#### Description of this paper

##### If a person of mass M simply moved forward with sp...

**Description**

Solution

**Question**

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? 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? What is the total kinetic energy due to both his forward motion and his rotation? What percentage of his kinetic energy is due to the rotation of his legs and arms?

Paper#13775 | Written in 18-Jul-2015

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