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CHAPTER 6
CIRCULAR MOTION AND GRAVITATION
Goals for Chapter 6
• To understand the dynamics of circular motion.
• To study the unique application of circular motion
as it applies to Newton’s Law of Gravitation.
• To study the motion of objects in orbit as a special
application of Newton’s Law of Gravitation.
Uniform circular motion is due to a centripetal acceleration
This aceleration is always pointing to the center
This aceleration is due to a net force
•
Period = the time for one revolution
1) Circular motion in horizontal plane:
- flat curve
- banked curve
- rotating object
2) Circular motion in vertical plane
Rounding a flat curve
• The centripetal force coming only from tire
friction.
Rounding a banked curve
• The centripetal force comes from friction and a
component of force from the car’s mass
Dynamics of a Ferris Wheel
The "Giant Swing" at a county fair consists of a vertical central
shaft with a number of horizontal arms attached at its upper
end. Each arm supports a seat suspended from a 5 m long
cable, the upper end of which is fastened to the arm at a point
3m from the central shaft.
Find the time of one revolution of the swing if the cable supporting the seat
makes an angle of 300 with the vertical.
GRAVITATION
Spherically symmetric
objects interact
gravitationally as though all
the mass of each were
concentrated at its center
Cavendish Balance
•The slight attraction of the masses causes a nearly imperceptible
rotation of the string supporting the masses connected to the
mirror.
•Use of the laser allows a point many meters away to move
through measurable distances as the angle allows the initial and
final positions to diverge.
Newton’s Law of Gravitation
• Always attractive.
• Directly proportional to the masses involved.
• Inversely proportional to the square of the
separation between the masses.
• Masses must be large to bring Fg to a size
even close to humanly perceptible forces.
A diagram of gravitational force
G = 6.674x10-11 N.m2/kg2
Each mass is 2 kg
Find the magnitude of the net gravitational force on each
mass and its direction
Each mass in the
figure below is 3 kg.
Find the force
(magnitude and
direction) on each
mass in the figure .
WEIGHT
Gravitational force falls off quickly
• If either m1 or m2 are small, the force decreases quickly
enough for humans to notice.
• In January 2005 the Huygens probe landed on Saturn's
moon Titan, the only satellite in the solar system having
a thick atmosphere. Titan's diameter is 5150 km, and its
mass is 1.35×1023 kg, The probe weighed 3120 N on
earth. What did it weigh on the surface of Titan?
Satellite Motion
What happens when velocity rises?
•When v is large enough, you achieve escape velocity.
The principle governing the motion of the satellite is Newton’s
second law; the force is F, and the acceleration is v2/r, so the
equation Fnet = ma becomes
GmmE/r
2
= mv 2/r
v = GmE/r
T= 2πr/v
= (2πr3/2)/
GmE
Larger orbits correspond to slower speeds and longer periods.
A 320 kg satellite experiences a gravitational force of 800 N.
What is the radius of the of the satellite’s orbit? What is its
altitude?
F = GmEmS/r 2
r 2 = GmEmS/ F
r 2 = (6.67 x 10 -11 N.m2/kg2) (5.98 x 10 24 kg) (320 kg ) / 800 N
r 2 = 1.595 x 1014 m2
r = 1.26 x 107 m
Altitude = 1.26 x 107 m – radius of the Earth
Altitude = 1.26 x 107 m – 0.637 x 107 = 0.623 x 107 m
We want to
place a
satellite into
circular orbit
300km
above the
earth
surface.
What speed,
period and
radial
acceleration
it must
have?