Physics : Lecture 09 (Chapter 8 Halliday )

1. Physics : Lecture 09(Chapter 8 Halliday) • Conservative forces & potential energy • Conservation of “total mechanical energy” • Example: pendulum • Non-conservative forces • friction • General work/energy theorem • Example problem

2. Conservative Forces: • In general, if the work done does not depend on the path taken (only depends the initial and final distances between objects), the force involved is said to be conservative. • Gravity is a conservativeforce: • Gravity near the Earth’s surface: • A spring produces a conservative force:

3. Conservative Forces: • We have seen that the work done by a conservative force does not depend on the path taken. W2 W1 = W2 W1 • Therefore the work done in a closed path is 0. W2 WNET = W1 - W2 = W1 - W1 = 0 W1

4. ò W=F.dr = -U U2 r2 r2 ò U = U2 - U1 = -W=-F.dr r1 U1 r1 Potential Energy • For any conservative force F we can define a potential energy function U in the following way: • The work done by a conservative force is equal and opposite to the change in the potential energy function. • This can be written as:

5. Gravitational Potential Energy • We have seen that the work done by gravity near the Earth’s surface when an object of mass m is lifted a distance y is Wg = -mg y • The change in potential energy of this object is therefore:U = -Wg= mg y j m y Wg = -mg y

6. Gravitational Potential Energy • So we see that the change in U near the Earth’s surface is:U = -Wg= mg y= mg(y2 -y1). • So U = mg y + U0 where U0 is an arbitrary constant. • Having an arbitrary constant U0 is equivalent to saying that we can choose they location where U = 0 to be anywhere we want to. y2 j m Wg = -mg y y1

7. Conservative Forces: • We have seen that the work done by gravity does not depend on the path taken. m R2 R1 M m h Wg = -mgh

8. (a) (b) (c) 2 Work & Energy • A rock is dropped from a distance RE above the surface of the earth, and is observed to have kinetic energy K1 when it hits the ground. An identical rock is dropped from twice the height (2RE) above the earth’s surface and has kinetic energy K2 when it hits. RE is the radius of the earth. • What is K2 / K1? 2RE RE RE

9. Where c = GMm is the same for both rocks Solution • Since energy is conserved, DK = WG. 2RE RE RE

10. For the second rock: So: 2RE Solution • For the first rock: RE RE

11. Conservative Forces: • We have seen that the work done by a conservative force does not depend on the path taken. W2 W1 = W2 W1 • Therefore the work done in a closed path is 0. W2 WNET = W1 - W2 = W1 - W1 = 0 W1

12. Conservative Forces • The pictures below show force vectors at different points in space for two forces. Which one is conservative ? (a)1(b)2(c)both y y x (1) x (2)

13. Lecture 11, Act 2Solution • Consider the work done by force when moving along different paths in each case: WA = WB WA > WB (1) (2)

14. W = 0 W = -5 J W = 0 • In fact, you could make money on type (2) if it ever existed: • Work done by this force in a “round trip” is > 0! • Free kinetic energy!! WNET = 10 J = DK W = 15 J Note: NO REAL FORCES OF THIS TYPE EXIST, SO FAR AS WE KNOW

15. Elastic Potential Energy Energy stored in a stretched or compressed spring. Work Done By Spring (Elastic) Force: Fe = -kx Choosing Uo = 0 for x = 0 (unstretched position), we have:

16. S2 U = U2 - U1 = -W=-F.dr S1 Potential Energy Recap: • For any conservative force we can define a potential energy function U such that: • The potential energy function U is always defined onlyup to an additive constant. • You can choose the location where U = 0 to be anywhere convenient.

17. Conservative Forces & Potential Energies Change in P.E U = U2 - U1 Work W(1-2) P.E. function U Force F ^ Fg = -mg j mg(y2-y1) -mg(y2-y1) mgy + C ^ Fg = r Fs = -kx (R is the center-to-center distance, x is the spring stretch)

18. Potential Energy • All springs and masses are identical. (Gravity acts down). • Which of the systems below has the most potential energy stored in its spring(s), relative to the relaxed position? (a)1 (b)2 (c)same (1) (2)

19. Solution • The displacement of (1) from equilibrium will be half of that of (2) (each spring exerts half of the force needed to balance mg) 0 d 2d (1) (2)

20. The potential energy stored in (1) is The potential energy stored in (2) is The spring P.E. is twice as big in (2) ! Solution 0 d 2d (1) (2)

21. Potential Energy Summary Can be defined only in a conservative field of forces. Mathematical definition: 2 ds F 1

22. Potential Energy

23. Potential Energy and Field Force For a conservative force in one dimension: dU = - W = - Fxdx Therefore, the force can be expressed as the negative derivative of the potential energy (- gradient of potential). The field force will act in the direction of the drop of potential energy with distance. Let’s illustrate for a spring-mass system:

24. Potential Energy and Equilibrium Let’s illustrate for a spring and mass system: First lets note that the slope is the derivative of the function. The force being the negative derivative of the potential energy U, is going to be also equal to the negative of the slope. At x = 0, Fx = 0 and the mass is in equilibrium, slope = 0.

25. Stable Equilibrium Curve has a minimum If x>0, slope>0, F<0 If x<0, slope<0, F>0 In either case the force is in the direction that will accelerate the object towards lower potential energy. A slight displaced from the equilibrium position x = 0, gives rise to a force directed back toward x = 0.

26. Unstable Equilibrium Curve has a maximum at x = 0 [example skier (Avram) at the top of the hill] F If x>0, slope<0, F>0 If x<0, slope>0, F<0 Again the force moves the object toward lower potential energy, but away from the equilibrium position.

27. Neutral Equilibrium Near x = 0 the region is flat. Small displacements near this location won’t produce any net force, and hence the particle is still in equilibrium

28. Exercise1 dU/dr = slope <0 Fr>0

29. Ex.1 mv2/2 = 3Uo-Uo = 2Uo ΔK=-ΔU = 0

30. Ex.1 2Uo/ro

31. Ex. 2

32. Ex. 2 Area =( 4Fo + Fo) x ro/2 = 5/2 (Foro) = 2.5 Foro= W ΔK = mv2/2 = 2.5Foro v = √5Foro/m F = k/r2 ΔK = mv2/2 = 2Foro v = √4Foro/m

33. Ex.2 Gravity

34. Conservation of Energy • If only conservative forces are present, the total kinetic plus potential energy of a systemis conserved, i.e. the total “mechanical energy” is conserved. • (note: E=Emechanical throughout this discussion) E = K + U is constant!!! • Both K and U can change, but E = K + U remainsconstant. • But we’ll see that if non-conservative forces act then energy can be dissipated into other modes (thermal,sound) E = K + U E = K + U = W + U = W + (-W) = 0 • using K = W • using U = -W

35. Example: The simple pendulum • Suppose we release a mass m from rest a distance h1 above its lowest possible point. • What is the maximum speed of the mass and wheredoes this happen? • To what height h2 does it rise on the other side? m h1 h2 v

36. Example: The simple pendulum • Kinetic+potential energy is conserved since gravity is a conservative force (E = K + U is constant) • Choose y = 0 at the bottom of the swing, and U = 0 at y = 0 (arbitrary choice)E = 1/2mv2 + mgy y h1 h2 y = 0 v

37. Example: The simple pendulum • E = 1/2mv2 + mgy. • Initially, y = h1and v = 0, so E = mgh1. • Since E = mgh1initially, E = mgh1always since energyisconserved. y y = 0

38. Example: The simple pendulum • 1/2mv2will be maximum at the bottom of the swing. • So at y = 0 1/2mv2 = mgh1v2 = 2gh1 y y = h1 h1 y = 0 v

39. Example: The simple pendulum • Since E = mgh1 =1/2mv2 + mgy it is clear that the maximum height on the other side will be at y = h1 = h2and v = 0. • The ball returns to its original height. y y = h1 = h2 y = 0

40. Example: The simple pendulum • The ball will oscillate back and forth. The limits on its height and speed are a consequence of the sharing of energy between K and U. E = 1/2mv2 + mgy= K + U = constant. y

41. Example: The simple pendulum • We can also solve this by choosing y = 0 to be at the original position of the mass, and U = 0 at y = 0.E = 1/2mv2 + mgy. y y = 0 h1 h2 v

42. Example: The simple pendulum • E = 1/2mv2 + mgy. • Initially, y = 0 and v = 0, so E = 0. • Since E = 0 initially, E = 0 always since energyisconserved. y y = 0

43. Example: The simple pendulum • 1/2mv2will be maximum at the bottom of the swing. • So at y = -h11/2mv2 = mgh1v2 = 2gh1 y Same as before! y = 0 h1 y = -h1 v

44. Example: The simple pendulum • Since 1/2mv2 - mgh = 0it is clear that the maximum height on the other side will be at y = 0 and v = 0. • The ball returns to its original height. y y = 0 Same as before!

45. Example: Airtrack & Glider • A glider of mass M is initially at rest on a horizontal frictionless track. A mass m is attached to it with a massless string hung over a massless pulley as shown. What is the speed v of M after m has fallen a distance d ? v M m d v

46. v M m d Example: Airtrack & Glider • Kinetic+potential energy is conserved since all forces are conservative. • Choose initial configuration to have U=0.K = -U

47. Problem: Hotwheel • A toy car slides on the frictionless track shown below. It starts at rest, drops a distance d, moves horizontally at speed v1, rises a distance h, and ends up moving horizontally with speed v2. • Find v1 and v2. v2 d h v1

48. Problem: Hotwheel... • K+U energy is conserved, so E = 0 K = - U • Moving down a distance d, U = -mgd, K = 1/2mv12 • Solving for the speed: d h v1

49. Problem: Hotwheel... • At the end, we are a distance d - h below our starting point. • U = -mg(d - h), K = 1/2mv22 • Solving for the speed: v2 d - h d h

50. Non-conservative Forces: • If the work done does not depend on the path taken, the force is said to be conservative. • If the work done does depend on the path taken, the force is said to be non-conservative. • An example of a non-conservative force is friction. • When pushing a box across the floor, the amount of work that is done by friction depends on the path taken. • Work done is proportional to the length of the path!