Physics 203 College Physics I Fall 2012

1. Physics 203College Physics IFall 2012 S. A. Yost Chapter 7Part 1 + Brief Review for Exam Momentum and Impulse

2. Announcements • The Exam on Chapters 4 – 6 will be Thursday. • Next Tuesday: Read Ch. 7, but you can skip sections 7.7 and 7.9. • A problem set HW07A is already open and due next Tuesday. It covers sections 1 – 3, 8, and 9 in Chapter 7. • A problem set HW07B on sections 4 – 6 (collisions) will be posted soon and due next Thursday.

3. Energy Conservation • When only conservative forces act on a system, energy is conserved. • All of the fundamental forces in the universe are conservative. • That means that if you keep track of the total energy in a closed system, it can never increase or decrease – it can only change form.

4. Energy Conservation • The total energy is the sum of the kinetic energy and the potential energy of an object. • The potential energy is the amount of work it took to put the object in its current position. It is normally written as U. • For example, the potential energy of a book of mass m on top of a cabinet of height h is U = mgh. • The potential energy of a spring compressed or stretched a distance x from equilibrium is U = ½ kx2.

5. Two water slides have the same length, but are shaped different. Who is going faster at the bottom of the slides? A) Paul B) Kathleen C) No difference Water Slide

6. Who gets to the bottom first? A) Paul B) Kathleen C) No difference Water Slide

7. Power • Power is the rate of doing work: P = W/t. • If the force F acts in the direction of motion, then • P = Fv (instantaneous) • These are consistent because x = v t is the distance traveled, so • P = F v = F x/t = W/t. • Horsepower: 1 hp = 746 W.

8. The Exam • Topics on Exam: • Chapter 4: Newton’s Laws • Free body diagrams, F= ma, … • Chapter 5: Circular motion, Universal gravitation. • ac = v2/R, F= ma, Fg = Gm1m2/R, orbits • Chapter 6: kinetic energy, work, W = DK, • potential energy, power → → → →

9. The Effect of a Force over Time → • We say that when a force F acts for time t, a mass acquires momentum • m v = m a t = F t. • If the force is changing, we can use the time-averaged force: • m v= Favg t . • The right-hand side of the equation is called the impulse. → → → → →

10. Impulse and Momentum → → • The momentum can be written asp = mv. • The impulse can be written J = Favgt. • Newton’s second lawimplies that the net impulse equals the change in momentum. • Dp = J → → → →

11. Impulsive Forces • Momentum can be used in any dynamical situation, but is especially useful for impulsive forces, which act over a short time. • The impulse is the area under the curve, geometrically.

12. The impulse is also equal to the average force times the time interval: (same area in blue) J(t) = FavgDt Impulse and Average Force → →

13. Conservation of Momentum • If there is no external force on a system of objects, then its total momentum cannot change, since there is no net impulse. • The total momentum of a system isolated from external forces is conserved.

14. Center of Mass • The center of mass a set of objects is the average position of their mass. For two objects in 1D: • x1 m1 + x2 m2 • m1 + m2 x cm = x1 x2 CM m2 m1 xcm

15. rcm= vcm= Center of Mass • In more dimensions you can use vectors to locate the CM: • m1r1 + m2r2 +m3r3 • m1 + m2 + m3 • It moves with velocity m2 m1 r2 r1 m2 CM r3 • m1v1 + m2v2 +m3v3 • m1 + m2 + m3 P = M vcm

16. Motion of the Center of Mass • If an external force F acts on an extended object or collection of objects of mass M, the acceleration of the CM is given by • F= Macm. • You can apply Newton’s 2nd Law as if it were a particle located at the CM, as far as the collective motion is concerned. • This says nothing about the relative motion, rotation, etc., about the CM. That comes up in chapter 8.

17. Motion of Extended Objects • The motion of extended objects or collections of particles is such that the CM obeys Newton’s 2nd Law.

18. Motion of the Center of Mass • The CM of a wrench sliding on a frictionless table will move in a straight line because there is no external force. In this sense, the wrench may be though of as a particle located at the CM. cm motion

19. Motion of the Center of Mass • For example, if a hammer is thrown, its CM follows a parabolic trajectory under the influence of gravity, as a point object would.

20. Motion of the Center of Mass • For example, if a hammer is thrown, its CM follows a parabolic trajectory under the influence of gravity, as a point object would.

21. Additional Slides • Problems for extra practice on chapter 6 follow.

22. Compressed Spring • When a box is set gently on a spring, it compresses it a distance d. • What would happen if I hold it at the uncompressed position, and then let go?

23. Compressed Spring • The spring will compress – how far? • DU = – mgh – ½ kh2 • = 0 • h = 2mg/k • = 2d.

24. Compressed Spring • Then what happens? • Where does the box attain its maximum speed? • v is maximum where U is a minimum. • This is the equilibrium position.

25. Compressed Spring • What is the maximum speed? • Start at the top: DK +DU = 0 • ½ mv2 – mgd + ½ kd2 = 0 • v2 = 2gd – (k/m)d2 • with k = mg/d. • v2 = 2gd – gd = gd. • v = √gd

26. Example • How long will it take 50 hp motor to pull a 120 kg sled 100 m up a hill, if the coefficient of kinetic friction is m = 0.10, and the elevation increases by 20 m on the way up? Assume a constant slope and speed.

27. Example • First find the work done by the motor. • Assume the net work is zero. • Wm + Wf = DU = mgh • = 2.35 ×104 J. h = 20 m

28. Example m= 0.10 m = 120 kg x = 98 m • Work done by friction? • Wf = –Ffd • Ff = mN, N = mg cosq • Ff = m mgcosq = – mmg (cosq) d = – mmg x = – 1.15 ×104 J → N d = 100 m → Ff h = 20 m θ → mg x =√ 1002 – 202 m θ

29. Example • Wm = mgh – Wf = 2.35 ×104 J – (– 1.15 ×104 J ) • = 3.50 ×104 J. • Time: t = Wm/Pm • Pm = 5.0 hp (746 W/hp) = 3730 W. • t = 9.4 s. h = 20 m