Download CHAPTER 5 DYNAMIC OF UNIFORM CIRCULAR MOTION

CHAPTER 5
DYNAMIC OF UNIFORM CIRCULAR MOTION
5.1 UNIFORM CIRCULAR MOTION:
Uniform circular motion is the motion of an object traveling at a constant or uniform speed on a
circular path.
If the period T is the time required for the object to complete one circular path i.e one revolution
, the relation between speed v , period T and the circumference of the circle 2  r is ;
2r
 Eq 5.1
T
where r is the radius for the circular path.
v
Constant speed v is equal to the
magnitude of tangential velocity v
which is also constant but changing in
direction.
r
The motion of an airplane flying at a constant speed on a horizontal circular path is an example
of uniform circular motion.
EXAMPLE 5.1:
How long does it take a plane traveling at a constant speed of 110 m/s to fly once around a circle
whose radius is 2850 m?
REASONING The speed of the plane is given by Equation 5.1: v  2 r / T , where T is the
period or the time required for the plane to complete one revolution.
SOLUTION Solving Equation 5.1 for T we have
T
2  r 2  ( 2850 m )

 160 s
v
110 m / s
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EXAMPLE 5.2:
The wheel of a car has a radius of r = 0.29 m and is being rotated at 830 revolution per minute
on a tyre-balancing machine . Determined the speed in m/s at which the outer edge of the wheel
is moving.
SOLUTION
1
= 1.2 x 10-3 min/revolution
830revolution / min
Therefore T = 1.2 x 10-3 min = 0.072 s
v
2r 2 (0.29m)
=
= 25 m/s
T
0.072s
5.2 CENTRIPETAL ACCELERATION
The centripetal acceleration of an object moving with a speed v on a circular path of radius r has
the magnitude as given by
v2
r
The direction of centripetal acceleration vector always points toward the center of the circle and
continually changes direction as the object moves.
ac 
The centripetal acceleration occurs from the changes of direction of velocity vector (magnitude
constant) during object moves in circular path.
The effect of radius on centripetal acceleration.
The larger the radius, the smaller the centripetal acceleration. Refer EXAMPLE 3 on page 127.
5.3 CENTRIPETAL FORCE
Centripetal force is the net force required to keep an object of mass m , moving at a speed v, on a
circular path of radius r, and it has a magnitude of
mv 2
v2
[ From  F= ma ,  F = Fc and a = ac 
]
r
r
The direction of centripetal force is always points toward the center of the circle and continually
changes direction as the object moves.
Fc =
 F = Fc can becomes any external Forces i.e Gravitational Force, Normal Force, Frictional
Force and Tension Force.
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APPLICATIONS INVOLVING UNIFORM CIRCULAR MOTION:
5.4 BANKED CURVES
A vehicle can negotiate a circular turn without relying on static friction to provide the centripetal
force, provided the turns is banked at an angle relative to the horizontal. The angle  at which a
friction-free curve must be banked is related to the speed v of the vehicle, the radius r of the
curve, and the magnitude g of the acceleration due to gravity is given by :
tan  
v2
rg
Refer example below on how to get equation 5.4.
A car travels on a frictionless banked
road with radius r from center c and
the banking angle
The forces acting on the car are its
weight mg and the normal force FN.
A component FN sin of the normal
force provides the centripetal force.
mv 2
 (1)
r
FN cos   mg
 (2)
Fc = FN sin  
(1)  (2)

FN sin 
mv 2 / r
=
FN cos 
mg
v2
tan  
rg
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EXAMPLE 5.3 :
For a car traveling with a speed v around a curve of radius, r, determine a formula for the angle
at which a road should be banked so that no friction is required. And what is this angle for an
expressway off-ramp curve of radius 50m at a design speed of 50km/h?
EXAMPLE 5.4:
How large must the coefficient of static friction be between the tires and the road if a car is to
round a level curve of radius 85m at a speed of 95km/h?
Solutions:
In the free-body diagram, the car is coming out of the paper at the reader, and the center of the circular
path is to the right of the car, in the plane of the paper. The vertical forces (gravity and normal force)
are of the same magnitude, because the car is not accelerating vertically. We assume that the force
of friction is the force causing the circular motion. If the car has its maximum speed, it would be on
the verge of slipping, and the force of static friction would be at its maximum value.
2
FR  Ffr

 1m s  
 95 km hr   3.6 km hr  
2
v

   0.84
 m v 2 r   s FN   s mg   s 

2
 85 m   9.8 m
rg
s

Notice that the result is independent of the car’s mass.
EXAMPLE 5.5
A 1200kg car rounds a curve of radius 67m banked at an angle of 120 . If the car is traveling at
95km/h, will a friction force be required? If so, how much and in what direction?
Solutions:
The car moves in a horizontal circle, and so there must be a net horizontal
centripetal force. The car is not accelerating vertically. Write Newton’s 2nd law for both the x and y
directions.
mg
 Fy  FN cos  mg  0  FN  cos
y

x
FN
F

F

F
sin


ma
 x  R N
x
The amount of centripetal force needed for the car to round the
curve is
mg

2

 1.0 m s  
 95 km h   3.6 km h  

   1.247  104 N
FR  m v 2 r  1200 kg  
67 m
.
The actual horizontal force available from the normal force is
mg
FN sin  
sin   mg tan   1200 kg  9.80 m s 2 tan12o  2.500 103 N .
cos 
Thus more force is necessary for the car to round the curve than can be
y

supplied by the normal force. That extra force will have to have a
x
FN


4
mg


Ffr
horizontal component to the right in order to provide the extra centripetal force. Accordingly, we
add a frictional force pointed down the plane. That corresponds to the car not being able to make the
curve without friction.
Again write Newton’s 2nd law for both directions, and again the y
acceleration is zero.
mg  F sin 
 Fy  FN cos  mg  Ffr sin   0  FN  cosfr
F
 FN sin   Ffr cos   m v 2 r
Substitute the expression for the normal force from the y equation into the x equation, and solve for
the friction force.
mg  Ffr sin 
v2
sin   Ffr cos   m v 2 r   mg  Ffr sin   sin   Ffr cos 2   m cos 
cos 
r
x
Ffr  m
v2
r




cos   mg sin   1.247  104 N cos12o  1200 kg  9.80 m s 2 sin12o
 9.752  103 N
So a frictional force of 9.8 103 N down the plane is needed to provide the necessary centripetal
force to round the curve at the specified speed.
5.5 VERTICAL CIRCULAR MOTION
Vertical circular motion occurs when an object move on a vertical circular path. The speed of the
object often varies from moment to moment, and so do the magnitudes of the centripetal
acceleration and centripetal force. Refer example below:
A vertical loop-the-loop
motorcycle stunt
The normal force FN and the weight mg
of the cycle and the rider are shown at
four location
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The magnitude of the centripetal force at each of the four points is given as follows in terms of
mg and FN :
mv1
(1) Fc1 = FN1 – mg =
r
2
mv3
(3) Fc3 = FN3 + mg =
r
2
2
2
mv2
mv4
(2) Fc2 = FN2 =
(4) Fc4 = FN4 =
r
r
EXAMPLE 5.6 :
At what minimum speed must a roller coaster be traveling when upside down at the top of a
circle so that the passengers will not fall out? Assume a radius of the curvature of 7.4m.
Solutions:
At the top of a circle, a free-body diagram for the passengers would be as shown, assuming the passengers
are upside down. Then the car’s normal force would be pushing DOWN on the passengers, as shown in
the
diagram. We assume no safety devices are present. Choose the positive direction to be down, and write
Newton’s 2nd law for the passengers.
F  F
N

 mg  ma  m v 2 r  FN  m v 2 r  g

We see from this expression that for a high speed, the normal force is positive, meaning the
passengers are in contact with the car. But as the speed decreases, the normal force also decreases. If
the normal force becomes 0, the passengers are no longer in contact with the car – they are in free
fall. The limiting condition is
2
vmin
r  g  0  vmin  rg 
9.8 m s   7.4 m   8.5 m s
2
5.6 HORIZONTAL CIRCULAR MOTION
Vertical circular motion occurs when an object move on a vertical circular path
Refer to example below:
An object with mass m whirled at the end of the string in a horizontal circle with radius r on a
smooth table. If the speed of the object is v , the centripetal force which enable the object to
move in a circle is provided by the tension in the string.
mv 2
T=
r
O
If the string snaps, no more centripetal
force acts on the object so that the object
will continue its motion along a straight line
in the direction of v, which is tangential
to the string.
T
v
m
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EXAMPLE 5.7:
An object at the end of the string of length move in a horizontal circle of radius r as figure below:
Given the speed of the object is 28m/s and the radius is 150m, find the value of θ.
If
T
is the angle the string make with vertical,
T cos
= mg
T sin
=
Tan
r
=
mg
EXAMPLE 5.8:
18. REASONING AND SOLUTION The centripetal acceleration of the block is
ac = v2/r = (28 m/s)2/(150 m) = 5.2 m/s2
The angle  can be obtained from Tan  =
v2
rg
5.7 KEPLER’S LAW
Three statements of Kepler’s Laws:
(i)
The law of orbits : All planets move in elliptical orbits with the sun at one of the focal
points.
(ii)
The law of areas : A line drawn from the sun to any planet sweeps out equal areas in
equal time intervals.
(iii)
The law of periods : The square of the orbital period of any planet is proportional to
the cube of the average distance from the planet to the Sun.
 4 2
T  
 GM S
2

 3
r  K S r 3

T r
2
3

Ks is a constant given by 2.97 x 10-19 s2/ m3
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EXAMPLE 5.9:
The asteroid belt that lies between Mars and Jupiter may be the debris of a planet that broke apart or that was not able to form as
a result of Jupiter’s strong gravitation. The asteroid belt has a period of about 5.0 years. Approximately how far from the Sun
would this “fifth” planet have been?
T = 5.0 y = (5.0)(365)(24)(3600 s) = 1.58  108 s.
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we have r =
T2
K =
3
T 2 = Kr 3,
From
(1.58  108 s)2
= 4.4  1011 m .
3.0  1019 s2/m
EXAMPLE 5.10:
The value of the Kepler constant is K 2.97  10
19 2
2

s / m3 , and T  2.97  10
2
19 2
3

3
s / m r . With
3
nonstandard units, K can be made equal to unity, and the equation can be written as T  r . Using data
on the Earth (from the tables on the inside back cover of the book), find the units of K that make it equal
to unity. [Hint: The average, or mean, distance of the Earth from the Sun is used in astronomy as a unit
called an astronomical unit (AU).]
T 2 = Kr 3.
So
For the Earth,
T = 1 year
(1 y)2 = K(1 AU)3,

and
r = 1 AU.
K = 1 y2/AU3 .

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