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The kinetic energy of walking

  1. Oct 22, 2012 #1
    1. The problem statement, all variables and given/known data
    If a person of mass M simply moved forward with speed V, his kinetic energy would be 0.5MV2. However, in addition to possessing a forward motion, various parts of his body (such as the arms and legs) undergo rotation. Therefore, his total kinetic energy is the sum of the energy from his forward motion plus the rotational kinetic energy of his arms and legs. The purpose of this problem is to see how much this rotational motion contributes to the person's kinetic energy. Biomedical measurements show that the arms and hands together typically make up 14.0% of a person's mass, while the legs and feet together account for 38.0% . For a rough (but reasonable) calculation, we can model the arms and legs as thin uniform bars pivoting about the shoulder and hip, respectively. In a brisk walk, the arms and legs each move through an angle of about +/- 30° (a total of 60 °) from the vertical in approximately 1 second. We shall assume that they are held straight, rather than being bent, which is not quite true. Let us consider a 74.0 kg person walking at 5.00 km/h having arms 66.0 cm long and legs 94.0 cm long.

    A)Using the average angular velocity from part A, calculate the amount of rotational kinetic energy in this person's arms and legs as he walks.




    2. Relevant equations
    I figured that the mass of the arm is 10.36 kg and the mass of the leg is 28.12 kg. In addition 5km/h converts to about 1.38889 m/s. I've found the angular velocity ω, t be 1.05 rad/s.
    Krot=.5MIω2 is the equation I'm assuming I should use but I'm not sure where to go from here.


    3. The attempt at a solution
    Krot=.5(10.36*.662+28.12*.942)(1.05)
     
  2. jcsd
  3. Oct 22, 2012 #2

    tiny-tim

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    hi flip290! :smile:
    (you mean Krot=.5Iω2 :wink:)

    ok, now just plug in the figures :smile:

    (but you'll need to look up the formula for the moment of inertia of a rod about one end)
     
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