1. Physics 203 – College Physics I Department of Physics – The Citadel Physics 203 College Physics I Fall 2012 S. A. Yost Chapter 5 Circular Motion, Universal Gravitation

2. Physics 203 – College Physics I Department of Physics – The Citadel Announcements Chapter 4’s homework, HW04, was due today. Chapter 5’s homework, HW05, is due next Tuesday. Read Chapter 5, except the last section “Nonuniform Circular Motion” for Thursday. Today we will do an inclined plane problem in class, then look at circular motion.

3. Physics 203 – College Physics I Department of Physics – The Citadel Quiz: Question 1 Which of the following is the frictional force on a sliding object always proportional to? (a) the contact surface area (b) its weight (c) the normal force (d) the contact pressure

4. Physics 203 – College Physics I Department of Physics – The Citadel Quiz: Question 2 For a given object on a surface, which is larger? (a) the coefficient of kinetic friction (b) the coefficient of static friction (c) Neither, they are equal. (d) Either could be larger, depending on the object.

5. Physics 203 – College Physics I Department of Physics – The Citadel Quiz: Question 3 A rocket of mass m is near earth, accelerating into space, as its engines produce a thrust of F th. The rocket is propelled upward by (a) the force of the rocket on the earth. (b) the force of the earth on the rocket. (c) the force of the rocket on its exhaust. (d) the force of the exhaust on the rocket. (e) the force of the rocket exhaust on empty space. a

6. Physics 203 – College Physics I Department of Physics – The Citadel Quiz: Question 4 A rocket of mass m is near earth, accelerating into space, as its engines produce a thrust of F th. The force of the rocket on the earth at this time is (a) F th downward (b) F th upward (c) mg downward (d) mg upward (e) zero a

7. Physics 203 – College Physics I Department of Physics – The Citadel Inclined Plane A block slides down the inclined plane shown, starting at the top. How long does it take to reach the bottom? 5 m 3 m 4 m

8. Physics 203 – College Physics I Department of Physics – The Citadel Inclined Plane F net = ma F net = mg sin  = 3/5 mg a = 3/5 g = 5.88 m/s 2 Time to reach bottom… x = ½ a t 2 5 m = (2.94 m/s 2 ) t 2 t = 1.3 s. N F net  y x sin  = 3/5 mgmg  a = g/5 = 1.96 m/s 2

9. Physics 203 – College Physics I Department of Physics – The Citadel Uniform Circular Motion In uniform circular motion, and object goes around the circumference at constant speed. Which way does the velocity vector point? It is always tangential to the path. R → v →

10. Physics 203 – College Physics I Department of Physics – The Citadel Uniform Circular Motion The circle has radius R. If the period is T, what is the magnitude of the velocity vector? It is the length the radius vector traces out divided by the period: v = 2  R/T. R → v →

11. Physics 203 – College Physics I Department of Physics – The Citadel Uniform Circular Motion What is the direction of the acceleration vector? The speed is constant, so the component of acceleration in the direction of motion must be zero. a is perpendicular to v. a points inward, since v turns that way. R → v → a → → a is called the centripetal acceleration. →

12. Physics 203 – College Physics I Department of Physics – The Citadel Uniform Circular Motion What is the magnitude of the acceleration? Each revolution, the velocity vector traces out a circle of radius 2  v. Then a = 2  v/T. Recall that T = 2  R/v. Then a = v 2 /R. R → v → a → v →

13. Physics 203 – College Physics I Department of Physics – The Citadel Force in Uniform Circular Motion From the fact that there is acceleration directed toward the center of a circle in uniform circular motion, we can infer that the net force on the object is F = ma, also directed toward the center of the circle. Note! This is not a new force due to the circular motion. It is the result of all of the forces acting on the object together. →

14. Physics 203 – College Physics I Department of Physics – The Citadel Circular Motion If you twirl an object around your head on a string and then let go, which way does it travel? view from above (a) (b) (c)

15. Physics 203 – College Physics I Department of Physics – The Citadel Roller Coaster A roller coaster car goes over a hill with radius of curvature R at speed v. How fast must it go for the riders to feel weightless at the top? R v Note: it is ok that the motion isn’t uniform – it’s the instantaneous speed and radius that matter.

16. Physics 203 – College Physics I Department of Physics – The Citadel Roller Coaster Feeling “weightless” means there is no contact force between the rider and car: F N = 0 Gravity must be responsible for the entire centripetal acceleration: mg = mv 2 /R g = v 2 /R. v = √ g/R mgmg FNFN → →

17. Physics 203 – College Physics I Department of Physics – The Citadel Conical Pendulum A 120 g mass hanging from a string of length 1.50 m swings in a circle of radius 24 cm. a) What is the direction of the net force on the mass? Toward the center. b) Find the tension in the string and the period of revolution. L = 1.50 m R = 0.24 m F net →

18. Physics 203 – College Physics I Department of Physics – The Citadel Conical Pendulum The net force is F = mv 2 /R. What forces are responsible for the net force? F net = T + mg. Use F = ma, separated into horizontal and vertical components. L = 1.50 m R = 0.24 m F net → → → → → FTFT → mgmg →

19. Physics 203 – College Physics I Department of Physics – The Citadel Conical Pendulum x: F Tx = F T sin  = ma = mv 2 /R y: F Ty = F T cos  = mg F T = mg / cos  = mg L/H H = √ L 2 – R 2 = 1.48 m F T = 1.16 N mg tan  = mv 2 / R v = √ Rg tan  = R√g/H = 0.618 m/s T = 2  R/v = 2.4 s L = 1.50 m R = 0.24 m FTFT → mgmg →  H = 1.48 m

20. Physics 203 – College Physics I Department of Physics – The Citadel Vertical Circular Motion A mass is rotated at constant speed in a vertical circle on the end of the rod. Which vector(s) could correctly show the force of the rod on the mass? A B C D

21. Physics 203 – College Physics I Department of Physics – The Citadel Vertical Circular Problem Suppose the mass is 1.5 kg and the rod is 35 cm long. If the rod is tilted at  = 30 o with respect to the horizontal, and it is spinning around at 200 rpm, what is the force of the rod at this instant? FrFr FcFc FgFg 

22. Physics 203 – College Physics I Department of Physics – The Citadel Vertical Circular Problem F c = mv 2 /R = m(2  R/T) 2 /R = mR(2  /T) 2 T = 1/(200 min -1 ) = 0.005 min = 0.30 s. F c = 230 N. F cx = -F c cos  = -199 N = F rx F cy = -F c sin  = -133 N = F ry – mg = F ry – 15 N F ry = -118 N. FrFr FcFc FgFg  F r = 168 N,  r = 211 o.