Transcript File - Mrs. Haug`s Website

Chapter 5
5.1 Uniform Circular Motion
Uniform circular motion is the motion of
an object traveling at a constant
(uniform) speed on a circular path.
 Since the path is in a circle, period (T)is
often used to express time.
 The distance around the circle is equal
to the circumference (2πr)

If the magnitude of velocity if
constant, what causes
acceleration?
2r
v
T
5.2 Centripetal Acceleration
Since the direction of the velocity vector
is constantly changing, the object must
be accelerating.
 Centripetal acceleration is a vector
quantity.
 The direction of acceleration is always
toward the center of the circle or arc.

Centripetal Acceleration Defined

Magnitude: The centripetal acceleration of
an object moving with a speed v on a circular
path of radius r has a magnitude ac given by:
2
v
ac 
r
Direction: The centripetal acceleration always
points toward the center of the circle and
continually changes direction as the object
moves.
5.3 Centripetal Force
Newton’s 2nd Law of Motion holds true
for objects moving with uniform circular
motion
 Remember F=ma
 Applying our equation for centripetal
acceleration to Newton’s 2nd Law gives
the equation

2
v
Fc  m
r
Points to Ponder
Centripetal means “directed toward the
center”
 Centripetal force is caused by other
forces such as friction, tension, normal
force, applied force, etc.
 You may need to use equations from
previous chapters to determine the
centripetal force
 Remember: NET FORCE!

5.4 Banked Curves
Roadways are banked as a means of
eliminating friction as the cause of
centripetal force.
 On a banked turn, the normal force is
perpendicular to the surface of the road.
The vertical component of the normal
force (FN sinθ) provides the centripetal
force.
 This will only work for a certain velocity at a
certain banking angle.

Useful Equations for Banked Curves
mv2
Fc  FN sin  
r
The vertical component of the normal
force is Fn cosθ and since the car
doesn’t accelerate in the vertical
direction, this component must
balance the weight (mg) of the car.
Therefore this component of the
normal force will equal the weight.
Using this relationship we derive
the equation….
mg
FN 
cos 
2
v
tan  
rg
v  rg tan 
5.5 Satellites in Circular Orbits
Gravitational force will act as the
centripetal force for satellites in orbit
about the earth.
 There is only one speed that a satellite
can have if the satellite is to remain in
an orbit with a fixed radius.
 For a given orbit, a satellite with a large
mass has exactly the same orbital
speed as a satellite with a small mass.

Equations for Satellites in
Remember:
Circular Orbit
Centripetal force
mM E mv
Fc  G 2 
r
r
GM E
v
r
2
comes from any
number of forces. As
long as the force is
directed toward the
center of an arc or
circle, it is considered
a centripetal force.
These equations apply to manmade earth satellites or to
natural satellites like the moon.
It also applies to circular orbits
about any astronomical object.
Replace the mass of the earth
with the mass of the object on
which the orbit is centered.
Relating Period to Centripetal
Force




The period of a satellite is the time required to
complete one orbital revolution.
Shown on the next slide is an equation that
can be used to find the period of planets in
nearly circular orbits.
Replace the mass of the earth with the mass
of the sun.
The period is proportional to three-halves
power of the orbital radius. This is Kepler’s
third law of planetary motion.
More Equations!
GM E 2r

r
T
The first equation sets velocity of a
satellite in orbit at a fixed radius
equal to the velocity of an object
with uniform circular motion. If we
rearrange this equation to solve for
T, the equation on the right will be
useful to determine the period of a
satellite with fixed orbital radius.
3
2r
T
GM E
This equation is useful for communications experts
launching “synchronous satellites” where the period
is equal to one day. Check out Example 11 on page
147!
2
5.6 Apparent Weightlessness and
Artificial Gravity





When a person is in an orbiting satellite, both
the person and the scale are experiencing
uniform circular motion.
Both will continually accelerate or “fall” toward
the center of the circle.
Since both fall at the same rate, the person
cannot push on the scale
The apparent weight in the satellite is zero!
Read through Example 13 on page 149 of
your text!
5.7 Vertical Circular Motion
Consider a motorcycle stunt driver
driving around a vertical loop.
 The magnitude of the normal force
changes because the speed changes
and because the weight does not have
the same effect at every point.
 In the “east and west” positions, the
weight is tangent to the circle and has
no effect on the centripetal force.

A Picture is worth a thousand words!
Great Websites with lots of Concept
Development!
http://library.thinkquest.org/2745/data/loops.htm
http://www.its-abouttime.com/htmls/apcoreselect/apcorech4_278.html