2. CIRCULAR MOTION AND GRAVITATION
CHAPTER 6
CIRCULAR MOTION AND GRAVITATION

3. 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.

4. 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

6. Period = the time for one revolution
Circular motion in horizontal plane:
- flat curve
- banked curve
- rotating object
2) Circular motion in vertical plane

7. Rounding a flat curve
The centripetal force coming only from tire friction.

8. Rounding a banked curve
The centripetal force comes from friction and a component of force from the car’s mass

9. Dynamics of a Ferris Wheel

10. 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.

11. GRAVITATION

12. Spherically symmetric objects interact gravitationally as though all the mass of each were concentrated at its center

13. 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.

14. 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.

15. A diagram of gravitational force
G = 6.674x10-11 N.m2/kg2

16. Each mass is 2 kg Find the magnitude of the net gravitational force on each mass and its direction

17. Each mass in the figure below is 3 kg
Each mass in the figure below is 3 kg. Find the force (magnitude and direction) on each mass in the figure .

18. WEIGHT

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

20. 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?

21. Satellite Motion

22. What happens when velocity rises?
When v is large enough, you achieve escape velocity.

23. v = GmE/r T= 2πr/v = (2πr3/2)/
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)/
Larger orbits correspond to slower speeds and longer periods.

24. r 2 = (6.67 x 10 -11 N.m2/kg2) (5.98 x 10 24 kg) (320 kg ) / 800 N
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 N.m2/kg2) (5.98 x kg) (320 kg ) / 800 N
r 2 = 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 – x 107 = x 107 m

25. We want to place a satellite into circular orbit 300km above the earth surface.
What speed, period and radial acceleration it must have?