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5.2 Uniform Circular Motion
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A force, , is directed
toward the center of the
circle
This force is associated
with an acceleration, ac
Applying Newton’s
Second Law along the
radial direction gives
Fig 5.8
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Uniform Circular Motion, cont
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A force causing a
centripetal acceleration
acts toward the center of
the circle
It causes a change in the
direction of the velocity
vector
If the force vanishes, the
object would move in a
straight-line path tangent to
the circle
Fig 5.9
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Centripetal Force
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The force causing the centripetal
acceleration is sometimes called the
centripetal force
This is not a new force, it is a new role
for a force
It is a force acting in the role of a force
that causes a circular motion
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Conical Pendulum
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The object is in
equilibrium in the
vertical direction and
undergoes uniform
circular motion in the
horizontal direction
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v is independent of m
Fig 5.11
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Banked Curve
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These are designed
with friction equaling
zero
There is a component
of the normal force that
supplies the centripetal
force
Fig 5.11
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Horizontal (Flat) Curve
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The force of static
friction supplies the
centripetal force
The maximum speed at
which the car can
negotiate the curve is
Note, this does not
depend on the mass of
the car
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Fig 5.13
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Loop-the-Loop
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This is an example
of a vertical circle
At the bottom of the
loop (b), the upward
force experienced
by the object is
greater than its
weight
Fig 5.14
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Loop-the-Loop, Part 2
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At the top of the
circle (c), the force
exerted on the
object is less than
its weight
Fig 5.14
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Non-Uniform Circular Motion
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The acceleration and
force have tangential
components
produces the
centripetal acceleration
produces the
tangential acceleration
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Fig 5.15
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Vertical Circle with NonUniform Speed
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The gravitational
force exerts a
tangential force on
the object
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Look at the
components of Fg
The tension at any
point can be found
Fig 5.17
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Top and Bottom of Circle
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The tension at the
bottom is a
maximum
The tension at the
top is a minimum
If Ttop = 0, then
Fig 5.17
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5.4 Motion with Resistive Forces
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Motion can be through a medium
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Either a liquid or a gas
The medium exerts a resistive force, , on an
object moving through the medium
The magnitude of depends on the medium
The direction of is opposite the direction of
motion of the object relative to the medium
nearly always increases with increasing
speed
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Motion with Resistive Forces,
cont
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The magnitude of can depend on the
speed in complex ways
We will discuss only two
is proportional to v
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Good approximation for slow motions or small
objects
is proportional to v2
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Good approximation for large objects
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R Proportional To v
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The resistive force can be expressed as
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b depends on the property of the
medium, and on the shape and
dimensions of the object
The negative sign indicates is in the
opposite direction to
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R Proportional To v, Example
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Analyzing the
motion results in
Fig 5.18(a)
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R Proportional To v, Example,
cont
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Initially, v = 0 and dv/dt = g
As t increases, R increases and a
decreases
The acceleration approaches 0 when R
 mg
At this point, v approaches the terminal
speed of the object
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Terminal Speed
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To find the terminal speed,
let a = 0
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Solving the differential
equation gives
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t is the time constant and t
= m/b
Fig 5.18(b)
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R Proportional To
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2
v
For objects moving at high speeds through air,
such as airplanes, sky divers and baseballs,
the resistive force is approximately
proportional to the square of the speed
R = 1/2 DrAv2
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D is a dimensionless empirical quantity that is
called the drag coefficient
r is the density of air
A is the cross-sectional area of the object
v is the speed of the object
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R Proportional To
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2
v,
example
Analysis of an object
falling through air
accounting for air
resistance
Fig 5.19
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R Proportional To
Speed
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2
v,
Terminal
The terminal speed
will occur when the
acceleration goes to
zero
Solving the equation
gives
Fig 5.19
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Some Terminal Speeds
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Exercises of chapter 5
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8, 13, 21, 26, 27, 29, 32, 39, 46, 55, 58
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Exercise 32
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Consider an object on which the net force is a
resistive force proportional to the square of its
speed. For example, assume that the resistive force
acting on a speed skater is , where k is a constant
and m is the skater’s mass. The skater crosses the
finish line of a straight-line race with speed v0 and
then slows down by coasting on his skates. Show
that the skater’s speed at any time t after crossing
the finish line is v(t )  v0 /(1  ktv0 ).
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