Physics 1710 Chapter 6—Circular Motion

1. 0 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

2. 0 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 = -ω2r 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.

3. vy vx 0 Physics 1710 Chapter 6—Circular Motion x = r cos θy= 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 θvy = +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 cosθ(d ω/dt)a = - ω2r + r (d vtangential /dt) θ Centripetal Acceleration:

4. vy vx 0 Physics 1710 Chapter 6—Circular Motion θ a centripetal= - v 2 / r atangential= r (d vtangential/dt) F = m a F centripetal = m a centripetal= - mv 2 / r Centripetal Acceleration:

5. m Fcentripetal v 0 Physics 1710 Chapter 6—Circular Motion r Demonstration: Ball on a String Fcentripetal = m v 2/r

6. Physics 1710 — e-Quiz Answer Now ! 0 0 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? • About 12. N • About 30. N • About 78. N • About 380. N

7. m Fcentripetal v 0 Physics 1710 Chapter 6—Circular Motion r Demonstration: Ball on a String 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

8. 0 Physics 1710 Chapter 6—Circular Motion Demonstration: Marble in a bottle Why does the marble stay up on the side?

9. No Talking! Think! Confer! 0 Physics 1710 Chapter 6—Circular Motion Why does the marble stay up on the side? Peer Instruction Time

10. 0 Physics 1710 Chapter 6—Circular Motion Satellites in Orbit Which orbit is most like that of the Space Shuttle?

11. 4 3 2 1 No Talking! Think! Confer! 0 Physics 1710 Unit 1—Review Peer Instruction Time

12. 0 Physics 1710 Chapter 6—Circular Motion Satellite Motion: h ~ 100 km , R⊕ ~ 6300 km Fg= G Mm/ r 2Gravity 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?”

13. 0 Physics 1710 Chapter 6—Circular Motion Satellites in Orbit Orbiting satellites are in free fall but miss the earth because it curves.

14. 0 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.