, so
the coaster would
leave the track
Circular Motion, Artificial Gravity
Example
• Circular motion can be used to
create “artificial gravity”
• The normal force acting on the
passengers due to the floor
would be
mv 2
N
r
• If N = mg it would feel like the
passengers are experiencing
normal Earth gravity
Tomorrow: (Ch. 5)

Circular Motion and Gravitation