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Chapter 5: Circular Motion Uniform circular motion Radial acceleration Unbanked turns (banked) Circular orbits: Kepler’s laws Non-uniform circular motion.

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Presentation on theme: "Chapter 5: Circular Motion Uniform circular motion Radial acceleration Unbanked turns (banked) Circular orbits: Kepler’s laws Non-uniform circular motion."— Presentation transcript:

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. 1

2 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.

3 3 Radians s = arc length r = radius

4 4 motion tangent to circle

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

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

7 7 Angular Equations of Motion Valid for constant-  only

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

9 Top View Back View ff Centripetal Force

10 Acceleration with Non-Uniform Circular Motion Total acceleration = tangential + centripetal = forward/backward + left/right a(total) = r  (F/B) + v 2 /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

11 11 Centripetal Force required for circular motion F c = ma c = mv 2 /r Example: 1.5kg moves in r = 2m circle v = 8m/s. a c = v 2 /r = 64/2 = 32m/s/s F c = ma c = (1.5kg)(32m/s/s) = 48N

12 12 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%

13 13 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 = mv 2 /r = (1)(2)(2)/(.15) = 26.7N

14 14 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 = mv 2 /r, v 2 = Tr/m v = sqrt{(9.8)(0.15)/(1)} = 1.21m/s

15 15 Kepler’s Laws

16 Kepler’s Laws of Orbits 1.Elliptical orbits 2.Equal areas in equal times (ang. Mom.) 3.Square of year ~ cube of radius

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

18 18 Summary s = r  v = r  a(tangential) = r . a(centripetal) = v 2 /r F(grav) = GMm/r 2 Kepler’s Laws, Energy, Angular Momentum

19 19 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.

20 20 rotational speeds rpm = rev/min frequency “f” = cycles/sec period “T” = sec/cycle = 1/f degrees/sec rad/sec  = 2  f

21 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

22 22 Rolling Motion v = v cm = R 

23 23 Example: Rolling A wheel with radius 0.25m is rolling at 18m/s. What is its rotational rate?

24 24 Example A car wheel angularly accelerates uniformly from 1.5rad/s with rate 3.0rad/s 2 for 5.0s. What is the final angular velocity? What angle is subtended during this time?

25 25 Ex: Changing Units

26 26 Rotational Motion vtvt atat acac acac vtvt r

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


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