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Physics 111: Lecture 12 Today's Agenda

Physics 111: Lecture 12 Today's Agenda. Problems using work/energy theorem Spring shot Escape velocity Loop the loop Vertical springs Definition of Power, with example. Problem: Spring Shot.

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Physics 111: Lecture 12 Today's Agenda

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  1. Physics 111: Lecture 12Today's Agenda • Problems using work/energy theorem • Spring shot • Escape velocity • Loop the loop • Vertical springs • Definition of Power, with example

  2. Problem: Spring Shot • A sling shot is made from a pair of springs each having spring constant k. The initial length of each spring is x0. A puck of mass m is placed at the point connecting the two springs and pulled back so that the length of each spring is x1. The puck is released. What is its speed v after leaving the springs? (The relaxed length of each spring is xr). xr x1 x0 m m m v

  3. Problem: Spring Shot • Only conservative forces are at work, so K+U energy is conserved. EI = EFK = -Us x1 x0 m m

  4. Problem: Spring Shot • Only conservative forces are at work, so K+U energy is conserved. EI = EFK = -Us m m at rest v

  5. m Problem: Spring Shot Spring Shot • Only conservative forces are at work, so K+U energy is conserved. EI = EFK = -Us m v

  6. Problem: How High? • A projectile of mass m is launched straight up from the surface of the earth with initial speedv0. What is the maximum distance from the center of the earth RMAX it reaches before falling back down. RMAX m RE v0 M

  7. All forces are conservative: WNC = 0 K = -U And we know: Problem: How High... RMAX m RE v0 hMAX M

  8. Problem: How High... RMAX m RE v0 hMAX M

  9. Escape Velocity • If we want the projectile to escape to infinity we need to make the denominator in the above equation zero: We call this value of v0 the escape velocity,vesc

  10. gp(m/s2) vesc(m/s) Rp(m) Mp(kg) Earth 6.378x106 5.976x1024 9.81 11.2x103 Moon 1.737x106 7.349x1022 1.62 2.38x103 Jupiter 7.149x107 1.900x1027 24.8 59.5x103 Sun 6.950x108 1.989x1030 275 618.x103 Escape Velocity • Remembering that we find the escape velocity froma planet of mass Mp and radius Rpto be:(where G = 6.67 x 10-11 m3 kg-1 s-2).

  11. Lecture 12, Act 1Escape Velocity • Two identical spaceships are awaiting launch on two planets with the same mass. Planet 1 is stationary, while Planet 2 is rotating with an angular velocity . • Which spaceship needs more fuel to escape to infinity? (a) 1 (b)2 (c)same •  (1) (2)

  12. Lecture 12, Act 1Solution • Both spaceships require the same escape velocity to reach infinity. • Thus, they require the same kinetic energy. • Both initially have the same potential energy. • Spaceship 2 already has some kinetic energy due to its rotational motion, so it requires less work (i.e. less fuel).

  13. r1 r2 K1 = m(r1)2 K2 = m(r2)2 > Lecture 12, Act 1Aside v = r • This is one of the reasons why all of the world’s spaceports are located as close to the equator as possible. r2 > r1

  14. d k Problem: Space Spring • A low budget space program decides to launch a 10,000 kg spaceship into space using a big spring. If the spaceship is to reach a height RE above the surface of the Earth, what distance d must the launching spring be compressed if it has a spring constant of 108 N/m.

  15. Problem: Space Spring... • Since gravity is a conservative force, energy is conserved. Since K = 0 both initially and at the maximum height (v = 0) we know: • Ubefore = Uafter • (US + UG )before = (UG )after

  16. d k Problem: Space Spring So we find • For the numbers given, d = 79.1 m • But don’t get too happy... • F = kd = ma • a = kd/m • a = 79.1 x 106 m/s2 • a = 791000 g • unhappy astronaut! a

  17. Problem: Loop the loop • A mass m starts at rest on a frictionless track a distance H above the floor. It slides down to the level of the floor where it encounters a loop of radius R. What is H if the mass just barely makes it around the loop without losing contact with the track. H R

  18. N v mg = v Rg Problem: Loop the loop • Draw a FBD of the mass at the top of the loop: • FTOT = -(mg+N) j • ma = -mv2/R j • If it “just” makes it, N = 0. • mg = mv2/R v j H R i

  19. 5 mg(H-2R) = 1/2 mRg = H R 2 Loop the Loop Problem: Loop the loop • Now notice that K+U energy is conserved. K = -U. • U = -mg(h) = -mg(H-2R) • K = 1/2 mv2 = 1/2 mRg h = H - 2R v H R

  20. Lecture 12, Act 2Energy Conservation • A mass starts at rest on a frictionless track a distance H above the floor. It slides down to the level of the floor where it encounters a loop of radius R. What is H if the normal force on the block by the track at the top of the loop is equal to the weight of the block ? (a)3R(b)3.5R (c)4R H R

  21. N v mg Lecture 12, Act 2Solution • Draw a FBD of the mass at the top of the loop: • FNET = -(mg+N) j • ma = -mv2/R j • In this case, N = mg. • 2mg = mv2/R v j H R i

  22. Lecture 12, Act 2Solution • Use the fact that K+U energy is conserved: DK = -DU. • DU = -mg(h) = -mg(H - 2R), DK = 1/2 mv2 = mRg • mg(H - 2R) = mRg h = H - 2R v H R

  23. m mg = -kye (ok since yeis a negative number) Vertical Springs (b) (a) • A spring is hung vertically. Its relaxed position is at y = 0(a). When a mass m is hung from its end, the new equilibrium position is ye(b). j k • Recall that the force of a spring is Fs = -kx. In case (b) Fs = mg and x = ye:-kye - mg = 0 (ye < 0) y = 0 y = ye -kye mg

  24. but mg = -kye Vertical Springs (b) (a) • The potential energy of the spring-mass system is: j k y = 0 y = ye choose C to make U=0 at y = ye: m

  25. Vertical Springs (b) (a) • So: j k y = 0 which can be written: y = ye m

  26. Vertical Springs (b) (a) • So if we define a new y coordinate system such that y = 0 is at the equilibrium position, ( y = y - ye ) then we get the simple result: j k y = 0 m 

  27. Vertical Springs (b) (a) • If we choose y = 0 to be at the equilibrium position of the mass hanging on the spring, we can define the potential in the simple form. • Notice that g does not appear in this expression!! • By choosing our coordinates and constants cleverly, we can hide the effects of gravity. j k y = 0 m

  28. 160 140 Uof Spring 120 U 100 US = 1/2ky2 80 60 40 20 0 -10 -8 -6 -4 -2 0 2 4 6 8 10 -20 y -40 -60

  29. 160 140 Uof Gravity 120 U 100 80 60 UG = mgy 40 20 0 -10 -8 -6 -4 -2 0 2 4 6 8 10 -20 y -40 -60

  30. 160 UNET = UG + US 140 Uof Spring + Gravity 120 U 100 US = 1/2ky2 80 60 UG = mgy 40 20 0 -10 -8 -6 -4 -2 0 2 4 6 8 10 -20 y -40 -60 ye 0 shift due to mgy term

  31. 160 Choose C such as to show that the new equilibrium position has zero potential energy: 140 Uof Spring + Gravity 120 U 100 80 60 40 UNET =UG + US + C 20 0 -10 -8 -6 -4 -2 0 2 4 6 8 10 -20 US = 1/2ky2 y -40 -60 ye 0 shift due to mgy term

  32. Lecture 12, Act 3Energy Conservation • In (1) a mass is hanging from a spring. In (2) an identical mass is held at the height of the end of the same spring in its relaxed position. • Which correctly describes the relation of the potential energies of the two cases? (a)U1 > U2(b)U1 < U2(c)U1 = U2 case 2 case 1 d

  33. In case 2 the total potential energy is then . relaxed y = d d y = 0, U1 = 0 Lecture 12, Act 3Solution • In case 1, it is simplest to choose the mass to have zero total potential energy (sum of spring and gravitational potential energies) at its equilibrium position. The answer is (b) U1 < U2.

  34. m We know: Vertical Springs:Example Problem • If we displace the mass a distance d from equilibrium and let it go, it will oscillate up & down. Relate the maximum speed of the mass v to d and the spring constant k. • Since all forces are conservative,E = K + U is constant. j k y = d v y = 0 y = -d

  35. m Vertical Springs:Example Problem Spring • At the initial stretched positionand K = 0 (since v=0). j • Since E=K+U is conserved,will always be true ! k y = d • Energy is shared between the K and U terms. • At y = d or -d the energy is all potential • At y = 0, the energy is all kinetic. v y = 0 y = -d

  36. Power is the “rate of doing work”: • If the force does not depend on time: dW/dt = F.dr/dt = F.v P = F.v Power Ladder • We have seen that W = F.r • This does not depend on time! F r • Units of power: J/sec = N-m/sec = Watts v

  37. y x v T winch  mg Power • A 2000 kg trolley is pulled up a 30 degreehill at 20 mi/hrby a winch at the top of thehill. How much power is thewinch providing? • The power is P = F.v = T.v • Since the trolley is not accelerating, the net force on it must be zero. In the x direction: • T - mg sin= 0 • T = mg sin

  38. y x v T winch  mg Power • P= T.v= Tvsince T is parallel to v • So P= mgv sin v = 20 mi/hr = 8.94 m/sg = 9.81 m/s2m = 2000 kgsin = sin(30o) = 0.5 and P = (2000 kg)(9.81 m/s2)(8.94 m/s)(0.5) = 87,700 W

  39. Recap of today’s lecture • Problems using work/energy theorem • Spring shot • Escape velocity • Loop the loop • Vertical springs • Definition of Power, with example (Text: 6-3) • Look at textbook problems Chapter 8: # 81, 85, 95 Chapter 11: # 49

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