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

3. 1′ Lecture: 1′ Lecture: The net force on a body executing circular motion is equal to the mass times the centripetal acceleration of the body. F centripetal = m a centripetal The net force on a body executing circular motion is equal to the mass times the centripetal acceleration of the body. F centripetal = m a centripetal a centripetal = v 2 / R [toward the center]; a = -ω 2 ra centripetal = v 2 / R [toward the center]; a = -ω 2 r In non-inertial frames of reference one may sense fictitious forces. 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. At terminal velocity the velocity-dependent resistive forces balance the accelerating forces so that no further acceleration occurs. Physics 1710 Chapter 6—Circular Motion

4. Centripetal Acceleration: Physics 1710 Chapter 6—Circular Motion x = r cos θ y = r sin θ v x = (dr/dt) cos θ- r sin θ(d θ/dt) v y = (dr/dt) sin θ+r cos θ(d θ/dt) Let (dr/dt) = 0v x = - r ω sin θ v y = +r ω cos θ a x = dv x /dt= d(- r ω sin θ)/dt = r ω 2 cos θ- r ω sin θ(d ω/dt) a y = dv y /dt= d(r ω cos θ)/dt = -r ω 2 sin θ+ r ω cos θ(d ω/dt) a = - ω 2 r + r (d v tangential /dt) θ vxvxvxvx vyvyvyvy

5. Centripetal Acceleration: Physics 1710 Chapter 6—Circular Motion a centripetal = - v 2 / r a tangential = r (d v tangential /dt) F = m a F centripetal = m a centripetal = - mv 2 / r θ vxvxvxvx vyvyvyvy

6. Demonstration: Ball on a String Physics 1710 Chapter 6—Circular Motion F centripetal rmv F centripetal = m v 2 /r

7. 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? Physics 1710 — e-Quiz Answer Now ! 1.About 12. N 2.About 30. N 3.About 78. N 4.About 380. N

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

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

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

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

12. No Talking! Think! Confer! Peer Instruction Time Physics 1710 Unit 1—Review 1234

13. Satellite Motion: h ~ 100 km, R ⊕ ~ 6300 km h ~ 100 km, R ⊕ ~ 6300 km F g = G Mm/ r 2 GravityF g = G Mm/ r 2 Gravity F g = F r = mv 2 /r F g = F r = mv 2 /r v = √[GM/r] = √[GM/(R ⊕ + h)]~ √[GM/R ⊕ ] = √[gR ⊕ ]= √[(9.8 m/s 2 )(6.3 x10 6 m) ]= 7.9x10 3 m/s ~17,600 mph v = √[GM/r] = √[GM/(R ⊕ + h)]~ √[GM/R ⊕ ] = √[gR ⊕ ]= √[(9.8 m/s 2 )(6.3 x10 6 m) ]= 7.9x10 3 m/s ~17,600 mph Why do the shuttle astronauts appear “weightless?” Why do the shuttle astronauts appear “weightless?” Physics 1710 Chapter 6—Circular Motion

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

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