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Physics 111: Lecture 11 Today’s Agenda

Physics 111: Lecture 11 Today’s Agenda. Review Work done by variable force in 3-D Newton’s gravitational force Conservative forces & potential energy Conservation of “total mechanical energy” Example: pendulum Non-conservative forces friction General work/energy theorem Example problem.

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Physics 111: Lecture 11 Today’s Agenda

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  1. Physics 111: Lecture 11Today’s Agenda • Review • Work done by variable force in 3-D • Newton’s gravitational force • Conservative forces & potential energy • Conservation of “total mechanical energy” • Example: pendulum • Non-conservative forces • friction • General work/energy theorem • Example problem

  2. F r Work by variable force in 3-D: • Work dWF of a force F acting • through an infinitesimal • displacement r is: • dW = F.r • The work of a big displacement through a variable force will be the integral of a set of infinitesimal displacements: • WTOT = F.r ò

  3. m Work by variable force in 3-D:Newton’s Gravitational Force • Work dWg done on an object by gravity in a displacement drisgiven by: • dWg = Fg.dr= (-GMm / R2r).(dR r + Rd)dWg = (-GMm / R2)dR (sincer.= 0, r.r = 1) ^ ^ ^ ^ ^ ^ ^ ^ dR  ^ r dr Rd Fg d R M

  4. R2 R2 R1 R1 m Work by variable force in 3-D:Newton’s Gravitational Force • Integrate dWg to find the total work done by gravity in a “big”displacement: • Wg = dWg= (-GMm / R2)dR = GMm (1/R2 - 1/R1) Fg(R2) R2 Fg(R1) R1 M

  5. Work by variable force in 3-D:Newton’s Gravitational Force • Work done depends only on R1 and R2, not on the path taken. m R2 R1 M

  6. (a) (b) (c) 2 Lecture 11, Act 1Work & 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

  7. Where c = GMm is the same for both rocks Lecture 11, Act 1Solution • Since energy is conserved, DK = WG. 2RE RE RE

  8. For the second rock: So: 2RE Lecture 11, Act 1Solution • For the first rock: RE RE

  9. Newton’s Gravitational ForceNear the Earth’s Surface: • Suppose R1 = REand R2 = RE + ybut we have learned thatSo: Wg = -mgy m RE+ y RE M

  10. 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

  11. Conservative Forces: • In general, if the work done does not depend on the path taken, the force involved is said to be conservative. • Gravity is a conservativeforce: • Gravity near the Earth’s surface: • A spring produces a conservative force:

  12. 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 = 0 W1 • The work done can be “reclaimed.”

  13. Lecture 11, Act 2Conservative 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)

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

  15. W = 0 W = -5 J W = 0 Lecture 11, Act 2 • 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

  16. 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:

  17. 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

  18. 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

  19. 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.

  20. Conservative Forces & Potential Energies (stuff you should know): 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

  21. Lecture 11, Act 3Potential 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)

  22. Lecture 11, Act 3Solution • 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)

  23. The potential energy stored in (1) is The potential energy stored in (2) is The spring P.E. is twice as big in (2) ! Lecture 11, Act 3Solution 0 d 2d (1) (2)

  24. Conservation of Energy • If only conservative forces are present, the total kinetic plus potential energy of a systemis conserved.E = K + U is constant!!! • Both K and U can change, but E = K + U remainsconstant. E = K + U E = K + U = W + U = W + (-W) = 0 using K = W using U = -W

  25. 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

  26. 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

  27. 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

  28. 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

  29. 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

  30. Example: The simple pendulum Bowling • 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

  31. 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

  32. 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

  33. 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

  34. Example: The simple pendulum Galileo’s 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!

  35. 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

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

  37. 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

  38. 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

  39. 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

  40. 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!

  41. Non-conservative Forces: Friction • Suppose you are pushing a box across a flat floor. The mass of the box is m and the coefficient of kinetic friction is k. • The work done in pushing it a distance D is given by:Wf = Ff• D = -kmgD. Ff = -kmg D

  42. Non-conservative Forces: Friction • Since the force is constant in magnitude and opposite in direction to the displacement, the work done in pushing the box through an arbitrary path of length L is justWf = -mgL. • Clearly, the work done depends on the path taken. • Wpath 2 > Wpath 1 B path 1 path 2 A

  43. Generalized Work/Energy Theorem: • Suppose FNET = FC + FNC (sum of conservative and non-conservative forces). • The total work done is: WNET = WC + WNC • The Work/Kinetic Energy theorem says that: WNET = K. • WNET = WC + WNC = K • WNC = K - WC • But WC = -U So WNC = K + U = E

  44. Generalized Work/Energy Theorem: • The change in kinetic+potential energy of a system is equal to the work done on it by non-conservative forces. E=K+U of system not conserved! • If all the forces are conservative, we know that K+U energy is conserved: K + U = E = 0 which says that WNC = 0, which makes sense. • If some non-conservative force (like friction) does work,K+Uenergy will not be conserved and WNC= E, which also makes sense. WNC = K + U = E

  45. Problem: Block Sliding with Friction • A block slides down a frictionless ramp. Suppose the horizontal (bottom) portion of the track is rough, such that the coefficient of kinetic friction between the block and the track is k. • How far, x, does the block go along the bottom portion of the track before stopping? d  k x

  46. Problem: Block Sliding with Friction... • Using WNC = K + U • As before, U = -mgd • WNC = work done by friction = -kmgx. • K = 0 since the block starts out and ends up at rest. • WNC = U -kmgx= -mgd x = d / k d k x

  47. Recap of today’s lecture • Work done by variable force in 3-D (Text: 6-1) • Newton’s gravitational force (Text: 11-2) • Conservative Forces & Potential energy (Text: 6-4) • Conservation of “Total Mechanical Energy” (Text: 7-1) • Examples: pendulum, airtrack, Hotwheel car • Non-conservative forces (Text: 6-4) • friction • General work/energy theorem (Text: 7-2) • Example problem • Look at Textbook problems Chapter 7: # 9, 21, 27, 51, 77

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