Circular Motion and Kepler's Laws - Understanding Uniform and Non-Uniform Circular Motion

1. Chapter 5: Circular Motion • Uniform circular motion • Radial acceleration • Unbanked turns (banked) • Circular orbits: Kepler’s laws • Non-uniform circular motion • Tangential & Angular acceleration • (apparent weight, artificial gravity) • Hk: CQ 1, 2. Prob: 5, 11, 15, 19, 39, 49.

2. angular measurement degrees (arbitrary numbering system, e.g. some systems use 400) radians (ratio of distances) e.g. distance traveled by object is product of angle and radius. 2

3. Radians r s r s = arc length r = radius 3

4. motion tangent to circle 4

5. Angular Motion radian/second (radian/second)/second 5

6. angular conversions Convert 30° to radians: Convert 15 rpm to radians/s 6

7. Angular Equations of Motion Valid for constant-a only 7

8. Centripetal Acceleration • Turning is an acceleration toward center of turn-radius and is called Centripetal Acceleration • Centripetal is left/right direction • a(centripetal) = v2/r • (v = speed, r = radius of turn) • Ex. V = 6m/s, r = 4m. a(centripetal) = 6^2/4 = 9 m/s/s

9. Planar Acceleration • Total acceleration = tangential + centripetal • = forward/backward + left/right • a(total) = ra (F/B) + v2/r (L/R) • Ex. Accelerating out of a turn; 4.0 m/s/s (F) + 3.0 m/s/s (L) • a(total) = 5.0 m/s/s

10. Centripetal Force • required for circular motion • Fc = mac = mv2/r • Example: • 1.5kg moves in r = 2m circle v = 8m/s. • ac = v2/r = 64/2 = 32m/s/s • Fc = mac = (1.5kg)(32m/s/s) = 48N

11. Rounding a Corner • How much horizontal force is required for a 2000kg car to round a corner, radius = 100m, at a speed of 25m/s? • Answer: F = mv2/r = (2000)(25)(25)/(100) = 12,500N • What percent is this force of the weight of the car? • Answer: % = 12,500/19,600 = 64%

12. Mass on Spring 1 • A 1kg mass attached to spring does r = 0.15m circular motion at a speed of 2m/s. What is the tension in the spring? • Answer: T = mv2/r = (1)(2)(2)/(.15) = 26.7N

13. Mass on Spring 2 • A 1kg mass attached to spring does r = 0.15m circular motion with a tension in the spring equal to 9.8N. What is the speed of the mass? • Answer: T = mv2/r, v2 = Tr/m • v = sqrt{(9.8)(0.15)/(1)} = 1.21m/s

14. Kepler’s Laws

15. Kepler’s Laws of Orbits • Elliptical orbits • Equal areas in equal times (ang. Mom.) • Square of year ~ cube of radius

16. Elliptical Orbits • One side slowing, one side speeding • Conservation of Mech. Energy • ellipse shape • simulated orbits

17. s = rq v = rw a(tangential) = ra. a(centripetal) = v2/r F(grav) = GMm/r2 Kepler’s Laws, Energy, Angular Momentum Summary

18. Centrifugal Force • The “apparent” force on an object, due to a net force, which is opposite in direction to the net force. • Ex. A moving car makes a sudden turn to the left. You feel forced to the right of the car. • Similarly, if a car accelerates forward, you feel pressed backward into the seat.

19. rotational speeds • rpm = rev/min • frequency “f” = cycles/sec • period “T” = sec/cycle = 1/f • degrees/sec • rad/sec w = 2pf

20. 7-43 • Merry go round: 24 rev in 3.0min. • W-avg: 0.83 rad/s • V = rw = (4m)(0.83rad/s) = 3.3m/s • V = rw = (5m)(0.83rad/s) = 4.2m/s

21. Rolling Motion v = vcm = Rw

22. Example: Rolling A wheel with radius 0.25m is rolling at 18m/s. What is its rotational rate?

23. Example A car wheel angularly accelerates uniformly from 1.5rad/s with rate 3.0rad/s2 for 5.0s. What is the final angular velocity? What angle is subtended during this time?

24. Ex: Changing Units

25. vt at ac vt ac Rotational Motion r

26. Convert 50 rpm into rad/s. • (50rev/min)(6.28rad/rev)(1min/60s) • 5.23rad/s