1. Center of MASS (COM) Definition--The Point located at an objectsaverage position of the mass In other words. The center of an objects mass Symmetrical objects, like a baseball the C of M would be in the exact center of object However other oddly shaped objects will findCOM in any number of positions, depending on weight distribution COM

2. C.O.M. When objects rotate freely they must rotate about an axis through the COM Basically treat the object as if all its weight is concentrated at that one pt. 3. C.O.M. --Balancing For an object to balance, and not topple support must be directly below C.O.M. 4. Where C.O.M. is located Generally found in the middleof all the weight Does not even have to be within, the object itself Ex. boomerang Will be located toward oneside of an object where most of its mass is focused Ex. WeeblesCOMgravity 5. Weebles Wobble, but they dont fall down??? Weebles have very low COM Whenever rolling it will roll to a stopwhen its COM is as low as possible This occurs when it is standing upright Also occurs for inflatable toy clowns Objects with a low COM are less likely to topple because of this principle Higher COM is, the easier to topple 6. Balancing Stuff Again, all that has to happen to balance, is fora support to be directly beneath COM 7. Advantage of low COM Athletic advantages wrestlingharder to takedown Football Both easier to drive power through their legs SUVs . Tip over all thetime b/c COM is too high Farmers tractors Much more control in allvehicles w/ low COM 8. Deadliest Catch Pots on deck and freezing ice makeboat top heavy more likely to roll and sink Ballast tanks at bottom help lower boats COM 9. Animals Low COMHigh COM 10. T. Rex& Tails 11. Humans-Where is our COM? Just below our bellybutton Notice, support always below COM Bipedalism?? Only mammals w/ thisability to walk on 2 legs Because of Evolution and how our legs changed to balance between steps is why we are only mammals to walk 12. Because our legs/hips evolved so that our support base (feet) were close together allows us to be bipedal Apes and our early ancestors hips were constructed differently with a wide set base. Impossible to walkbipedalHip protruding from joint. Creates inward angled femurs Which makes feet close together. Providing a stable/efficient base for walking upright 13. Definition The center of mass (or mass center) is the mean location of all the mass in a system Marked with: 14. Examples The center of mass of a two-particle system lies on the line connecting the particles (or, more precisely, their individual centers of mass). The center of mass is closer to the more massive object The center of mass of a ring is at the center ofthe ring (in the air). 15. More Examples The center of mass of a solid triangle lies on all three medians and therefore at the centroid, which is also the average of the three vertices. 16. Applications Objects rotate around their center of mass In a uniform gravitational field, the center of mass and center of gravity are the same. A projectiles center of mass will follow a parabolic path. If an objects center of mass is outside its base of support, it will topple. An applied force that is not through an objects center of mass will cause rotation. 17. CoM of Symmetrical Object The CoM of any symmetrical objectlies on an axis of symmetry and on any plane of symmetry. 18. Toppling Rule of Thumb If the CoM of the object isabove the area of support, the object will remain upright. If the CoM is outside the area of support the object will topple. 19. Another look at Stability Stable equilibrium: when for a balanced objecta displacement raises the CoM Unstable equilibrium: when for a balancedobject a displacement lowers the CoM . Neutral equilibrium: when the height of theCoM does not change with displacement. 20. Stability 21. Center of Mass of: System of Particles Extended Object 22. Center of Mass of a System of Particles in motionvCM =Smivi SM mi is the mass of each particle M is the sum of the masses of all particles MomentumvCM =Spcm SM 23. MOTION of center of mass(CoM) Refer pg 37 of course book the grey out area. 1. Consider just the single motion of the center of mass(CoM) 2. Motion of the vcm does not change in magnitude or direction 3. ptotal is always conserved(pbefore = pafter = ptotal4. CoM momentum and velocity is unaffected by interactions and collisions)=pcm ) 24. Level 3 COURSE book solutions page 38Page 38 number 2 and 4 2a p =mv = 260 x 3.2 = 832kgm/s b vcm = 0m/s CoM velocity and momentum does not change. c 2nd dogem must stop as well 3a 100m/s right b. 100m/s left *** NO. 4 VERY IMPORTANT TO grasp CoM 4a vcm = (m1v +m2v /(m1 +m2) = 15000x 5 / (10000+15000) = 3m/s left b 3.0m.s 25. Level 3 COURSE book solutions page 38pg 38 number 1 at the bottom and pg 39 no 2 1d period T = =0.25 x component of vcm = 2cm/0.25 = 8cm/s e pt = pcm = 0.08 x 0.3 = 0.024 kg m/s 43 = 122a8x2=16 122 + 162 = 20b 20kgm/s c vcm = Spcm = 20 = 1.7m/s SM 12 d tan = 16/12 = 53 to x axis 26. Level 3 COURSE book solutions page 39pg 39 no 3 3a100002 + 120002 =800x15=12 000 100010 = 10 000b 15600kgm/sc vcm = Spcm = 15600 SM 1800 d 8.7m/s e tan = 12000/10000= 8.7m/s = 50 to x axis 27. Example 1: Center of Mass in one Dimension Find the CM of a system of four particles thathave a mass of 2 kg each. Two are located 3cm and 5 cm from the origin on the + x-axis and two are 2 and 4 cm from the origin on the x-axis Answer: 0.5cm 28. Level 3 Text book solutions for activity 7D page 101number 1 a CoM(90 x X) 90X 200X X= = = ={110 x (15-X)} 1650 -110X 1650 1650/200 = 8.3mbi CoM has not moved ii 8.25/ 15 x (15-5.5) =5.2m from Aiii t =8.5s d =3.1MX 90X 90X 200X XOR= m(d-X) = 110(9.5-X) = 1045 110X = 1045 = 5.2m from Av = d/t 3.1/8.5 = 0.36m/siv pa = mv = 90 x 0.36 = 32.4kgm/s pb = 32.4kgm/s = 90 x vbvb= 32.4/90 = 0.29m/s 29. Level 3 Text book solutions for activity 7D page 101number 4 aio m/sii5mbi use rotational torque Ta = Fa x d = (12.6x9.8) x 3 = 370.44Nm Tc = 370.44Nm Fc = Ta d= 370.44 2 = 185.22Nm = 185.22 9.8 = 19kg bii d = 5m v = 2m/s t = d/v 5/2 = 2.5s iii zero no external forces therefore vcm = 0m/s *** ivcenter of mass moved 3m in 2.5s therefore vcm = 3/2.5 = 1.2m/s 30. Level 3 Text book solutions for activity 7D page 101number 5 avcm = 3.4m/sb mm = mass of Moanamj = mass of Joebefore: pcm = (mm + mj ) x 3.4 = 3.4 mm + 3.4 mj After: pm = mv = 3.1 x mmpj = 4.0 x mj pcm = (3.1 x mm + 4.0 x mj ) pcm = 3.4 mm + 3.4 mj (3.1 x mm + 4.0 x mj ) = 3.4 mm + 3.4 mj 4.0 mj - 3.4 mj = 3.4 mm - 3.1 mm 0.6 mj=0.3mmmm = 0.6/0.3 mj = 2mjMass of Moana = 2 mass of Joe