L20 - Finding the CoM

• What makes for a successful high jump performance?

  • The best jumpers will have the following characteristics:
  • Decreased Fg - Less mass means less force is required to accelerate the body upward (Law 2)
  • Largest vertical velocity - They can direct force precisely to redirect the hz velocity into vt velocity (Law 3)
• What makes for a successful high jump performance?

  • The best jumpers will have the following characteristics:
  • Decreased Fg - Less mass means less force is required to accelerate the body upward (Law 2)
  • Goal is jumping high against gravity so notice that they are all fairly thin
• What makes for a successful high jump performance?

  • The best jumpers will have the following characteristics:
  • Largest vertical velocity - They can direct force precisely to redirect the hz velocity into vt velocity (Law 3)
  • Have to produce torque on body to cause body to rotate in the air
  • Want large hz velocity to transfer into vt velocity - so relative impulse important
• What makes for a successful high jump performance?

  • Create 2 torques on body:

  1. Take a J turn when running to bar creates rotation along long axis
  2. To hitch kick when take off, is unbalances also creates rotation
• What makes for a successful high jump performance?

  • To get precise movement to get over bar
  • As doing this location of CoM changing continuously
  • Is the defining factor if they are successful or not
• But there is something else that is important:

  • Javier Scotomayer 2.45 m (8.05 ft) 1993; 1.95 m tall & 82 kg
  • Mutaz Barashim 2.42 m (7.99 ft) 2014; 1.89 m tall & 65 kg
• The science behind the Fosbury Flop:

  • We have used an off centre force which created a rotation of the body with our push off
  • In the air we can redistribute our segments/limbs to manipulate the location of the body, keeping the CoM as near to the bar as possible
  • Apply same force but jump higher
  • Prev CoM had to clear bar, but now can be under bar with same force
• Flight phase of jump - goal:

  • To get over the bar as efficiently as possible
  • Get CoM as close to bar as possible
  • Athlete can't really change the flight path of the CoM but it’s possible to redistribute body parts (limbs in air) to maximise position over the bar & take advantage of action reaction principle
  • Must create rotation about/around the CoM
  • To keep mass at highest point possible
• Weightlifting:

  • Using location of mass to use greatest advantage to get bar moving
• Basketball Dunk/Volleyball Spike:

  • Gain height but maintain it long enough to dunk/spike = hang time; to hold self in air for long period of time, redistribute body mass to stay in air longer (increasing peak)
• What is the Centre of Mass?

  • The weight of the body is a function of the mass of each particle & the acceleration due to gravity
  • F = ma
  • The point around which all the particle are evenly distributed is the point at which we can draw a single weight vector (Fw)
• What is the Centre of Mass?

  • Point of balance
  • Usually have weight vector acting
  • F = ma → different at CoM (check)
  • eg merry go round, centre easier to hold on compared to outside, more force acting on body further away from axis of rotation
• Centre of Gravity (CoM):

  • The centre of gravity is the point about which the resultant torque of all the forces acting on the body is 0
  • ∑T = 0
  • ∑T = Fw1 * d1 - Fw2 * d2
• Centre of Gravity (CoM):

  • If mass distributed equally around that point (neg & pos so sum = 0)
  • Know as going to balance thru that point then the sum of the torques must be 0
  • Find balance point find CoM
• How do we locate the Centre of Mass?

  • Quantitative method to locate CM:
  • rcm = ∑ri mi / ∑mi
  • rcm = (0.1 * 20 + 1.1 * 80) / 100
  • ∑T = (Fw1 * d1) - (Fw2 * d2)
  • ∑T = (20 * 0.8) - (80 * 0.2) = 0
  • >> confirm the definition of CM
• How do we locate the Centre of Mass?

  • Point in body related to sum of all the points/masses on the body; take position of all limbs centres & find avg position divide mass of person = gives location of mass
  • 0 = origin point to measure from
  • rcm = where CoM is located; is the sum of our masses/torque
  • Moment arm of force mass divide by total mass = 0.9 m
  • CoM are axis of rotation: so 0.8 (1.1 - 0.9) for 80 kg
• How do we locate the Centre of Mass of an athlete in action:

  • Stop it a different locations & calculate CoM in that frame
  • Want to know mass & location (x & y axis - use grid)
  • Tables for this
  • 0.007 = 7% of BM
• How do we locate the Centre of Mass?

  • 3 segments/chunks:

  1. Head: m = 7%BM, r = [0.98 i + 1.03 j] m
  2. Torso & arms: m = 61%BM, r = [1.18 i + 0.83 j] m
  3. Legs: m = 32%BM, r = [0.85 i + 0.8 j] m

  • rCM = ∑ri mi / ∑mi = {0.07 BM [0.98 i + 1.03 j] → Head
  • 0.61 BM [1.18 i + 0.83 j] → Torso + arms
  • 0.32 BM [0.85 i + 0.8 j]} / BM → Legs
  • = BM [0.0686 i + 0.0721 j]
  • + BM [0.7198 i + 0.5063 j]
  • + BM [0.272 i + 0.256 j]} / BM → remove from equation
  • = [1.0604 i + 0.8343 j] m
  • rCM = [1.0604 i + 0.8343 j] m
• How do we locate the Centre of Mass?

  • % BM x by position for each segment then divide by BM
  • Do we need to know what is body mass is can be solve it without his actual mass
  • Yes can solve it as BM isn’t changing
  • Location of CoM is going to be multiplying thru percentages
  • Can use this technique to locate CoM & tells us the rotation
• The science behind the Fosbury flop:

  • How does this help us understand how someone can jump higher?
  • First, we have to do work on our CoM to give it kinetic energy for the jump
  • We have used an off centre force which created a rotation of the body with our push off
  • In the air we can redistribute our segments to manipulate the location of the CoM, to control the rotation of the body, keeping the CoM as near to the bar as possible
• The science behind the Fosbury flop:

  • To redistribute body masses to keep neutralising torque around that point to get higher in the jump