Chapter 5 Circular Motion Uniform circular motion Radial

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

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

Radians s = arc length r = radius 3

motion tangent to circle 4

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

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

Angular Equations of Motion Valid for constant-a only 7

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 = 6 m/s, r = 4 m. a(centripetal) = 6^2/4 = 9 m/s/s 8

Centripetal Force f f Top View Back View

Acceleration with Non-Uniform Circular Motion • Total acceleration = tangential + centripetal • = forward/backward + left/right • a(total) = ra (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

Centripetal Force • required for circular motion • Fc = mac = mv 2/r • • Example: 1. 5 kg moves in r = 2 m circle v = 8 m/s. ac = v 2/r = 64/2 = 32 m/s/s Fc = mac = (1. 5 kg)(32 m/s/s) = 48 N 11

Rounding a Corner • How much horizontal force is required for a 2000 kg car to round a corner, radius = 100 m, at a speed of 25 m/s? • Answer: F = mv 2/r = (2000)(25)/(100) = 12, 500 N • What percent is this force of the weight of the car? • Answer: % = 12, 500/19, 600 = 64% 12

Mass on Spring 1 • A 1 kg mass attached to spring does r = 0. 15 m circular motion at a speed of 2 m/s. What is the tension in the spring? • Answer: T = mv 2/r = (1)(2)(2)/(. 15) = 26. 7 N 13

Mass on Spring 2 • A 1 kg mass attached to spring does r = 0. 15 m circular motion with a tension in the spring equal to 9. 8 N. What is the speed of the mass? • Answer: T = mv 2/r, v 2 = Tr/m • v = sqrt{(9. 8)(0. 15)/(1)} = 1. 21 m/s 14

Kepler’s Laws 15

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

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

Summary • • • s = rq v = rw a(tangential) = ra. a(centripetal) = v 2/r F(grav) = GMm/r 2 Kepler’s Laws, Energy, Angular Momentum 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 = 2 pf 20

7 -43 • • Merry go round: 24 rev in 3. 0 min. W-avg: 0. 83 rad/s V = rw = (4 m)(0. 83 rad/s) = 3. 3 m/s V = rw = (5 m)(0. 83 rad/s) = 4. 2 m/s

Rolling Motion v = vcm = Rw 22

Example: Rolling A wheel with radius 0. 25 m is rolling at 18 m/s. What is its rotational rate? 23

Example A car wheel angularly accelerates uniformly from 1. 5 rad/s with rate 3. 0 rad/s 2 for 5. 0 s. What is the final angular velocity? What angle is subtended during this time? 24

Ex: Changing Units 25

Rotational Motion vt at ac r vt ac 26

Convert 50 rpm into rad/s. • (50 rev/min)(6. 28 rad/rev)(1 min/60 s) • 5. 23 rad/s