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2. Circular Motion • Your body cannot sense constant velocity. (Earth, windowless car) But, your body is very good at feeling acceleration(punch gas pedal or brake) • Acceleration is change in velocity. Up to now, that has been on one direction. We will see that one can have constant speed and acceleration at the same time. • A constant speed and changing direction is also acceleration.

3. Period… • Period is the time needed for one complete cycle. • T is the symbol for period

4. … • When something is in a curve, it changes its direction every instant. • When any object is in a curve or revolving,we say it is acceleration (even if speed remains the same) • We call this acceleration CENTRIPETAL ACCELERATION Ac

5. Centripetal Force • Centripetal Force is the force that keeps an object moving in a circular path. Remove the force and the object will NOT move in a circle. • Ball & String • Car & Road • Earth & Gravity

6. When Fc is removed, the net force has been removed and the object travels in a straight line Centripetal is Center Seeking “Trip to center”

7. When you observe objects moving away from the center of curvature, they do so b/c they do NOT have enough Fc

8. a = ΔV/ Δt ac = ΔV/ Δt ac = v2/ r Velocity for something turning V = d/t = circumference/ period V = 2πr/ T Substitute V into our ac equation ac = (2πr/ T)2 r ac = 4π2r / T2

9. Newton’s 2nd law says f= ma • Therefore centripetal force = mac Fc = mac Fc = mV2/ r Fc = m(4π2r / T2)

10. Example Problem p.49 • A 0.013kg rubber stopper is attached to a 0.93m string. The stopper is swung in a horizontal circle, making 1 revolution in 1.18s. • Find the speed for the stopper • Find the stopper’s centripetal acceleration • Find the force the string exerts on the stopper

11. Given m = 0.013kg r = 0.93m T = 1.18sec V = ? ac = ? Fc = ? Formula V = 2πr/ T ac = V2/r Fc = Mac • Solution • 2(3.14)(.93m)/ 1.18 = 5m/s • (5m/s)2/ .93 = 26.9m/s2 • Fc = (.013)(26.9) = .35N inward

12. Class Work 1. Sue is swinging a yo-yo around her head. What happens to the size of the ac if the mass of the yo-yo is doubled without changing the period or length of string? 2. Imagine you are in a car seat tossing a ball straight up in the air. • If the car is moving with constant velocity, will the ball land in front, behind or in your hand? • If the car is decelerating in a straight line , will the ball land in front, behind, or in your hand? • If the car rounds a curve at constant speed, where will the ball land?

13. 3. Geraldo whirls a 20g rubber stopper above his head. The string that holds the stopper is 1.5m long. The rate of spin is 3 revolutions every second. • Find ac of the stopper • Find linear velocity • Find tension in string (Fc) 4. A 100kg pilot is in an airplane doing a loop de loop. The jet has a velocity of 88m/s. The radius of the loop is 200m • What is the pilot’s weight? • what is the pilots ac? • What is the pilots Fc? • compare the pilots Fc to his weight

14. Class work Solutions • Nothing, ac is independent of mass • A= In your Hand B= In front of your hand C= To the side of your hand. The outside of the curve

15. Given m = .02kg r= 1.5m 3 revs/sec T= 1/3 = .33sec #3 Formula (A) ac = 4π2r/ T2 (B)V = 2πr/ T (C) Fc = mac Solution (B) 2(3.14)1.5 .33 V = 28.5m/s Solution (C) .02 (543.3) =10.9N Solution (A) 39.44(1.5m) (.33)2 ac = 543. 3m/s2

16. #4 Given m = 100kg V = 88m/s r = 200m Formula W = mg ac = V2/r Fc = Mac Solution (B) (88)2/ 200 ac = 38.7m/s2 Solution(C) 100(38.7) =3872N Solution (A) 100(9.8) = 980 N compare: 3872/ 980 = 3.95 Fc is 3.95x greater – aka a 4g turn

17. ?? • An early objection to the idea that Earth is spinning on its axis was that Earth would turn so fast at the equator that people would be thrown off into space. Show the error in this logic by calculating… • The speed of a 97kg person at the equator, the radius of Earth is 6400Km • The centripetal Force on the person • The weight of the person

18. Formula • V = 2πr/T • Fc = m(v2/r) • W = mg Given r = 6,400,000m M = 97kg Solution (A) 2(3.14)(6,400,000) 24hr(3600s) V= 465 m/s Solution (B) Fc = 97(4652/ 6,400,000) Fc = 3.28N

19. Solution (C) 97kg(9.8m/s2) = 950.6 N 3.3/ 950.6 = Fc is only .3% of the person’s weight -too small to worry about

20. Simple Harmonic Motion • A playground swing, vibrating spring, guitar string & pendulum, are all examples of simple harmonic motion. • SHM occurs if the restoring force varies linearly with the displacement.

21. Uniform Circular Motion: Period Object repeatedly finds itself back where it started. The time it takes to travel one “cycle” is the “period”.

26. Tension Can Yield a Centripetal Acceleration: If the person doubles the speed of the airplane, what happens to the tension in the cable? Doubling the speed, quadruples the force (i.e. tension) required to keep the plane in uniform circular motion.

27. Centripetal Force: Question A car travels at a constant speed around two curves. Where is the car most likely to skid? Why? Smaller radius: larger force required to keep it in uniform circular motion.

28. Q: Why exit ramps in highways are banked?

30. Vertical Circular Motion