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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 dur 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
A tetherball problem
Dynamics of a Ferris Wheel
HOMEWORK CH6
3; 5; 8; 13; 15; 19; 22; 29; 36; 40
• Each stride is taken as one in a series of arcs
Spherically symmetric
objects interact
gravitationally as though all
the mass of each were
concentrated at its center
A diagram of gravitational force
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.
•Use Newton’s Law of Universal Gravitation with the
specific masses and separation.
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.
WEIGHT
Gravitational force falls off quickly
• If either m1 or m2 are small, the force decreases quickly
enough for humans to notice.
What happens when velocity rises?
• Eventually, Fg balances and you have orbit.
• When v is large enough, you achieve escape velocity.
Satellite Motion
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
Larger orbits correspond to slower speeds and longer periods.
We want to
place a
satellite into
circular orbit
300km
above the
earth
surface.
What speed,
period and
radial
acceleration
it must
have?
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