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## Ch5. Uniform Circular Motion

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Title: Ch5. Uniform Circular Motion

1
Ch5. Uniform Circular Motion
Uniform circular motion is the motion of an
object traveling at a constant (uniform) speed on
a circular path. Period T is the time required
to travel once around the circle, that is, to
make one complete revolution.
r
2
Example 1 A Tire-Balancing Machine
The wheel of a car has a radius of r 0.29m and
is being rotated at 830 revolutions per minute
(rpm) on a tire-balancing machine. Determine the
speed (in m/s) at which the outer edge of the
wheel is moving. The speed v can be obtained
directly from , but first the
period T is needed. It must be expressed in
seconds.
3
830 revolutions in one minute
T1.210-3 min, which corresponds to 0.072s
4
Uniform circular motion emphasizes that 1. The
speed, or the magnitude of the velocity vector,
is constant. 2. Direction of the vector is not
constant. 3. Change in direction, means
acceleration 4. Centripetal acceleration , it
points toward the center of the circle.
5
Centripetal Acceleration
Magnitude ac of the centripetal acceleration
depends on the speed v of the object and the
radius r of the circular path. acv2/r
6
in velocity divided by the elapsed time
or a /
7
COP is an isosceles triangle. Both triangles have
equal apex angles .
acv2/r
The direction is toward the center of the circle.
8
Conceptual Example 2 Which way will the object
go?
An object on a guideline is in uniform circular
motion. The object is symbolized by a dot, and at
point O it is release suddenly from its circular
path.
If the guideline is cut suddenly, will the object
move along OA or OP ?
9
Newtons first law of motion guides our
reasoning. An object continues in a state of rest
or in a state of motion at a constant speed along
a straight line unless compelled to changes that
state by a net force. When the object is suddenly
released from its circular path, there is no
longer a net force being applied to the object.
In the case of a model airplane, the guideline
cannot apply a force, since it is cut. Gravity
certainly acts on the plane, but the wings
provide a lift force that balances the weight of
the plane.
10
In the absence of a net force, then, the plane or
any object would continue to move at a constant
speed along a straight line in the direction it
had at the time of release. This speed and
direction are given in Figure 5.4 by the velocity
vector v.
As a result, the object would move along the
straight line between points O and A, not on the
circular arc between points O and P.
11
Example 3 The Effect of Radius on Centripetal
Acceleration
The bobsled track at the 1994 Olympics in
Lillehammer, Norway, contained turns with radii
of 33 m and 24 m, as the figure illustrates. Find
the centripetal acceleration at each turn for a
speed of 34 m/s, a speed that was achieved in the
two-man event. Express the answers as multiples
of g9.8m/s2.
12
From acv2/r it follows that
13
Conceptual Example 4 Uniform Circular Motion and
Equilibrium
A car moves at a constant speed, and there are
three parts to the motion. It moves along a
straight line toward a circular turn, goes around
the turn, and then moves away along a straight
line. In each of three parts, is the car in
equilibrium?
14
An object in equilibrium has no acceleration,
according to the definition given in Section
4.11. As the car approaches the turn, both the
speed and direction of the motion are constant.
Thus, the velocity vector does not change, and
there is no acceleration. The same is true as the
car moves away from the turn. For these parts of
the motion, then, the car is in equilibrium. As
the car goes around the turn, however, the
direction of travel changes, so the car has a
centripetal acceleration that is characteristic
of uniform circular motion. Because of this
acceleration, the car is not in equilibrium
during the turn.
In general, an object that is in uniform circular
motion can never be in equilibrium.
15
The car in the drawing is moving clockwise around
a circular section of road at a constant speed.
What are the directions of its velocity and
acceleration at (a) position 1 and (b) position
2?
16
(a) The velocity is due south, and the
acceleration is due west.
(b) The velocity is due west, and the
acceleration is due north.
17
Centripetal Force
18
Concepts at a glance Newtons second law
indicates that whenever an object accelerates,
there must be a net force to create the
acceleration. Thus, in uniform circular motion
there must be a net force to produce the
centripetal acceleration. As the
Concept-at-a-glance chart, the second law gives
this net force as the product of the objects
mass m and its acceleration v2/r. This chart is
an expanded version of the chart shown previously
in Figure 4.9. The net force causing the
centripetal acceleration is called the
centripetal force FC and points in the same
direction as the acceleration- that is, toward
the center of the circle.
19
centripetal force does not denote a new and
separate force created by nature.
20
Example 5 The Effect of Speed on Centripetal
Force
The model airplane has a mass of 0.90 kg and
moves at a constant speed on a circle that is
parallel to the ground. The path of the airplane
and its guideline lie in the same horizontal
plane, because the weight of the plane is
balanced by the lift generated by its wings. Find
the tension T in the guideline(length17m) for
speeds of 19 and 38m/s.
21
Equation 5.3 gives the tension directly
FCTmv2/r
Speed 19m/s
22
Conceptual Example 6 A Trapeze Act
In a circus, a man hangs upside down from a
trapeze, legs bent over the bar and arms
downward, holding his partner. Is it harder for
the man to hold his partner when the partner
hangs straight down and is stationary or when the
partner is swinging through the straight-down
position?
23
Reasoning and Solution When the man and his
partner are stationary, the mans arms must
support his partners weight. When the two are
swinging, however, the mans arms must do an
additional job. Then the partner is moving on a
circular arc and has a centripetal acceleration.
The mans arms must exert and additional pull so
that there will be sufficient centripetal force
to produce this acceleration.
Because of the additional pull, it is harder for
the man to hold his partner while swinging than
while stationary.
24
Example 7 Centripetal Force and Safe Driving
Compare the maximum speeds at which a car can
safely negotiate an unbanked turn (r 50.0m)
25
The car does not accelerate ,
FN mg 0 FN mg.
26
As expected, the dry road allows the greater
maximum speed.
27
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28
• A car is traveling in uniform circular motion on
slippery, and the car is just on the verge of
sliding.
• If the cars speed was doubled, what would have
to be the smallest radius in order that the car
r.
• What would be your answer to part (a) if the car
were replaced by one that weighted twice as much?

29
(a) 4r
(b) 4r
30
Banked Curves
31
A car is going around a friction-free banked
curve. The radius of the curve is r.
FN sin that points toward the center C
FN cos and, since the car does not
accelerate in the vertical direction, this
component must balance the weight mg of the car.
32
At a speed that is too small for a given , a
car would slide down a frictionless banked curve
at a speed that is too large, a car would slide
off the top.
33
Example 8The Daytona 500
The Daytona 500 is the major event of the NASCAR
(National Association for Stock Car Auto Racing)
season. It is held at the Daytona International
Speedway in Daytona, Florida. The turns in this
oval track have a maximum radius(at the top)of
r316m and are banked steeply, with
. Suppose these maximum-radius turns were
frictionless. At what speed would the cars have
to travel around them?
34
From Equation 5.4, it follows that
(96 mph)
Drivers actually negotiate the turns at speeds up
to 195 mph, however, which requires a greater
centripetal force than that implied by Equation
5.4 for frictionless turns.
35
Satellites in Circular Orbits
36
If the satellite is to remain in an orbit of
radius r, the speed must have precisely this
value.
The closer the satellite is to the earth, the
smaller is the value for r and the greater the
orbital speed must be.
Mass m of the satellite does not appear. For a
given orbit, a satellite with a large mass has
exactly the same orbital speed as a satellite
with a small mass.
37
Example 9 Orbital Speed of the Hubble Space
Telescope
Determine the speed of the Hubble Space Telescope
orbiting at a height of 598 km above the earths
surface.
Orbital radius r must be determined relative to
the center of the earth. The radius of the earth
is approximately 6.38106m, r6.98106m
38
The orbital speed is
39
Global Positioning System(GPS)
40
Example 10 A Super-massive Black Hole
The Hubble Telescope has detected the light being
emitted from different regions of galaxy M87. The
black circle identifies the center of the galaxy.
From the characteristics of this light,
astronomers have determined an orbiting speed of
7.5105m/s for matter located at a distance of
5.71017m from the center. Find the mass M of the
object located at the galactic center.
41
Replacing ME with M
4.81039kg
42
The ratio of this incredibly large mass to the
mass of our sun is (4.81039kg)/(2.01030kg)2.41
09. Matter equivalent to 2.4 billion suns is
located at the center of galaxy M87. The volume
of space in which this matter is located contains
relatively few visible star. There are strong
evidences for the existence of a super-massive
black hole. black hole tremendous
mass prevents even light from escaping. The light
that forms the image comes not from the black
hole itself, but from matter that surrounds it.
43
Period T of a satellite is the time required for
one orbital revolution.
44
Period is proportional to the three-halves power
of the orbital radius is know as Keplers third
law. (Johannes Kepler, 1571-1630). Keplers third
law also holds for elliptical orbits.
synchronous satellites orbital period is
chosen to be one day, the time it takes for the
earth to turn once about its axis. Satellites
move around their orbits in a way that is
synchronized with the rotation of the earth,
appearing in fixed positions in the sky and can
serve as stationary.
45
Example 11 The Orbital Radius for Synchronous
Satellites
The period T of a synchronous satellite is one
day. Find the distance r from the center of the
earth and the height H of the satellite above the
earths surface. The earth itself has a radius of
6.38106m.
T8.64104s
46
r 4.23107m
H4.23107m-0.64107m3.59107m (22300 mi)
47
Two satellites are placed in orbit, one about
Mars and the other about Jupiter, such that the
orbital speeds are the same. Mars has the smaller
mass. Is the radius of the satellite in orbit
about Mars less than, greater than, or equal to
the radius of the satellite orbiting Jupiter?
Less than.
48
Apparent Weightlessness and Artificial Gravity
The idea of life on board an orbiting satellite
conjures up visions of astronauts floating around
in a state of weightlessness
49
Conceptual Example 12 Apparent Weightlessness
and Free-Fall
Objects in uniform circular motion continually
accelerate or fall toward the center of the
circle, in order to remain on the circular path.
The only difference between the satellite and the
elevator is that the satellite moves on a circle,
so that its falling does not bring it closer to
the earth. True weight is the gravitational force
(FGmME/r2) that the earth exerts on an object
and is not zero.
50
Example 13 Artificial Gravity
At what speed must the surface of the space
station (r1700m) move in the figure, so that the
astronaut at point P experiences a push on his
feet that equals his earth weight?
51
FCmv2/r
Earth weight of the astronaut (massm) is mg.
FCmgmv2/r
52
Example 14 A Rotating Space Laboratory
A space laboratory is rotating to create
artificial gravity. Its period of rotation is
chosen so the outer ring (r02150m) simulates the
acceleration due to gravity on earth (9.80 m/s2).
What should be the radius r1 of the inner ring,
so it simulates the acceleration due to gravity
on the surface of Mars (3.72 m/s2)?
The outer ring (radiusr0) simulates gravity on
earth, while the inner ring (radiusr1) simulates
gravity on Mars
53
Centripetal acceleration acv2/r, speed v
T is the period of the motion. The laboratory is
rigid. All points on a rigid object make one
revolution in the same time. Both rings have the
same period.
54
Inner ring
Dividing the inner ring expression by the outer
ring expression,
r1 816 m
55
• The acceleration due to gravity on the moon is
one-sixth that on earth.
• Is the true weight of a person on the moon less
than, greater than, or equal to the true weight
of the same person on the earth?
• Is the apparent weight of a person in orbit about
the moon less than, greater than, or equal to the
apparent weight of the same person in orbit about
the earth?

(a) Less than
(b) Equal to
56
Vertical Circular Motion
Usually, the speed varies in this stunt.
non-uniform
57
The magnitude of the normal force changes,
because the speed changes and the weight does not
have the same effect at every point.
58
The weight is tangent to the circle at points 2
and 4 and has no component pointing toward the
center. If the speed at each of the four places
is known, along with the mass and radius, the
normal forces can be determined.
They must have at least a minimum speed at the
top of the circle to remain on the track. v3 is a
minimum when FN3 is zero.
Weight mg provides all the centripetal force. The
rider experiences an apparent weightlessness.
59
Concepts Calculation Examples 15 Acceleration
At time t0 s, automobile A is traveling at a
speed of 18 m/s along a straight road and its
picking up speed with an acceleration that has a
magnitude of 3.5 m/s2. At time t0 s, automobile
A is traveling at a speed of 18 m/s in uniform
circular motion as it negotiates a turn. It has a
centripetal acceleration whose magnitude is also
3.5 m/s2. Determine the speed of each automobile
when t2.0 s.
60
Which automobile has a constant acceleration?
Both its magnitude and direction must be
constant. A has constant acceleration, a
constant magnitude of 3.5 m/s2 and its direction
always points forward along the straight road. B
has an acceleration with a constant magnitude of
3.5 m/s2, a centripetal acceleration, which
points toward the center of the circle at every
instant.
Which automobile do the equations of kinematics
apply?
Apply for automobile A.
61
Speed of automobile A at t2.0 s vv0at18
m/s(3.5 m/s2)(2.0 s)25 m/s
B is in uniform circular motion and goes around
the turn. At a time of t2.0 s, its speed is the
same as it was at t0 s, i.e., v18 m/s.
62
Concepts Calculation Example 16 Centripetal
Force
Ball A is attached to one end of a rigid
mass-less rod, while an identical ball B is
attached to the center of the rod. Each ball has
a mass of m0.50kg, and the length of each half
of the rod is L0.40m. This arrangement is held
by the empty end and is whirled around in a
horizontal circle at a constant rate, so each
ball is in uniform circular motion. Ball A
travels at a constant speed of vA5.0m/s. Find
the tension in each half of the rod.
63
How many tension forces contribute to the
centripetal force that acts on ball A?
A single tension force of magnitude TA acts on
ball A, due to the tension in the rod between the
two balls. This force alone provides the
centripetal force keeping ball A on its circular
How many tension forces contribute to the
centripetal force that acts on ball B?
Two tension forces act on ball B. TB-TA
Is the speed of ball B the same as the speed of
ball A?
No, it is not. Because A travels farther than B
in the same time. A travels a distance equal to
the circumference of its path, which is
(2L). B is only L . Speed of ball B is one
half the speed of ball A , or vB2.5 m/s
64
Ball A
Ball B
Centripetal force FC
Centripetal force FC
23N
65
Problem 4
R 3.6m
OA ?
66
Problem 4
REASONING AND SOLUTION Since the speed of the
object on and off the circle remains constant at
the same value, the object always travels the
same distance in equal time intervals, both on
and off the circle. Furthermore since the object
travels the distance OA in the same time it would
have moved from O to P on the circle, we know
that the distance OA is equal to the distance
along the arc of the circle from O to P.
67
Circumference
360o 22.6m
1o (22.6/360)m
25o (22.6/360)25m
and, from the argument given above, we conclude
that the distance OA is 1.6m.
68
Problem 43
REASONING In Example 3, it was shown that the
magnitudes of the centripetal acceleration for
the two cases are
According to Newton's second law, the centripetal
force is (see Equation
5.3).
69
SOLUTION a. Therefore, when the sled undergoes
the turn of radius 33 m,
b. Similarly, when the radius of the turn is 24
m,
70
Problem 46
REASONING AND SOLUTION The force P supplied by
the man will be largest when the partner is at
the lowest point in the swing. The diagram at
the right shows the forces acting on the partner
in this situation. The centripetal force
necessary to keep the partner swinging along the
arc of a circle is provided by the resultant of
the force supplied by the man and the weight of
the partner.
71
From the figure
Therefore
Since the weight of the partner, W, is equal to
mg, it follows that m (W/g) and