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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

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
        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

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

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.
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
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
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
         v = 2540cm/s = 25.4 m/s.
             Angular Velocity
Velocity can be defined in terms of multiples of
  the radius, called radians
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

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
                      2= 2 rad =  rad/s = 1.6 rad/s
                      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
          Rotational Kinematics
Rotational velocity, displacement, and
 acceleration all follow the linear forms, just
 substituting the rotational values into the
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

The wheel on a moving car slows uniformly from 70 rads/s to
42 rads/s in 4.2 s. If its radius is 0.32 m:
      a. Find a b. Find q       c. How far does the car go?
a. a= Dw = (42-70) rads/s = -6.7 rads/s2
      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
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?
 q = 2 revs = 2(2) radians = 4 rads       wf=0 rads/s wo= 0.21 rads/s
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
   Homework: Read
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
     Gravitation between earth and moon
     Electromagnetic force between protons and electrons in an
     Friction on the tires of a car rounding a curve
  Equation is Fc=mac = mv2/r
Approximately how much force does the earth
   exert on the moon? Moon’s period is 27.3
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
 Fc= mac = m 42r
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
  “Centrifugal force effect” makes a
   rotating object fly off in straight line if
   centripetal force fails
  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
We may equate these two force expressions and solve for the period T:
mg =m42r                              750m
     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
  radius at any linear speed
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
Determine q from
  q = tan-1 (v2/ g r)
   = tan-1 (66.72/9.81(975))= 24.9o
   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
              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
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
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
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
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
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
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
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
In this case the attractive force is the gravitational force
   between the earth and moon             F = G ME m
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
 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
                                   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
  above the earth's surface.Set   T = gR        2   = g
                                                    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
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.
  Greater radius causes greater period
      Stationary orbiting satellite with period 24hrs
      has radius approx. 23,000 miles
                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
       vesc =               according to the
                            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
            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
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f            T=2     g
  v=2r/T = 2rf

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