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Chapter 9

Chapter 9. Systems of Particles. Center of Mass. Real objects are not point particles. How do I define the position of the cow? What is h for U = mgh?. Take average position of mass: “Center of mass”. Treat extended objects as point particles at CM. (Easy for spherical objects). r 2.

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Chapter 9

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  1. Chapter 9 Systems of Particles

  2. Center of Mass Real objects are not point particles How do I define the position of the cow? What is h for U = mgh? Take average position of mass: “Center of mass” Treat extended objects as point particles at CM (Easy for spherical objects)

  3. r2 r3 r1 Center of Mass (Discrete) To find the center of mass for a bunch of point masses: m1, m2, m3,… at r1, r2, r3,… rCM

  4. r Center of Mass (Continuous) For an extended object, break it into point masses (mass = dm) Note: This equation assumes that density is constant

  5. Center of Mass (Demo) 3 4 To balance, rcm = 0 m1 0.5 kg m2 ???

  6. Center of Mass (Bigger Demo) 1 14 Again, to balance: rcm = 0 m2 ??? m1 14.7 kg As before: Why didn’t it work this time?!

  7. Center of Mass (Bigger Demo) 1 14 We forgot about the bar! mbar=0.745 kg m1 14.7 kg m2 ??? CM in center of bar: 0.5 7

  8. Center of Mass (Bigger Demo) 1 14 m1 14.7 kg m2 ???

  9. rCM Center of Mass Example m1 at position (1,1) meter m2 at position (2,0) meter m1 = 3 kg Mtotal = 8 kg m2 = 5 kg

  10. rCM m2 = 5 kg Center of Mass for 2D objects Often don’t need to do integral: Use symmetry m1 = 3 kg Can also add symmetric objects together

  11. Another Way to Determine CM

  12. Brick falls if CM not supported Why Do We Want Center of Mass? Can treat extended objects or groups of objects as points aCM = Ftot /Mtot Gravity pulls on CM: h = hCM for Ugrav

  13. Mystery Hill Gravity pulls down on the CM of the system…

  14. Example: (Problem 9.9) A cylindrical can with mass M, height H, and uniform density is initially filled with soda/pop of mass m. We then punch a hole in the top and bottom so that the liquid can drain out. Find: • The center of mass of the can when full • The center of mass of the can when empty • What happens to the height of the CM while draining? • Find the height of the soda/pop when the CM reaches its lowest point This one is pretty tough!

  15. Ftot = dp/dt Momentum: New fundamental quantity (like force,energy,..) For point particle p = mv For group of particles: ptotal=m1v1+m2v2+... For extended object: pCM = mvCM Relation to Force: Ftot = ma = m dv/dt = d[mv]/dt

  16. Ftot = dp/dt pinitial = pfinal Conservation of Momentum If Ftot = 0, then momentum is constant For an isolated system (no external forces): True also for groups of particles: If Fexternal = 0, then pCM = constant Even if there are internal forces

  17. PSmith PPulse Momentum of a Noisy Cricket Will Smith: 200 lbs ~91 kg Thrown back ~3m in 1s  3 m/s

  18. Momentum of a Noisy Cricket So how much energy is this? But light is massless… Looking ahead to Phys 214: Recall: A car at 60 mph has ~ 105 J!

  19. The Celebrated Jumping Frog of Calaveras County 3 m A 2 kg frog sits on the end of a motionless, floating 50 kg log (3 meters long). She jumps to the other end. How far does the log move due to the jump?

  20. Celebrated Jumping Frog (cont.) Initial: x Final: Consider frog and log together as a system of objects No external forces  momentum of system doesn’t change pinitial = pfinal = 0 Center of mass of system does not move

  21. Celebrated Jumping Frog (cont.) xCM = 1.53 m x x = 0 x = 1.5m x = 3m CM does not move, log moves to keep it in same place! 1.53 m From geometry, we see that: initial 1m 1.53 m x final

  22. vB vG Gun and Bullet Both are at rest when fired, no net external force Total Momentum: pinitial = pfinal = 0 What does this say about the motion after the gun is fired?

  23. vB vG Gun and Bullet (continued) mB << mG so vG << vB The gun is heavier so it moves slower!

  24. vB vG Gun and Bullet (Question 1) Does the Center of Mass of the system move? No, because pinitial = pfinal = 0 for the system!

  25. vB vG Gun and Bullet (Question 2) If the momenta are the same, why do you want to hold the gun rather than catch the bullet? Consider the kinetic energy: K = ½mv2 = p2/(2m) |pG| = |pB| and mB << mG  KG << KB

  26. Spaceship Example A spaceship in outer space, free from any interactions with other objects, has mship=34,000 kg and vship,i=1,000 m/s i. The ship launches a satellite msat=1,000 kg which moves with vsat,f=(1,500m/s)i + (400 m/s)j. What is the velocity of the spaceship after the satellite launch?

  27. Spaceship (continued) No external forces, so momentum is conserved: Solve for vship,f

  28. Spaceship (question) Is Kinetic Energy conserved? Kinitial = Kship = ½mtotal(vinitial)2 = ½(3.5 x 104 kg)(1000 m/s)2 = 1.75 x 1010 J Kfinal = Kship + Ksatellite = ½mshipvship2 +½msatvsat2 = ½(3.4 x 104 kg)[(985 m/s)2 + (11.8 m/s)2] + ½(1000 kg)[(1500 m/s)2 + (400 m/s)2] = 1.77 x 1010 J NO! Extra energy added when satellite ejected.

  29. Rockets A gun shooting a bullet is a discrete process We can generalize this to continuous processes Rockets operate by shooting outa continuous stream of gas Note: Rocket propulsion was originally thought to be impossible… Nothing to push against!!

  30. Rockets

  31. Rockets Δm 0 Note: dm = -dM Dividing both sides by dt Let dM/dt = -R, where R is the rate of fuel consumption

  32. Rockets To find the velocity of the rocket we must integrate Note: Velocity is a function of mass loss, not time

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