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This is a useful tool for studying collisions, explosions and other forms of motion. Centre of mass = point where the to

This is a useful tool for studying collisions, explosions and other forms of motion. Centre of mass = point where the total mass acts. Another name for it is centre of gravity . This is of paramount importance in architecture:. CoM. CoM. Centre of mass acts through base building stable.

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This is a useful tool for studying collisions, explosions and other forms of motion. Centre of mass = point where the to

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  1. This is a useful tool for studying collisions, explosions and other forms of motion. Centre of mass = point where the total mass acts. Another name for it is centre of gravity. This is of paramount importance in architecture: CoM CoM Centre of mass acts through base building stable. Centre of mass is not supported by the base, result is building falls E.g. the leaning tower of Pisa
  2. The foundations subsided which caused the tower to lean. However, because the CoM is over the base the tower will not topple over. As people walk up to the top of the tower the CoM shifts over time. As a result the CoM moves further along until eventually the base will not support it and it will fall. CoM
  3. MgX = mgx MX = mx M/m = x/X MX = m(d-X) 4kg 1kg CoM is four times closer to the 4kg mass than the 1kg mass. If the distance between the two masses is 12m then the CoM is 2.4m from the 4kg mass.
  4. MgX = mgx MX = mx M/m = x/X MX = m(d-X) 4x = 1 ( 12 – x) 4x = 12 – 1x 5x = 12 x = 2.4 m from 4kg 4kg 1kg 12m
  5. Example 1: The Earth and Moon are like a two particle system, if we imagine all their mass as concentrated at their CoM. Calculate the distance of the CoM of the Earth-Moon system from the centre of the Earth. Mass of Earth = 600 x 1022 kg Mass of moon = 7.3 x 1022 kg Distance between Earth & moon = 3.8 x 105 km SOLUTION: Distance between Earth & Moon = 3.8 x 108 m Earth: 600x1022kg X Moon: 7.3x1022kg CoM
  6. d = X + x MX = m(d – X) X is the distance from the centre of the earth X = md/(M + m) Rearranging X = 7.3x1022 x 3.8x108 / (600x1022 + 7.3x1022) X = 4.6 x 106 m As the radius of the earth is 6.4 x 106m, the centre of mass of the earth-Moon system is beneath the earth’s surface. While it is usually said that the Moon orbits the earth, it would be more correct to say that the Moon orbits the earth each orbit around their common CoM.
  7. CoM & Conservation of Momentum Any form of calculation involving a collision, as youlearnt last year, can be solved using momentum.  = mv The law of conservation of momentum can also be used for isolated systems. The behaviour of the system can be analysed by looking at what happens to its CoM. system = msystemx vCoM system = (M + m)x vCoM If the system is isolated then the momentum will not change. The masses of the individual particles also remain the same thus the vCoM must therefore remain constant.
  8. The momentum of a system is also the total momentum of the particles that make up the system. system = Σparticles of system system = (M + m)x vCoM (M + m) x vCoM = MvM + mvm As two objects approaching each other collide they can change both their speed and direction, but their CoM moves with a constant velocity. The velocity of the CoM is unchanged by the collision.
  9. Example 2: A 2000kg truck moving at 8.0ms-1 hits a stationary car of mass 1200kg. After the accident, the two vehicles are locked together. Calculate the velocity of the truck and car after the collision. BEFORE AFTER vms-1 2000kg 8ms-1 2000kg 1200kg 1200kg SOLUTION: The total momentum before the collision = (2000 x 8.0) + (1200 x 0) = 16000 kgms-1 The velocity of the CoM before the collision is given by: vCoM = total momentum / total mass vCoM = 16000 / (2000 + 1200) = 5.0 ms-1 This will also be the velocity of the CoM after the collision. Since the vehicles are locked together their velocity after the collision will also be 5.0 ms-1.
  10. Why can’t it stop easily ?? HARD TO STOP 2. Itis FAST 1.It is MASSIVE IT has a lot of MOMENTUM
  11. Momentum is an especially useful tool to measure collisions / explosions.  = mv Momentum is a vector quantity having direction and velocity. An unbalanced force acting on an object causes a change in velocity, so it also changes the momentum of the object. This change in momentum can be calculate from:  = f - i Conservation of momentum: In collisions the total momentum of the objects involved is conserved as long as there are no external forces acting. This is very useful for solving problems involving collisions.
  12. Example 3: Conservation of  in 2-dimensions The diagram shows a radioactive nucleus, which was initially at rest, immediately after it has decayed. An emitted electron moves off in a northerly direction, a neutrino moves off in an easterly direction, and the decayed nucleus moves off in a different direction. Electron Decayed nucleus Neutrino The momentum of the electron is 1.2 x 10-22 kgms-1, and that of the neutrino is 6.4 x 10-23 kgms-1. Calculate the size of the momentum of the nucleus after the decay.
  13. SOLUTION: Σ (before) = Σ(after) Σ (before) = 0 Σ(after) = 1.2 x 10-22kgms-1  + 6.4 x 10-23kgms-1  + (nucleus) (nucleus) = 0 – 1.2 x 10-22kgms-1  - 6.4 x 10-23kgms-1   (nucleus) (nucleus) = (1.2x10-22)2 + (6.4x10-23)2 = 1.4 x 10-22kgms-1 1.2x10-22kgms-1 6.4x10-23kgms-1
  14. F = ma a = v/ t F = mv/ t Substituting both OR F t = mv F t =  
  15. Example 4: Impulse in 2-dimensions A ball of mass 100g is moving north at 5.0ms-1. Raj hits the ball with his stick so that it travels west at 12ms-1. The diagram is a view from above and shows the position of the stick at the instant the ball is hit. Find the size and direction of the change in momentum of the ball. If the stick is in contact with the ball for 0.020s, what is the magnitude and direction of the average force acting on the ball. 12ms-1 5ms-1
  16. SOLUTION: Initial i= mv = 0.10 x 5.0 = 0.50 kgms-1 Final f = mv = 0.10 x 12 = 1.2 kgms-1  = f - I = 1.2 kgms-1 - 0.50 kgms-1 1.2kgms-1  0.50kgms-1  Using trigonometry: tan  = 0.50 / 1.2  = 23o  =  1.22 + 0.502 = 1.3 kgms-1 23o south-west
  17. b. The average force F on the ball can be calculated from: Ft =  F x 0.020 = 1.3 F = 1.3 / 0.020 F = 65N (in the same direction as )
  18. BEFORE AFTER y3 x2 x1 A+B y1 y2 x3 B A HORIZONTALLY: x1+ x2 = x3 VERTICALLY: y1 + y2 = y3 RESULTANT = x3 + y3 Use trigonometry
  19. Kinetic energy Ek = ½mv2 Gravitational potential energy Ep = mgh Work done W = Fd Power P = W/t Remember if all energy is conserved then the collision is ELASTIC. If energy is lost then it is INELASTIC. Paddys Note: C drive/PYX/Vol2/Mechanics/Forces
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