Physics 1710 Chapter 6—Circular Motion Answer: a = ∆v/∆t a = (112 m/s)/(2.5 sec) = 44.8 m/s2 Thrust = force F = ma = (1800. kg)(44.8 m/s 2 ) ~ 81 kN

Physics 1710 Chapter 6—Circular Motion 1′ Lecture: The net force on a body executing circular motion is equal to the mass times the centripetal acceleration of the body. Fcentripetal = m acentripetal acentripetal = v 2/ R [toward the center]; a = -ω2 r In non-inertial frames of reference one may sense fictitious forces. At terminal velocity the velocity-dependent resistive forces balance the accelerating forces so that no further acceleration occurs.

Physics 1710 Chapter 6—Circular Motion Centripetal Acceleration: vx vy x = r cos θ y = r sin θ vx = (dr/dt) cos θ- r sin θ(d θ/dt) vy = (dr/dt) sin θ+r cos θ(d θ/dt) Let (dr/dt) = 0 vx = - r ω sin θ vy = +r ω cos θ ax = dvx /dt= d(- r ω sin θ)/dt = r ω2cos θ- r ω sin θ(d ω/dt) ay = dvy /dt= d(r ω cos θ)/dt = -r ω2sin θ+ r ω cos θ(d ω/dt) a = - ω2 r + r (d vtangential /dt) θ

Physics 1710 Chapter 6—Circular Motion Centripetal Acceleration: vx vy θ a centripetal = - v 2 / r a tangential = r (d vtangential /dt) F = m a F centripetal = m a centripetal = - mv 2 / r

Physics 1710 Chapter 6—Circular Motion Demonstration: Ball on a String m v r Fcentripetal Fcentripetal = m v 2/r

Physics 1710 — e-Quiz An athlete swings a 8 kg “hammer” around on a 1.2 m long cable at an angular frequency of 1.0 revolution per second. How much force must he exert on the hammer throw handle? Answer Now ! About 12. N About 30. N About 78. N About 380. N 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

Physics 1710 Chapter 6—Circular Motion Demonstration: Ball on a String m v r Fcentripetal v = 2π r/T = 2(3.14) (1.2 m)/(1.0 sec) = 7.5 m/s Fcentripetal = m v 2/r = (8.0 kg)(7.5 m/s)2/(1.2 m) = 378 N ~ 380 N

Physics 1710 Chapter 6—Circular Motion Demonstration: Marble in a bottle Why does the marble stay up on the side?

Physics 1710 Chapter 6—Circular Motion Why does the marble stay up on the side? Think! Confer! No Talking! Peer Instruction Time

Physics 1710 Chapter 6—Circular Motion Satellites in Orbit Which orbit is most like that of the Space Shuttle?

Physics 1710 Unit 1—Review 1 2 3 4 Think! Confer! No Talking! 1 2 3 4 Think! Confer! No Talking! Peer Instruction Time

Physics 1710 Chapter 6—Circular Motion Satellite Motion: h ~ 100 km , R⊕ ~ 6300 km Fg= G Mm/ r 2 Gravity Fg = Fr = mv 2/r v = √[GM/r] = √[GM/(R⊕ + h)]~ √[GM/R⊕ ] = √[gR⊕ ]= √[(9.8 m/s2)(6.3 x106m) ]= 7.9x103 m/s ~17,600 mph Why do the shuttle astronauts appear “weightless?”

Physics 1710 Chapter 6—Circular Motion Satellites in Orbit Orbiting satellites are in free fall but miss the earth because it curves.

Physics 1710 Chapter 6—Circular Motion Summary The net force on a body executing circular motion is equal to the mass times the centripetal acceleration of the body. acentripedal = v 2/ R [toward the center] The “centrifugal” force is a fictitious force due to a non-inertial frame of reference.