## Related questions with answers

If a person of mass M simply moved forward with speed V, his kinetic energy would be $\frac{1}{2} M V^{2}$ 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 13% of a person’s mass, while the legs and feet together account for 37%. 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 $\pm 30^{\circ}$ about (a total of $60^{\circ}$) from the vertical in approximately 1 second. Assume that they are held straight, rather than being bent, which is not quite true. Consider a 75-kg person walking at 5.0 km/h, having arms 70 cm long and legs 90 cm long. What is the total kinetic energy due to both his forward motion and his rotation?

Solution

Verified## C.

The total kinetic energy of a walking person is constituted by both the rotational kinetic energy of his arms and legs, and his translational kinetic energy given as equation $(4)$. Hence,

$\begin{aligned} K_{total}&=K_{rotational}+K_{translational}\\ &=4.9814~\mathrm{J}+\dfrac{1}{2}(75)\left[5.0~\mathrm{\dfrac{km}{\cancel{hr}}}\left(\dfrac{1~\mathrm{\cancel{hr}}}{3600~\mathrm{sec}}\right)\left(\dfrac{1000~\mathrm{m}}{1~\mathrm{km}}\right)\right]\\ &=\boxed{77.3194~\mathrm{J}}\\ \end{aligned}$

Hence, the total kinetic energy of the walking person is $\mathbf{77.3194~\mathrm{J}}$

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