+ All Categories
Transcript
Page 1: CHAPTER 6 CIRCULAR MOTION AND GRAVITATION
Page 2: CHAPTER 6 CIRCULAR MOTION AND GRAVITATION

CHAPTER 6CIRCULAR MOTION AND GRAVITATION

Page 3: 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.

Page 4: CHAPTER 6 CIRCULAR MOTION AND GRAVITATION

Uniform circular motion is due to a centripetal accelerationThis aceleration is always pointing to the centerThis aceleration is dur to a net force

Page 5: CHAPTER 6 CIRCULAR MOTION AND GRAVITATION

• Period = the time for one revolution

1) Circular motion in horizontal plane: - flat curve - banked curve - rotating object2) Circular motion in vertical plane

Page 6: CHAPTER 6 CIRCULAR MOTION AND GRAVITATION

Rounding a flat curve

• The centripetal force coming only from tire friction.

Page 7: CHAPTER 6 CIRCULAR MOTION AND GRAVITATION

Rounding a banked curve

• The centripetal force comes from friction and a component of force from the car’s mass

Page 8: CHAPTER 6 CIRCULAR MOTION AND GRAVITATION

A tetherball problem

Page 9: CHAPTER 6 CIRCULAR MOTION AND GRAVITATION

Dynamics of a Ferris Wheel

Page 10: CHAPTER 6 CIRCULAR MOTION AND GRAVITATION

HOMEWORK CH6

3; 5; 8; 13; 15; 19; 22; 29; 36; 40

Page 11: CHAPTER 6 CIRCULAR MOTION AND GRAVITATION

• Each stride is taken as one in a series of arcs

Page 12: CHAPTER 6 CIRCULAR MOTION AND GRAVITATION

Spherically symmetric objects interact

gravitationally as though all the mass of each were concentrated at its center

Page 13: CHAPTER 6 CIRCULAR MOTION AND GRAVITATION

A diagram of gravitational force

Page 14: CHAPTER 6 CIRCULAR MOTION AND GRAVITATION

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.

Page 15: CHAPTER 6 CIRCULAR MOTION AND GRAVITATION

•Use Newton’s Law of Universal Gravitation with the specific masses and separation.

Page 16: CHAPTER 6 CIRCULAR MOTION AND GRAVITATION

•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.

Cavendish Balance

Page 17: CHAPTER 6 CIRCULAR MOTION AND GRAVITATION

WEIGHT

Page 18: CHAPTER 6 CIRCULAR MOTION AND GRAVITATION

Gravitational force falls off quickly• If either m1 or m2 are small, the force decreases quickly enough for humans to notice.

Page 19: CHAPTER 6 CIRCULAR MOTION AND GRAVITATION

What happens when velocity rises? • Eventually, Fg balances and you have orbit.

• When v is large enough, you achieve escape velocity.

Page 20: CHAPTER 6 CIRCULAR MOTION AND GRAVITATION

Satellite Motion

Page 21: CHAPTER 6 CIRCULAR MOTION AND GRAVITATION

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.

Page 22: CHAPTER 6 CIRCULAR MOTION AND GRAVITATION

We want to place a satellite into circular orbit 300km above the earth surface.

What speed, period and radial acceleration it must have?

Page 23: CHAPTER 6 CIRCULAR MOTION AND GRAVITATION

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


Top Related