Circular Motion and Gravitation

1
Circular Motion and Gravitation
2
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

3
Examples

• 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

4
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
• 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

5
Period

• Another important measure in UCM is period, the
  time for 1 rotation or revolution
• Since xv0t , this implies that vT 2?r and
  thus T 2pr/ v
• Rearranging differently, v 2pr/ T and then
  inserting it into the acceleration equation
• ac v2/r 4p2r/T2

6
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.
• ac 4p2r 4p2(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

7
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
• ac4p2f 2r

8
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, 2pr, 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 2prf
• v (2?)(25.4cm/2)(1910/1 min)(1min/60s)
• v 2540cm/s 25.4 m/s.

9
Angular Velocity

• Velocity can be defined in terms of multiples of
  the radius, called radians
• There are 2p radians in a circle, and so the
  angular velocity w v/r
• In terms of period w 2p/T
• In terms of frequency w2pf

10
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
• 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 ?2r (?/2 rad/s)2(7.7m) 19m/s2.
• Notice that this is almost twice the acceleration
  of a body in free fall.

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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

12
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
• wfw0 at q(w0 wf) t/ 2
• q x/r a a/r w v/r

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Example 9.1
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
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Example 9.2
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(2p) radians 4p rads wf0 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(4p) a -(0.21)2 -1.8 x10-3 rad/s2
2(4p)
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Homework p210ff 1,3,4,7,10,11,15
16
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 Fcmac mv2/r

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Example

• Approximately how much force does the earth exert
  on the moon? Moons 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
• Fc2.005 x 1020 N.

18
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

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Forces in Circular Motion

• Centrifugal Force Can Mimic Gravity
• Bug in can-can pushes up on bug due to
  centripetal force
• Bug pushes down on can due to inertia
• Space station could simulate gravity by
  rotation in same manner
• Rotational speed to simulate gravity depends on
  radius of the station
• 1 RPM would require 1 kilometer radius
• People have difficulty adjusting to rotational
  speeds over 2 RPM
• Stations would have to be large to compensate for
  proper gravity

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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 gravity 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
• T2 T2? 2p 55s 0.92 min.

21
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

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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

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Homework!!p211ff17,21,23,25,27
24
Law of Universal Gravitation

• Newtons 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

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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

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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

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Example

• Consider a mass m falling near the earth's
  surface. Find its acceleration in 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 earths surface,
  so replace r with the average radius of the earth
  RE
• g GME
• RE2

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Example

• Show that Keplers third law follows from the law
  of universal gravitation. Keplers 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
• FcGmM 4p2mr Rearranging gives T2
  4p2
• r2 T2 r3 GM

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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 4/3pRE3 r
• 4/3pRE3
• The equation for g can then be rewritten in terms
  of the density as
• g G(4/3pRE3 r) 4/3GpRE r
• RE2

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Density Example (cont)

• Upon rearranging, we find the density to be
• r 3g
• 4pREG
• Inserting the numerical values, we get
• r 3(9.81 m/s2)
• 4p(6.38 x 106 m)(6.67 x 10-11 N m2/ kg2
• r 5.50 x 103 kg/m3

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Moon Period Example

• Calculate the period of the moons 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 4p2mr
• 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.

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Period of Moon Example

• The centripetal force is provided by the
  gravitational force, so that
• GME m 4p2mr
• r2 T2
• Solving for T gives
• T
• T2.37x 106 s, or T 27.4 days.

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Period of a Satellite Example

• T
• Use the result g RE2 GME, the above expression
  for the period becomes
• T
• Notice that the period depends only on the
  radius of the earth and the acceleration of
  gravity. Insert the approximate values of p2 10,
• g 10 m/s2, and RE 6.4 x 106m,
• T 5100 s 85 min

p
2
4
R
m
E
2
T

• Estimate the period of an artificial
  earth-orbiting satellite that passes just above
  the earth's surface.Set the force required to
  give a circular orbit--the centripetal
  force--equal to the gravitational force. The mass
  of the satellite m. The mass of the earth ME,
  the radius of the orbit RE, and the satellite's
  period T

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Homework!!P21232,34,36,37,39
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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 GF/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.
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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

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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
• according to the
• celestial object involved

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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

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Black Holes

• Gravity is great due to small distance with huge
  mass
• Gravity only great near the object, at distance
  gravity is no different

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Homework!!P21343,47,52,53,62,64
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Major Equations!!
T2 4p2 r3 GM qtan-1 (v2/gr) GGM/r2 Ug-G
Mm/r
acv2/r ac4p2r/T2 ac4p2rf2 ac
w2r wv/r w2p/T w2pf v2pr/T 2prf
f1/T Fcmac Fcmv2/r FgGMm/r2 gGM/r2 r3g/4pRG
T2p