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```					       Uniform Circular Motion
Motion in a circular path at
constant speed
Speed constant, velocity
changing continually
Velocity changing direction, so
there is acceleration
Called centripetal acceleration,
since it is toward the center
of the circle, along the radius
Value can be calculated by
many formulas, first is
ac = v2/r
Example

A bicycle racer rides with constant speed around
a circular track 25 m in diameter. What is the
acceleration of the bicycle toward the center
of the track if its speed is 6.0 m/s?

ac = v2 =__(6.0 m/s)2 = 36 (m/s)2 = 2.9 m/s2
r    12.5 m       12.5 m
Rotation and Revolution
Rotation-Around an Internal Axis-Earth
rotates 24 hours for a complete turn
Linear (tangential) versus rotational
speed
Linear is greater on outside of
disk or merry-go-round, more
distance per rotation
Linear is smaller in middle of
disk, less distance per rotation.
Rotational speed is equal for both
Rotations per minute (RPM)
Linear speed is proportional to
both rotational speed and
distance from the center
Rotation and Revolution
Revolution-Around an
External Axis-Earth
revolves 365.25 days per trip
around sun
Same relationship
between linear and
revolutional speeds
as with rotational
Planets do not revolve
at the same
revolutional speeds
around the sun
Period

Another important measure in UCM is period,
the time for 1 rotation or revolution
Since x=v0t , this implies that vT = 2r and thus
T= 2r/ v
Rearranging differently, v= 2r/ T and then
inserting it into the acceleration equation
ac = v2/r = 42r/T2
Example

Determine the centripetal acceleration of the moon as it circles the earth,
and compare that acceleration with the acceleration of bodies falling
on the earth. The period of the moon's orbit is 27.3 days.
According to Newton's first law, the moon would move with constant
velocity in a straight line unless it were acted on by a force. We can
infer the presence of a force from the fact that the moon moves with
approximately uniform circular motion about the earth. The mean
center-to-center earth-moon distance is 3.84 x 108 m.
Example
ac= 42r = 42(3.84 x 108)       T= 27.3 da (24 hr/da)(3600 s/hr) = 2.36 x 106 s
T2         (2.36 x 106)2
ac = 2.72 x 10-3 m/s2

The ratio of the moon's acceleration to that of an object falling near the
earth is          ac = 2.72 x l0-3 m/s2 =            1
g      9.8 m/s2                 3600
Frequency
The number of revolutions per time unit
Value is the inverse of the period, 1/T
Units are sec-1 or Hertz (Hz)
Inserting frequency into the ac equation
ac=42f 2r
Example
An industrial grinding wheel with a 25.4-cm
diameter spins at a rate of 1910 rotations
per minute. What is the linear speed of a
point on the rim?
The speed of a point on the rim is the distance
traveled, 2r, divided by T, the time for
one revolution. However, the period is the
reciprocal of the frequency, so the speed of
a point on the rim, a distance r from the axis
of rotation, is
v = 2rf
v= (2)(25.4cm/2)(1910/1
min)(1min/60s)
v = 2540cm/s = 25.4 m/s.
Angular Velocity
Velocity can be defined in terms of multiples of
There are 2 radians in a circle, and so the
angular velocity w = v/r
In terms of period     w= 2/T
In terms of frequency      w=2f
Example

At the Six Flags amusement park near Atlanta. The Wheelie carries passengers in a
circular path with a radius of 7.7 m. The ride makes a complete rotation every 4.0 s.
(a) What is a passenger's angular velocity due to the circular motion? (b) What
acceleration does a passenger experience?
a) The ride has a period T = 4.0 s. We can use it to compute the angular velocity as
T      4.0 s            2.0
(b) Because the riders travel in a circle, they undergo a centripetal acceleration
given by
ac= w2r = (/2 rad/s)2(7.7m) = 19m/s2.
Notice that this is almost twice the acceleration of a body in free fall.
Angular Velocity and Acceleration
Any real object that has a definite shape can be made
to rotate – solid, unchanging shape
Angular displacement -- q -- Radians around circular path
Angular velocity -- w --radians per second, angle between
fixed axis and point on wheel changes with time
Angular acceleration -- a -- increase of w , when angular
velocity of the rigid body changes, radians per seconds
squared
Rotational Kinematics
Rotational velocity, displacement, and
acceleration all follow the linear forms, just
substituting the rotational values into the
equations:
q= wot + 1/2 at2        wf2 = wo2 + 2aq
wf=w0 +at               q=(w0 +wf) t/ 2
q= x/r         a= a/r   w= v/r
Example

The wheel on a moving car slows uniformly from 70 rads/s to
a. Find a b. Find q       c. How far does the car go?
Dt       4.2 s
b. q= wot + 1/2 at2 = (70)(4.2) + 1/2(-6.7)(4.2)2 = 235 rads
c. q = x / r in rads so x = q r = (0.32)(235) = 75 m
Example
A bicycle wheel turning at 0.21 rads/s is brought to rest by the brakes in
exactly 2 revolutions. What is its angular acceleration?
Use angular equivalent of vf2 = vo2 + 2ax which is
wf2 = wo2 + 2aq
(0)2 = (0.21)2 + 2a(4)
a = -(0.21)2 = -1.8 x10-3 rad/s2
2(4)
pp.898-903 Practice
Problems 7A, 7B
Forces in Circular Motion
Centripetal Force
Force toward the center from an object, holding it in
circular motion
At right angle to the path of motion, not along its
distance, therefore does NO work on object
Examples
Gravitation between earth and moon
Electromagnetic force between protons and electrons in an
atom
Friction on the tires of a car rounding a curve
Equation is Fc=mac = mv2/r
Example
Approximately how much force does the earth
exert on the moon? Moon’s period is 27.3
days
Assume the moon's orbit to be circular about a
stationary earth. The force can be found from
F = ma. The mass of the moon is 7.35 x 1022
kg.
Fc= mac = m 42r
T2
Fc = (7.35 x 1022 kg)42(3.84 x 108m)
((27.3 days)(24 hr/day)(3600 s/hr))2
Fc=2.005 x 1020 N.
Forces in Circular Motion
Centrifugal “Force”
Not a true force, but really the result of
inertia
“Centrifugal force effect” makes a
rotating object fly off in straight line if
centripetal force fails
Example
Imagine a giant donut-shaped space station located so far from all
heavenly bodies that the force of gravity may be neglected. To enable
the occupants to live a “normal” life, the donut rotates and the
inhabitants live on the part of the donut farthest from the center. If the
outside diameter of the space station is 1.5km, what must be its period
of rotation so that the passengers at the periphery will perceive an
artificial gravity equal to the normal ravity at the earth's surface?
The weight of a person of mass m on the earth is a force F = mg.
The centripetal force required to carry the person around a circle of
radius r is F =mac = m 42r
T2
We may equate these two force expressions and solve for the period T:
mg =m42r                              750m
r
T2        T=2    g    =2      9.81m / s 2   =55s = 0.92 min.
Banked Curves
“Banking” road curves makes turns without skidding possible
For angle q, there is a component of the normal force toward the
center of the curve, thus supplying the centripetal force. The
other component balances the weight force.
FN sin q = mv2/r        FN cos q = mg     tan q= v2/gr
thusly q = tan-1 (v2/gr)
This equation can give the proper angle for banking a curve of any
Banked Curve Example
A race track designed for average speeds of
240 km/h (66.7 m/s) is to have a turn with
a radius of 975 m. To what angle must the
track be banked so that cars traveling
240km/h have no tendency to slip
sideways?
Determine q from
q = tan-1 (v2/ g r)
= tan-1 (66.72/9.81(975))= 24.9o
Homework!!
Law of Universal Gravitation
Newton’s first initiative for the Principia was
investigating gravity
From his 3rd law, he proposed that each object
would pull on any other object
He likewise noted differences due to distance
His final relationship was that Force was
proportional to masses and inversely proportional
to distance squared
Using a constant Fg = Gm1m2
r2
Center of Gravity
Newton found that his law would only work when
measuring from the center of both objects
This idea is called the center of gravity
Sometimes it is at the exact center of the object
Sometimes it may not be in the object at all
All forces must be from the CG of one object to the
CG of the other object
Universal Gravitation Constant
G was elusive to find since gravity is a weak force if
masses are small
Cavendish developed a device which made
measurement of G possible
The value of G is 6.67 x 10-11 N m2
This puts Fg in Newtons      kg2
G can be used then to find values of many
astronomical properties
Example m falling near the earth's surface. Find its acceleration in
Consider a mass
terms of the universal gravitational constant G. The gravitational
force on the body is        F = GmME
r2
ME = mass of the earth        r = the distance of the mass from the center
of the earth, essentially the earth's radius.
The gravitational force on a body at the earth's surface is F = mg.
mg= GmME         or    g =GME
r2                  r2
Both G and ME are constant, and r does not change significantly for
small variations in height near the surface of the earth. The right-hand
side of this equation does not change appreciably with position on the
earth’s surface, so replace r with the average radius of the earth RE
g = GME
RE2
Example third law follows from the law of universal
Show that Kepler’s
gravitation. Kepler’s third law states that for all planets the ratio
(period)2/ (distance from sun)3 is the same.
Make the approximation that the orbits of the planets are circles and that
the orbital speed is constant.
The sun's gravitational force on any planet of mass m is
F= GmM
r2
M =the mass of the sun. Because the mass of the sun is so much larger
than the mass of the planet, we can assume, as Kepler did, that the sun
lies at the center of the planetary orbit. The circular orbit implies a
centripetal force. This net force for circular motion is provided by the
gravitational force. Equating these two forces, we get
Fc=GmM = 42mr              Rearranging gives T2 = 42
r2         T2                              r3     GM
Example
Use the law of universal gravitation and the measured value of the
acceleration of gravity g to determine the average density of the earth.
The density, r of an object is defined as its mass per unit volume: r
= m/V where m is the mass of the object whose volume is V
From a previous example       g = GME
RE2
Substitute for M an expression involving r, r = ME/V.
If we take the earth to be a sphere of radius RE. Then
r =       ME         and ME= 4/3RE3 r
4/3RE3
The equation for g can then be rewritten in terms of the density as
g = G(4/3RE3 r) = 4/3GRE r
R E2
Density Example (cont)
Upon rearranging, we find the density to be
r=         3g
4REG
Inserting the numerical values, we get
r     =            3(9.81 m/s2)
4(6.38 x 106 m)(6.67 x 10-11 N m2/ kg2

r = 5.50 x 103 kg/m3
Moon Period Example
Calculate the period of the moon’s orbit about the earth,
assuming a constant distance r = 3.84 x 108 m.
The magnitude of the attractive force must equal the
centripetal force.
Fc = 42mr
T2
In this case the attractive force is the gravitational force
between the earth and moon             F = G ME m
r2
where ME is the mass of the earth (5.98 x 1024 kg), m is the
mass of the moon, and r is the earth-moon distance. We can
equate these forces, solve for T and substitute the numerical
values.
Period of Moon Example
The centripetal force is provided by the gravitational
force, so that
GME m = 42mr
r2       T2
Solving for T gives
T = 4 2 r 3     =             4 2 (3.84x108 m)3
-11    2    2      24
GM E            (6.67 x10 Nm / kg )(5.98x10 kg)
T=2.37x 106 s, or      T = 27.4 days.
Period of a Satellite Example
m4 2 RE = GmM E               T=   4 2 RE
3
2
T 2       RE                       GM E
Estimate the period of an
artificial earth-orbiting     Use the result g RE2 = GME, the above
satellite that passes just      expression for the period becomes
4 R
2       3
4 R
2
above the earth's surface.Set   T = gR        2   = g
E       E

E
the force required to give a    Notice that the period depends only
circular orbit--the             on the radius of the earth and the
centripetal force--equal to     acceleration of gravity. Insert the
the gravitational force. The    approximate values of 2 =10,
mass of the satellite = m.
g =10 m/s2, and RE = 6.4 x 106m,
The mass of the earth ME,
the radius of the orbit RE,         4(10)(6.4 x106 m)
and the satellite's period T  T=        10m / s 2     = 5100 s = 85 min
Homework!!
Gravitational Field Strength
Gravity works at a distance, and distance limits its strength
At any point in space, the strength of the field would be
G=F/m0 , where m0 is the test mass
Substituting, we get G = GMm0 = GM
r2m0         r2
This picture illustrates
that far from a body,
the field lines are far
apart and thus its
strength is reduced.
Gravitation Considerations
Orbital Speed--if an object is projected
horizontally with enough speed, it remains
in orbit around any celestial object
For Earth, this is 8000 m/s
This causes satellites to orbit every 90 min.
Stationary orbiting satellite with period 24hrs
Escape Velocity
Earth spacecraft must get entirely away from the earth
to go on to other planets
This requires giving a spacecraft enough energy to
overcome the gravitational potential energy of earth
This gives an equation such that
Where M and R vary
2GM
vesc =               according to the
R
celestial object involved
Black Holes
If the escape velocity is equal to the speed of
light, gravity will keep even light from
escaping--the idea behind the black hole
Conjecture due to observations from space
Theory is a supergiant star collapses in on
itself creating super strong gravity at a
small point
Black Holes
Gravity is great due
to small distance
with huge mass
Gravity only great
near the object, at
distance gravity is
no different
Keplers Laws
First Law: Each planet travels ina an
elliptical path around the sun, and the sun
is at one of the focal points
Keplers Laws
Second Law: An
imaginary line is
drawn from the
sun to any planet
sweeps out equal
areas in equal time
intervals.
Keplers Laws

Third Law: The
square of a planet’s
orbital period (T2) is
proportional to the
cube of the average
distance (r3)
between the planet
and the sun or
T2 ∝ r3
Period and speed of an object in
circular orbit
Homework!!
Major Equations!!
ac=v2/r          f=1/T        T2 = 42
ac=42r/T2       Fc=mac       r3 GM

ac=42rf2        Fc=mv2/r
ac= w2r
w=v/r            Fg=GMm/r2
w=2/T
r
w=2f            T=2     g
v=2r/T = 2rf

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