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If a person of mass M simply moved forward with speed V, his kinetic energy would be \frac{1}{2} MV^{2}.
If a person of mass M simply moved forward with speed V, his kinetic energy would be frac{1}{2} MV^{2}. 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 12.0 % 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 pm 30;^circ (a total of 60;^circ) 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 79.0 {rm kg} person walking at 5.00 {rm km/h} having arms 68.0 {rm cm} long and legs 92.0 {rm cm} long.What is the average angular velocity of his arms and legs?rad/sUsing the average angular velocity from part A, calculate the amount of rotational kinetic energy in this person's arms and legs as he walks?joulesWhat is the total kinetic energy due to both his forward motion and his rotation?joulesWhat percentage of his kinetic energy is due to the rotation of his legs and arms?%
Answer to 7300532If a person of mass M simply moved forward with speed V, his kinetic energy would be \frac{1}{2} MV^{2}.However, in addition to possessing a forward motion, various parts of his...