Transcript Ch5. Uniform Circular Motion

Ch5. Uniform Circular Motion
Uniform circular motion is the
motion of an object traveling at a
constant (uniform) speed on a
circular path.
r
Period T is the time required to
travel once around the circle, that
is, to make one complete
revolution.
2r
v
T
1
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.
2 r
The speed v can be obtained directly from v 
,
T
but first the period T is needed. It must be
expressed in seconds.
2
830 revolutions in one minute
1
 1.2 103 min/revolution
830 revolutions / min
T=1.2*10-3 min, which corresponds to 0.072s
2r 2 (0.29 m)
v

 25m / s
T
0.072 s
3
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.
4
Centripetal Acceleration
Magnitude ac of the centripetal acceleration
depends on the speed v of the object and the
radius r of the circular path. ac=v2/r
5
v in velocity divided by the elapsed time t
or a= v / t
Sector of the circle COP.
t is very small, the arc
length OP is approximately
a straight line whose length
is the distance v t
traveled by the object.
6
COP is an isosceles triangle. Both triangles
have equal apex angles  .
v vt

v
r
v / t
ac=v2/r
The direction is toward the center of the circle.
7
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 ?
8
Newton’s 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.
9
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.
10
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
g=9.8m/s2.
11
From ac=v2/r it follows that
Radius=33m
2
(34m / s)
ac 
 35m / s 2  3.6 g
33m
Radius=24m
2
(34m / s)
ac 
 48m / s 2  4.9 g
24m
12
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?
13
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.
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Check your understanding 1
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?
15
(a) The velocity is due south, and the acceleration
is due west.
(b) The velocity is due west, and the acceleration
is due north.
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Centripetal Force
17
Concepts at a glance: Newton’s 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-aglance chart, the second law gives this net force as the
product of the object’s 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.
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2
mv
FC 
r
“centripetal force” does not denote a new and
separate force created by nature.
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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(length=17m) for
speeds of 19 and 38m/s.
20
Equation 5.3 gives the tension directly:
FC=T=mv2/r
Speed =19m/s
(0.90kg )(19m / s) 2
T
 19N
17m
Speed =38m/s
(0.90kg )(38m / s) 2
T
 76N
17m
21
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?
22
Reasoning and Solution: When the man and his
partner are stationary, the man’s arms must
support his partner’s weight. When the two are
swinging, however, the man’s arms must do an
additional job. Then the partner is moving on a
circular arc and has a centripetal acceleration. The
man’s 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.
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Example 7: Centripetal Force and
Safe Driving
Compare the maximum
speeds at which a car can
safely negotiate an
unbanked turn (r= 50.0m)
S
S
S  dry  0.9
N
FS
FS
mg
f
--dry = 0.9
--icy = 0.1
s
max
 s N
24
f
MAX
s
 S FN
N
fS
The car does not accelerate ,
FN – mg = 0

2
v
s g 
r
mg
FN = mg.
v   s gr
25
Dry road (  s =0.900)
v  (0.900 )( 9.80 m / s )( 50.0m)  21.0m / s
2
Icy road (
 s =0.100)
v  (0.100 )( 9.80 m / s )( 50.0m)  7.00 m / s
2
As expected, the dry road allows the greater
maximum speed.
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Upward on the wing surfaces with a net lifting force L, the
plane is banked at an angle  , a component L sin  of the
lifting force is directed toward the center of the turn.
Greater speeds and/or tighter turns require greater centripetal
forces. Banking into a turn also has an application in the
construction of high-speed roadways.
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Check your understanding 2
A car is traveling in uniform circular motion on a
section of road whose radius is r. The road is
slippery, and the car is just on the verge of sliding.
(a) If the car’s speed was doubled, what would have
to be the smallest radius in order that the car does
not slide? Express your answer in terms of r.
(b)What would be your answer to part (a) if the car
were replaced by one that weighted twice as
much?
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(a) 4r
(b) 4r
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Banked Curves
30
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
2
mv
FC  FN sin  
r
FN cos  and, since the car does not accelerate in the
vertical direction, this component must balance the
weight mg of the car.
FN cos  mg
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FN sin  m v / r

FN cos
mg
2
2
v
tan 
rg
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.
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Example 8:The 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 r=316m and are banked
steeply, with   31 . Suppose these maximumradius turns were frictionless. At what speed
would the cars have to travel around them?
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From Equation 5.4, it follows that
v  rg tan   (316 m)( 9.80 m / s 2 ) tan 31   43m / s
(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.
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Satellites in Circular Orbits
2
m ME m v
FC  G 2 
r
r
GM E
v
r
35
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.
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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
earth’s surface.
Orbital radius r must be determined relative to the
center of the earth. The radius of the earth is
approximately 6.38*106m, r=6.98*106m
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The orbital speed is
GM E
(6.671011 N .m 2 / kg 2 )(5.981024 kg)
v

r
6.98106 m
v  7.5610 m / s(16900mi/ h)
3
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Global Positioning System(GPS)
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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 7.5*105m/s for matter located at a distance of
5.7*1017m from the center. Find the mass M of
the object located at the galactic center.
40
Replacing ME with M
GM
v
r
v r (7.5 10 m / s) (5.7 10 m)
M

11
2
2
G
6.6710 N .m / kg
2
5
2
17
=4.8*1039kg
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The ratio of this incredibly large mass to the mass
of our sun is (4.8*1039kg)/(2.0*1030kg)=2.4*109.
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.
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Period T of a satellite is the time required for
one orbital revolution.
v  2r / T
GM E 2r

r
T
2r
T
GM E
3/ 2
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Period is proportional to the three-halves power
of the orbital radius is know as Kepler’s third
law. (Johannes Kepler, 1571-1630). Kepler’s 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”.
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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 earth’s
surface. The earth itself has a radius of 6.38*106m.
2r
T
GM E
3/ 2
T=8.64*104s
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4
11
2
2
24
T
GM
(8.6410 s) (6.6710 N.m / kg )(5.9810 kg)
E
3/ 2
r 

2
2
r = 4.23*107m
H=4.23*107m-0.64*107m=3.59*107m (22300 mi)
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Check your understanding 3
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.
GM
v
r
MMars ---smaller
Since v is the same, M small -----r small.
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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”
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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
(F=GmME/r2) that the earth exerts on an object
and is not zero.
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Example 13: Artificial Gravity
At what speed must the
surface of the space
station (r=1700m) move
in the figure, so that the
astronaut at point P
experiences a push on his
feet that equals his earth
weight?
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FC=mv2/r
Earth weight of the astronaut (mass=m) is mg.
FC=mg=mv2/r
v  rg  (1700 m)( 9.80 m / s )  130 m / s
2
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Example 14: A Rotating Space
Laboratory
The outer ring (radius=r0)
simulates gravity on earth, while
the inner ring (radius=r1) simulates
gravity on Mars
A space laboratory is rotating
to create artificial gravity. Its
period of rotation is chosen so
the outer ring (r0=2150m)
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)?
52
Centripetal acceleration:
speed v and radius r:
ac=v2/r,
v  2r / T
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.
 2r 


2
2
v
4

r
T 

ac 

 2
r
r
T
2
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4 (2150m)
9.80m / s 
T2
2
2
2
4

r1
2
3.72m / s 
T2
Inner ring
Outer ring
Dividing the inner ring expression by the outer ring expression,
2
r1
3.72m / s

2
2150m
9.80m / s
r1 = 816 m
54
Check your understanding 4
The acceleration due to gravity on the moon is onesixth that on earth.
(a) 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?
(b)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
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Vertical Circular Motion
Usually, the speed varies in this stunt.
“non-uniform”
56
m v1
(1) F  m g 
N1
r
=FC1
(2) FN 2
=FC2
m v22

r
2
2
3
mv
(3) FN 3  m g 
r
=FC3
(4)
FN 4
m v42

r
=FC4
The magnitude of the normal force changes, because
the speed changes and the weight does not have the
same effect at every point.
57
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. v3  rg
Weight mg provides all the centripetal force. The
rider experiences an apparent weightlessness.
58
Concepts & Calculation
Examples 15: Acceleration
At time t=0 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 t=0 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 t=2.0 s.
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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.
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Speed of automobile A at t=2.0 s
v=v0+at=18 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 t=2.0 s, its speed
is the same as it was at t=0 s, i.e., v=18 m/s.
61
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 m=0.50kg, and the length
of each half of the rod is L=0.40m.
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 vA=5.0m/s. Find
the tension in each half of the rod.
62
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 path of radius 2L
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 2 (2L). B is only2 L . Speed of ball B is
one half the speed of ball A , or vB=2.5 m/s
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m vA2
TA 
2L
Ball A
Centripetal force FC
m vB2
Ball B TB  TA 
L
Centripetal force FC
m vA2 (0.50kg)(5.0m / s 2 )
TA 

 16N
2L
2(0.40m)
2
B
2
A
2
2
mv mv
(0.50kg)(2.5m / s ) (0.50kg)(5.0m / s )
TB 



L
2L
0.40m
2(0.40m)
=23N
64
Problem 4
R = 3.6m

= 25
OA = ?
65
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.
66
Circumference =
360o
1o
25o
2  r  2  (3.6 m) = 22.6 m
22.6m
(22.6/360)m
(22.6/360)*25m
and, from the argument given above, we conclude that the
distance OA is 1.6m.
67
Problem 43
REASONING In Example 3, it was shown that the
magnitudes of the centripetal acceleration for the
two cases are
Radius  33 m
a C  35 m / s
Radius  24 m
a C  48 m/ s2
According to Newton's second law, the centripetal
force is FC  ma C (see Equation 5.3).
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2
SOLUTION
a. Therefore, when the sled undergoes the turn of
radius 33 m,
FC  ma C  ( 350 kg)(35 m / s 2 )  1.2  10 4 N
b. Similarly, when the radius of the turn is 24 m,
FC  ma C  ( 350 kg)(48 m / s 2 )  1.7  10 4 N
69
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.
70
From the figure
Therefore
mv 2
P  mg 
r
mv 2
P
 mg
r
Since the weight of the partner, W, is equal to mg, it
follows that m = (W/g) and
(W/g )v2
[(475 N)/(9.80 m/s2 )] (4.00 m/s)2
P
W 
 (475 N) = 594 N
r
(6.50 m)
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